Transcript for:
Class 11 Physical Chemistry Marathon Notes

for hey guys good morning and welcome back again to your an academy n English Channel I hope all of you doing amazing all of you doing great so people quickly let me know in the chats if all of you can hear me if I'm perfectly audible visible to every one of you quickly let me know in the chats everyone everyone good morning morning people good morning good morning good morning good morning guys good morning good morning and welcome back good morning and welcome back so what's up how are you all doing I'm all good guys thank you so much I'm all good perfect perfect perfect so I hope I'm perfectly audible visible to everyone so my dear students you all must be knowing today it's going to be the mega marathon of your class 11th physical chemistry right and let me exactly show you the chapters which are there in class 11th physical chemistry please have a look mole concept redox reactions atomic structure thermodynamics chemical equilibrium ionic equilibrium right so these are in total some six chapters right these are in total some six chapters and let me first of all tell you how many marks you guys are going to get from each chapter my de students Mo concept you can expect two questions redox reactions you can expect two questions atomic structure you can expect two questions thermodynamics you can expect three questions chical equilibrium you can expect one question ionic equilibrium you can expect two questions right so in total 2 4 6 9 10 11 12 so 12 questions you will be getting from your class 11th physical chemistry right perfect well people to cover all these six chapters since these are Big chapters basically I would require two marathons and the first marathon is is today which will be there for 10 to 12 hours perfect I've already given you the schedule I've told you class 11 physical chemistry will be done in two marathons because I want to cover each and everything I do not want to skip anything among these chapters perfect that is the reason why have I have dedicated two marathons for all these six chapters perfect perfect guys so can you let me know have you studied these chapters before I just want to know have you studied these chter before see I'm going to teach every single thing that two from the basics only right that two from the basics only you need not to worry about anything you need not to worry about anything and once we are done with the marathons I mean once we are done with the two marathons of your class 11 physical chemistry after that there'll be extra problem practice session as well and I believe all of you have already seen the schedule which I gave you earlier in the uh like before 2 days perfect I gave you the schedule till uh till 30th of April yeah perfect so are you all ready are you all ready to be with me till the end are you all ready to be with me till the end whatever I'm going to teach you today it is going to be more than sufficient when it comes to your class 11th physical chemistry whatever I'll be teaching teaching you apart from that nothing is going to be asked when it comes to your neat examination from these chapters okay perfect perfect people exactly practical chemistry will be also done there'll be a separate one 2hour session for that right that's not something huge practical chemistry will be also done well while I'll be teaching you can you can you can make the short notes with me right you can make the short notes with me you need not to write everything but be ready with your pen and paper right and take a note take a note of all the important things whatever I'll be mentioning on the screen perfect and whatever questions I'll be giving you try to solve them with me okay and at the end of the session right I'll be sharing the session PDF with you well I could not share the inorganic PDF with you right you know what is the reason perfect that got deleted because of the electricity cut right okay but right after today's session I'll be sharing this PDF with you for sure because I've already saved this because I've already saved this so should we start I would want some fire fire fire fire fire in the chats smoke in there yeah I would want some fire in the chats I would want some fire in the chats before starting the session let the Jo be high let the Jo be high because this is not going to be the session for 1 hour 2 hour this is the session for complete 10 12 hours right so I want you guys to stay with me till the end with the same Jo the way you're starting right now and in between I'll be giving you the breaks as well you know it right after every 2 to 3 three hours I'll be giving you like 15 to 20 minutes breaks for sure perfect guys so without wasting any sort of a time let's get going and let's get started with the first chapter of your class 11th physical chemistry that is the mole concept this particular chapter it forms the foundation of your physical chemistry if this particular chapter is clear then and then your complete physical chemistry can get clear otherwise if there are issues in this particular chapter there'll be issues in your physical chemistry throughout so make sure make sure whatever I'll be teaching you in the mole concept you understand each and everything in detail because this chapter means if this chapter is strong that actually means your physical chemistry automatically is getting stronger okay perfect all right guys so let's get going let's get started let's get started guys let's get started just give me a second let's get started just a second perfect my dear students over here I'm writing a term that is 1 AMU that is 1 AM U which is also what I call as 1 U which is also what I call as 1D now what this AMU stands for what this U stands for and what this d stands for let me tell you this AMU which I wrote over here it stands for the atomic mass unit U stands for the unified mass and this D over here which I wrote it stands for Del it stands for Del now what this 1 AMU exactly means let me tell you this one AMU which is what you call as one U which is what you call as 1 D it is defined as it is defined as it is defined as 1 12th of the mass of carbon toal atom it is defined as 1 12th of the mass of carbon 12 atom imagine you have taken the carbon carbon 12 atom imagine you have taken the carbon 12 atom you have divided the carbon 12 atom into 12 equal parts after dividing the carbon 12 atom into 12 equal parts the smaller part which you take right whatever is the mass of that smaller part that is something which you call as 1 AMU 1 AMU is basically defined as 1 12th of the mass of whole carbon toot 1 12th of the mass of whole carbon toot let me tell you let me tell you the mass of one carbon toal atom the mass of one whole carbon toal atom how much is that that is 1.99 into 10^ - 23 g if you know the mass of one whole carbon 12 atom if you divide that number by 12 you'll be getting the value exactly as 1.66 into 10^ -24 G okay so what exactly we did we took a carbon to at right mass of whole carbon to atom is how much 1.99 into 10-23 G and I divided that carbon to atom into 12 equal parts and then I took one smaller part the mass of that smaller part the mass of one 12th part of the carbon is how much 1.66 10 -24 G and my students I can write it as 1 / avagadro's number as well because when you divide one by the avagadro number when you divide one by by 6.022 into 10^ 23 you'll be getting the value exactly as how much 1.66 10 power minus 24 this is the first statement which I would want every one of you to remember this is the first statement which I would want every one of you to remember okay 1 AMU which is what you call as 1 U which is what you call as 1D it is defined as 1 12th of the mass of carbon 12 atom mass of whole carbon 12 atom is this much if you divide this number by 12 you'll be getting the value exactly this much which can be written as 1 / na as well where na is something which you call as your avagadro's number perfect now guys what this 1 AMU is let me tell you this 1 AMU over here which is what you call as one U which is called as 1D it is basically a standard 1 AMU it is basically a standard it is basically the reference it is a standard it is a reference with which we shall be comparing the masses of different atoms or molecules 1 AMU or one U or 1 D it is nothing but it is a standard it a reference it is a standard it's a reference with which we shall be comparing the masses of different atoms or molecules right with which we shall be comparing the mass of different atoms or molecules now how exactly for that have a look try to understand exactly what I'm going to say my dear students try to understand carefully what exactly I'm going to say over here I'm dividing this board into three parts try to understand exactly what I'm going to say I'm going to Define certain terminologies here one is called as atomic mass one more term is what you call as molecular mass and one more term will be called as the molar mass these are the three basic terminologies which I'm defining with all of you in the beginning itself try to understand what these terms are this forms the basics of the mole concept first thing how do we Define the at atomic mass atomic mass is nothing it is defined as the mass of atomic mass is defined as the mass of mass of one atom of an element atomic mass is defined as the mass of one atom of an element for example I'm asking you what is the atomic mass of oxygen what does that mean that means I'm asking you what is the mass of one atom of oxygen let's say I'm asking you what is the atomic mass of Florine that means I'm asking you what is the mass of one atom of Florine so so in short if I want to define the ter atomic mass it simply means the mass of one atom of an element now my dear students for example if I talk about oxygen if I talk about oxygen you know what is the atomic mass of oxygen you know it is 16 AMU it is 16 AMU what does it mean what does it mean I can divide the 16 AMU into two parts I can write it as 16 multiplied by 1 AMU I can do that and you know this 1 Amu is nothing it is 1 12th of the mass of carbon toal atom yes 1 AMU is also 1 AMU is also equal to 1.66 into 10^ -4 G this 1 AMU is also equal to 1 / na G now why did I do all these things I just took an element oxygen right and I'm telling you the atomic mass of is 16 AMU what does it mean it means one atom of oxygen one atom of oxygen is 16 times one atom of oxygen is 16 times heavier than 1 AMU one atom of oxygen is 16 times heavier than 1/ 12 of the mass of carbon to atom this is the mass of one atom of oxygen in G even this is the mass of one atom of oxygen in grams as simple as that as simple as that for example you're defining I mean I mean I'm I'm asking you what is the atomic mass of nitrogen you will say atomic mass of nitrogen is 14 AMU what does that mean that means one atom of nitrogen is 14 times heavier than 1 AMU one atom of nitrogen is 14 times heavier than 1 12th of the mass of carbon to atom right if I ask you what is the mass of one nitrogen atom in G you will say 14 * 1.6 6 10^ -4 G or you will say 14 * 1/ avagadro's number in G yes as simple as that as simple as that perfect so what is atomic mass it is nothing but mass of one atom of an element my dear students did you understand one thing I told you in the beginning 1 AMU is a standard 1 AMU is the reference with which we compare the masses of different atoms did you see I told you one atom of oxygen is 16 times heavier than 1 U so this 1 U over here is the reference it is the standard with which I'm comparing the masses of different atoms as simple as that yes I believe this is clear I believe this is clear right now if I talk about the term molecular mass look at the name molecular mass molecular mass simply means mass of one molecule right mass of one molecule right mass of one molecule how do you calculate it you know it already how do you calculate it it is it is calculate as the sum of atomic masses the sum of atomic masses of all the elements present in a molecular compound right for example for example for example I'm taking water I'm asking you what is the molecular mass of water I'm asking you what is the molecular mass of water how do you calculate it how do we calculate it sum of atomic masses hydrogen its atomic mass is one multiplied by how many hydrogen atoms two plus oxygen its atomic mass is 16 AMU how many oxygen atoms one when you solve this term what is the value which you'll getting you'll get the value exactly as 18 now in case of molecules preferably in case of molecules preferably we write u in place of AMU meaning same AMU u d they are same right but in case of molecules right we prefer the term U instead of what instead of AMU now my dear students if I ask you what it means we got the molecular mass of water as 18 U what does that mean we got the molecular mass of water as 18 U can I break this 18 U as 18 multipli 1 U I can do that instead of one U can I write 1 12th of the mass of carbon to atom absolutely I can do do that instead of 1 U can I write 1.66 into 10us 24 G absolutely can do that instead of one U can I write 1/ avagadro's number in grams absolutely I can do that perfect so what did we get right understand we got the molecular mass of water it means that one molecule of water one water molecule is 18 times heavier than one U one molecule of water is 18 times heavier than one 12 of the mass of carbon to2 this is the mass of this is the mass of one water molecule in G even this is the mass of one water molecule in grams I believe this is super clear to everyone I believe this is super clear to everyone now my dear students the third terminology which we have what do we have the third terminology it is the molar mass molar mass what is molar mass molar mass is simply defined as the mass of the mass of one mole of a substance the mass of one mole of a substance the mass of 1 mole of a substance or I can write it like this mass of 6.022 into 10^ 23 particles of the substance right mass of 6.022 into 10 part 23 particles of the substar or I can write it like this Mass mass of one particle of a substance mass of one particle of the substance multiplied by avad number simple right molar mass mass of one mole of the substance you know one mole of the substance contains 6.022 into 10 23 particles so instead of mass of one mole of the substance I can write mass of 6.022 into 10 23 particles of the substance I can write it I can write it as mass of particle of a substance multiplied by number all these things are absolutely the same right for example I want to define the M mass of marker what is meant by the term M mass of marker it means I'm asking you to calculate the mass of one mole marker one mole marker means 6.022 into 10 23 markers so indirectly I'm asking you calculate the mass of 6.022 into 10 23 markers how do I get that if I just calculate the mass of one marker and multiply that term with avagadro number I'll be getting the mass of 6.022 into 10^ 23 markers which is what I'll be calling as M mass of marker am I clear with this am I clear with this say yes or no in the chats quickly say yes or no in the charts quickly understand people let's say I want to calculate molar mass of molar mass of for example for example molar mass of for example for example oxygen oxygen I have taken in atomic form o I've not written O2 what is meant by m mass of oxygen M mass of oxygen it means mass of one oxygen atom multiplied by aad number what is the mass of one oxygen atom mass of one oxygen atom is 16 AMU mass of one oxygen atom is 16 AMU and I'm multiplying it with what avagadro's number my de students you already know 1 AMU I can write as 1/ na G right multiplied by this na so na na cancel the value came out to be 16 G so what is the 16 G what is the 16 G it is the molar mass of oxygen this is the mass of 1 mole of oxygen this is the mass of 6.022 into 10^ 23 oxygen atoms yes right people if you look carefully what was the atomic mass of oxygen 16 AMU what is the mol mass of oxygen 16 G so magnitude wise they are same right 16 16 same it is just units are different it is just units are different perfect for example sodium its atomic mass is 23 AMU that means mass of one atom of sodium is 23 AMU now if I'm asking you what is the molar mass of sodium instead of AMU you just have to write grams that is the molar masse of sodium that is the mass of 1 mole of sodium that is the mass of 6.022 into 10 23 sodium atoms right right people so in short what do you have to remember atomic mass molar mass difference is in the units here the unit is AMU here the unit is grams right okay one more thing one more thing water its molecular mass is 18 U its molecular mass is 18 U that means mass of one molecule of water is 18 U now if I ask you what is the molar mass of water instead of you you'll write grams that's it 18 G will be the molar mass of water as simple as that right right people I hope I'm clear with all these things yeah now to quickly summarize all these things try to understand what exactly I'm going to say try to understand what exactly I'm going to say I'm quickly summarizing all these things which is super important my dear students here I'm writing O oxygen in the atomic form oxygen is in the atomic form here I'm writing O2 so oxygen hair is in molecular form oxygen hair is in the molecular form now try to understand perfectly I'm writing few terminologies related to this o one I'm writing 16 AMU another terminology only and only I'm writing 16 without the units third terminology instead of one AMU I'm writing 1.66 into 10 power - 24 G right and at the end I'm just writing 16 G I'm just writing 16 G try to understand carefully what I say similarly O2 first of all I'm writing 32 U then I'm only writing 32 then I'm writing 32 multiplied 1.66 into 10 ra^ minus 24 g at the end I'm just writing 32 G 32 G now let me tell you the meaning of all these things right perfect perfect try to understand guys try to understand carefully first of all oxygen here I've taken in atomic form so this is what I call as atomic mass of oxygen right this is what I call as atomic mass of oxygen only 16 this is what I call as a relative atomic mass of oxygen this what they call as a relative atomic mass of oxygen right this particular term this particular term this I call as actual mass of O this is what you call as actual mass of O and this particular term you call this as molar mass of O you call this as molar mass of O right or instead of the term molar mass you can write one more thing see oxygen here is in atomic form so instead of molar mass you can write gram atomic mass you can write gam at atomic mass of oxygen as well this is what you call as gram atomic mass of oxygen as well but you should know the meaning you should know the meaning of all these things this is the mass of one oxygen atom right perfect this is the atomic mass of oxygen perfect this 16 G this is not the mass of one atom of oxygen this is the mass of one mole of oxygen this is the mass of 6.022 into 10^ 23 oxygen atoms as simple as that right now look at all these terminologies oxygen hair is in molecular form so this particular term I'll be calling as molecular mass of oxygen molecular mass of oxygen this particular term without the units you'll be calling it as relative molecular mass of oxygen you'll be calling it as a relative molecular mass of oxygen this particular term you'll be calling this particular term as the actual mass of o2 actual mass of o2 and this particular term first of all you'll be calling it as mass of o2 mol mass of o2 or or or since oxygen is in molecular form you can call it as G molecular mass of oxygen as well you can call it as the gram molecular mass of oxygen as well am I clear with all these things say yes or no in the chats quickly quickly guys everyone I want everyone to say it are all these things clear to you are all these things clear to you say it in the chats quickly yes perfect one statement which I would want every one of you to remember from now onwards I'm taking a note of that statement I'm taking the note of that particular statement what is the standard which I've have chosen I have chosen 1 AMU as the standard and which we defined as 1 12th of the mass of carbon to atom which we defined as 1 12th of the mass of carbon to atom my dear students remember and directly remember no need of explanation etc etc remember directly on changing the standard on changing the reference on changing the standard on changing the reference on changing the standard on changing the reference for example if I'm not defining AMU as 112th of mass of carbon if I'm defining it as 1 16 of mass of something 124th of mass of something that mean I'm changing the standard when you change the standard when you change the standard remember remember directly actual mass of the substance does not change actual mass of the substance remains unchanged remains unchanged remains unchanged do remember this particular point from now onwards right perfect perfect now my dear students the things which I've told you till now let's see what kind of question can be asked out of these things the first type of the question which can be asked is on your screen look at the question carefully and let's try to solve this the question is saying if one U one u means 1 AMU if one U is defined as 124th of the mass of carbon to atom that means we are changing the standard we are changing the standard calculate the molecular mass of carbon dioxide on new scale understand how exactly you'll be approaching to these sort of questions remember them directly my dear students if you remember we had defined one U as 12th of the mass of carbon to Atom 1 12th of the mass of carbon to atom this is how we had defined one U right as for the question how you are defining one U which I'm calling as one U Dash as for the question one U is defined as 124th 124th of the mass of carbon Toom right yes now people understand this scale which I've chosen over here this is the old scale conventional scale which we have defined and this particular scale is the new scale that is given to us as an equation that is given to us as the question question what are we supposed to calculate we are supposed to calculate the molecular mass we are supposed to calculate the molecular mass of carbon dioxide on new scale we have to calculate molecular mass of carbon dioxide on new scale my dear students the first thing which I'll be doing always I'll be dividing these two equations why do I need to divide these two equations I just need to get the relation between U and U Dash I just need to get the relation between U and U Dash perfect I'll be dividing these two equations equation one divide by equation two so it becomes U divid by U Dash uid by U Das is equal mass of carbon 12 mass of carbon 12 cancel so 1X 12 divid 1 by4 the value comes out be two so U is nothing but 2 * u- U is nothing but U * 2 * U Dash Guys these are the standard questions these are the standard questions that can be asked the first thing I got the relation between U and U Dash I got the relation between old scale and new scale perfect number one number two number two if I ask you if I ask you what is the molecular mass of carbon dioxide on Old scale molecular mass of carbon dioxide which we know 12 * 1 + 16 multiped by 2 that is 44 U this is something which we know but do we have to get the answer in terms of U no we have to get the answer in terms of U Dash in terms of U Dash so I can write it as 44 * 1 U and 1 U Already is equal to what it's equal to * u- the value comes out to be 88 u- so what is this 88 u- this is the molecular mass of carbon dioxide on new scale this is the molecular mass of carbon dioxide on new scale right this is something which we were supposed to calculate done and dusted right now if I ask you one thing can you tell me can you tell me the actual mass of carbon dioxide on new scale I'm just asking you this question directly can you can you tell me the actual mass of carbon dioxide on new scale quickly actual mass of carbon dioxide on a new scale I told you already I told you already I told you already actual mass does not change when you change the standard actual mass will still remain the same what was the actual Mass actual mass was 44 * 1.66 into 10^ - 24 G this was the actual mass of carbon dioxide on Old scale on new scale it will be the same you did not to check anything it remains unchanged it remains unchanged my dear students quickly tell me the answer of this particular question quickly quickly we don't have a lot of time quickly tell me the answer of this particular question be very quick be very quick be very quick be very quick and I'm damn sure you can easily solve this now be very quick you also solve it I'll also solve it as per the question first of all we know already 1 U is nothing but 1 12th of the mass of carbon to this is something which we already know this are old scale now as per the question the new scale one u-h is defined as 1 16 of the mass of carbon to Atom 1 six of the mass of carbon to atom this is your new scale first of all I need to divide them up so u/ U Dash uid U Das this divid by this will be equal 1id 2 or U will be equal u-/ 2 this is the relation which I got between U and u-h done understood now after this after this after this calculate the molecular mass of NH3 molecular mass of NH3 on Old scale is what 17 U but I do not need to write the answer in terms of uu I have to write the answer in terms of U Dash so 17 multiplied by instead of U I can write U Dash divid by 2 so this is 8.5 u-h this is the molecular mass of NH3 on the new scale is it done and dusted yes is it done and dusted people say it in the chats say it in the chats perfectly done moving ahead moving ahead can you let me know the answer of this simple question which is on your screen the mass of one atom of X we are given with the mass of one atom of some element X mass of one atom is given calculate its molar mass can you can't you solve it I told you already molar mass of the substance molar mass of the substance and substance here is X molar mass of X can be written as can be written as if you remember I told you mass of one atom of X mass of one atom of X multiplied by aatos number multiplied by aad number which I'm roughly writing as 6 into 10^ 22 6.022 into 10 23 right I told you I told you that already molar mass of substance is equal mass of one atom of substance multiplied by the G number now what is the mass of one atom of X it's given to me as 4 into 10 this - 23 here by the way 4 and 10us 23 g multiplied by what multiplied by 6 into 10^ 23 so this term this term cancel 6 4 are 24 the value comes out to be 24 G this is the m mass of X now if in the question if in the question let's assume they are not asking you directly the molar mass of X they're asking you identify the element X and your options are sodium your options are magnesium your options are chlorine your options are is carbon which is going to be the correct answer let's say they are not asking you calculate the M mass of X they asking you identify X you calculated mol mass of X and you know 24 G that is basically the M mass of magnesium 24 G that is the m mass of magnesium so with like this also the question can be asked pretty much simple yeah perfectly done so be careful with all these things mov ahead mov ahead have you heard about something called as Isotopes have you heard about something called as isot otop what are isotopes my dear students what are isotopes you should be knowing it Isotopes these are the atoms these are the atoms of what these are the atoms of same element these are the atoms of same element having same atomic number having same atomic number but different but different atomic masses but different atomic masses yes something which you all must be knowing same atomic number different atomic masses for example for example for example if I talk about chlorine if I talk about chlorine you should be knowing chlorine chlorine has got two atoms chlorine has got two atoms which have got same atomic number but different atomic masses 35 37 chlorine has got two atoms which have got same atomic number but different atomic masses so should I be calling these two as the Isotopes of chlorine absolutely these two are called as isotopes of chlorine right hydrogen it has three Isotopes right h11 h12 h13 cium duum titium right you should be knowing now my de students in general if I write something like this try to understand in general I'm writing something like this let's say you have the element X let's say you the element X or let me write it in the middle imagine that you got the element X I'm assuming this x has got three Isotopes this x has got three isotopes that means this x has got three atoms which have got same atomic number but their mass number is different this is M1 this is M2 this is M3 right so I should be calling these three as the Isotopes of this X and let me tell you whenever element has got Isotopes whenever an element exists in the form of Isotopes at that point of time what do we do we do not calculate the normal atomic mass of element we calculate the average atomic mass of the element so when do we Define the average atomic of the element when do we Define it it is defined when the element exists in the form of Isotopes right now people be careful with what I say we have got the element X it has got Isotopes whenever element exists in the form of Isotopes we do not calculate the normal atomic mass of the element what do we calculate we calculate the average atomic mass now in order to calculate the average atomic mass of the element what do we need what do we need remember we need the percentage abundances we need the percentage abundances of these Isotopes what are percentage abundances what is meant by that you'll get the idea in some time first you remember in order to calculate average atopic Mass what do we need we need we need percentage abundances of these Isotopes or I can say we need mole percentage of these isotopes we need Mo percentage of these Isotopes I'm assuming the percentage abundances of these Isotopes are nothing but a% b% and c% once you know the values of a b and c you can get the formula over here and the formula is pretty much simple M1 a + M2 b + M3 C divided by what divided by a + B+ C this is one General result by means of which you can calculate the average atomic mass of an element which exists in the form of Isotopes as simple as that yeah right and people do you remember one thing the sum of percentage abundances the sum of percentage abundances of all the Isotopes of a particular element that's always equal to what that's always equal to 100 now let's try to solve an example so that you understand what it means one very simple and basic example I'm giving you try to understand carefully my dear students for example I'm taking the element Chlorine I'm taking the element Chlorine understand carefully I'm taking the element Chlorine chlorine you know it has got two isotopes one is cl35 one is CL 37 one is CL 37 perfect I'm am telling you that the percentage abundances of cl35 and cl 37 it is 75 and 25 respectively 75 and 25 respectively right now I'm asking you to calculate the average atomic mass of chlorine can you do that can you do that you should be in a position to do it it's pretty much simple see this 35 is basically the value of M1 this is the value of M2 this is the value of a this is the value of B all the terms are in front of you done understood right done and dusted so you'll be using the result M1 a + M2 B / a plus b that's always 100 right M1 is given M2 is given a is given B is given when you solve this the value comes out to be 35.5 Amu so what is the 35.5 U this is the average atomic mass of chlorine this is the average atomic mass of chlorine right yes am I clear am I clear people am I clear people say yes or no in the chats perfectly done all right all right people got it perfect this is the question which has been asked many a times this was asked once in J Main's examination see what kind of question was asked don't you kill it can't you kill it people boron has two isotopes this and this what is the average atomic weight of boron this is something which we are supposed to calculate is it it simple right students Boron it has got two isotopes it has got two isotopes one is your Boron 10 one is your Boron 11 percentage abundances are given given to us this is 19 and this is 81 perfect so M1 is given M2 is given a is given B is given I just need to write the formula and formula you know yeah and formula you know so I'll write average atomic mass of boron is nothing but M1 a + M2 B divide what divide by a plus b all these things are given when you solve it the value comes out to be 10.81 AMU the answer of this particular question is going to be option A I believe all these sort of questions you can solve easily from now on yeah perfectly done perfectly done people now how do we calculate the number of moles that's important how do we calculate the number of moles how do we calculate the number of moles if I ask you my dear students the answer is again very simple you know the formulas already I just need to write the formulas and I'll show you the application try to understand carefully number of moles of the substance which I'm representing with n number of moles of the substance which I'm representing with n or let me write it number of moles of a substance number of moles of a substance I'm representing this term with n how do we calculate the first Formula which we have given mass of the substance in GRS given mass of the substance in Gs divided by what divided by molar mass of the substance which is expressed either in grams or you can say G per mole this is the first Formula which we have to calculate number of moles given Mass divide by m mass number one number two number of moles of the substance is always equal given number of particles given number of particles given number of particles of the substance divided by avagadro's number divided by avagadro number second formula third formula which is only valid for gases which is only valid for gases be careful with this this only valid for gases and the gases should be present at STP standard temperature pressure and the formula is it is given volume of the gas in lers given volume of the gas in liters divided by 22.4 l these are the three basic formulas which we have to calculate number of moles or sometimes you can use ideal gas equation to calculate number of moles of the gas which is n is equal pv/ RT right this can be also sometimes used so these are some basic results by means of which we calculate number of moles of a particular substance for example I'm just giving you certain basic questions first of all slowly will rise the level right for example these are the basic questions which we have can't you solve them how do you solve them people calculate the following calculate the number of moles in 160 G of no look carefully I'm given with mass of no I'm supposed to calculate moles I'm supposed to calculate moles that will be equal given mass of Na in G divided by molar mass of n when you calculate 23 + 16 + 1 value comes out to be 40 and 40 G per mole the value comes out to be four so that means 160 G of NA means what 160 of no means 4 moles of no done and dusted right look at the second one 67.2 L of carbon dioxide at STP we are given with the volume of the gas we are given with the volume of the gas we are supposed to calculate moles of the gas how do we calculate moles when volume is given given volume of the gas in liters divided by 22.4 right the value comes out be three so 67.2 L of carbon dioxide at STP means that we are given with three moles of carbon dioxide as simple as that yeah look at the next one number of moles in 24 into 10^ 23 atoms of carbon look carefully we are given with the particles we are given with the particles we are supposed to calculate moles so I'll say number of moles of carbon is equal given number of particles given number of particles divide by what divide by avagadro's number given number of particles divide by avagadro's number the value is four so basically 24 into 10 23 atoms of carbon means that we have got four moles of carbon as simple as that done understood right look at the next one people number of moles in these many molecules of water again we are given with particles again we are given with particles so calculate moles moles of water will be equal given number of particles given number of particles divide by what divide by avad number right the value comes out to be 68 this is 80 perfect these are the simple results three four basic results by means of which you can calculate moles of particular substance right am I clear with all these things I should be yeah right people now now understand now understand over here I'm writing a term number of moles I'm writing a term number of moles which is represent by n i students how many ways do we have how many ways do we have to calculate number of moles of a particular substance you tell me how many wavs do we have either I'll be given with mass of substance or I'll be given with particles of the substance or I'll be given with volume of the gas there are three ways by means of which I can calculate moles right yes now people let me tell you in order to calculate moles either mass of the substance should be given to us either mass of the substance should be given to us number one number two or number of particles of the substance should be given to us number of particles of the substance should be given to us or or or I would say volume of the gas volume of the gas in liters at STP should be given to us perfect in order to calculate moles these are the things which should be given to us either mass or volume or particles right now you know already now you know already if we are given the mass of the substance and we are supposed to calculate moles if we are given the mass of the substance and we are supposed to calculate moles what do we need to do given mass of the substance divided by m mass of the substance what do we get we get moles if I reverse the same if I reverse the same if moles of the substance are given if moles of the substance are given and we are supposed to calculate mass of the same substance what do we need to do moles of the substance multiplied by m mass of the same substance I'll be getting the mass I'll show its application in some time now if number of particles are given if number of particles are given if number of particles are given and you are supposed to calculate moles given number of particles divided by aad number or if you reverse it if moles are given and you are supposed to calculate particles multip by aad number simple right if volume is given you are supposed to calculate moles divide by 22.4 L right if moles are given and you are supposed to calculate volume multiplied by 22.4 L this is the simple mole conversion diagram which you need to remember now people now people now people if I ask you something like this understand very carefully for example you are given with the mass of substance you are given with the mass of substance and you are supposed to calculate volume you are given with mass let's say you are given with mass you're supposed to calculate volume of the substance is there any direct result between mass and volume no what do we need to do what do we need to do if mass of the substance is given and we are supposed to calculate volume there is no result so what do we need to do we need to follow the arrows we need to follow the arrows which take us from Mass till volume we need to follow the arrows which take us from Mass till volume now which all arrows take us from mass to volume so first I'll be following this particular Arrow given mass divid by m Mass I'll be getting moles then I'll be following this Arrow moles multiplied 22.4 I'll be getting volume as simple as that right for example there's a question in which mass is given and we are supposed to calculate particles mass is given and particles are to be calculated for example mass is given particles are to be calculated there is no direct Arrow what do I need to do I need to follow all the arrows which take me from Mass still particles and which all arrows are those this is the first Arrow so given Mass divide by m Mass gives me moles then once you calculate moles multiply by that gives me particles I believe you can do that I believe you can do equations on this simple for example for example for example one two questions I'll solve with you with you for the related to this diagram let me see if we can solve this or not can you do this calculate the mass of 67.2 L we are given with volume we are supposed to calculate mass we are given with volume we are supposed to calculate mass we are given with volume we are given with volume and we are supposed to calculate mass right so this is given this to be calculated so I'll start from here and I'll follow all the arrows which take me from volume till Mass so if I ask you which all arrows I need to follow my dear students this is the first Arrow which I'll be following and this is the second Arrow which will be following so question will be done and dusted with that right so what is the given volume what is the given volume 67.2 l so 67.2 L divide by 22.4 the value comes out be three so I got number of moles as three three multiplied by m mass m mass of what M mass of carbon dioxide so 3 multiplied by mol mass of carbon dioxide is how much 44 G so 3 multi 44 comes out to be 132 G so 132 G is the answer of the question perfect so I would say the mass of 67.2 L of carbon dioxide ATP is how much 132 G as simple as that done understood right yeah now similarly if I ask you calculate the molecules present in 440 G 440 G of carbon dioxide my dear students in this question mass is given particles are to be calculated mass is given particles are to be calculated so you'll start from Mass and you'll reach till particles so first step will be number of moles of carbon dioxide will be equal given mass of carbon dioxide in Gams divide by mol mass of carbon dioxide the value comes out be 10 so you got the moles right now number of molecules of carbon dioxide number of molecules of carbon dioxide the second step is moles are to be multiplied by the avad number moles are to be multiplied by the avad number moles are 10 multiplied by avad number so these many molecules are present in 440 G of carbon dioxide am I clear 10 na is the final answer of the question I believe you can do it I believe you can do it from now on easily guys I want you to solve this question look at the question carefully once look at the question carefully once can you solve it just say it just say it just say it can you solve it I just want to know can you solve it yes or no yes or no yes or no people yes or no yeah you should be in a position to solve it the question is pretty much simple calculate the number of moles present in calculate the number of moles present in 96 kg of X so we are given with mass of X we are supposed to calculate moles we are given with mass of X we are supposed to calculate moles so number of moles of X is equal mass of X in G which is 96,000 G given Mass divide by mol mass of X divide by m mass of X now people if I ask you are we somewhere given with the m mass of X I don't think so molar mass of X is not given to us but but but mass of one atom of X is given to us if you know the mass of one atom of X just multiply it with avagadro's just multiply this term 4 into 10-23 G just multiply it with avagadro's number what do I get what do I get the value comes out be 24 G 24 G is a m mass of X something which I told you already right so divide this 96,000 with what with 24 G per mole the value comes out be how much quick quick people are saying 4,000 absolutely you guys are perfectly right done and dusted right done underst dusted done and dusted guys if I ask you a question like this have you seen this question this question was asked in J long back but it's a proper neat standard question can you kill this one how many moles of electron so you supposed to calculate moles how many moles of electron V is 1 kg so you are given with mass you have to calculate moles you are given with mass you have to calculate moles so number of moles of electrons is equal mass of electrons in Gs divide divided by m mass of electron divided by mass of electron divided by mass of electron mass of electron in Gams is 1 kg which means 1,000 G molar mass of electron means mass of one electron which is 9.1 into 10us 28 G multiplied 6.022 into 10^ 23 this becomes G per mole just the matter of calculation and you guys are done just the matter of calculation and you guys are done yeah perfectly done people perfectly done now one more type of question that will be asked one more type of question that will be asked calculation of atoms electrons protons neutrons ions this you would have seen frequently how many electrons are present in 67.2 L of carbon dioxide how many valence electrons are present in 4.2 G of ENT negative which was asked 3 years back in meat right these sort of questions you would have I mean you would have seen questions based on these atoms electrons protons Etc now in order to solve this set of questions in order to solve this set of questions I would want to give you one diagram which you will be remembering that's all one diagram and we are done and we are done understand properly what exactly I'm going to say my dear students here I'm writing something as given quantity given quantity given quantity what whatever will be given to us as for the question be it mass of the substance be it particles be it volume whatever is given to us you'll be calling that as given quantity whatever is given in the question your first step always is to convert the given quantity into number of moles your first step always is going to you will be converting the given quantity into number of moles right you'll be converting the given quantity number of moles once we get the moles we'll be multiplying moles by the avagadro's number will be multiplying moles by the avad number the substance whose quantity is given the substance whose quantity is given if that substance is in molecular form if that substance is in molecular form then at that point of time moles multiplied by avagadro's number it gives me the number of molecules it gives me the number of molecules once we get the molecules we are done we can calculate atoms we can calculate electrons we can calculate protons we can calculate neutrons etc etc this diagram I would want every one of you to remember and follow directly from now onwards now people how do we apply this form apply this diagram yeah how do we how do we apply it how do we apply it for that I'll be taking one simple question which will make you understand all these things properly for example one basic question one basic question calculate the total number of atoms in 44 sorry 448 L calculate the total number of atoms in 448 L of carbon dioxide at what at STP this the question what are we supposed to calculate be careful calculate the atoms calculate the atoms and whenever we are supposed to calculate atoms we are supposed to follow this diagram the first step convert the given quantity into moles moles multipli by's number you get the molecules once you get the molecules you can calculate atoms done and dusted see see people the first step this is my given quantity I'll be converting the given quantity in mol so number of moles of carbon dioxide is equal to given volume of carbon dioxide in liters divided by 22.4 perfect the value comes out to be 20 perfect I got the moles my first step is done now the given substance is in molecular form given substance is in molecular form Second Step moles multiplied by V number moles multiplied by V number that gives me molecules right so number of molecules number of molecules of what of carbon dioxide will be equal 20 multipli G number these many molecules we have so these many molecules of carbon dioxide we have you getting it we are given with 448 L of carbon dioxide I'll say in 448 lers of carbon dioxide we'll have these many molecules of carbon dioxide now what are we supposed to calculate atoms if I ask you tell me how many atoms are present in one molecule of carbon dioxide in one molecule of carbon dioxide there are three atoms present right you know it in one molecule of carbon dioxide there are three atoms present now people we are done one molecule of carbon dioxide contains three atoms two molecules will contain multipli by two three molecules multiplied by three four molecules multiplied by four how many molecules do we have 20 na so 20 na molecules of carbon dioxide will contain 3 * 20 na the value comes 60 na atoms so there are in total 60 na atoms present in 448 L of carbon dioxide am I clear am I clear people am I clear am I clear am I clear say it say it in the chats say it in the chats quickly say it in the chats quickly yeah this your ask question calculate the valence electrons calculate the valence electrons in 4.2 G of Entry negative be careful with what I say ENT TR negative is it in molecular form or ionic form is it in molecular form or ionic form this is in ionic form this is an ionic form so once I calculate its moles If I multiply those moles by aatos number will I get molecules or I'll get ions directly you say it you say it first of all I'll be calculating it moles once I calculate moles once I multiply those moles by number what will I get will I get molecules or ions I directly get the I'll directly get the I right so my first point convert the given quantity into moles so number of moles of nry negative is equal given mass of ent2 divided by m mass of ENT ne2 right the value will be 0.3 perfect so you got the moles now moles multipli by the G number that gives me number of ions directly number of ions of Entry negative will be equal moles multiply by so these many ions we have these many ions we have now comes one point what do we have to calculate valence electrons valence electrons is tell me one thing if I talk about nitrogen whose atomic number is seven one atom of nitrogen will contain how many electrons seven electrons one atom of nitrogen contains seven electrons but this is TR negative so it has gained three more so in total there are 10 electrons present in one and tri negative in one and tri negative there are in total 10 electrons present now out of these 10 electrons can I say two electrons will be in the first shell eight electrons will be in the second shell perfect two electrons in the first one eight in the second one so this is my valence shell where in eight electrons are present this is my valence shell wherein eight electrons are present so if I ask you if I ask you in one ENT Tri negative in one andry negative how many valence electrons are present there are eight valence electrons present in one entry negative so if you have two entry negative multip by two three entry negative multip by 3 how many do we have 0.3 na so in 0.3 na andry negatives how many valence electrons will be there 8 multiplied by 0.3 na valence electrons which will come out to be how much 2.4 na valence electrons so these are the total valence electrons which are present in 4.2 G of nry negative yes right people right am I clear say it in the chats quickly am I clear am I clear it's just give me a second just give me a second perfect guys perfect perfect perfect I believe all these things said okay one question wherein one question wherein you might face some problems one simple question these are three questions basic questions these are but I believe there'll be some students who will fall in trap these are one-step questions these are one-step questions calculate the number of atoms in 16 G of heat heli atoms we have to calculate atoms so I'll be following this diagram which I gave you first step convert the given quantity into moles the number of moles of helium is equal given mass of helium in grams divided by m mass of helium right the value comes out to be four right now helium is in atomic form helium is in atomic form so moles of helium multip by aad number what that gives us that does not give us molecules that gives us atoms because this is an atomic form right so number of atoms of helium number of atoms of helium will be equal to what it'll be four multiplied by av's number four multip done this is something which I was supposed to calculate number of atoms in 16 G of helium done underst dusted right the second one the second one atoms in 16 U of helium 16 U 16 U 16 U 16 U it is not 16 G here so you cannot use the result given mass in grams divide M Mass because this is 16 U how do we do this understand 16 U we are given with 16 U of helium right tell me one thing what is the atomic mass of helium 4 U 4 U 4 U is basically I would I would say 4 U is the mass of one atom of helium correct 4 U is the mass of one atom of helium now tell me one U will be 1X 4 atoms of helium I just divided this equation by four 1 U is equal to 1x 4 atoms of helium am I given with 1 U or 16 U 16 U 16 U will be 1X 4 * 16 right which will be four atoms of helium perfect so 16 U is nothing but four atoms of helium it's simple 4 U is equal one atom so multip four 16 U is equal 4 atoms correct no need to thank a lot okay molecules in one 26 U of hno3 molecules in 126 U of hno3 now you know 63 U is the mass of one molecule of hno3 is the mass of one molecule so if I multiply this equation by two it becomes 126 U will be the mass of two molecules of hno3 right so molecules 126 u means two molecules of H3 am I clear say it am I clear am I clear people quickly quickly say it I want you guys to say it quickly quickly quickly yeah right perfectly done let's go ahead and let's talk about something called as stomry right something which you hate something which you hate I believe do you no I'll make it simpler for you try to understand what exactly I'm going to do my dear students the toeter it is almost used everywhere when it comes to your physical chemistry right this is used everywhere now the real mole concept starts yeah exactly okay story symetry first of all in order to make you understand what the stoeter is how to apply this and what kind of questions are asked I'm going to write a general equation N2 gas plus H2 gas what does it give it gives NH3 G this is a general chemical equation right general chemical equation if I ask you is this reaction balanced or not is the reaction balanced or not the reaction is not at all balanced so first of all first of all whenever whenever a question is asked in St symetry the first thing which we need to do is to balance the reaction is to balance the reaction so balance the reaction this has to be three times this has to be two times and this is one balanced my dear students after balancing the reaction after balancing the reaction this one three and two these are called as tetric coefficients these are called as tetric coefficients now after balancing the reaction I can treat these tetric coefficients as moles and how exactly I can say as for the stet coefficients of reactants and products we can say from stoeter we can say 1 mole N2 completely reacts with 3 moles of H2 and produces 2 mol of NH3 and produces 2 mol of NH3 this is something this statement this information we got from the stet coefficients we got to know one Mo of N2 in this reaction one mole of N2 completely reacts with three moles of H2 and produces what produces two moles of NH3 right now if I ask you if one mole of N2 is reacting with three moles tell me what about two moles of N2 can I say two moles of N2 will react with six moles of H2 and will produce four moles of NH3 it is simple as if I multipli this equation by two right simple simple guys yes right Point number one point number two point number two here in this particular statement I wrote all the terms in terms of moles so basically which method I'm using right now I'm using the mole method in sty symmetry I'm using the mole method number one number two my dear students if I ask you what is the m mass of N2 molar mass of N2 will be 28 G what does that mean that means mass of one mole of N2 is 28 G am I given with one mole of N2 only yes so what is the mass of 1 mole of N2 you know it already it's 28 G so instead of 1 Mo of N2 I wrote 28 G of N2 same now what is the m mass of H2 2 g m mass of H2 is 2 G that means mass of one mole of H2 is 2 G but I'm get with three moles so what is the mass of three moles of H2 3 2 are 6 so I'll write 6 G of H2 gives what is the molar mass of NH3 17 G that means mass of one mole of NH3 is 17 G therefore mass of two moles of NH3 will be 34 G so 34 G of NH3 this particular statement I wrote only in terms of mass so I can say I'm using mass method here in STO material right now yeah now people I can treat these stoet coefficients directly as volume as well I can say 1 L of N2 completely reacts with 3 l of H2 and produces what produces 2 L of NH3 it is not necessary that you take volume in liters you can take volume in ml as well you can say 1 ml of N2 completely reacts with 3 ml of H2 and produces 2 ml of NH3 done IND dusted right so first of all you should be in a position you should be in a position to write all these three statements for a particular equation whenever you are given with a chemical equation perfect you should be in a position to write all these three statements be it in terms of moles be it in terms of mass be it in terms of volume and my dear students always always will be trying to solve the questions with the help of mole method no need to remember all the methods just remember mole method and how do we solve the equations with the help of mole method before letting you know how do we solve the equations with the help of mole method I would want to give you certain reactions which are frequently Asked certain reactions number one number one remember it K3 K cl3 when it is heated it gets converted into KCl with that you get O2 gas this one reaction which you have to remember number two calcium carbonate upon heating gives calcium oxide plus carbon dioxide number three number three acid plus base acid plus base what does it give it gives salt and water it gives salt and water okay it gives salt and water number four whenever you have got the hydrocarbon like this CX Hy CX it can be methane CH4 c26 whatever right when you do its combustion it produces always carbon dioxide and water there is one General way to balance this particular equation see this is X this can be X here this is y it has to be y by 2 and the stric efficient of oxygen has to be x + y by 4 x + y by 4 right similarly similarly if you have got a compound like this CX H yo Z like glucose C6 h206 if you do it's combustion it will again lead to the formation of carbon dioxide and water only how do we balance this type of the reaction this is X here so make it X this is y here so make it y by 2 and make this as x + y by 4us Z by 2us Z by 2 this is how you balance this particular reaction this is how you balance this particular reaction okay perfect similarly guys I want you to remember one more thing sodium bicarbonate nco3 when you heat it up it gives sodium carbonate plus water plus carbon dioxide this reaction also you would have seen somewhere right similarly one more reaction uh bac2 Plus na3 po4 plus na3 P4 what does it give it gives ba3 P4 whole twice plus na this one more reaction which you need to remember which you need to remember now my dear students let's try to see the questions which can be asked I'll be starting from the basic question from zero I'm starting basic I'll be trying to solve all the equations with the help of which method I'll be trying to solve all the equations with the help of mole method the equation is calculate the mass of o2 produced on heating 12.25 G of kcl3 this is the question so imagine that you got a container and in that container you have kept 12.25 G of K3 in that container you have kept 12.25 G of K3 now you are heating that you heating 12.25 G of kcl3 calculate the mass of o2 that will be produced since you know since you know when K3 is heated when kcl3 is heated it gives you what it gives you KCl and with KCl you get O2 gas right so O2 gas is produced through this particular reaction you know it yeah now I'll be solving all the equations with the help of mole method in the mole method the first thing which you need to do is to balance the reaction so try to balance the reaction try to balance the reaction reaction is balanced after balancing the reaction convert the given quantity into number of moles convert the given quantity number of moles I am given with the mass of K3 convert the mass of K3 into into moles of K3 moles of K3 is equal given mass of K3 in G divided by m mass of K3 divided by m mass of K3 the value comes out to be 0.1 so as per the question as per the question we are heating 12.25 G of K3 that means we are heating 0.1 moles of K3 we are hitting 0.1 moles of K3 now guys now guys this is the one whose data is given this is the one whose data is to be calculated the one whose data is given the one whose data is calculated I just have to do I just have to relate these two I've got nothing to do with KCl here I can say as per stoet I can say from stoet I got to know 2 mol of K3 upon heating 2 moles of K3 upon heating produces 3 moles of o2 produces 3 moles of o2 2 mol of K3 upon heating produces 3 mol of o2 from stoet therefore 1 Mo K3 1 Mo K3 upon heating 1 mole K3 upon heating produces 3x2 mol of o2 if I use the unitary method 1 mole of K3 upon heating produces these many moles of o2 but as per the question do I have to heat 1 mole of K3 no I need to heat 0.1 mol of K3 so I can say 0.1 mol of K cl3 upon heating reduces 3x2 multipli by 0.1 the value comes out to be 0.15 moles of o2 so these many moles of o2 are produced these many moles of o2 are produced right these many moles of o2 are produced but am I supposed to calculate moles of o2 no I'm supposed to calculate mass of o2 how do you convert moles into mass mass of o2 will be equal moles of o2 multipli by m mass of o2 as simple as that this is 0.15 and this is 32 the value comes out to be 4.8 G so I would say 4.8 G of o2 are produced when we do the decomposition of 12.25 G of kcl3 am I clear am I clear people St it in the chats am I clear am I clear one more thing one more thing remember it directly one more thing one more thing people I wrote a statement over here in this format the one whose data is given and the one whose data is to be calculated they are to be related they are to be related and remember the one whose data is to be calculated the one whose data is to be calculated that has to be on the right side of the statement always the one whose data is to be calculated that has to be on the right side of the statement always always okay in this question I was supposed to calculate the mass of oxygen so O2 is the one whose data was to be calculated that's why it's on right side if I was supposed to calculate something about kcl3 then I was supposed to reverse the statement I believe it's clear I believe it's clear one more question simple question understand calculate the volume of carbon dioxide produced at STP by complete combustion of 570 G of octane now tell me the formula of octane first of all C8 h18 this your octane you are doing the combustion of octane that means you are making it to react with what you are making it to react with O2 it will lead to the formation of carbon dioxide and with carbon dioxide there is water with carbon dioxide there is water with carbon dioxide there is water you know it now my dear students tell me one thing tell me one thing tell me one thing which method we are going to use mole method in the the first thing which we do in the mole method we balance the reaction and I've already told you how do we balance it this is X x is 8 so this is 8 Y is 18 y by 2 makes it 9 x + y by 4 makes it 25 by2 simple after this convert the given quantity into number of moles what is my given quantity check it I'm given with mass of octane I'm given with mass of octane so I'll be calculating number of moles of octane which will be equal mass of octane in Gs divided by molar mass of octane the value comes out to be five so as for the question as for the question we are doing the combustion of 570 G of octane what does that mean that means we are doing the combustion of five moles of octane we are doing the combustion of five moles of octane that means we have to do the combustion of five moles of octane basically now the one whose data is given and the one whose data is to be calculated they are to be related that's it they are to be related perfect and the whose data is to be calculated that has to be on the right side of the statement I'll write the statement from stoet I know one mole octane when it under goes combustion it produces 8 moles of carbon dioxide it produces 8 moles of carbon dioxide as for the question do I have to do the combustion of one mole of octane no I have to do the combustion of five moles of octane so when five moles of octane under goes combustion when 5 moles of octane under goes combustion I'll say 8 multipli 5 40 mol of carbon dioxide are produced 40 mol of carbon dioxide are produced but am I supposed to calculate moles of carbon dioxide produced no I'm supposed to calculate volume of carbon dioxide how do we convert moles into volume volume of carbon dioxide at STP is equal moles multiplied by 22.4 L the value comes out to be 896 l so I would say 896 L of carbon dioxide will be produced when we do the combustion of 570 G of octane when we do the combustion of 5 moles of octane if I'm clear let me know once in the chats everyone everyone people Everyone quickly yeah done I want you to say it in the chats guys please and please do not sleep I do not want you guys to sleep I do not want you guys to sleep one more question with this and then we'll move ahead can you give this question a try can you give this question a try guys look at the question carefully simple question again again one simple question you are given with the reaction and the reaction is given in the balance format how many GRS of NO2 how many GRS of NO2 how many grams of NO2 are required to produce 25.2 G of hno3 the reaction is balanced the first thing convert the given quantity into number of moles right we are given the mass of H3 so we'll calculate the moles of H3 first which will be equal given mass of H3 in G divided by mol mass of H3 that is 63 G per mole what is the value coming out after solving this how much it comes out be how much it comes out be I think it'll be 0.4 approximately right it'll be approximately 0.4 right I hope so it is 0.4 correct it'll be 0.4 so I converted the given quantity to moles right now now I just have to relate these two things and the one whose data is to be calculated that has to be on the right side so I'll write the statement like this I'll say from the storage symmetry 2 moles of hno3 are produced from 3 moles of NO2 2 moles of hno3 are produced are produced from 3 moles of NO2 therefore 1 mole hno3 1 mole hno3 is produced from 3x 2 mol of NO2 therefore 0.4 mol of hno3 as per the equation 0.4 mol of H3 will be produced from 3x2 multiplied 0.4 mol of NO2 the value comes out be how much 0.6 mol of NO2 so as for the question we got to know 0.6 mol of NO2 are required to produce 0.4 mol of H3 now am I supposed to calculate moles of NO2 no what what am I supposed to calculate I'm supposed to calculate mass of NO2 how do you convert moles into Mass how do you convert moles into mass mass of NO2 is equal moles of NO2 multipli by mol mass of NO2 which will be how much 46 right is this going to be 46 46 multiplied 0.6 approximately it will be 27.6 g 26.7 g of NO2 26 27.6 G of NO2 are required to produce 25.2 G of H I hope I'm clear someone is feeling sleepy right do the same in the examination Hall as well yeah it'll be good for you I hope people I'm clear with all these things yeah yeah someone was feeling sleepy of course they should sleep right they should know exactly from 2: to 5:20 yeah you need to sleep do you have anything else to say if you're feeling sleepy is there anyone else who's feeling sleepy is there anyone else who is feeling sleepy is there anyone else who is ah someone is going to sleep after need that's good that's nice yes there'll be breaks in between just wait guys we just have studied like I mean we have started one and a half hour before even after Nate I I2 will sleep so let's have a look on one more important concept that is limiting reagent right this again one very important concept which you need to know what is limiting Regent first of all how do you define the term limiting Regent my dear students limiting Regent is basically defined as the reactant the reactant which gets the reactant which gets completely consumed the reactant which gets completely cons consumed earlier in the reaction completely consumed earlier in the reaction is what you call as limiting reagent and and the other one is called as excess reagent and the other one is called as the excess reagent now what is meant by that try to understand carefully what is limiting region because this is something which is very very very important my dear students for example for example you have got a balanced chemical equation like this A + B gives C plus d it is one balanced chemical equation imagine that okay let's say I'm writing at time T is equal to0 time T is equal to0 means the reaction has not started time T is equal to0 means reaction has not started reaction has not started means products products have not got formed yet products have not got formed yet right so initially at time T is equal to Z what do I have I'll be only having reactants with me imagine I've got a container in which this reaction has to take place at time T is equal Z before the start of the reaction there will be only reactant present you know it right let's say you have got 10 moles of reactant a and 5 moles of reactant B in the container initially and right now I'm assuming reaction has not started so this is zero this is zero whenever in a question you are given with the amounts of all the reactants whenever in equation you are given with the amounts of all the reactants at that point of time you'll be using the concept of limiting reagent okay here here you given with the amounts of all the reactants so over here I'll be using the concept of limiting region now my dear students my dear students when the reaction will start when the reaction will start when these reactants will start reacting and forming the products what will happen to the moles of reactants with time moles of reactant with time will decrease right can I see can I say out of these two reactants there will be one reactant which would have got completely consumed I'm assuming this B is the reactant which got completely consumed earlier in the reaction so you are left with zero moles of B that reactant which gets completely consumed earlier in the reaction that's something which you call as limiting reagent other one is what you call as excess reagent right and people it's clear it's clear when one of the reactant is consumed completely right this a this what I call as excess reagent that means out of 10 moles of a there will be some moles of a still left in the container similarly some moles of c and some moles of D would have got formed as well but tell me one thing when one of the reactant is completely consumed can I say at that point of the time the reaction would have stopped absolutely the reaction would have stopped absolutely the reaction would have stopped that particular reactor which gets completely consumed earlier in the reaction that's called as limiting reent other one is called as excess reagent excess reagent does not get completely consumed there will be some amount of excess reagent still left in the container right and there would be some products which which would have got formed as simple as that now how do we use this concept in the equations that's the point that is the point my dear students one question and every single thing will be done in that particular question for example for example for example I'm given with a reaction N2 + H2 gives 2 * NH3 understand what I say understand what I say see what kinds of questions can be asked imagine at time T is equal to Z I'm given with 10 moles of N2 I'm given with 10 moles of H2 and this is zero so first of all which concept do I need to use here I'm given with the amounts of all these reactants I'm given with the amounts of all these reacts I need to use the concept of limiting Regent and my dear students whenever we need to use the limiting Regent concept the first thing is balance the reaction reaction is already balanced reaction is already balanced right after balancing the reaction we are supposed to identify the limiting Regent we are supposed to identify the limiting Regent how do we identify the limiting Regent have a look see one simple drink one simple trick you have got N2 you have got s2c is one here is three how many moles of N2 are given 10 how many moles of H2 are given 10 divide the given moles of reactance divide the given moles of reactance by their respective stric coents by their respect to stric coefficients the one which gets the Lesser value that's always your limiting rate Regent so this is getting the Lesser value so H2 is my limiting Regent I hope you got to know how to identify right so H2 is my limiting Regent if H2 is my limiting Regent I'll be left with nothing of H2 the reaction would have stopped okay this is what you call as excess reagent perfect so there will be some moles of N2 which will be present in the container there will be some moles of NH3 which would have got formed now how to get to know how many moles of NH3 have got formed how many moles of N2 are still in the container how do we check that understand all the data all the data all the data will be given by the limiting region people have a look when I relate limiting reagent with NH3 I'll say p moles of limiting reagent produces 2 moles of NH3 limiting Regent is the one which decides everything three moles of limiting Regent produces two moles of NH3 therefore one mole of limiting gradient produces 2x3 moles of NH3 2x3 moles of NH3 one mole of limiting Regent produces these many moles of NH3 therefore 10 moles of limiting Regent therefore 10 moles of limiting Regent produces 2x3 * 10 the value is 6.66 moles of NH3 so I got to know at the end of the reaction 6.66 moles of NH3 would have got formed this is point number one point number two out of 10 moles of excess reagent how many moles of excess reagent will be left in the container how do we check that that see guys again limiting Regent is going to decide that relate the limiting Regent with N2 now right and how do I relate it I really like this I'll say three moles of limiting Regent three moles of limiting Regent reacts with reacts with one mole of N2 reacts with one mole of N2 so one mole of limiting reagent one mole of limiting Regent reacts with 1X 3 moles of N2 therefore 10 moles of limiting Regent 10 moles of limiting regent would have reacted with 1X 3 multiplied by 10 the value is 3.33 moles of N2 so I got to know 10 moles of limiting reagent have reacted with have reacted with 3.33 moles of an so 3.33 moles of I to have reacted how many did I take I took 10 out of 10 how many are reacting 3.33 so 10 minus 3.33 will be 6.66 so these are the moles of N2 which will be still left in the container as such right these are the moles of N2 which will be still left in the container as such so you got to know the moles of NH3 form you got to know moles of N2 left now if they do not ask you the data about moles let's say they're asking you calculate the molecules of NH3 formed calculate the molecules of NH3 formed you got the moles multip by number that will give you the molecules let's say they are asking you calculate the mass of NH3 formed you got the moles multiplied by m mass of NH3 that is mass of formed what is the mass of excess Regent left these are the moles of excess Regent left multiplied by its M Mass you'll be getting the mass of excess Regent left how many molecules of N2 are left how many molecules of N2 moles of N2 left multip number anything can be asked am I clear am I clear people am I clear people quick am I clear now I'm going to give you one question which you'll be solving and that's it and that's it tell me the answer of this question very simple and basic and I have given the reaction already in the balanced format quick be very quick now the reaction that's given that is 6 * LH + 8 * BF3 what does it give it gives 6 * l i BF4 plus what plus b26 plus B2 H6 as for the equation at time T is equal to Z we are starting with two moles of each of reactants we are starting with two moles of each of reactants how many moles of B2 H6 can be produced how many moles of B2 H6 can be produced the first thing is the reaction balanced yes the reaction is balanced now identify the limiting Regent how divide the given moles by the respective stet coefficients 2ide by 6 2ide by 8 whose value will come out to be less is 2x 6 less or 2 by 8 less I'll say 2 by 8 is less so this is the limiting region this is the one which is going to decide everything right this is the one which is going to decide everything what do we have to calculate moles of B2 H6 formed so just relate limiting Regent with b26 so I'll say 8 moles of limiting Regent 8 moles of limiting region gives 1 mole b26 8 moles of limiting region gives 1 M b26 therefore one mole limiting Regent gives what gives 1 by 8 moles of b26 1 by 8 moles of b26 therefore 2 moles of limiting Regent 2 moles of limiting Regent gives gives 1 by 8 multiplied by 2 the value is 1X 4 which will be 0.25 moles of b26 so these many moles of b26 would have got formed at the end of the reaction this is something which I was supposed to calculate am I clear am I clear people am I clear people say in the chats quickly everyone everyone in the chats everyone yeah so I hope you can do the questions when it comes to the topic limiting created I hope it's super clear to you right I hope it is super clear to you guys one question I want you guys to solve right now can you do it can you do this question I do not know whether I'll be doing thermodynamics in this Marathon or in the next marathon right let's see how much we can finish today and the rest of the things we'll be finishing the next marathon you need not to worry okay first of all can you do this question on your own can you do this question on your own quick done see guys we are given with 10 moles each of reactants 10 10 0 0 now identify the limiting Regent 10/ 3 10id 2 this is limiting what do we have to calculate calculate the moles of B3 P4 whole twice so relate limiting region with this three moles gives 1 mole 1 mole gives 1 by3 so 10 moles gives 10 by3 so the answer is 3.33 3.33 moles of barium phosphate would have got formed it is very simple and basic okay now people comes one more topic that is percentage yield have you heard about the term percentage yield this again important percentage yield before doing the questions I'll first of all give you the result calculate percentage yield result by means of which you can calculate the percentage yield remember the result then I'll show you how it's to be done see guys percentage yield is always equal experimental amount or the actual amount experimental amount or the actual amount divided by expected amount divided by expected amount multiplied by 100 this is the formula by means of which we calculate the percentage yield of the reaction okay this is the formula by means of which we calculate the percentage yield of the reaction remember this experimental yield it'll be most of the times given to you in the question itself right and expected amount we always calculate with the help of stomry expected amount we calculated with the help of stomry this will be given to us in the question right this we calculate with the help of stomry once we get the two amounts multiply with 100 we get the percentage yield for example for example for example this is the question this is the question we are given with a reaction that is calcium carbonate calcium carbonate upon heating reduces calcium oxide plus carbon dioxide this is the reaction which we have this is the reaction which we have as as per the question is concerned on heating 40 G of calcium carbonate 20 G of calcium oxide is produced this is something given to us in the question on heating 40 G of calcium carbonate 20 G of calcium oxide is produced so it is given to us in the question and in the question which data is given experimental data so it has been seen experimentally that when you heat 40 G of calcium carbonate 20 G of calcium carbon calcium oxide are produced so this has been observed experimentally so this 20 G it was observed 20 G of calcium oxide is produced as per the equation so your experimental amount is equal to what 20 G so I got one term experimental amount what do I need now what do I need now what do I need now I need expected amount how do I get the expected amount with the help of stet understand reaction is balanced reaction is balanced now let me use the mass method in stet tell me one thing what is the mass of 1 Mo of calcium carbonate the mass of 1 Mo of calcium carbonate means M mass of calcium carbonate which is how much 100 G what is the mass of 1 mole of calcium oxide that means M mass of calcium oxide 56 G so can I say as per St symmetry 100 G of calcium oxide sorry 100 G of calcium carbonate when you heat 100 G of calcium carbonate as per stet 56 G of calcium oxide should get formed yes as per tretry 100 G of calcium carbonate when heated should produce 56 G of calcium oxide so 1 G calcium carbonate upon heating should produce how much 56 ided 100 G of calcium oxide therefore 40 G of calcium carbonate 40 G of calcium carbonate upon heating should produce how much should produce how much these many G of calcium oxide this will come out to be 22.4 G of calcium oxide side so my dear students I used stomry and I got to know with the help of stomry that when you heat 40 G of calcium carbonate these many grams of calcium oxide should produce these many grams of calcium oxide should produce this is something which I got with the help of stet right so this is the amount of calcium oxide which I was expecting to get formed with the help of St chetry so this is something which I call as the expected amount correct now if I want to calculate percentage yield which will be experimental amount divide by expected amount multip 100 the value will be 89% so this is the percentage yield of the reaction this is the percentage yield of the reaction right my dear students try to understand try to understand one thing if the yield of the reaction was 100 if the yield of the reaction was 100% at that point of time whatever amount of calcium oxide you would have expected with the help of Storer same would have been the actual amount of calcium oxide formed in the lab if the yield of the reaction was 100% okay the yield of the reaction is 89% that means whatever you expected to get formed not the entire amount will get formed in the lab only 89% of this amount will get formed in the lab only 89% of this amount will get formed in the lab I hope I'm clear I hope I am clear people yes right let's solve one more question with the help of which it will be completely clear to you completely clear to you okay one more question on this one more question on this as for the question is concerned my dear students there's a reaction given and it's said that 15.6 G of benzene produces 18 G of nitro Benzene something which is given to us in the question in the question what do we get in the equation we get the experimental amount so it has been observed that when you heat 15.6 G of benzene you get 18 G of nitro Benzene so this is something which has been observed in the lab so I'll be calling this particular amount as experimental amount now what do I need I need expected amount I need expected amount what do I do for that I'll be using St symetry I'll be using St symetry to get the expected amount so this is one this is one the mass of one mole of benzene is how much 78 G which is the molar mass what is the mass of 1 Mo of nitro Benzene that means M mass of nitro Benzene I think it will be 123 G so with the help of stoet I can say 78 G of benzene should produce should produce 123 G of nitrobenzene therefore 1 G of benzene should produce how much 1 G of benzene should produce 123 divided 78 G of nitro Benzene and therefore 15.6 G of benzene 15.6 G of benzene should produce how much should produce how much multipli by 15.6 G of nitro benzine right this value will come out to be 24 G so as for stoet I got to know that 24 G of nitro Benzene should get produced should get produced so this is something which I was expecting to get formed this my expected amount how much is actually getting produced only 18 G of nitro Benzene now calculate the yield yield is experimental divide by expected multipli 100 solve it I think the value comes out be 7375 something like that I hope this is clear to everyone I hope this is clear to everyone right I hope this is clear to everyone now comes one more topic that is percentage Purity yes if the yield of the reaction is 100% at that point of time experimental and expected will be the same yeah all right percentage Purity percentage purity Purity let me first of all write the formula percentage purity of a compound in an impure sample in an impure sample is equal to is equal mass of the pure compound mass of the pure compound in grams divided by divided by the mass of impure sample mass of impure sample multiplied 100 now let me make you understand first of all what this percentage Purity is all about it's an important concept from which a question was asked last year as well right a question was asked last year as well guys try to understand what exactly I'm going to say try to understand carefully imagine that you got a container the is a container in this particular container for example you have kept calcium carbonate right you have kept calcium carbonate let's say with calcium carbonate there is something else present as well a plus b plus C I don't know what this a plus b plus C is this a plus b plus C I'm calling as impurities a plus b plus C I'm calling as impurities now tell me in the container do we have pure calcium carbonate or we have got impure sample of calcium carbonate we have got impure sample of calcium carbonate I'm assuming that the mass of the whole sample the mass of the impure sample of calcium carbonate is for example 50 g right now in the 50 g of complete sample in the 50 g of complete sample let's assume that 25 G is the mass of pure calcium carbonate 25 G is the mass of pure calcium carbonate 25 G is the mass of pure calcium carbonate 25 G is the mass of pure calcium carbonate if I need to calculate the percentage purity of calcium carbonate in this particular sample in the sample what do I need to write percentage purity of calcium carbonate in the sample is equal mass of pure calcium carbonate how much is that 25 G divided by mass of impure sample of calcium carbonate how much is that 50 g multipli what multip 100 multip 100 okay multipli 100 perfect how much this value will come out be this one this two This Is 50 so percentage purity of calcium carbonate in this sample is 50 what does that mean that means as per definition as per definition that means 50 g of pure calcium carbonate 50 g of pure calcium carbonate is present in 100 G of the impure sample right so percentage purity of a compound in a sample what does it mean it means that that mass of pure compound let me write the definition that mass of the pure compound that mass of the pure compound which is present in 100 G of the impure s which is present in 100 G of the impure sample I hope you got to know the meaning of all this stuff right number one number two number two before solving the equation remember in the reactions in the reactions only pure substance participates in the reactions only pure substance participates pure substance participates impurities they do not participate in the reaction impurities do not participate in the reaction remember this okay now now I'm going to give you one statement you need to decode that statement you need to decode that statement see guys I'm writing something like this I'm writing 40 G of 80% pure calcium carbonate 40 G of 80% pure calcium carbonate what it means what it means 40 G of 80% pure calcium carbonate if you see a statement written like this how do we decode this statement see this 40 G this is the mass of the whole impure sample of calcium carbonate this is the mass of whole impure sample of calcium carbonate now now out of these 40 G how much is pure calcium carbonate 80% 80% of these 40 G which will come out to be 32 so 32 G is basically the mass of pure calcium carbonate the mass of pure calcium carb so in total in total the whole sample is 40 G out of 40 G 32 G is the mass of pure calcium carbonate so 8 G will be the impurities right 8 G will be the impurities I hope this clear perfect right now guys understand understand try to solve this question when can we when we start next lesson sir once we complete this then only we can go to the second one right I don't want to skip any topic guys okay 50 of 80% pure calcium carbonate is heated calculate the volume of carbon dioxide get produced at STP so basically tell me one thing when calcium carbonate is heated when calcium carbonate is heated what do we get we get calcium oxide plus carbon dioxide what do we need to calculate volume of carbon dioxide tell me one thing people 50 g of 80% pure calcium carbonate tell me what is the mass of pure calcium carbonate that's given to us what is the mass of pure calcium carbonate that's given to us that is 80% of 50 g so the sample which we have in that sample we have got 40 G of pure calcium carbonate right 40 G of pure calcium carbonate so how many pure moles of calcium carbonate do we have pure moles of calcium carbonate pure mass of calcium carbonate divide M mass of calcium carbonate so these are the pure moles of calcium carbonate these are the pure moles of calcium carbonate in the reaction only pure substance participates impurities do not participate now we are done impurities do not participate I can say as per the sto symmetry 1 mole calcium carbonate when you heat it up 1 mole carbon dioxide is produced 1 mole carbon dioxide is produced therefore therefore 0.4 moles of calcium carbonate upon heating will produce 0.4 moles of carbon dioxide so we got the moles of carbon dioxide produced we got the moles of carbon dioxide produced and my dear students was I supposed to calculate moles of carbon dioxide produced no I'm supposed to calculate volume so volume of carbon dioxide produced will be moles of carbon dioxide produced multiplied by 22.4 L the value will be 8.96 l so 8.96 lers of carbon dioxide will be produced am I clear am I clear people am I clear quickly is this percentage Purity concept clear how many equations will be asked from physical chemistry from class 11 physical chemistry minimum 10 11 questions will be asked yeah perfect guys I hope you can easily solve the questions which are based on what which are based on which are based on the percentage Purity okay perfect now now comes one of the easiest topics and the topic from which questions are frequently asked what is that empirical molecular formula have you heard about this one empirical moleular formula empirical molecular formula let's have a look people what kind of questions are asked from empirical molecular formula first of all let's understand what empirical molecular formula is all about see guys first of all let's say I'm talking about glucose I'm talking about glucose do you know any formula for glucose do you know you should be knowing it guys C6 h126 right okay if I ask you what is the ratio over here 6 is 12 is 6 this is the ratio this is the ratio of the atoms this a normal ratio or is it the simplest ratio can you simplify it further you can simplify it further when you simplify it if you divide through out by six this becomes 1 is to 2 is to 1 this ratio over here I'm calling a simplest ratio this ratio over here I'll be call calling is actual ratio right as per actual ratio the formula is this the formula is this perfect so this formula I'll be calling as molecular formula this formula I'll be calling as molecular formula and as per simplest ratio the formula has to be C1 h201 this is what I call as empirical formula this is what I call as emperical formula okay the one which gives the simplest ratio that's empirical formula the one which gives the actual ratio that is the molecular formula number one number one so one is your empirical formula one is your molecular formula now now now with the help of molecular formula you can calculate molecular mass you can calculate molecular mass with the help of what with the help of molecular formula right calculate the molecular mass 12 * 6 + 1 * 12 + 16 multiplied by six the value comes out be 180 U this is the molecular mass of glucose with the help of empirical formula you can calculate empirical Mass how much is that 12 * 1 + 1 * 2 + 16 * 1 the value comes out be 30 U so you got the molecular mass as well you got the empirical mass as well once you get molecular and empirical Mass my dear students I'm going to I'm going to write a number n what is n n is just the ratio of it is the ratio of molecular mass to that of empirical mass and is just the ratio of molecular mass to that of iCal Mass if I ask you what is the n value for glucose you'll say molecular mass of glucose is 180 divide 30 the value of n is 6 and once you get n value my dear students remember molecular formula is always equal empirical formula multipli by n empirical formula multiplied by n now guys just these things you have to remember we are sorted now how do we apply all this how do we get questions on this before that there's one more result which I would want you guys to remember do remember molar mass is equal to two times the vapor density sometimes in equation they give you the vapor density of a compound sometimes in equation they give you Vapor density of a compound from Vapor density you just have to multiply that by two you'll be getting the molar mass okay remember this one as well remember this one as well for example let's see what kind of questions are asked let's see what kind of questions are asked okay and I'll just tell you simple things by means of which you can solve the equations in lesser time see guys directly I'm taking equation try to understand calculate the empirical formula of the compound that contains 26.6 percentage of potassium this much percentage of chromium this much percentage of oxygen by we so you have got a compound you have got a compound which contains three elements what are the elements potassium chromium oxygen you have got a compound which has got three elements potassium chromium oxygen the mass percentage of these is given mass percentage of potassium in the compound is this much mass percentage of chromium in the in the compound is this much mass percentage of oxygen in the compound is this much right what do we have to get we have to calculate empirical formula how do we calculate empirical formula there is a general way of getting the empirical formula first of all you should know the atomic masses of all these elements first of all you should know the atomic mass of all these elements okay first of all first of all you should know the atomic mass of all these elements once you know the atomic mass of all these elements what is the second thing which you need to do whatever percentage is given to you divide the percentage by the atomic mass divide this percentage by its atomic mass divide this percentage by its atomic mass on dividing what do we get we get we get Atomic ratio you call this as atomic ratio you call that as atomic ratio you call this as atomic ratio so what am I supposed to do given percentage divide by atomic mass what do we get Atomic ratio now tell me one thing which one is the least number among these three numbers 0.68 I'll be dividing all these three numbers by that least number okay what do I get I got the least Atomic ratio this was Atomic ratio now I'll be calling this as least Atomic ratio now the atomic ratio here is coming out be 1 is to 1 is to 3.5 Atomic ratio here is coming out b 1 is to 1 is to 3 3.5 no doubt it's the least Atomic ratio but is it the simplest ratio is it the whole number ratio I mean this not the whole number ratio so convert it into whole number ratio multiplying through by two this becomes 2 is 2 is 7 this particular ratio should I be calling this as the simplest whole number ratio yes right and formula as for this will be K2 cr27 right and this particular formula you'll be calling us empirical formula am I done right people are we done with this one more question one more question a compound contains two elements A and B A and P right perfect two elements A and B combined to yield the percentage by mass percentage of a is given as 75.8 therefore percentage of B will be 100 minus this which will be how much almost 24 almost 24 something 2492 something I believe right perfect these are the percentage that that are given to us now atomic masses of A and B are given to us what is the atomic mass of a that is 75 what is atomic mass of B that's 16 now what do we get Atomic ratio how this divide by this value is one this divide by this the value is how much 1.5 I think yeah 24 divid 16 that be approximately 1.5 only okay we got the atomic ratio we got the atomic ratio since we got the atomic ratio what do we need to calculate now least Atomic ratio so divide them by the simplest number by the least number least number is one the value here is one the value here is 1.5 perfect the value here is one the value here is 1.5 now is this 1 is to 1.5 should I be calling this as simplest whole number ratio no I'll be dividing this by two this becomes two this becomes three so the formula becomes A2 B3 right and this is what you call as empirical formula here are we done are we done with this are we done with this quick are we done with this guys are we done with this look at one more type of the question look at one more type of the question you have got a compound that contains hydrogen carbon chlorine it is molar mass is this what will be the molecular formula of the compound what will be the molecular formula of the compound my dear students you know in order to get the molecular formula what do we need we need empirical formula once we get empirical formula multiply that with n that is the point basically so first of all my first point is to calculate what empirical formula and you know empirical formula stuff is same right you do the process you get the empirical formula empirical formula over here you got as H2 C1 cl1 H2 C1 cl1 this is the empirical formula this is the empirical formula right from the empirical formula what can I calculate I can calculate empirical Mass what will be empirical Mass 1 into 2 + 12 into 1 + 35.5 into 1 that will be approximately how much approximately it will be 49 G approximately it'll be sorry proximately it will be 49 U this is empirical Mass right my dear students since you got empirical Mass can I calculate molecular mass molecular mass is given to us as with the equation right 98.96% two if you got the N value we are done guys molecular formula is equal empirical formula was H2 C1 cl1 multiplied by n that's two so it is at the end C2 H4 cl2 this is the molecular formula of the compound I hope you can solve this sort of equation right perfect guys perfectly done yeah now now have you heard about something called as concentration terms the last topic of the chapter concentration terms this is the last topic of the mole concept chapter this is the last topic of the mole concept chapter laws and all all that is outdated perfect do not study gas laws I mean I mean I mean these laws of chemical combination that's outdated concentration terms my dear students first of all if I talk about solution if I talk about a solution just tell me in the chats how many components a solution has you tell me that how many components the solution has two components one is what you call a solute one is what you call a solvent perfect salute you are going to represent by number one solvent you are going to represent by number two okay now now now how do you define salute it is that component of the solution which is present in lesser amount and this is that component of the solution which is present in larger amount which is present in larger amount okay which is present in larger amount first of all do remember mass of the whole solution is always equal mass of solvent plus mass of solute mass of solvent plus mass of solute perfect do remember whatever is the phase of solvent phase of solvent is equal to the final phase of the solution Final Phase of the solution remember this sometimes you'll be given with something as binary solution what is a binary solution binary solution is the one which has got two components out of which one solute and one solvent one solute one solvent right erary solution what is that what is that two solutes and one solvent three components quary solution four components three solute components and one solvent components so that particular solution which has got two components is your binary right you know it perfect binary tary quary sometimes they'll mention a term aquous solution aquous solution is the one where in the solvent used is water aquous solution is the one where in the solvent used is water okay solvent used is water now guys the first concentration term which we have that is what you call as mass percentage of solute or you call it as weight by weight percentage of solute weight by weight percentage of solute now how do you define this particular term how do you define it let me not Define it how do you calculate it how do you calculate Weight by weight percentage of solute it is equal mass of solute in G divided by directly I'm going to give the formulas you have to remember all these formulas okay mass of solute in G divided by mass of solution in G multipli what multip 100 this is how you calculate Weight by weight percentage of solute this is how you calculate Weight by weight percentage of solute number one number two how do we calculate Weight by volume percentage of solute we by volume percentage this also asked I'll show you its application in some time but before that we have to write all the okay weight by volume percentage of solute how do you do that mass of solute in G divided by volume of solution in ml multiplied 100 multiplied 100 okay right the third one that is volume by volume percentage of solute how do we calculate the volume by volume percentage of solute it is equal volume of solute in ml divided by volume of solution in ml multipli 100 multiplied 100 similarly guys there is one result that connects this and this directly weight by volume percentage of solute is always equal weight by weight percentage of solute multiplied by density of solution in G per ml remember this one as well remember this one as well I'll show you the application in some time first of all you have to remember all these results to calculate all of them to calculate all of them okay the fourth one sorry the fifth one mole fraction of solute is always equal number of moles of solute divided by total moles present in the solution mole fraction of solvent number of moles of solvent divide by total moles present in the solution divide by total moles present in the solution right mity mity which is represent by capital M how do we Define marity by the way number of moles of solute present in 1 lit of solution how do you calculate it that's important that's important number of moles of solute divide by volume of solution in liters number of moles of solute divide by volume of solution in liters okay right or or you can write it like this number of moles of solute means mass of solute divide by m mass of solute volume of solution in liters means volume of solution you can take it in ml and you can multiply here with thand this one more result to calculate marity there is one more to calculate marity directly do remember that as well it is equal 10 * weight by weight percentage of solute multiplied by density of solution in G per ml divided by molar mass of solute this one more result by means of it you can calculate marity by means of it you can calculate marity directly right directly this marity you must be knowing it now marity is temperature dependent term this is temperature dependent term so when you change the temperature of the solution marity always change right perfect yeah right people so you have to remember these results now there are few more results then I'll show you certain questions by means of it you'll understand when to use which formula hey guys the seventh one that is mity that is mity mity is represented by m which is equal number of moles of solute divided by mass of solvent in kg or you can write it as mass of solute divided by m mass of solute mass of solvent in G and multip by th000 Perfect similarly guys there is one more result mity is equal th000 multipli marity of the solution ID th000 multipli density of solution minus marity multi M mass of salute this is one more result this is one more result right and at the same time mity can be calculated by one more result that is mole fraction of solute multip th000 divided by 1 minus mole fraction of solute multiplied M mass of salt these are few results which are asked nothing apart from these results is asked nothing apart from these results is asked and guys remember mity this is temperature independent this is temperature independent okay let me give you a few questions so that you understand what kind of questions are asked from these concentration terms understand first question let's say the question is like this calculate the mole fraction calculate the mole fraction of solute in 10% weight by weight of NaOH Aqua Solution NaOH Aqua Solution can you solve this question quickly quickly quickly it's a simple one calculate the mole fraction of solute in 10% weight by weight of NaOH aquous solution quick first of all what is meant by this 10% weight by weight of no Aqua Solution it means that 10 G of No 10 G of solute are present in 100 G of solution 10 G of solute are present in 100 G of solution okay so my dear students 10 G of solute that means mass of solute I calculate as 10 G mass of solvent I calculate as 100 - 10 which is 90 G I calculate mass of solute and mass of solvent I calculate mass of solute mass of solvent now can I calculate moles of solute and moles of solvent moles of solute is equal mass of solute divide by m mass of solute this is mass of solvent divide M mass of solvent this system W2 by M2 mass of solute is 10 g m mass of solute M mass of n that's 40 the value comes out to be 1X 4 mass of solvent is 90 G molar Mouse of solvent solvent here is water because it's Aquis so 18 the value comes be five so you got your N1 as well as N2 if you got your N1 as well as N2 what am I supposed to calculate mole fraction of solute how do I do it mole fraction of solute is equal number of moles of solute divid by total moles present in the solution now n 2 you have N1 you have when you solve this the value comes out be 1 upon 21 this is the mole fraction of solute which we were supposed to calculate am I clear with this people quickly I'll give you the break after like 15 minutes I'll give the break after 15 minutes just wait for sometime am I clear with this number one for example you get a question like this for example you get the question like this calculate calculate the mass of water in Gams that has to be added that has to be added to 16 G of methanol to 16 G of methanol to make it small fraction to make its small fraction as 0.25 can you solve this one can you solve this one in the last question 18 was the molar mass of solvent M mass of water guys this again a one simple question and see exactly how do you deal with this particular question this is simple as per the question imagine that you have got 16 G of methanol in the container you have got 16 G of methanol in the container as per the question how much water has to be added let's say we are adding some x g of water after the addition of x g of water in the container I'm assuming the mole fraction of methanol is becoming 0.25 mole fraction of methanol is becoming 0.25 after the addition of x g of water now if I ask you how many moles of methanol are there in the container mass of methanol divide by m mass of methanol .5 if I ask you how many moles of water are there in the container mass of water divide by m mass of water right I can say these are the moles of water which has been added to make the mole fraction of methanol 0.25 now tell me mole fraction of methanol means number of moles of methanol divided by total moles present in the solution total moles present in the solution that is equal 0.25 now moles of methanol is how much 0.5 divided by moles of methanol 0.5 moles of water xid 18 and this value is equal 0.25 which means 1 by 4 so when you calculate X from here x will come out be 27 G so 27 G of methanol sorry 27 G of water has to be added then only the mole fraction of methanol will come out to be how much then only the mole fraction of methanol will come out to be 0.25 am I clear am I clear yes am I clear let me give you one more question and with that we'll take a break and right after the break we'll start one more chapter right right after the break we'll start one more chapter but before taking the break one question which you'll be solving right now okay for example I'm giving equation like this dissolving 120 g of Ura dissolving 120 g of UA in 1,000 G of water gives the solution gives the solution of density 1.15 G per ml calculate molarity of that solution calculate marity of that solution quick quick can you give it a try it's a simple question guys so you have got one 20 G of Ura which you are dissolving in th000 G of water so first of all you identified your solute and solvent perfect the mass of solute that is given to me that is 120 g mass of solvent that's given to me how much that is 1,000 G therefore mass of solution will be how much mass of solution will be this plus this that's 1120 G perfect now what is the m mass of solute what is the m mass of UIA remember molar mass of U is 60 G per mole right what I suppose calculate marity of the solution which will be nothing but mass of solute multiplied th000 divided by m mass of solute and this is volume of solution and ml this is the formula which we'll be using now if I ask you am I given a W2 Yes 120 multip what multip th000 divided by m mass of solute yes but volume of solution is not given but somehow we have to calculate volume of solution somehow we have to calculate volume of solution how exactly how exactly we are given with the density of solution density of solution is 1.15 G per ml can I say density of solution is nothing it is just mass of solution divide by volume of solution which is equal to 1.15 G per what is mass of solution 1120 G divided by 1.15 G per ml that will give you what that will give you volume of solution and when you solve this this will come out to be 973 ml this is the volume of solution which we calculated from density if you got the volume of solution so put it here put the volume of solution here and when you put the volume of solution here what do I get the value final value will come out to be 2.05 M so this is the final marity of the solution this is the final marity of the solution and guys there will be one question which will be asked from this particular result as well right sometimes there will be a question given in which mity will be given and you are supposed to calculate marity use this result directly sometimes mole fraction will be given and you will be supposed to calculate mity use this particular result so these are some direct results which you can remember from now onwards okay am I clear and these are the only types of the questions my dear students which can be asked from the chapter from the chapter mole concept right now we shall be mov into the chapter redox reaction but before starting the redox reaction I know you need a break so I'm giving you the break this is the Lune Break by the way right it is the lunge break till till till uh it's 12:33 Lune break till one okay so be back on time be back on time and then we shall be starting the chapter redox reactions okay okay guys till then I'll also be back I'll also have some lunch but be back on time okay be back on time is it a promise because I don't trust you guys okay I would require everyone everyone to be here perfect then only I'll continue the session otherwise you're gone hello see you see you after the break e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e what's up people Everyone is back is everyone [Music] back call up call up everyone call up everyone let's start the chapter number two call up everybody quickly so guys now we shall be starting one more chapter which is very important that is the redox reactions okay are you guys done with your lunch are you guys done with your lunch quickly let me know in the chats are you guys done with your lunch okay all right people all right all right people are still doing it huh they are not giving us lunch who is not giving you the lunch who is not giving you the lunch people are still joining in let let people join then we shall be starting this chapter no I did not have my lunch yet no no because I don't want to to feel sleepy during the session that is the reason why I did not have the lunch yeah yes exactly whatever I'm teaching you this much is enough this much is enough people are people are joining in yes let them join in like 1 minute then we shall start let them join in okay whatever I discussed till now is every single thing clear you can honestly say yes or no honestly I would want honest answers from you whatever topics I discussed with you till now is every single thing clear yeah perfect that's great that's great people perfect then so let's start the next chapter that is your redox reaction again I must tell you this is one of the important chapters right which is there in the physical chemistry so my dear students the first topic which we are going to cover here that is going to be the oxidation state now first of all how do we Define the term oxidation state this is the first and the basic terminology over here how do we Define the term oxidation state oxidation state is defined as the charge the charge present on an atom of an element it is defined as the charge present on an atom of an element in a given compound what is meant by that you'll get the idea in sometime but before that I'm going to write few terms over here oxidation state it is defined as the charge present on an atom of an element in a given compound this charge can be real or virtual this charge can be real or virtual in case of the ionic compounds in case of the ionic compounds the charge is real and in case of the calent compounds in case of the calent compounds the charge is virtual the charge is virtual now let's get to know one by one what is the meaning of all these things because they are really important see guys for example for example if I take an ionic compound over here that is na this example I have taken many a times okay let's say I got an ionic compound that is NaCl but students in case of Na if you see how this NAC exactly gets formed in case of Na I would say sodium loses its outermost electron gets converted into na positive perfect and that one electron is gained by chlorine and after gaining this one electron it gets converted into CL Nega so what happens basically na positive and cl Nega is getting formed here na positive CL negative gets formed here and once na positive CL negative gets formed after that there will be force of attraction between them and that force of attraction you already know that's what you call as ionic bond that's what you call as ionic bond perfect now my de students over here if you look carefully sodium loses its one electron completely chlorine gains that one electron completely so can I say overhead during the formation of ionic compound can I say complete loss and gain of electron is taking place absolutely complete loss and gain of electrons is taking place now people if I ask you in case of NaCl what is the charge present on sodium you will say the charge present on sodium is plus one what is the charge present on chlorine the charge present on chlorine is minus one the charge present on an atom of an element is what you call as oxidation state that means if I ask you what is the oxidation state of sodium over here you will say + one what is the oxidation state of chlorine over here you will say minus1 perfect and my dear students in case of ionic compounds there is complete loss and gain of electrons happening and whenever charges arise on the atoms whenever charges arise on the atoms due to complete loss and Gain of electrons those charges are what you call as a real charge so this plus one over here I'm saying oxidation state of sodium is plus one it means that sodium has lost its one electron completely chlorine has gained its one chlorine has gained one electron completely so the charges which you see over here on sodium and chlorine these are the real charges these are the real charges so remember in case of the ionic compounds the charges present are real now in case of coent compounds we say the charges are virtual the charge are virtual for example let's say I have a coent compound over here perfect there's a bond between A and B I'm assuming that electro negativity of B is greater than that of electro negativity of a this something which I'm assuming if B is more Electro negative can I say B has got more tendency to attract the bonded pair towards itself due to which b gets Delta negative and a gets Delta positive if you ask me what is the oxidation state of a over here I'll say it's plus one what is the oxidation St of B over here it is minus one but but over here if you see lifting of electron pair is happening displacement of electron pair is happening displacement of the bonded pair is happening due to which charges are coming on these atoms the charges on these atoms are not due to complete loss and gain of electrons the charge on these atoms is due to what displacement of electron pair and whenever charges arise on the atoms due to the displacement of electron pair these charges are not the real charges these are just the virtual charges right so I would say this plus one minus one these are not the real ones these are the virtual ones right I hope this is clear for example there's a double bond between a and b and I believe B is for example more electr negative than a so B will attract both the electron pairs towards itself due to which it carries minus 2 Delta this carries plus2 Delta if I ask you what is the oxidation of a here you'll say plus two what is the oxidation SE of B -2 but does that mean a has lost two electron completely no does it mean B has gained two electron completely no it is just the two bonded pair of electrons have been shifted towards B so remember in case of ionic compounds due to the complete loss and gain of electrons the charges on the atoms are real but in case of coent compounds in case of covalent compounds the charges are not real they are virtual right they are not the real charges these are virtual charges I believe this is clear to everyone right and my dear students I hope that you already know on mov from left to right what happens to electro negativity electr negativity increases on moving from top to the bottom in the periodic table electro negativity decreases right in the periodic table Flor Florine is considered to be maximum electr negative followed by oxygen followed by nitrogen this is the electro negativity order which you have to remember remember these things remember these things they are very important okay remember these things they are very important perfect now guys there is one more thing there is one more thing which I want you guys to remember then I'll move ahead this is just the basics which needs to be understood okay for example between A and B let's assume there's a coordinate Bond there's a coordinate Bond okay if B is more Electro negative than a if B is more electronegative than a if B is guys just give me a second just give me a second just a second just a second guys just give me a second okay okay so what I was saying is if there's a coordinate bond between a and b and I'm assuming B is more Electro negative than a okay if this is the scenario then this a is assigned the charge plus two this B is assigned the charge minus two right but if the scenario is like this if the scenario is like this for example for example a is more electronegative than b then the contribution of this Bond towards the charges is going to be zero we do not con their charges this is the rule which you need to remember this is the rule which you need to remember okay remember this particular point this is for the coordinate bonds this is for the coordinate bonds from less electr negative to more electr negative yes you have to consider from more electr negative to less electr you do not have to consider remember these two things as well now I'm going to jump into the rules by means of which you can calculate the oxidation States you can calculate the oxidation States the first rule says that oxidation state of an element in its standard State oxidation state of an element in its standard state is taken as zero now what is the standard State how how do we Define the standard state in short I would say it is that state of the element standard state or you can call it as the most stable State most stable State as per thermodynamics right most stable State as per thermodynamics for example if I talk about hydrogen hydrogen in nature exists in the form of H2 gas chlorine in nature exists in the form of cl2 gas bromine in nature exists in the form of br2 liquid right perfect iron in nature exists as Fe solid copper exists as copper solid perfect so these are what I call as NATO States standard states of these elements and whenever an element is present in its standard State its oxidation state is taken as zero whenever the element is present in its standard State its oxidation state is taken as zero right its oxidation state is taken zero point number one point number two my dear students whenever you see an element which has got elot trops if the element has got elot drops lrop one elop two right elot drop one lrop two that elop which is thermodynamically more stable that elop which is thermodynamically more stable will be considered as the standard state of the element for example lrop 2 is thermodynamically more stable is thermodynamically more stable right lrop 2 let's assume it is thermodynamically more stable therefore lrop 2 will be considered as the standard state of the element a for example if you talk about carbon carbon has got two main elot trops one is Diamond one is graphite one is diamond and one is graphite thermodynamically graphite is considered to be more stable thermodynamically graphite is considered to be more stable and if graphite is thermodynamically more stable that means the standard state of carbon I'll be writing like this I'll be writing carbon graphite this is the standard state of carbon over here and whenever the element is present in its standard State it's oxidation state is taken as zero this is rule number one rule number two rule number two is for your group first and group second elements group first elements your alkal matters group second elements your alkaline earth matters right my dear students whenever in any compound you see alkal metal present its oxidation state directly you'll be taking as plus one whenever you see alkaline earth metal present its oxidation state directly you'll be taking as plus two okay oxidation state of your group first and group second elements they are taken as plus one and plus two respectively remember it directly number three number three number three is going to be for hydrogen number three is going to be for hydrogen my dear students whenever you see hydrogen hydrogen attached to an element whenever you see hydrogen attached to an element which is more electr negative than hydrogen whenever hydrogen is attached to an element which is more electronegative than hydrogen at that point of time the oxidation state of hydrogen is Tak as plus one whenever you see hydrogen attached to less electron negative element than itself right the oxidation state of hydrogen will be taken as minus one and whenever hydrogen will be present in its standard State its oxidation state will be taken as zero these are the three things which you have to remember about hydrogen these are the three things which you need to remember about hydrogen that's all that's all okay these are the three things which you need to remember about hydrogen now there are few things which you need to remember about oxygen as well there are few things which you need to remember about oxygen as well there are few things which you need to remember about oxygen as well the first point about oxygen whenever you see oxygen attach to the less electr negative element than itself whenever you see oxygen attached to less electr negative element than itself generally oxidation state of oxygen at that point of time is taken as minus two for example for example you have got MGO oxygen attached to less electr negative element than itself its oxidation state is minus two for example you have got water oxygen attached to less Electro negative element than itself its oxidation state again is minus two perfect so whenever you see oxygen attached to less electr negative element than itself generally its oxidation state will be taken asus2 minus two now my dear students now my dear students there are few special compounds which oxygen forms one is what you call as peroxides one is what you call as super oxides super oxides and one is what you call as oxy fluorides oxy fluorides these are some some special compounds which oxygen forms whenever you see oxygen in the form of o2 dtive whenever in a compound oxygen will be present in the form of o2 DGA understand that's a peroxide there is a peroxide right whenever you see oxygen present in the form of o2 negative O2 negative remember that a super oxide for example for example you know about your hydrogen peroxide H2O2 you know about your hydrogen peroxide H22 do you know its structure this is O single Bond o this for example H this is H this is the structure of your H2O2 now you tell me one thing if you look at this particular bond between oxygen and hydrogen which one is more Electro negative oxygen so this bonded pair will be shifted towards oxygen so oxygen carries minus one this carries plus one what about this particular bond this oxygen will carry minus one this will carry plus one what about this particular Bond on both the sides you have got oxygen so electron pair is exactly at the center so what is the charge present on one oxygen atom minus one the charge present on one atom of an element is what you call as oxidation state so I would say I would say oxidation state of oxygen here is minus1 so remember whenever oxygen forms the peroxide whenever the oxygen forms the peroxide right oxidation state of oxygen is taken as minus one oxidation state is basically charge present on one atom right no doubt you writing O2 D negative here O2 DG here but I have to check how much charge is present on one atom of oxygen and charge present on one atom of oxygen is minus one so that is the oxidation state of oxygen in case of peroxide right in case of peroxide and my dear students this particular linkage a single bonded linkage between two oxygen atoms a single bonded linkage between two oxygen atoms that's what you call as peroxide linkage this is what you call as peroxide linkage this is what you call as per oxide linkage perfect now in case of super oxides you will find oxygen in the form of o2 negative now if I ask you the charge present on two oxygen atoms here is minus1 therefore the charge present on one oxygen atom will be minus 1 by2 and the charge present on one atom of an element is what you call as oxidation state so in case of super oxides the oxidation state is minus 1 by2 right perfect I hope this is clear now in case of oxy fluorides what you'll find you will find oxygen show positive oxidation States in case of oxy fluorides you will find oxygen showing positive oxidation States for example for example I'm going to take two compounds over here one is O F2 and one is O2 F2 these are the two compounds which I've taken o F2 and O2 F2 first of all if you look at the O F2 structure this is O this is f and this is f right O2 F2 this is the structure of your O2 F2 perfect look at these structures carefully look at these structures carefully guys first of all if you look at this particular single bond between oxygen and Florine which one is more Electro negative chorine is more Electro negative so Bond pair is shifted towards Florine due to which this will have minus one this will have plus one right similarly look at this particular bond between Florine and oxygen which one is more electr negative Florine so minus one plus one so what is the total charge present in oxygen here plus two so oxidation state of oxygen hair is plus2 similarly O2 F2 look at this particular bond between oxygen and Florine which one is more electr negative Florine so minus one + one due to this particular Bond minus one plus one due to this Bond nothing is going to happen because on both the sides you got oxygen the charge present on one oxygen atom here is plus one and the charge present on one atom of an element is something which you call as oxidation state I would say oxidation state of oxygen here is plus one the charge present on one oxygen atom is plus one which is what you call as oxidation state yes I believe it's clear I believe it's clear someone is saying majority from Kashmir is that guys can you let me know in the chats where are you guys from can you write your States I want to see from which place majority of the state majority of the students are watching me right now see I'm live from Kashmir where are you guys from quickly write your states in the chats write your states in the chats let me see the majority majority is not from sh majority is not from Kashmir yes majority is from Tamil Nadu Karnataka okay from Kashmir also few people are watching nice what about the rest oh someone is watching from Nepal Andra people watching yes great all right so basically students are watching from all over the country right okay so I hope you got to know about oxygen and these a few things which you have to remember about oxygen yeah okay now guys now guys uh one more thing which I would want to discuss here that is about halogens about halogens about halogens okay you know halogens they are your group 17 elements my dear students hogen you represent with X right whenever you see hogen attached to a less electr negative element attached I mean whenever hogen is attached to less elect negative element than itself the oxidation state of hogen will be taken as minus one the oxidation state of hogen will be taken as minus one for example you have got HCL you got HCL perfect you got HF perfect you got HF right you got HF yeah so over here this is hogen this is hogen attached to less Electro negative element than itself so oxidation state directly here is minus one now the point is what about halogens do they show positive oxidation state anytime do they show positive oxidation state my de students there are few compounds where in halogens they show positive oxidation state for example you have got interhalogens you have got interhalogens you have got interhalogens for example let's say I'm taking one compound clf3 clf3 right it's an interhalogen perfect you got brf5 interhalogen you have got if7 interhalogen compound if I ask you what about the oxidation state of these underlined of these underlined halogens what you'll say you'll say oxidation state here will be plus three because it's attached to three Florin atoms here is going to be plus 5 here is going to be plus 7 so basically halogens can show oxidation state positive as well they can show positive oxidation States as well similarly for example you know there's a compound cl2 there's a compound cl2 chlorine here is the hogen if you look at its structure this is O for example this is CL and this is CL correct this is O2 if you look at this particular bond between oxygen and chlorine which one is more electr negative oxygen so due to this Bond minus one + one due to this Bond minus1 + one so again you can see the charge present on one chlorine is+ one so I'll say hogen over here is showing plus one oxidation state so basically my point is halogens can show positive oxidation state as well right exactly it's all about electr negativity I believe this is clear I believe this is clear it's all about the electro negativity okay now people Now understand there is one important rule which you need to know one important rule which you need to know what is that the sum of oxidation States the sum of oxidation states of all the elements the sum of oxidation states of all the elements in a compound in a compound is equal to is equal to the net charge present on the compound is equal to the net charge present this is something which every one of you would have heard I'm I'm pretty much sure about that right so people the sum of oxidation states of all the elements present in a compound it's always equal to what it is always equal the net charge present on the compound just give me a second perfect see guys see guys see what is the meaning of this particular statement I have taken over here two compounds right and I'm underlining certain elements over here I'm asking you to calculate that oxidation States how exactly you are going to do it how exactly it's simple let the oxidation state of chromium be X we have got two chromium atoms so multiplied by two plus oxygen attach to less electr negative than itself so oxygen generally shows minus to oxygen State there are seven oxygen atoms is equal to net charge is minus 2 when you solve x x will come out to be plus 6 so plus 6 is the oxidation state of chromium here that means plus 6 is the charge present on one chromium atom as simple as that right sulfur assume its oxidation state as X so which will be 2x oxygen will show min-2 oxygen State there are three oxygen atoms net charge is min-2 so the value of x will be here plus2 plus2 is the oxidation state of sulfur here correct let the oxidation of nitrogen be X hydrogen shows plus one there are four hyd atoms net charge is plus one the value of x when you solve it will come out to be minus 3 simple C4 negative right similarly when you solve it this is x + oxygen min-2 there are four oxygen atoms net charge minus one it is + 7 right this is plus 6 perfect so this is how you calculate oxidation states of different elements for example you got Fe2O3 Fe2O3 the oxidation state of iron when you calculate it will be+ 3 here it will be + 8id 3 plus 8id by 3 and whenever you see oxidation state coming be fractional whenever you see oxidation state coming out be fractional remember that is the average oxidation state remember that is the average oxidation state right that is the average oxidation state for example fe304 fe304 it actually is a mixed oxide it is a mixed oxide it is a mixed oxide and it exists in the form of feo do fe203 it exists in this format it exists in this format and if you look carefully this iron this will be x - 2 is equal 0 so iron this iron is in plus2 and this iron is in + 3 so if I calculate average oxidation state of iron over here see the first iron is in plus two this iron is in + three this iron is in plus three perfect and these are two iron atoms in total there are 2 + 1 three iron atoms when you calculate it this value will come out be plus 8 by 3 so this Plus 8 by3 is basically the average oxidation state of iron ha in case of fe304 so similarly I believe you would have studied the coordination compounds or if not there will be a marathon for that you know this CN CN it is a negative charged liant I hope you remember it's a negatively charged Li it carries charge minus one so if this is X right CN will carry minus1 there are 6 CN net charge is minus 3 so x value when you solve it will come out be plus three this carbonal is a neutral liant neutral liant so zero perfect similarly you can calculate here also perfect now look at these two comp comps nh4 NO2 nh4 NO2 let me tell you this nh4 NO2 it exists in the form of nh4 positive and NO2 negative and the oxidation state of these two nitrogen atoms air will be different check it out let's say the oxidation state of this nitrogen is X hydrogen will show plus one there are four hydrogen atoms the charge here is Plus One X value when you sol it it will be minus 3 so charge present on this nitrogen is minus 3 right look at this particular nitrogen this is X oxygen will show minus two there are two oxygen atoms net charge is minus one X will come out be what it will come out be plus three so the charge present on this nitrogen is plus three if I ask you to calculate average oxidation of nitrogen on the first one it is minus three on the second one it is plus three total nitrogen atoms are two right average oxidation state of nitrogen here comes out be zero it comes out be zero right now talk about K3 talk about K3 see potassium it belongs to what potassium it belongs to what potassium belongs to group for elements and group first elements what do they show they show plus one oxidation state now let's assume the oxidation state of iodine here is X perfect so it is + 1 + 3x net charge is zero so the value of x comes out be - 1 by 3 so - 1X 3 is the oxidation state of iodine over here I hope I'm clear I hope I'm clear I hope I'm clear now guys there is one very strong point which you have to remember and take a note of it take a note of of it let me tell you the maximum oxidation state of an element the maximum oxidation state of an element can never be can never be the maximum oxidation state of an element can never be greater than plus 7 can never be greater than plus 7 except except osmium in oso4 where osmium shows Plus 8 oxidation state the maximum oxidation state of an element it can never be greater than plus 7 there is one exception that is osmium osmium in osmium tetroxide it shows the oxidation state of Plus 8 okay it shows the oxidation state of Plus 8 now guys can you tell me one thing can you tell me one thing I'm writing a compound over here C5 can you let me know the oxidation St of chromium here quickly can you let me know the oxidation of ch in C5 quick the rules which we have discussed as per those rules the rules which we have discussed as per those rules as per those rules this has to be X this has to be min-2 multiplied by five net charge is zero so the value of x has to be+ 10 as per the rules we discussed but can the oxidation state of element be greater than plus 7 no that means we have done some mistake it is incorrect whenever whenever with the help of formula with the help of formula oxidation state of an element comes out to be greater than plus 7 whenever with the help of formula oxidation state of an element comes out to be greater than plus 7 at that point of time you have to consult the structure of that particular compound you have to consult the structure of that particular compound right so there are some compounds my dear students there are some 11 to 12 compounds who structure you need to remember on priority now what are those compounds what are those structures let's have a look let's have a look people let's have a look the first compound which I have that is C5 cr5 perfect let me tell you the C5 we say it has got a butterfly like structure which is not true though right butterfly like structure let me tell you it is deep blue in color as well it is deep blue in color as well now the point is what is its structure this is chromium double bond O oxygen oxygen single Bond and here oxygen oxygen single bond this is the actual structure of your C5 this is actual structure of your C5 now look at this particular double bond between chromium and oxygen between chromium and oxygen oxygen is more electr negative so due to double bond it will be minus two this chromium will be plus two due to this single Bond due to this single bond this oxygen is minus one this is plus one due to this single bond this minus one this plus one due to this single Bond minus one + one due to the single Bond minus one and + one so if I ask you what is the total charge present on chromium here total charge present on chromium it will be how much plus 2 + 3 + 4 plus 5 plus 6 so plus 6 is the actual oxidation state of chromium in case of cr5 in case of C5 now people if I ask you how many peroxide linkage do we have how many oxide linkage peroxide linkage means oxygen single B oxygen this is the oxygen single B oxygen this oxygen single B oxygen so in case of C5 the number of peroxide linkage is one number of peroxide linkage is one okay right at the same time if they ask you how many oxygen atoms are there with minus1 oxidation state how many oxygen atoms are there with minus1 oxygen state right what you have to do number of peroxide linkage multiplied by two number of peroxide linkage multiplied by two number of peroxide linkage multiplied by two the value comes up be two this was your first structure that was C5 right that was your C5 let's have a look on the second structure cr6 T6 right this has been asked in your neat as well C6 right this is CR oxygen oxygen peroxide oxygen oxygen peroxide oxygen oxygen peroxide now you tell me quickly what is the oxidation state of this chromium it is going to be plus six because six single bonds with oxygen atoms right now if I ask you how many peroxide linkage are there there are three peroxide linkage how many oxygen atoms are there with minus one oxidation state 3 2 are 6 done understood right there is something which you call as caros acid h205 you call it as you call it as the caros acid perfect what about the structure of this caros acid you need to remember this as well this as double bond o for example Right double bond o or attached with hydrogen oxygen oxygen and hydrogen H2 s and five oxygen 1 2 3 4 5 yes perfect after having a look at this particular structure can you let me know the oxidations of sulfur see double bond double bond single Bond single Bond so this sulfur will have plus 6 oxidation state because plus two due to this double bond plus two due to this double bond plus one due to this single Bond plus one due to this single Bond so plus six right how many peroxide linkage do we have peroxide linkages this is a peroxide linkage one if I ask you how many oxygen atoms will be there withus one oxidation state per oxide linkage multiplied by two done now you have got the fourth one fourth one is going to be your this was your uh H2 so5 okay now we have got H2 s27 you call it as olum you call it as olum H2 s27 perfect H2 s207 so this s this is O this is s double bond o double bond o double bond o double bond O oxygen attached with hydrogen and this oxygen attached with hydrogen so this is the structure of your H2 s27 check how we used all the seven oxygen 1 2 3 1 2 3 yes right this is the structure of your h227 now if I ask you what the oxidations of this sulfur and this sulfur check it out there's a double bond so plus two from this double bond plus two from this double bond plus one from this single Bond plus one from this single Bond so plus six similarly this will be plus 6 right plus 6 plus 6 now if I ask you the average of oxidation state of sulfur over here the first atom is plus 6 second is plus 6 total number of sulfur 2 so plus 6 is also the average oxidation state of sulfur here in case ofum believe it's clear right I believe it's clear is it clear people quickly perfect sir butterfly structure peroxide linkage two what did I write oh guys there's a mistake which I have done over here oh thank you so much there are two peroxide linkage right so peroxide linkage are two how many oxygen atoms are there with minus one oxidation state 2 2 are four right I hope you would have corrected a L I said the same but I wrote different yeah right this is auto correct corrected perfect this was your Oleum H2 s207 now M ahead M ahead have you heard about Marshal's acid H2 s28 this is your H2 s28 if you look at this if you look at this the oxidation state of this particular sulfur will come out be plus 6 the oxidation state of this particular sulfur will come out be plus 6 now this there's a peroxide linkage here so number of peroxide linkage is one and number of oxygen atoms with minus one oxidation state is two correct it is just if you remember the structure you are done right the sixth one the sixth one the sixth one this has been frequently asked in many examination na2 s46 many examinations in J two times the same question has been asked na2 s46 right and remember its structure there's the structure this is the structure of your na2 s406 now if you want to calculate the oxidation states of different sulfur atoms this sulfur is having uh + 2 + 4 + 5 so + 5 this will be zero this will be 0 and this will be + 5 so the correct set of oxidation state here is + 5 0 0 + 5 now if you ask me the average oxidation St of sulfur here it will be + 5 second one carries 0 third one carries Z fourth one plus 5 total sulfur four the value comes out be plus 2.5 plus 2.5 is the average oxidation state of sulfur in case of na2 s406 okay this was your six structure now the seventh one uh your K3 C8 K3 C8 remember it structure as well so this is CR this is oxygen oxygen single Bond oxygen oxygen oxygen then you have got these peroxide linkage and you have got potassium potassium potassium this is K3 CR 1 2 3 4 5 6 7 8 yes perfect now guys if I ask you about all these things what is the oxidation state of chromium here you will say plus 5 if I ask you how many peroxide linkage do we have first peroxide second peroxide third one one and fourth one four peroxide linkage how many oxygen atoms with oxygen State minus one it is eight done this was your K3 C8 this was the seventh one now eth one eth structure eth structure is your c302 c302 C32 this is your c302 if you calculate the oxidation States this carbon is having plus two this carbon zero because on both SI you got carbon this is +2 if you ask me average oxidation of carbon here the first one plus two second one 0o third one plus2 total carbon 3 the value is plus 4x3 the value is plus 4x3 right the value is plus 4x3 this was your c302 now the ninth one that is br38 br38 you need to remember its structure as well this is BR this is BR this is BR Right double bond double bond double bond double bond double bond double bond everywhere you have our double bonds like this okay and if you remember the structure now you can easily calculate the oxidation states of different elements easily you can do that now if you talk about this particular bromine plus 2 plus 2 plus 2 that mean plus 6 this one + 2 + 2 + 4 + 2 + 2 + 2 this is plus 6 so if you ask me the average oxidation state of bromine over here it will be plus 6 followed by + 4 followed by plus 6 total three bromine 66 12 4 12 4 16 so + 16 by 3 is the average oxidation state of bromine here right perfect this was your br38 br38 now the 10th one bleaching powder c cl2 c cl2 this is CA this is O and this is CL directly okay now if you look at the oxidation states of different elements here calcium and chlorine chlorine is more electr negative so minus one + one oxygen is more electr negative minus1 + one oxygen is more Electro negative minus one + one right so oxidation St of this calcium is plus2 this oxygen is min-2 this chlorine is + one this chlorine is minus one if you ask me the average oxidation St of chlorine here one is having plus one the second one is having minus one two chlorine atoms in total so the average oxidation state of chlorine here comes out be zero right perfect guys right this was the 10th one now few more three four structures 11th 11th b a s CN hold twice you will find I mean this particular compound why am I giving you because uh after this we have to discuss n Factor calculation in N Factor calculation equ question is asked from this as well BN hold tce you need to know its structure as well s c triple bond right s CLE Bond now people understand understand first of all there's a triple bond between carbon and nitrogen which one is more Electro negative nitrogen so minus 3 and + three single bond between carbon and sulfur sulfur is more electr negative minus1 + one right because you know sulfur is group 16 carbon is group 14 on Mo from left to right electr negativity increases now barium and sulfur sufur is more electus one + one due to this Bond min-1 + one perfect due to this Bond -1 + one due to this triple bond - 3 + 3 can you calculate the oxidation states of all these elements here can we calculate the oxidation states of all these elements here right see guys this nitrogen is having -3 oxygen State this carbon is having +4 oxidation state -2 + 2 -2 + 4 and minus 3 okay I believe all the things are clear I believe all the things are clear now H3 po2 and H3 po3 H3 po4 and H3 PO5 you need to remember their structures as well you need to remember their structures as well now guys let's have a look what their structures are first of all everywhere you keep p double Bondo keep p double bond everywhere keep p double bond everywhere now one oxygen directly attached with hydrogen one oxygen directly attached with hydrogen right so we have got H3 P4 so this is your uh what did I do here just a second guys uh I had to write H3 P2 I made for H3 P4 so this is H3 P4 here this is H3 h3p 1 2 3 4 okay uh here I'll make H3 P2 here I'll make H3 P2 p double bond o double bond O oxygen attached with hydrogen oh wait guys how do we make the structure for H3 po2 I forgot hello let's make for H3 po3 oxygen attached with hydrogen okay there's a peroxide linkage here no peroxide linkage is here H3 P5 oxygen attached with hydrogen oxygen attached with hydrogen this oxygen oxygen and hydrogen this your H3 P1 2 3 45 H3 P5 okay let's talk about these two now H3 P3 h3p H3 P O3 okay this is p dou Bond o perfect this oxygen attach with hydrogen we have three hydrogen atoms right uh this oxygen attached with hydrogen and this will be one hydrogen directly here so this will be H3 P1 2 3 perfect H3 P2 this is p double bond o this is oxygen attached with hydrogen Perfect Two oxgen atoms so this hydrogen and this hydrogen so these are the structures of these H3 po2 2 H3 po3 h3po4 right yeah exactly brain is not braining anymore done so Guys these are the structures which you need to remember these are the structures which you need to remember these are the structures which you need to remember I hope I am perfectly clear to everyone yeah is it is it all clear till now let me know once in the chats I would want some fire emojis guys do not sleep please otherwise I'll sleep e I want I want from everyone is it like clear properly clear honestly you can say whatever structures I gave you just remember these structures whatever structures I gave you just remember these structures that's all that's all perfect if the structures are clear now let's move ahead oh by the way structures were already made here why did I why did I make them again structures were already on the slides perfect now guys if I ask you how do you define the term oxidation there many ways by means of which you can Define the term oxidation reduction right perfect there are many ways by means of which you can Define the term oxidation and reduction oxidation involves loss of electrons number one oxidation involves loss of electrons oxidation involves removal of hydrogen something which you have seen in class 8th 9th addition of oxygen is also called as oxidation addition of more electr negative element is also called as oxidation right but I would just want you guys to remember this one addition loss of electrons just remember this one loss of electrons gain of electrons is what you call as a reduction addition of hydrogen is called as a reduction removal of oxygen is called as a reduction removal of more electr negative element is called as a reduction for example for example try to understand carefully if you look at the first one if you look at the first one magnesium is losing two electrons magnesium is losing two electrons getting convert into mg positive so this process involves loss of electrons I would say magnesium is undergoing oxidation here magnesium is undergoing oxidation here right over here over here removal of hydrogen is happening removal from water you are removing hydrogen removal of hydrogen is happening removal of hydrogen is also what you call as oxidation over here addition of O oxygen is happening addition of oxygen is also what you call as oxidation and in this particular case addition of more electronegative element is happening right this also what you call as oxidation if you see there is one thing common in everyone all these are oxidation processes all these processes over here they are oxidation processes but if you perfectly analyze magnesium its oxidation state here is zero here the oxidation state is plus2 so the oxidation state of magnesium is increasing here correct oxygen its oxidation state here is min-2 here it is zero so - 2 to 0 again the oxidation state is increasing perfect calcium oxidation state St 0 calcium oxidation state plus2 again the oxidation state is increasing carbon it's oxidation state zero here the oxidation state is plus4 oxidation state again increasing all these process were oxidation processes right and there's one thing common oxidation state of an element is increasing so in short forget everything and remember one thing increase in the oxidation state is what you call as oxidation whenever from now onwards if you see if you see oxidation state of an element increase in whenever from now onwards if you see oxidation state of an element increasing remember that particular element is undergoing oxidation in the similar way decrease in the oxidation state is what you call as reduction so that particular element which under goes decrease in the oxidation state which under goes decrease in the oxidation state we say that spey under go reduction and the one which under goes oxidation is what you call as reducing agent the one which under goes reduction is what you call as oxidizing agent you should know it the one which under go oxidation the one which under goes oxidation is what you call as a reducing agent the one which under goes reduction that's what you call as oxidizing agent right now now now let's try to analyze all these things properly for example I'm giving you a reaction I'm asking you to identify your oxidizing agent and reducing agent you should be able to do it you should be able to do it you should be able to do it see guys first of all I would want you guys to calculate the oxidation state of every speci over here oxidation state of every specie over here so you calculate the oxidation state of every specie number one now check is the oxidation state of zinc increasing or decreasing 0 to plus2 increasing the oxid State increase in the oxidation state is something which you call as oxidation so I would say zinc is undergoing oxidation if zinc is undergoing oxidation definitely it'll be losing electrons it'll be losing electrons right and the one which is undergoing oxidation that's what you call as reducing agent so zinc here X as a reducing agent right similarly copper plus 2 to 0 + 2 to 0 decrease in the oxidation state decrease in the oxidation state is what you call as reduction and reduction involves gain of electrons so your copper duls to hair will be gaining electron right the one which undergo reduction that's what you'll be calling as oxidizing agent simple yeah is it simple people for example I'm giving you one more I'm giving you one more calculate the oxidation state of every everything over here calculate the oxidation state of everything over here iron in plus two oxygen state magnes plus 7 oxidation state iron plus three oxygen State magnes plus two oxygen state right you calculate the oxidation States you calculated the oxidation States now people try to understand plus 2 to+ 3 increase plus 7 to +2 decrease increase in the oxidation state is what you call as oxidation so I would say this feo is undergoing oxidation the one which under goes oxidation that is a reducing agent plus 7 to Plus decrease in the oxidation state decrease in the oxidation state is something which you call as reduction so here I can say M4 negative is undergoing reduction the one which under goes reduction is termed as the oxidizing agent so I believe from now onwards you can easily identify which species is your oxidizing agent and which species is your reducing agent yeah am I clear people am I clear I hope from now onwards you can easily Define your oxidizing agent and reducing agent now comes the most important part of the chapter the most important part of the chapter the most important part of the chapter the most important part of the chapter what is that equivalent Mass how do we exactly Define the term equivalent Mass how do we do it see guys equivalent Mass it is defined as that mass of the substance in grams that mass of the substance in Gs which can which can combine or displace which can combine or displace either either which can combine or displace either 1 G of hydrogen or 8 G of oxygen or 35.5 G of chlorine or 108 G of silver or 80 g of bromine or 80 g of bromine what does it mean try to understand my dear students for example for example I'm writing a reaction over here N2 + 3 * H2 it gives NH3 it gives 2 * NH3 this is the reaction which I wrote over here if you look carefully if you look carefully my dear students your nitrogen is being combined nitrogen is combining with what nitrogen is combined with hydrogen I mean nitrogen it's combining with hydrogen in this particular case nitrogen is combining with hydrogen let's assume that I want to calculate the equivalent weight of nitrogen let's assume that I want to calculate the equivalent weight of nitrogen how do we Define the ter equivalent weight it is that mass of the substance it is that mass of the substance which can combine or displace either 1 G of hydrogen or this or this or this and in this particular case your nitrogen is combining with hydrogen so no doubt I have to calculate equalent mass of nitrogen but here in this particular case I'll be calculating equivalent mass of nitrogen by taking hydrogen into reference by taking hydrogen into reference so I will exactly check how many G of nitrogen are combined with 1 G of hydrogen that's something which I need to check how many grams of nitrogen I combined with 1 G of hydrogen how exactly I'm going to do that understand what is the mass of 3 mol of hydrogen 3 mol of hydrogen means 6 G so can I say 6 G of hydrogen can I say six as per stoet 6 G of hydrogen combines with combines with 28 G of nitrogen here from the stoet I got to know 6 G of hydrogen combines with 28 G of nitrogen therefore I can say 1 G of hydrogen combines with 1 G of hydrogen combines with 28 / 6 which comes out to be 14 ID 3 G of nitrogen if you look carefully can I say this is the mass of nitrogen can I say this is the mass of nitrogen that has combined with 1 G of hydrogen that mass of substance which combines with 1 G of hydrogen that's what you call as equivalent Mass so this is something which I'll be calling as equivalent mass of your nitrogen so this is the mass of nitrogen which has combined with 1 G of hydrogen that's what I'll be calling as equalent mass of nitrogen I hope I'm clear I hope I'm clear yes yes perfect let me give you one more example so that it makes sense to you I'm writing a reaction magnesium plus O2 it gives MGO if I balance it this has to be two times this has be two times now it's a balanced chemical equation my dear students let's assume that I need to calculate the equal mass of magnesium equalent mass of magnesium but over here if you look carefully magnesium is combining with O2 magnesium is combining with oxygen so no doubt I will be calculating equal mass of magnesium but with oxygen taking into reference with oxygen taking into reference with oxygen taking into reference okay now how exactly understand what is the mass of 1 Mo of oxygen 32 G 32 G what is the mass of 2 moles of magnesium 1 mole of magnesium means 24 G so it's 48 G as per sto symetry I got to know 32 G of oxygen combines with 48 G of magnesium so 1 G of oxygen combines with 48 / 32 G of magnesium therefore 8 G of oxygen 8 G of oxygen combines with these many grams of magnes right 8184 the value comes out be 12 G so can you quickly check what did I calculate I calculated basically that mass of magnesium I basically calculated that mass of magnesium which is combining with 8 G of oxygen and that mass of substance which either combines with 1 G of hydrogen or 8 G of oxygen that's something which you call as equivalent Mass so this is what you will be calling as equivalent mass of magnesium this is the general definition this is how you calculate the equalent mass of difference species as per their definition as per their definition right as per their definition now there is one direct formula also by means of which we can calculate equivalent mass and what is that direct formula remember that directly right remember equivalent mass of the substance is always equal equalent mass of the substance is always equal molar mass of the substance divided by n factor of the same substance n factor is also called as valency Factor molar mass of the substance divided by n Factor now the molar mass of every substance you already know or you can calculate but the point is how do we calculate n Factor how do we calculate n Factor now comes the actual topic that is the n Factor calculation okay now let's have a look on the N Factor calculations n Factor calculations n Factor calculations first of all I'm starting with first of all I'm starting with n factor of an element n factor of an element how do we calculate n factor of an element it's always equal the valency of an element n factor of an element is what n factor of an element is basically equal to what the valency of an element for example take certain elements let's say I'm taking nitrogen let's say I'm taking magnesium right perfect let's say I'm taking sodium these are few Elements which I took now people if I ask you what is the valency of nitrogen I'm asking you about the valency of nitrogen what is the valency of nitrogen you tell me quickly what is the valency of nitrogen it's three what is the valency of magnesium it's two what is the valency of sodium it's one whatever is the valency of an element that's something which you call as n factor of an element so this is the n factor of your nitrogen this is the n factor of your magnesium this is the n factor of your sodium now equivalent weights of all these all these species so equivalent weight of nitrogen will be equal molar mass of nitrogen divide by n Factor equivalent mass of magnesium is equal to M mass of magnesium divide by n Factor equalent mass of sodium is equal to M mass of sodium IDE n Factor so this is how you calculate n factor of an element I mean this is how you calate equalent mass of an element old mass by n Factor n factor of an element is nothing but n factor of an element is nothing but someone is saying this part is deleted who said this part is deleted who said so just make me talk to him or her whosoever it is that's why you should not watch random videos right this stick only one channel someone is saying for nitrogen valency is five wow those are valence electrons valence electrons are five 8 - 5 is three valency something which you study in class 8th okay someone is asking certain questions why can't we take 16 G of oxygen for finding equalent Mass I'll ask you one more question why can't we take 10 G of hydrogen to calculate equalent mass these are all definitions which you need to follow right okay I have not given this somebody gave it like hundreds of years before that's what I'm teaching yeah so remember it remember it so so the first part is done n factor of an element is basically what n factor of an element is the valency of the element once we get the valency that's something which you call as the N factor and you're done this is point number one point number two point number two that is n factor of a Salt N factor of a Salt N factor of the salt is equal LCM of LCM of LCM of charge on K comma the charge on an LCM of charge on K comma charge on N I hope you remember how to calculate LCM I wish and hope at the same time for example for example try to understand people let's say let's say I'm taking certain salts first one is your NAC which you all must be knowing right let's say I'm taking mgcl2 right let's say I'm taking al2 s so4 whole tce these are certain salts which I will take yeah now first of all this is one this is one this one goes a charge here this one will be the charge here that crisscross thing which you studied earlier this two is a charge here this one is a charge here this three is the charge here this two is the charge here so you got the charges on LC charges on K and an since you got the charges on K and an now you have to take the LCM LCM of 1 and 1 is 1 LCM of 2 2 and 1 is 2 LCM of 3 and 2 is 6 so LCM of charge on C and anion that gives you the N Factor right so this is the n factor of n this is the n factor of mgcl2 this is the n factor of al2 so4 whole thce right since you calculated the N factors you can easily calculate the equivalent weights M mass of n divide by n Factor M mass of this divide by n Factor M mass of this divide by n factor that gives you the equalent mass of these salts it gives you the equivalent weight of these salts this was second point the third point this is important n factor of acids n factor of acids how do we calculate n factor of aets n factor of acid is nothing but basicity of the acid n factor of an acid is nothing but basicity of the acid now how do we Define the term basicity of an acid there are many ways I means of which you can Define basicity of the acid number one number one number of H positives donated number of H positives which an acid donates number two number of O negatives which an acid accepts or number three the number of Lone pairs which an asset donates perfect for example for example let me take certain acids like simple simple acids HCL h204 H3 po4 these are some simple acids which you already know when you when you when you introduce this HCL into water or when you introduce this h204 into water when you introduce this H3 P4 into water tell me how many H positives it's going to donate it will give away 1 H positive this will give away 2 H positives this will give away 3 H positives and you know number of H positives and acid donates number of H positives and acid donates that defines its n Factor n factor of HCL here is one n factor of h204 here is 2 N factor of H3 po4 here is three if you got the N factors of these acids you can easily calculate the equalent equalent weights right how molar mass of HCL divide by n Factor M mass of this divide by n Factor M this divide by Factor correct right now guys here I'm going to write one statement be careful with that I'm writing an acid an acid donates only that H positive and acid donates only that H positive which is connected and acid donates only that H positive which is connected to more electr negative element the more electr negative element like Like chorine Oxygen Etc right why am I writing this particular statement there's a point for that you you'll have a point over here I students for example I'm writing h3p 2 here I'm writing H3 po3 here I'm writing H3 P4 and here I'm writing H3 PO5 H3 P5 if I ask you what is the n factor of this this this this what is the factor of all these four species what do you think all these are acids what do you think what do you think what is the n factor of all these species which I took quick it looks like this will give three h positives this will give three h positives this will give this will give three so n factor of all of them should be three right n factor of all of them should be three but I told you an acid donates only that edge positive which is connected to more electr negative element like Flor and oxygen right in this particular case there's only one if you look at its structure which I gave you few minutes back there is only one hydrogen which is attached to more electronegative element Like Oxygen it will donate only that hydrogen right over here in this particular structure two hydrogens are attached with oxygen so it will donate those two hydrogen this will donate three hydrogens this will also donate three hydrogen so their n factors are respectively 1 2 3 and 3 and if you got the N factors you're done right if you got their n factors if you got their n factors you can easily calculate the equivalent weights I hope this is clear to everyone I hope this is clear to everyone yeah but for that my dear students I I gave you the structures of these compounds you need to remember those structures first of all you need to remember those structures first of all then only you can calculate the N factors I hope this is clear now there is one more point which is related to acids only that is it is the boric acid have you heard about boric acid H3 bo3 right boric acid sometimes in the examination they might give you boric acid in this format B whole th this also the boric acid you can write it like this as well this again h33 right this again h33 boric acid perfect this again boric acid now when you introduce boric acid into water what happens exactly my dear students this is h33 when you introduce this boric acid into water you get a reaction like this B o4 Nega plus h positive B4 negative plus h positive let me write it like this this was your B whole Price Right plus water gives this gives this gives this now my students if you look at this particular reaction carefully how many o did we have here three now how many do we have four so when this reaction is happening can I say this acid is accepting 1oh negative it is accepting 1 negative right this acid is accepting one O Negative the number of O negatives and acid excepts that also defines its Bas and whatever is the basicity of the acid that's its n Factor so do remember n factor of boric acid over here will be one because it's accepting one oh negative let me tell you this boric acid is a it is a weak lowest acid by the way it's a weak lowest acid and do remember its n factor is one and if you got its n Factor you can easily calculate its equivalent weight molecular weight divid by m mass divid by n Factor done yeah perfect now there are few things which you need to know about assets as well I'm writing few reactions here let's say I'm writing h204 plus NaOH h2so4 plus NaOH for example it gives NS so4 plus water this one reaction which I wrote h204 plus 2 * nooh 2 * no it gives na2 s04 plus 2 * water these are the two reactions which I wrote over here now now in these two reactions if I ask you in these two reactions what is the n factor of this h204 and this h204 understand understand my dear students both the reactions are first of all balanced you know one when one H positive reacts with 1 O Negative what do we get we get one mole of water when two H positives combine with two o neges you get two moles of water perfect in the first case one mole of water is being formed if one mole of water is being formed say from here one mole of H posi would have come from here one mole of O Negative would have come they would have interacted and formed one mole of water so this acid is donating only one H positive the number of H positives and acid donates that defines its n Factor n factor of h204 in this reaction is one right if you look at the second one if you look at the second one two moles of water being formed two moles of water so for the formation of two moles of water I would say this would have given 2 moles of H positive this would have given 2 moles of O Negative they would have interacted and formed 2 moles of water so since this is giving two moles of H positive so it's n Factor has to be two so n factor of h204 I mean n factor of an acid it can change when you change the type of the reaction my point is that when you change the type of the reaction n factor of an acid can change okay n factor of an acid chains on changing the type of the reaction in which it is participating is that clear this was all about acids which you need to remember I hope all the things are clear till now is it is it guys now similarly similarly we can talk about the N factor of bases as well we can talk about the N factor of bases as well n factor of bases how do you exactly Define the N factor of bases what is the n factor of a base exactly n factor of the base is the acidity of the base how do I Define the acidity of the base number of O negatives a base donates number of O negatives are base donates second number of H positives a base excepts a base excepts you can't remember just these two this is more than sufficient this is more than sufficient see guys for example if I talk about some simple simple bases let let's say you got n you got no tell me when you put this no into water how many o negatives it's going to give it's going to give you one O Negative the number of O negatives donated by a base that defines its n Factor n factor of NaOH here is one all these simp simple simple ones you can calculate I believe you can do that on your own for example I'm writing NH3 I'm writing NH3 I'm introducing this NH3 into water I'm introducing this NH3 into water now what's going to happen here what's going to happen here this nitrogen has got the lone pair perfect this nitrogen has got the lone pair it will take H positive from here and it will get converted into what nh4 positive over here three hydrogens four so can you see this NH3 it has accepted one H positive this NH3 has accepted one H positive the number of hge positives a base accepts that defines it n Factor n factor of NH3 here is one because it's accepting 1 H positive it is accepting 1 H positive for example you have got something like this nh2 ch2 ch2 nh2 now here you have got two nitrogen atoms here one lone pair here one lone pair so it will accept how many H positives in water it will accept two H positives it will accept two h positives right so n Factor ha will be two n Factor hair will be two perfect for example I'm giving you reactions like this zinc hydroxide plus HCL let's say it gives ZN o CL plus water similarly I'm writing one reaction like this plus 2 * HCL it gives ZN cl2 + 2 * water can you tell me the N factor of zinc hydroxide in both these reactions is it same or different what do you think is it same or different what do you think same or different same or different quick see guys over here one mole of water is formed so for the formation of one mole of water this would have given one Mo of H O Negative right so it is donating 1 O Negative so it's n Factor here is one and over here it will be two perfect so if you got the N Factor you can easily calculate the equivalent weights so this was how you calculate the N factor of the basis okay now comes the most important thing the most important thing what is that that is n factor of oxidizing and reducing agents how do we exactly calculate the N factors of oxidizing and reducing agents this is the most important part this is the most important part I hope all these things whatever I discuss till now I hope every single thing is clear n factor of oxidizing and reducing agents guys understand the things which I'm telling you understand it in the similar manner the way I'm going to tell you okay perfect see guys here to make you understand all the things I'm going to write three reactions very general reactions with the help of which you'll understand first of all how do we get the N factors of oxidizing reducing agents see guys I'm writing a reaction for example a general reaction a plus b gives a positive plus b Nega this is the reaction which I wrote understand carefully first of all the oxidation of a is 0 this is zero this is + one this is minus one now tell me what is happening to the oxidation state of a it is increasing increase in the oxidation state is oxidation the one which under goes oxidation that is the reducing agent 0 to minus one decrease in the oxidation state reduction so it is the oxidizing agent first of all I identified which one is my oxidizing agent and which one is my reducing agent okay now how do I calculate n factors of these oxidizing and reducing agents first of all it is always equal number of moles of electrons lost gained or exchanged or exchanged by by one mole of substance and how do we get this how do we get this that is the point I'm writing here n factor of a which is the reducing agent n factor of a is equal n factor of a is equal what is the change in the oxidation state of a 0 to + one 0 to plus one calculate the change change will be final minus initial 1 - 0 that comes out to be one this is the change of the oxidation state for one atom of a how many atoms of a do we have only one so I'm multiplying it with one n factor of a here is one simple right n factor of a here is one what is the n factor of B here what is the n factor of B here 0 to minus one 0 to minus one calculate the change final minus initial final is minus1 initial is zero take the mod the value comes out be one one is the change for one atom how many we have only one this is the n factor of B simple simple guys yeah correct for example I'm writing something like this A2 + B2 it gives 2 * a positive + 2 * B negative this is the reaction which I wrote A2 + B2 gives 2 * A positive plus 2 * B negative try to understand oxidation at 0 0 here we have got + one here we have got minus one if I ask you what is the n factor of A2 here n factor of A2 it will be equal A2 0 to + one change final minus initial the value comes out to be one one is the change for one at but I have two so one multi by two the value comes out be two so two is n factor of A2 n factor of B2 will be equal 0 to minus one change is one one is a change for one atom but I have here two the value comes are2 so this is how you calculate n factors of different species right right people and if you got the N factors you can calculate the equalent weights molecular weight divid by n Factor perfect someone is asking question that comes from this particular topic this is the question right equivalent weight calculate the equivalent weight of this speci calculate the equivalent of that speci I hope all these things are absolutely clear to you okay I hope all these things are absolutely clear to you now guys something really important right something really important which I'm going to let you know you have got two important oxidizing agents here one is called as km4 one is called as K2 cr207 K4 and K2 cr27 my dear students this K4 or you can write M4 negative you can write M4 Nega in acidic medium it gets converted into m& positive your K4 or you can call it as M4 negative M4 Nega remember you have to remember these things in acidic medium it gets converted into MN di positive in which medium acidic medium acidic medium right in neutral in neutral or slightly alkaline medium in neutral or slightly alkaline medium the K4 it gets converted into what M2 and let me tell you in strongly alkaline medium in strongly alkaline medium the same K4 it gets converted to what M4 D negative these three things you have to remember first of all am4 in acidic medium gets converted to m& positive right in strong in in neutral or slightly alkaline medium M2 in strongly alkaline medium M4 D negative now what is the n factor of M4 negative in all these three mediums how do I do that how do I do that see oxidation St of magnes here is plus 7 here is plus2 so in acidic medium plus 7 to plus2 what is the change five five is the change for one atom of magnes I have one ATM only so n factor of K4 in acidic medium is just five remember it directly from now onwards right similarly magnes +4 so+ 7 to +4 what is the change three so n factor of K4 in neutral or slightly alkaline medium it's three similarly here it's plus 6 plus 7 to plus 6 change is one n factor in the third case is one n factor in the third case is one okay right right people is it clear K4 which is a strong oxidizing agent which is a strong oxidizing agent in acidic medium it shows n factor of five in neutral medium it shows three in strongly alkaline medium it shows one yeah okay and now people since this C4 it's a strong oxidizing agent it a strong oxidizing agent what it does in acidic medium that is again very important that is again very important what it exactly does in acidic medium first of all in acidic medium you know your K4 or you can say m for negative it gets converted into m& D positive in which medium I'm talking about acidic medium now what it does what it does to others let's have a look the first thing K4 in acidic medium converts Fe di positive into Fe Tri positive the first thing which you have to remember it converts Fe di positive into Fe Tri positive number two it converts SN di positive into SN +4 again in acidic medium right it converts oxalate ion into carbon dioxide it converts H2O2 into O2 H22 into O2 it converts H2S into sulfur it converts H2S into sulfur it converts I ne into I2 it converts I ne into I2 these things you have to remember guys K4 in acidic medium converts Fe diosi into Tri posi SN plus2 into SN plus4 oxalate into carbon dioxide H22 into O2 H2S into sulfur I2 I2 if you just remember this part you can easily write the reactions how you can easily write the reactions you need not to mug up the reactions which you see in D and block elements for k14 and K2 C2 you can check that here only how right to understand the first reaction your K4 your K4 converts Fe D postive in which medium acidic medium into what Fe triol 2 itself gets converted into M and Di positive and with that you just start write water this is the reaction done and yeah similarly your K4 your K4 look at the second reac ction it converts SN D positive in which medium acidic medium into what SN +4 itself gets converted into m& D positive and with that you'll be writing water as well that can you similarly write all the reactions can you similarly write all the reactions say yes or no in the chats I don't want to waste time in order to write all these reactions now can you write all these reactions on your own say it quickly perfect now if you talk about another one K2 cr27 K2 cr207 this is also a strong oxidizing agent and in acidic medium it gets converted into chromium tripositive in acidic medium it gets converted into chromium tripositive if I ask you first of all what is the n factor of K2 cr27 here quickly chromium here is in plus 6 oxidation state so plus 6 to plus three change that is three three is the change for one atom of chromium but I have two so 3 to are six n factor of K2 cr207 here is 6 okay now since K2 cr207 it's again a strong oxidizing agent so it is going to show the similar reactions which your k04 showed first one it converts Fe di positive into F Tri positive number two it converts SN +2 into SN +4 number three it converts I into I2 number four it converts oxalate ion into carbon dioxide to carbon dioxide number five what was the fifth one H2S to sulfur number six number six H2O2 to2 it shows the similar reactions now you should be able to write all these reactions on your own I'll just show you one or two rest you can do on your own see guys so first reaction will be your K2 cr27 it converts what Fe di positive in which medium acidic medium into what into Fe Tri positive itself gets converted into CR Tri positive with that you'll be writing water as well so this is the reaction done this is the reaction let's say I need to write this particular reaction K2 cr27 K2 cr27 convert s oxalate in which medium acidic medium into what carbon dioxide itself gets converted into chromium trios and with that you'll be writing water as well done in the similar way you can write all these reactions in the similar way you can write all these reaction someone is say I'm seeing you for the first time in the chat yeah suddenly you enter the session how will you get watch it from from the beginning then you'll get okay I believe all these reactions you can easily write from now on yeah am I clear people say yes or no in the chats am I clear am I clear am I clear this chapter will take still 1 hour one and a half hour one one and a half hour this chapter will take more yeah perfect now guys try to understand try to understand try to understand just a second just a second let me see whether you can calculate the N factors of these oxidizing and reducing agent okay the first question calculate the N factor of this I2 here can you do it n factor of I2 yes lania no need to worry can you calculate the N factor of I2 here what is it quickly see guys first of all the oxidation of iodine here is zero here it's + 5 so 0 to + 5 what is the change final minus initial the value is five five is the change for One ATM but I have two so 5 2 are 10 n factor for I2 here is how much it is 10 correct it is 10 yeah okay by chance I'm asking you to calculate the N factor of io3 negative what you'll do io3 negative what you'll do you'll just reverse the same process reverse now you'll be calling this as initial this as final so final minus initial is five five is the change for one atom and only one atom of iodine so n factor of io3 negative here will be five yeah just reverse the stuff that's all just reverse the stuff that's all okay this though you can easily do quickly tell me the N factor of I4 negative here io4 negative quickly see iodine over here is in plus 7 oxidation state this is zero so plus 7 to 0 plus 7 to 0 initial final final minus initial 7 s is the change for one atom I only one atom so n Factor here is seven now I2 0 to plus 7 0 to plus 7 final minus initial that's 7 7 is the change for one atom but I have two so 7 2 are 14 this is the n factor of I2 yeah I hope you can solve these sort of questions can you get the N factor of this particular speci can you get the N factor of this particular speci quickly n factor of this particular species can you get it okay guys see first of all if we calculate the oxidation set of iron here let's say it's X multiplied by 0.94 plus oxygen shows min-2 net charge is zero so the value of x will come out to be + 200 94 + 200id 94 here iron is in plus three oxidation state so this to this calculate the N Factor n factor is going to be equal final oxidation state minus initial oxidation state final minus initial this is the change in the oxidation state for one atom of iron but how many we have 94 so multiplying it with 9 for whatever value you get that's something which will give you n factor of this nonmetric compound yeah perfect right okay tell me the N factor of this oxalate here tell me the N factor of this oxalate ion here quickly quickly people tell me the N factor of oxalate ion quick quick quick very quick done done people you should be able to solve this oxidation of carbon here is plus three here it's plus4 so plus 3 to plus4 changes one plus 3 to plus 4 Change is one one is the change for one atom but I have two atoms of carbon so n factor of this particular spey is two here right so I believe you can easily solve the N Factor stuff yeah I believe you can do that now guys till now till now till now till now what I was doing till now if you observed we were calculating the N factors of those oxidizing or reducing agents wherein wherein wherein oxidation state of only one element was changing and we were calculating change in the oxidation state of that element right till now if you observed I gave you only those oxidizing and reducing agents wherein oxidation state of only one atom was changing and we were calculating that change in the oxidation state what if there's a specii what if there's the oxidizing or reducing agents wherein the oxidation state of more than one element is changing simultaneously how do we calculate this n Factor remember directly if you have got a specie in which oxidation state of let's say two elements is increasing or decreasing let's say you got a specie wherein oxidation state of two elements is is increasing or decreasing how do we calculate n factor of that particular speci it is simply going to be N1 plus N2 let's say you got a speci in which oxidation state of one element is increasing another one is decreasing it is going to be mod of N1 minus N2 just remember it directly just remember it directly whenever you see oxidation state of more than one element simultaneous increasing or decreasing simultaneously oxidation state of both elements is increasing or simultaneously oxidation state of both elements is decreasing at that point of time N1 plus N2 if one increases one decreases N1 minus N2 now what is this N1 what is N2 try to understand see guys if you look at this particular case if you look at this particular case carefully iron ha is in plus two oxidation State carbon here is in plus three oxidation state because it's oxalate right here iron is in plus three here carbon is in plus4 if you check carefully plus 2 to+ 3 increase + 3 to + 4 increase oxidation state of more than one element is increasing simultaneously so what do I do how do I calculate it n Factor its n Factor will be calculated as N1 plus N2 now the point is how do I get N1 and how do I get N2 that is something which we already discussed have a look + 2 to + three change is one one is the change for one atom of iron and I've got only one N1 is one plus three to+ 4 Change is one one is a change for one atom of carbon but I have two so N2 is two so N1 is 1 N2 is two so 2 + 1 comes out be three so three is the overall n factor of this particular speci I believe this is clear to everyone let me give one more example let me give one more example people all right look at this particular example first thing oxidation state of copper here is plus one sulfur here is min-2 copper here is +2 sulfur here is +4 so + 1 to +2 is increase -2 to +4 is also increase oxidation state of more than one element is increasing simultaneously so how do we do it N1 plus N2 plus 1 to plus 2 changes one one is the change for one atom but I have two so N1 is 2 - 2 to + 4 - 2 to Plus 4 Change is six right change is six and there's one atom only N2 is 6 6 + 2 is 8 6 + 2 is 8 yeah right people now let's see one of the element wherein oxidation state of one element increases another element decreases let's see what happens in that just a second let's take one example of that as well look at this particular case look at this particular case if you look carefully iron has in plus three oxygen State sulfur is in min-2 Iron here is in plus2 right and sulfur here is in plus4 my dear students if you look carefully which case it is which case it is + 3 to +2 decreas minus 2 to + 4 increase oxidation state of one element is decreasing another one is increasing simultaneously how do I do it N1 + N2 or N1 minus N2 I'll be calculating its n Factor as mod of N1 minus N2 now get the N1 get the N2 + 3 to + 2 change is one one is the change for one atom of iron but I have two N1 is two similarly - 2 to + 4- 2 to plus 4 Change six final minus initial change is six six is a change for one at but have three so 6 3 are 18 right so the value overall value comes out be 16 I hope this is clear I hope this is clear now the final case of the N factors the final case of the N factors the final case of the N factors how do we calculate n factor for the substance how do we calculate n factor for the substance that under goes disproportionation n Factor of substance that under goes disproportionation that under goes disproportionation n factor of that substance which under goes disproportionation this is the final case of the INF factors yeah no that was not disproportionation the one which I gave you see what is disproportionation let's say you got the reaction like this A2 it gives a positive plus a negative let's say this is the reaction okay oxidation state is zero here is Plus + 1 it is minus1 0 to +1 increase in the oxidation state 0 to minus1 decrease in the oxidation state so I can say oxidation state of same element is simultaneously increasing and decreasing there it was not the same element those were two different elements speci was same but elements were different here the element is same yeah right people the oxidation state of one element is increasing I mean oxidation state of an element is increasing as well as increasing simultaneously right how do I get it n Factor its n factor is calculated as N1 N2 / N1 + N2 N1 N2 ided by N1 + N2 yeah right people it is basically oxidation as well as reduction of same species is happening of same element is happening right right people now get the N1 get the N2 we are done we have sorted see 0 to minus one change is one one is the change for one at have two so N1 value here is two N1 value here is two 0 to minus one change is one one is a change for one at have two so N2 Valu is again two so 2 * 2 / 2 + 2 the value comes out to be 1 1 is the n factor of this A2 overhead I'm giving you a question it'll be the answer of this question br2 gives BR Nega plus br3 Nega tell me the N factor of this br2 you you are in a position to answer this one trust me oxidation state is zero here is minus one here is + 5 0 to minus one decrease 0 to + 5 increase get its n Factor it is N1 N2 divide by N1 plus N2 N1 N2 / N1 + N2 now 0 to minus1 change is one one is the change for one atom but I have two N1 value is two 0 to + 5 change is five five is the change for one atom but I have two so 5 2 is are 10 right so this is going to be 10 * 2 / 10 + 2 the value is 20 / 12 so this is 45 43 so n factor of br2 here comes out to be 5x3 yeah there was one question asked long back in as H2O2 gives H2O Plus O2 I'm giving you the oxidation stat this min-1 this min-2 this is 0o can you tell me it's n Factor right now can you tell me it's n Factor right now n factor of H22 quickly quickly people quickly quickly quickly quickly everyone everyone everyone people are saying it's one are you sure are you sure are you sure I feel you are sure about it yeah I feel you're sure about it so let's see -1 to -2 so decrease in the oxidation state -1 to 0 increase in the oxidation state again oxidation state of the same element is increasing as well as decreasing simultaneously right right people so how do we calculate n Factor N1 N2 div N1 + N2 now min-1 to-2 change is one one is a change for one atom but I have two so N1 is equal to 2 minus1 to 0 what is the change one one is the change for one atom but I have two so N2 is also 2 so 2 * 2 divided by 2 + 2 the value comes out be 1 n factor of H22 here in this disproportionation reaction it is it is one be careful with this right I have seen a lot of teachers think n factor of H2O2 is equal to two in this disproportionation but it is one remember this yeah remember that perfect guys perfect so with this your n Factor calculation is done and dusted now balancing won't take more than 15 minutes complete balancing guys wait what break break you are saying what break do you deserve the break or should do I deserve the break I deserve the break right more than you but I I'm not asking for it nor I want it we we have to complete a lot of stuff a lot of stuff okay so first of all since you are not going to write any board's examination you just have to write need examination so so so there is no need to go for all the methods of balancing it is just you should know how to balance the reaction that's enough right it is just you know you should know how to balance the reaction whether you are balancing it with this method or this method one and the same thing my our point is just to balance the reaction so we are not going to discuss all these balancing stuff we are just going to discuss one method which is going to be easy for you by means of which you can balance the reactions okay and that that method is your oxidation number method oxidation number method right oxidation number method so first of all let's discuss oxidation number method in acidic medium see how easy it is right see how easy it is see how easy it is for example my de students I'm writing a reaction like this M4 negative plus Fe e di positive it gives F Tri positive plus MN di positive this reaction let's assume you will have you have to balance in acidic medium how you guys are going to do it what are the step steps to be followed understand in less than 40 seconds you can balance the reaction if you understand the concept behind it see guys first of all get the oxidation states of these species + 7 + 2 + 3 + 2 got the oxidation States right calculate the N factor of this particular speci + 7 to +2 changes five so it's n Factor five plus 2 to+ three changes 1 it's n factor is 1 the first thing what I did over here is I calculated the N factors I calculated the N factors of all the reactants I calculated the N factors of all the reactants right got the N factors after getting the N factors just crisscross them make this one as a coefficient here make this five as the coefficient here first get the N factors after getting the N factors just crisscross them okay number two number three balance all the atoms except hydrogen oxygen balance B all the atoms except hydrogen oxygen so one magnes one magnes five iron five iron balance all the atoms except first one get the N factors of reactants second crisscross them third balance all the atoms except hydrogen oxygen now you are balancing the reaction in which medium in acidic medium now in acidic medium what do we need to do we'll be balancing oxygen with the help of water molecules and then we'll be balancing hydrogen with the help of H positives Now understand first you're going to balance oxygen how many oxygen atoms do we have on left side only four is there any Oxygen on right side no so can I say right side is oxygen deficient yes how many oxygen deficiency is there four oxygen deficiency and O oxygen deficiency is removed by water molecules so I'll be adding four water molecules on this side so that oxygen becomes equal on both the sides so oxygen is balanced now balance hydrogen how many hydren on this side 4 2 are 8 any hydrogen here no right so this side is hydrogen deficient and that side which is hydrogen deficient add H positives on that side eight hydrogen's no hydrogen so add h8 H positives here now this is a balanced chemical equation am I clear you are not supposed to do anything extra okay just follow these marathons follow the question practice which we do extra 500 stuff perfect that's more than sufficient nothing else you have to do don't don't waste your time done I'll take one more example with that we'll get over I'll take one more example CH for you guys I'll take one more let's say let's say let's say we have got the reaction like this okay cr27 dga2 plus oxalate ion it gives CR Tri positive plus carbon dioxide balance this reaction in acidic medium again the similar approach again people the similar approach I'll be using here get the N factors you know how to do it it's n Factor 6 it's n Factor plus 3 + 4 is 1 1 two okay okay you got the enactors now after getting the enactors after getting the enactors crisc cross them so make this two as a stoet coent here make this six as a stoet coent here done balance all the atoms except hydrogen oxygen except hydrogen oxygen four chromium four chromium 12 carbon 12 carbon now balance oxygen with the help of H2O molecules how many oxygen on left side 7 to 14 6 4 24 so 24 + 14 comes out be 38 38 oxygen atoms 24 oxygen atoms which side is oxygen deficient right side how many oxygen deficiency 14 oxygen deficiency 14 oxygen deficiency means add 14 water molecules yeah now talk about hydrogen 28 hydrogen atoms no hydrogen atoms so add 28 H positives on this side right to balance hydrogen now this is a balanced chemical equation no need to do anything yeah done people is everything clear this was how do you balance the reactions in acidic medium okay right I want everyone to say it in the chats is every single thing clear till here right okay now comes your which medium basic medium in basic medium I'll give you one example that's all in basic medium let me give you one example that's enough that's enough anyways balancing is least important okay how do we balance the redox reactions in in basic medium uh okay look at this particular reaction carefully again the first F steps are same the first few steps are same oxidation state of chlorine here is zero iodine here is + 5 chlorine here is minus1 iodine is + 7 get the N factors 0 to minus one change is one one is the change for one at two so it's n factor is 2 plus 5 to plus 7 it's n factor is also two so you got the N factors you got the N factors now you got the N factors now chrisc cross the N factors chrisc cross the N factors nothing else Chris Cross the N factors make the two coefficient here here as well you have crisscrossed the N factors balance all the atoms except hydrogen oxygen so four chlorine four chlorine two iodines two iodine so balanced all the atoms except hydrogen oxygen right now since the reaction is happening in basic medium till here everything is same but now in basic medium what do we need to do we need to balance the negative charge first by adding o negatives on that side which is negative charge deficient we need to balance in acidic media we used to balance oxygen first by adding water molecules but here I'll be balancing the negative charge first by adding o negatives on that side which is negative charge deficient and after that I'll balance oxygen with the help of water that's all that's all now have a look no charge here if you look at this one io3 molecule carries minus one charge so two will carry min-2 okay one cl carries minus1 charge so four will carry -4 one io4 carries minus1 2 will carry -2 so -4 - 2 comes out be Min - 6 so you got the total negative charge on left side total negative charge on right side okay tell me people - 2 - 6 which side is a negative charge deficient left side or right side which side is negative charge deficient left side or right side quickly left side or right side quickly left side is negative charge deficient how much negative charge deficiency is on left side four and that negative charge deficiency is compensated by o negatives so how many o negatives I would add here four o negatives because there was the deficiency of four negative charges right done negative charge balanced after that after that I have to go to oxygen how many oxygen on left side 3 2 are 6 6 4 10 so 10 oxygen atoms on left side and 4 2 are 8 eight oxygen atoms on right side 10 8 which side is oxygen deficient which side is oxygen deficient 10 8 right side is oxygen deficient how much oxygen deficiency is there two oxygen deficiency that means oxygen deficien is compensated by water molecules so how many water molecules do I have to add on right side I'll say I'll be adding two water molecules on right side perfect the reaction automatically gets balanced perfect right people am I absolutely clear to everyone you can let me know quickly in the chats this was your balancing which is done and dusted quick quick people done quickly everyone in the chats no the chapter is not complete yet wait chapter is not complete completed yet there are few more important things which you need to know I hope all these things are absolutely clear to everyone of you let's move ahead guys let's move ahead let's move ahead well there was one concentration term which I did not discuss with you that was normality so let's try to cover this normality and with normality we'll end this chapter okay with normality we'll end this chapter okay so guys before talking about normality before talking about normality before talking about normality I need to tell you about the number of gram equivalents the number of gram equivalents of any substance the way you calculate number of moles of any substance in the similar way we can calculate number of gam equivalents of any substance the way we calculate number of moles of any substance okay the way we calculate number of moles of any substance in the similar way we can calculate number of gram equivalents of any substance okay and how do we calculate number of gram equivalence of any substance it is equal mass of the substance in grams divided by equivalent mass of the substance divided by equivalent mass of the substance divided by equivalent mass of the substance I'll write it mass of the substance in Gs divide by what divided by equivalent mass of the same substance this is the first Formula by means of which we can calculate number of gram equivalent of particular substance it is also equal number of moles of the substance multiplied by n factor of the substance right number of moles of the substance multiplied by n factor of the substance okay now similarly there's a term called as number of M equivalents number of M equivalents of the substance it is also it's equal given mass of the substance in grams divided by equivalent mass of the substance multiplied by th000 or instead of moles you can write number of Mill moles of the substance multip by n Factor first of all remember these two results by means of which you can calculate the number of gam equivalents of the substance and number of M equivalents of the substance yeah perfect perfect guys okay I'll be discussing tiation also just wait just wait so my dear students after remembering these formulas after remembering these formulas now I'm going to move on to something called as normality normality which is represented by capital N understand normality which is repr present by capital N how do you define the term normality first of all how do we Define the term normality first of all normality is defined as the number of the number of gram equivalents of solute the number of gam equivalent of solute present in present in 1 lit of solution how do you define marity number of moles of solute present in 1 lit of solution right in place of moles what do we use gram equivalents number of gam equal of solute present in 1 lit of solution how do you calculate normality simple formula number of gam equivalents of solute divided by volume of solution in liters gam equal of solute means mass of solute in Gams divide equivalent mass of solute volume of solution in liters means if I take it in ml in the numerator I'll have to multiply that with th right perfect and there is one more direct result between normality and marity normality is equal to marity of the solution multiplied by n factor of solute these are the three results which I would want you guys to remember as well these are the three results which I want you guys to remember yeah perfect guys there is one more thing do you see normality is equal number of gam equivalence of solute divide by volume of solution in liters so can you say gam equivalence is equal normality multiplied by volume in liters gram equivalence is equal this multiplied by this I can say that right so so gram equivalence I can calculate like this as well normality multipli by volume of solution in liters M equivalence I can calculate like this normality multiplied with volume of solution in ml this is the just this is the only difference this is the only difference if I take volume in ml it gives you m equence if I take volume in liters it gives you gram equence that's all that's all that's all people if this is clear if this is clear if this is clear just tell me the answer of this very simple and basic question quickly what do you think is the answer of this one are we done with this are we done with this people quickly quickly are we done with this okay the gram equivalence in 50 mL of 0.5 normal NaOH guys I told you already number of gam equivalents is equal normality multipli volume of solution in liters normality is given as 0.5 volume of solution is given as ml 50 ml so divide it with th000 then only you'll be getting liters just solve it and get the gam equal of solute in the solution right perfect now comes the actual stuff and that is the last topic of the chapter that is redox titration that is redox titration okay that is redox titration or I'll be calling it as equivalent concept as well I'll be calling it as equivalent concept as well I'll be calling it as equivalent concept as well this is important have a proper eye on this right there can be questions asked from this particular Topic in this particular examination as well okay equivalent concept my dear students for example for example for example I have got a general chemical reaction a plus b it gives C plus d okay equivalent concept says that you are not supposed to balance the reaction equivalent concept says that no need to balance the reaction okay no need to balance the reaction simple equivalent concept says that that g equivalent of a gram equivalence of a reacted in the reaction has to be same as that of gram equivalence of B reacted which has to be same as that of G equivalence of C formed which has to be same same as that of gam equivalence of D formed one simple statement you need to write as per equalent concept whatever gram equal of a would have reacted same gram equivalence of B also would have reacted same gram equals of C would have formed same gram equals of D also would have formed well you can write the same statement in terms of M equivalence as well you can say mlie equalent of a reacted has to be equal m equivalence of B reacted has to be equal m equivalence of C formed mlie equivalence of D formed m equal of D formed this is the concept which I'll be using in the questions number one number one number two if this was case one let's say this is case two now this was case one now I'm going to give you case two for example you see a scenario like this let's say this is a container and in this container we have got a B and C it's a mixture of a b and c it's a mixture of a B and C for example I'm adding one more substance D over here let's say I'm adding one more substance D over here and for example the D which I'm adding it is reacting with a it is reacting with B but it's not reacting with C let's see it's reacting with a reacting with v but for example not reacting with C now how do you write the statement over here in the case second you'll write it like this gam equivalence of a reacted plus gram equivalence of B reacted it has to be equal directly G equivalence of D reacted I'll have got nothing to do with the c over here I've got nothing to do with the c over here nothing to do with C over perfect so remember these two statements remember these two statements okay these two statements are super important my dear students so remember them directly yeah remember them directly now how do we use them how do we use them before that before that there are few n factors which I want you guys to remember directly few n factors which I want you guys to remember directly for example for example K4 in acidic medium tell me it's n Factor we have discussed in acidic medium its n factor is five K2 cr207 K2 cr207 in acidic medium it's n factor is six okay now people for example you have got feo4 you have got feo4 let's say in acid AIC medium it's reacting with K4 let's say in acidic medium it's reacting with K4 so can I say this K4 it will convert this Fe D positive into Fe Tri positive Fe D positive is being converted into Fe Tri positive so change is one so if I ask you what is the n factor of feso4 here when feo4 when feo4 will be reacting with K4 or K2 cr207 in acidic medium its n Factor will be taken as one its n Factor will be taken as one correct now similarly if you have for example fc24 feros oxalate if it reacts with K4 or K2 c27 in acidic medium it's n Factor will be three just remember these few of n factors directly right perfect then then then you have got Fe c24 whole thce peric oxalate it's n Factor you'll be taking a six okay what else can we need here more salt more salt if it reacts with K4 or K2 cl27 in in and I mean if it reacts with k14 or K22 acidic Med effect is one more salt remember directly then you have got cu2s six cus 8 right C 8 sodium oxalate na2 c204 or potassium oxalate K2 c204 when it combines with your k14 or K2 C2 in acidic medium it's n factoral beating us to perfect remember these n factors and if you remember these n factors now we are in a position to solve a lot of questions we are in a position to solve lot of questions but before solving the questions I would want you guys to remember these formulas which I gave you how many formulas are there to calculate gram equivalence three formulas one is mass by equivalent mass given mass by equivalent Mass second moles multip by n Factor third normality multiplied by n factor sorry normality multiped by volume of the solution in liters similarly given Mass divide by equalent mass multip th milles multip n Factor third is normality m by volume of solution in milliliters right these are the results which we have you know it these are the results which we have now people it is the time it is the high time to use these results and do the questions based on titration the first question which is on your screen the first question which is on your screen the first question which is on your screen quickly quick quick quick can you do this see guys how exactly we are going to solve these questions see exactly how we are going to solve these questions first of all the actual procedure of solving these questions is you're supposed to write the reaction first then you are supposed to balance it then etc etc takes a lot of time okay uh did I write WR the N factor of Cu 2s reverse guys n factor of cu2s is 8 n factor of C is 6 if I interchanged by any mistake please change that c2s it is 8 C it is six okay all right now look at this particular question carefully look at the question carefully what volume of 0.1 m k04 is required to oxidize 100 ml of 0.3 M of peris oxalate in acidic medium so try to understand which all things are reacting over here as for the question your K104 is reacting with what your K4 is reacting with Fe c204 peris oxalate in which medium in acidic medium okay if I represent this by number one I'll be representing this by number two okay if I ask you what is the n factor of K4 in acidic medium you should know it five what is the n factor of feris oxalate here it will be already discussed already discussed what volume of 0.1 M K4 so marity of K4 is given 0.1 marity of FIS oxalate is also given 0.3 what volume of this what volume of this is required to oxidize 100 ml of this its volume is given correct people Now understand now understand and analyze things tell me one thing tell me one thing people you know I have already given you one result which converts I mean which connects normality and marity over here if you look carefully marity we know it's n Factor we know so marity multiplied by n Factor gives me normality so normality of 1 is equal to 0.5 correct mity multiplied n Factor here gives me normality so normality of two gives is equal to 0.9 done now tell me one thing tell me one thing tell me one thing tell me one thing volume is given in ml volume is given in ml the statement which I'll be using should I be using the statement in terms of gram equivalence or mil equivalence volume is an ml volume ml means mil equivalence I'll directly say mil equivalence of K4 reacted has to be equal what M equivalence of peris oxalate reacted right M equivalence of one means normality of 1 multiplied with volume of 1 in ml this is normality of two multipli with volume of 2 in ml what is normality of 1 0.5 volume of 1 we need to calculate normality of 2 0.9 volume of 2 is 100 V1 comes out to be 180 ML so what is meant by this 180 ML people what is meant by it it means that 180 ML of 0.1 m km4 is required to oxidize 100 ml of 0.3 M peris oxalate in acidic medium yes am I clear with this am I clear with this this is the first type of the question which can be asked in the titration first type of the question the second type I'm going to show you second type second type how many moles of K4 how many moles of K4 will be required to react completely with one Mo of K22 as per the question is concerned over here if you look carefully K4 is reacting with what potassium oxalate in acidic medium I'm calling this as number one this I'm calling as number two n factor of K4 in acidic medium is five n factor of this over here will be two you know it already discussed now people now how many moles of K4 how many moles of K will be required to react completely with one mole of potassium oxalate one mole of potassium oxalate perfect now tell me since we are given with the data for moles not moles moles means gram equals if it was Mill moles then mil equals I'll write the statement in terms of gram equal I'll say gam equivalence of 1 has to be equal gam equivalence of two now how many formulas do we have to calculate gram equivalence one was mass result second was moles result third one was normality right I'll use moles result moles of one multip by factor of one is equal moles of two multip factor of two what is moles of one needs to be calculated this is five this is one this is two so N1 comes out be directly 0.4 I would say 0.4 moles of k04 are required react with one mole of potassium oxalate in acidic medium am I clear am I loud and clear people am I loud and clear say it similar type of question similar type of question just relate the gram equivalence you can do it you can do it N1 multi n Factor one is equal to this this this okay similar type of question similar type of question look at this one look at this one six moles of a solution of positive the question is six moles of a solution of positive 6 moles of a solution of an positive 6 moles of a solution of a positive requires 2 moles of cr27 DGA requires 2 moles of cr207 D negative for the oxidation of a positive to A3 negative for the oxidation of positive to A3 negative and itself in acidic medium it gets converted into CR Tri positive what do we have to calculate get the value of n get the value of this look carefully Guys these two are reacting so equate their gram equivalence directly first of all what is the n factor of this n positive plus n to + 5 plus n to + 5 change change is final minus initial right and we have got only one atom what about the N factor of cr27 D negative plus 6 to plus 3 changes three 3 is the change for one at two so 3 2 6 equate the gram equal gram equ of first has to be equal to gram equ of second N1 multi n Factor 1 is equal N2 * n Factor 2 what is N1 6 what is this 5 - n is equal 2 * 6 6 6 cancel so n value comes out be how much n value comes out be three what is this n n was a charge present on a so basically your a is present in a tri positive state am I clear am I clear people am I clear am I clear am I clear am I clear yes perfect and that's all what can be asked from the redox okay is the redox reaction chapter clear is the redox reaction chapter clear Which chapter do you want to start the next Which chapter do you want to start the next Which chapter do you want to start the next whatever chapter we are starting the next that will take us 4 hours because now these are Big chapters right that's why I told you I have kept two marathons for class 11th physical chemistry by the way do you think I listen to you yeah I'm just making you happy by asking you asking your suggestions etc etc huh do you really think do do you really think I'll listen to you huh just tell me that huh are are you that important I'll ask you huh what do we say I'm I'm the king of my own Kingdom right wait let me see what majority asks I'll create a poll here you just tell me okay then we'll start that chapter only after the break Buton just a second I'll create a poll just just be like this how to create a pole now I'm technologically unsound I know that uh just a second just a second start the poll Which chapter next options atomic structure number two is Thermo Dynamics number three is equilibrium po has been started okay you just fill it and let's take a break after the break I'll see what majority would have said and I'll go with that yeah it is 340 right now so break till have a good break okay have some tea okay so first let's take a break am she listens to you huh that's why I don't like her she listens to everybody right she's not supposed to do that I'll see you exactly at 4:50 now you can have your tea you can fresh yourself and then once we get back I'll see what majority said and we can go ahead right we can go ahead with what majority says hello guys I'll see you after the break H take care take care guys see you e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e so guys is everyone back is everyone back yes all results let me check the results of the poll uh just a second oh it is a tie 40% atomic structure 40% thermodynamics huh 40% atomic structure and 40% thermodynamics so now which one to start quick quick quick quick quick quick say it in the chats I'll do that say it in the chats I'll do that I think majority wants to start Thermo right okay Thermo Thermo Thermo Thermo Thermo Thermo thermodynamics all right hello then I'll listen to you only all right so I'll listen to you only let's start with hermo only yeah okay uh so tell me once in the chats like whatever things I discuss till now is every single thing clear Perfect People Perfect all right then let's go ahead with the thermodynamics let me check the slides once where is the thermodynamics where is the thermodynamics okay here we have got the thermodynamics just a second just a second just a second just a second just a second B you can call up everyone till then all right can you let me know in the chats like how many of you have studied thermodynamics before honestly you can say yes or no how many of you have studied thermodynamics before okay anyways I'll be starting from the basics only you need not to worry you need not to worry just I'm just asking you I'm just asking you I'm just asking you all right soell then let's get started let's get started people first of all let's have a look on the basic terminologies which are involved in the chapter thermodynamics the basic terminologies which are involved in the chapter thermodynamics and I believe all of you would be already knowing the basic terminologies which involves your system your surroundings your boundary now what is the meaning of all these particular things okay what is the meaning of all these particular things my dear students if you look at this particular definition system is basically that part of the universe that part of the universe which is under thermodynamic investigation that part of the universe which is under thermodynamic investigation what is meant by that just let me make you understand that quickly for example this is the container let's say this is the container in this container my dear students for example you have kept water you have kept water inside this particular container okay there's water in this particular container for example I have to raise the temperature of this water let's say I have to raise the temperature of this water by 1° Centigrade let's say I have to increase the temperature of the water by 1° Cen what do I have to do I have to supply heat I have to supply heat okay I have to supply heat to increase the temperature of this water by 1° C integr my dear students for example I have to do one investigation I have to do one investigation like how much heat is required this is the investigation which I need to do how much heat is required to raise the temperature of this water by one degree okay I have to do one thermodynamic investigation related to this particular water which is there in the container I have to check how much heat is required to raise the temperature of this water by 1° to raise the temperature of this water by 1° can I say this water which is inside this particular container right now can I say this water is a right now under thermodynamic investigation yes this water right now is under thermodynamic investigation correct that part of the universe which will be under thermodynamic investigation what do you call that as you call that a system so this water which is there in the container it is right now under thermodynamic investigation so I'll be calling this water as my system this is my system right now this is the part of the universe this water in the container is under investigation right now I'm calling this as the system my dear students apart from this water whatever is there in the universe apart from this water whatever is there in the universe I'll be calling that as surrounding I'll be calling that a surrounding okay and anything that separates system from the surroundings anything that separates system from the surroundings that is something which you call as boundary as you can see these walls of the container these walls of the container they are separating your system from the surroundings so you'll be calling it as you'll be calling it as boundary you'll be calling it as boundary right so three terminologies one is system one is surrounding one is boundary that part of the universe which will be under thermodynamic investigation that is your system anything in the universe except system that is your surrounding anything that separates system from the surroundings that is your boundary clear now how many different types of boundaries do we have how many different types of boundaries do we have myar students broadly speaking I can say in this particular chapter we are going to use these types of boundaries number one real boundary real boundary is the one which can be seen by the eyes which can be seen by the eyes that boundary which can be seen by the eyes imaginary boundary it cannot be seen it cannot be seen by the eyes right dimic boundary it is that boundary through which heat can pass through which heat can flow right through which heat can flow okay that boundary which allows the passage of heat which allows the passage of heat which allows the passage of heat either from system to surrounding are from surrounding to system atic boundary it does not allow the passage of heat it does not allow the passage of heat it does not allow the passage of heat either from system to surroundings or from surrounding to system right perfect which does not allow the passage of heat either from system to surroundings or from surrounding to system so one is real which can be seen by the eyes imaginary cannot be seen dimic is the one which allows the passage of heat either from system to surrounding or from surrounding to system adab batic is the one which does not allow the passage of heat either from system to surroundings or from surrounding to system right now people now people these are some basic basic things which I'm telling you these are some basic basic things I did not start the chapter eight what are the types of the system which we have what are the types of the system which we have my dear students we have got three types of system over here one is called as open system one is called as closed system and one is called as isolated system what is an open system what is closed and what is an isolated system let's have a look open system is the one open system is the one which allows the exchange of which allows the exchange of energy as well as matter with the surroundings which allows the exchange of energy as well as matter with the surroundings close system is the one which allows only the exchange of energy with the surroundings not the matter and isolated system is the one which neither allows which neither allows the exchange of energy nor the matter with the surroundings now what does it mean what does it mean let's try to understand for example I have got an open container let's say this container is open on one side okay there is water in this particular container there is water in this particular container H2O liquid perfect this water is my system right now it is under investigation it is under investigation imagine that the walls here are dimic imagine that the walls here are dimic that means through these walls heat can pass heat can either go from system to surroundings or from surrounding to system heat can pass right because I've used the walls as di now people understand if I Supply some heat to this system I'm supplying some heat to the system okay I'm supplying heat to the system can I say between system and surroundings exchange of energy is happening yes exchange of energy is happening now people what is going to happen when you supply heat to this water what will happen evaporation will start water will start getting convert into Vapors those Vapors will go into the surroundings those Vapors will go into the surroundings can I say between the system and surroundings exchange of matter is also taking place whenever you see such kind of the system which allows the exchange of energy as well as matter with the surroundings exchange of energy as well as matter with the surroundings that's what you call as an open system that's what you call as an open system simple right now people if you close this if you close this if you close this will there be exchange of matter between system and surroundings there won't be exchange of matter with between system and surroundings but exchange of energy will be there any such system which allows the exchange of energy not the matter with the surroundings what do you call that as you call that as the closed system and any such system which neither allows exchange of energy nor the matter with the surroundings you call that isolated system and the example of the perfectly isolated system is the universe if you take entire universe under thermodynamic investigation if you take entire universe under thermodynamic investigation if you make entire universe as the system then it becomes the example of the perfectly isolated system am I clear with this this was just the basics this was just the terminologies right let's not talk about them in detail the detail discussion has to be for extensive and intensive properties over here it needs a detailed discussion so people What are extensive and what are intensive properties What are extensive and what are intensive properties let me tell you extensive properties are the ones extensive properties are the ones extensive properties are the ones whose value extensive properties are the ones just a second extensive properties are the ones whose value chains whose value chains on changing on changing the size of the system on changing the size of the system or I can say whose value change on changing the amount of substance present in the system whose value chains on changing the size of the system or the amount of substance in the system amount of substance present in the system I'll write comma otherwise intensive otherwise intensive now what is meant by all this try to understand people what exactly is meant by all this okay try to understand for example this is a block it's an block for example let's say this it's an iron block this iron block imagine this iron block is under investigation right now it is under thermodynamic investigation so I should be calling this iron block as my system right now so this is my system this iron block is my system right now its mass is M its volume is V its temperature is T right mass of this iron block is M volume is V temperature is Ste now what exactly I'm going to do I'll be dividing this iron block into two parts I'll be dividing this iron block into two parts right after dividing this iron block into two parts if I take the smaller part if I take this smaller part if I take the smaller part what will be the mass of the smaller part initially the mass of whole iron block was M now it's going to be M by2 its volume will be V by 2 what about its temperature its temperature is still going to be T okay so what am I doing am I changing the size of the system yes initially this was the system now only the smaller part is the system I changed the size of the system I changed the size of the system now my dear students if you look carefully how many properties did I Define for this system I defined its mass m volume v temperature T when you changed the size of the system mass of the system changed from M to M by2 volume of the system changed from V2 vx2 temperature did not change those properties of the system who value change on changing the size of the system are what you call as extensive properties and the ones whose value does not change you call them as the intensive properties so I defined three properties here out of these three properties mass and volume these are your extensive properties and temperature is your intensive property right temperature is your intensive property are you guys clear with what are extensive and what are intensive properties quickly are you guys clear with this yes are you clear with this perfect perfect guys now there are some examples of extensive intensive properties which I would want you guys to remember directly questions are asked from these okay these are the examples of some extensive properties and over here I mentioned some of the examples of the intensive properties perfect from these examples questions are asked now there are few things which you need to know about extensive and intensive properties few important things number one if I talk about extensive properties extensive properties are additive in nature extensive properties are additive in nature are additive intens properties are non- additive what does that mean extensive properties you can you can add or subtract extensive properties directly you can add or subtract extensive properties directly but you cannot add or subtract intensive properties directly what does that mean extensive properties are additive whereas whereas intensive properties are non- additive intensive properties are nonadditive are non additive what it means let's try to understand this let's try to analyze this for example this is a container this is one more container okay but students this container is filled with water even this container is filled with water okay imagine its mass is M volume is V temperature T mass is M volume is V temperature is T of the water present in the container right now what am I going to do I'm going to add them up I'm going to add them up and put it here in the final container I'm just going to mix them up in the final container so this particular final container will be containing water if I ask you what is the mass of water present in the final container this was M this was M this will be 2 m now what is the volume of this water now V plus v makes it 2 V but if I ask you what is the temperature of this water it is simple guys you have one glass of water at temperature 25° Centigrade one more glass of water at temperature 25° Centigrade you're mixing them in a bucket in a bigger bucket right so what will be the temperature of the resultant water it will be 25 only so temperature of the water inside the container right now will be T also how many properties I defined for the system three mass volume temperature out of these three properties these two are extensive these two are extensive and if you look carefully if you look carefully when I added them up m+ M made it 2 m v + V made it 2 V but t plus T did not make it 2 t so can you say over here extensive properties are being directly added but intensive properties are not right so can I say intensive properties are non- additive if intensive properties were additive then temperature of this water should be 2T but it's t only this is first point extensive properties are additive in nature whereas intensive properties are non additive in nature number one number two number two division of division of two extensive properties division of two extensive properties division of two extensive properties makes the new property as intensive makes the new property as intensive makes the new property as you what does it mean for example you have got I'm defining two things X and Y let's say this x is your extensive property let's say your Y is also your extensive property perfect X and Y imagine these are two extensive properties for example I'm dividing these two extensive properties after dividing these two extensive properties I'm getting one more property Zed when you divide to extens of properties the new property which comes into play that will be intensive in nature that will be intensive in nature for example for example m stands for Mass v stands for volume mass is extensive volume is extensive mass by volume is what you call as density density you'll be calling as intensive density you'll be calling as intensive as simple as that right so division of two extensive properties makes the new property as intensive yeah makes the new property as intensive okay third Point third point which is again important my dear students an extensive property let me write it in short form if x is an extensive property if x is an extensive property if you express this extensive property per unit mole per unit mass or per unit volume right if you express this extensive property either per unit moles per unit mass or per unit volume the new property that comes into play that automatically becomes intensive for example there's a term which we have to Define in some time heat capacity heat capacity which is represented by C heat capacity which is represented by C it is an extensive property if you expresses per unit Mass it becomes specific heat capacity it becomes specific heat capacity specific heat capacity is intensive in nature it is intensive in nature heat capacity which is extensive if you Define it per unit mole it becomes molar heat capacity mol heat capacity is intensive so remember in short whenever an extensive property expressed either per unit mass per unit mole or per unit volume the new property that arises that is your intensive property that becomes your intensive property am I clear with this am I clear with what are extensive and what are intensive properties I hope I'm clear I hope I'm clear all of you say yes or no in the chats quick quickly yes all clear perfect now people Now understand now understand there is something called as state functions and path functions State functions what are State functions what are State functions exactly right from this also a question is asked State functions State functions are the ones State functions are the ones which only only and only depends on which only only and only depends on initial and final state of the system which only depends on initial and final state of the system initial and final state of the system they only depend on initial and final state of the system they're path independent they path independent they are path independent on the opposite side if I Define path function what's a path function what's the path function first of all path function also depends on path function also depends on initial and final it also depends on initial and final position initial and final state of the system but at the same time it's path dependent it is path dependent these are path dependent these are path dependent right what are the examples of the state functions your pressure volume temperature internal energy enthalpy entropy Gibbs free energy these are all your state functions these are all your state functions these are all your state functions what about the path functions heat and work heat and work these are the examples of your path functions these are the examples of your path functions these are the examples of your path function heat and work yes heat and work are the examples of your path function my dear students how can you remember what is State function what is path function you can relate it with one simple thing you can relate it with one simple thing for example you at Point a and you want to go to point B from A to B you want to go from A to B how many paths are available in order to go from A to B I can choose infinite paths this is one of the path and this is for example one more path which I've chosen to go from A to B this particular one I'll be calling as you know it displacement this is what distance now tell me displacement only depends on initial and final state it path independent so you can relate it with your state function distance it is path dependent you can relate it with your path function you can relate it with your path function yeah you got it there are two things which I would want to which I would want you guys to know about State functions and path functions remember them properly remember them guys for example a is a state function a is a state function B is a path function a is a state function is a path function understand what I'm going to say a is a state function B is a path function okay if I multiply this a by D it becomes da and this becomes DB I just multiplied a as well as b b with d right da DB what is this da da is what I call as small change in a DA is what I call a small change in a DB I won't be calling this a small change in B this is what we'll be calling as small amount on B this is what we'll be calling a small amount of B if I integrate this da and if I integrate this DB what do we get what do we get integral of small change gives you the large change I'll be either calling it as large change in a or total total change in a integral of small amount gives you total amount it is total amount of B small amount total amount integral of DB integral of Da over here I'll be representing with Delta a but integral of DB I'll be representing simply with capital b b i mean remember these two things these are important for example for example for example one term we which we have to discuss in some time that is internal energy internal energy is the example of what it is the example of a state function okay it is the example of the state function for example Q heat heat is the example of what it is the example of a path function it is the example of the path function correct my dear students if you multiply this U with d and over here also you write DQ what is the meaning of both these I mean what is the meaning of these two terms du and DQ what is du du is basically small change in internal energy du is basically small change in internal energy what is DQ DQ I won't be calling a small change in heat that does not make any sense this a small amount of heat there a small amount of heat now people if you integrate them if you integrate this one I mean if you integrate du as well as DQ if you integrate both of them over here this becomes Delta U and this becomes normal Q this Delta U is what I call as total change in internal energy or large change in internal energy this is total change in internal energy this is what you call as small amount integral is total amount of heat this is total amount of heat are these two things clear to you are these two things clear to you people yes clear people I hope these two basic things are absolutely clear to you right perfect now now comes one more term that is that is State variables State variables or you call them as state parameters what are State parameters or state variables what are State parameters or state variables let me tell you these are the variables which these are the variables which which defines these are the variables which defines the state of the system these are the variables which actually defines the state of the system which actually defines the state of the system these are the variables which actually defines the state of the system state of the system what does that mean what does that mean before that before letting you know the meaning of this particular statement I'm going to write one more I'm going to write one more on changing on changing the state parameters on changing the state parameters in the bracket I'm writing minimum one minimum one on changing the state parameters minimum one on changing the state parameters the state of the system automatically Chang the state of the system automatically chains let's try to understand let's try to understand what these State parameters exactly are my dear students for example for example for example here we have got a container okay here we have got a container in this container let's say you have got an ideal gas this is the ideal gas in the container okay let's say the pressure of ideal gas is P1 volume of this ideal gas is V1 temperature of this ideal gas is T1 right moles are N1 Perfect Right the ideal guas which is in the container whose pressure is P1 volume is V1 temperature is T1 moles are N1 I can say this ideal gas for example is present at its initial State these are basically the parameters these are the variables which are letting you know whether the system is at initial State whether the system is at final State whether the system is at intermediate state or whatever so basically these are the parameters which actually Define which actually defines the state of the system these parameters exactly let you know right what is the state of the system okay now people remember whenever you want to change the state of the system whenever you want to change the state of the system what do you have to do you just have to change the value of these State parameters in order to change the state of the system if you want this this system to go from initial state to final State what do you have do if you want to change the state of the system you just have to change the value of State parameters minimum minimum if you change only one state parameter if you change only one state parameter the state of the system automatically Chang for example if I just change pressure rest everything I keep same if I just change the pressure from P1 to P2 rest everything I'm keeping same rest everything I'm keeping same I'll say the state of the system has automatically changed so remember whenever you want to change the state of the system you have to change the value of State parameters minimum if you change one state parameter the state of the system automatically Chang okay am I clear am I clear now let's move on my dear students there's a term called as thermodynamic process I'm just giving you the introduction of the things first thermodynamic process how do we Define the thermodynamic process what is a thermodynamic process what is a thermodynamic process first of all how do you define the thermodynamic process it is simply defined as the path followed it is simply defined as the path followed to change the path followed to change the state of the system it is a path followed to change the state of the system now what that means what that means try to understand this is important and I I'm just giving you the introduction of the things I did not go into the details yet I did not go into the details yet details I'll let you know in some time first of all you have to understand the overall scenario like what all things are to be discussed in this chapter okay see guys for example for example we have got a gas in the container for example we have got the gas in the container okay let's say the pressure of this gas is P1 volume is V1 temperature is T1 moles are N1 for example I'm assuming that this system is present at its initial State the system is present at its initial State now my dear students now my dear students what exactly am I going to do I am going to change the state of the system how do we change the state of the system by changing the value of State parameters right by changing the value of State parameters let's say let's say the final State parameters of the system are P1 V2 P2 and N2 what did I do what did I do people what did I do we changed the state of the system how exactly we changed the value of State parameters we change the value of State parameter since we change the state parameter value automatically the state of the system will automatically change you tell me one thing you tell me one thing no doubt I changed the state of the system but if you notice one thing pressure of the gas here was P1 here also the pressure of the gas is P1 no doubt I changed the state of the gas I changed the state of the gas I changed the state of the system but during the change in the state of the system I made sure that pressure of the system does not change pressure press of the gas does not change no doubt I changed the state of the gas by keeping its pressure constant I changed the pressure of the gas I change the pressure of the system by keeping the pressure of system I Chang the state of the gas by keeping its pressure constant I could have changed the state of the system by keeping its volume constant as well I could have changed the state of the system by keeping its temperature constant as well so can I basically say there are lot of paths there are lot of paths available right to change the state of the system you can change the state of the system by keeping the pressure of the system constant you can change the state of the system by keeping volume constant you can change the state of the system by keeping temperature constant etc etc so there are a lot of paths available to change the state of the system and the path followed among these paths which path you are following to change the state of the system that's what you call as thermodynamic process so what is a thermodynamic process it is a path followed to change the state of the system right it is a path followed to change the state of the system I hope I'm clear so over here first of all I'm writing isocoric process isocoric process what is an isocoric process tell me quickly what is an isor process when do I say okay first of all let me tell you one more thing from now onwards whenever I use the term system understand I've taken a container in that container I have kept the ideal gas and that ideal gas is under investigation that ideal gas is my system so from now onwards whenever I say system understand I'm talking about the ideal gas which is there in the container okay understand I'm talking about the ideal gas which is there in the container perfect now for example I'm saying that the ideal gas is undergoing isocoric process what does that mean that means we are changing the state of the system but we are making sure during the change in the state volume of the system is remaining constant we are not going to let the volume of the system to change no doubt we are changing the state but during the change in the state we are not letting the volume of the system to change okay perfect during isocoric process volume of the system does not change if volume of the system does not change Delta V change in volume will be zero change in volume will be zero for example how to plot a graph how to plot a graph between pressure of the gas versus volume of the gas how the graph will look like since volume is not supposed to change it has to be a straight line and this particular straight line I'll be calling as isocore this something which I'll be calling as isocore right ISO code perfect now similarly guys I'm using the term isobaric process what is meant by isobaric process during in isobaric process when the ideal gas undergo isobaric process I'll say pressure of the gas remains unchanged pressure does not change so change in pressure Delta P it will be zero Delta P will be zero right Delta P will be zero I'm plotting the graph between pressure and volume I'm plotting the graph between pressure and volume the graph is like this and you call this as isobar you call it as isobar simple right similarly guys if I talk about isothermal processes I'm just giving the introduction of the things isothermal process what is an isothermal process when do we say that ideal gas is undergoing isothermal process whenever ideal gas under goes a change in state but during the change in state temperature of the gas does not change if temperature of the gas if temperature of the system does not change we say the gas is undergoing isothermal process right temperature constant if temperature is constant delta T is z delta T is z change in temperature is zero okay for an isothermal process delta T will be zero right if you would have studied the gas laws gas laws which gas law used to be there in which we used to keep the temper temp of the ideal gas constant I hope you remember that was your boils law right perfect so the ideal gas whose temperature is kept constant the ideal gas whose temperature is kept constant can I say this gas whose temperature is kept constant it will be following the boils law right e is equal constant PV is equal constant this is what I'll be calling as equation of State for isothermal processes this what I'll be calling is equation of State for isothermal process equation of State for isothermal process right now people if I want to plot a graph between pressure and volume for the ideal gas which is undergoing isothermal process you can write P equal K by V you can write p is equal K by V pressure and volume here are inversely proportional to each other and you know whenever we need to plot a curve between two which are inversely proportional to each other the nature of the curve is hyperbola right this curve is something which we call as isotherm this is something which we call as ISM perfect if you look at the first process isocoric ideal gas is kept at constant volume is kept at constant volume ideal gas is kept at constant volume ideal gas is kept at constant pressure ideal gas is kept at constant temperature perfect perfect guys now there's one more process which we we have to discuss all these things we have to discuss in detail by the way okay there is one more process that's called as ad diotic process when do we say that ideal gas is undergoing tic process yeah just tell me that whenever an ideal gas undergoes a change in state whenever an ideal gas undergoes a change in state but during the change in state if there is no heat exchange between system and surroundings if there is no heat exchange between system and surroundings during the change in the state if there is no heat exchange between the system and surroundings then and only we say that your ideal gas is undergoing ad diabatic process perfect now what is the equation of State in case of adaba process pv^ gamma is equal to constant this is basically the equation of State for reversible adaba that's something which I'll explain you after some time when I'll discuss reversible irreversible process with you okay this is the equation of State the way PV is equal K was the equation of State for isothermal similarly this is the equation of State for adaba right reverse B over here if you want to plot a graph between pressure and volume if you want to plot a graph between pressure and volume okay you know p is equal k / V gamma pressure and volume again they are inversely proportional and whenever you have got two variables which are inversely proportional to each other the nature of the curve is your rectangular hyperbola okay and this curve is something which we call as adabat this curve is what we call as adabat my dear students if you look carefully over here you got the hyperbola in this case also you got the hyperbola in ISM hyperbola right in your adab batic curve you got hyperbola now how do you identify how do you check whether the hyperbola is adab batic or isotherm how do we check that how do we check that how do we check that one simple trick I would want you to remember that's all one simple trick I would want you to remember that's all let's say this pressure this is volume okay this is for example one hyperbola this is the second one these are the two hyperbolas that that are given to me okay I'm keeping the arrow towards right I'm keeping the arrow towards right I'm keeping the arrow towards right towards right volume is increasing towards right volume is increasing right volume is increasing increase in the volume is what you call as expansion both these graphs they represent what expansion they represent expansion now which graph is for isothermal which graph is for tic what do we do what do we do simple this is the common point this is our common point I'll place my thumb at the common point I'll place the thumb at the common Point Cur your fingers in a clockwise direction that curve which touches the fingers first that's always isothermal but this is isothermal this isothermal so automatically this this becomes automatically this becomes adamatic right perfect similarly guys similarly let's say you going to curve like this this is pressure and this is volume pressure versus volume okay pressure versus volume I'm starting from here I'm starting from here let's say this one hyperbola this is one more hyperbola Arrow towards left Arrow towards left towards left volume is decreasing decrease in the volume is something which we call as compression and both these curves they represent compression both these curves they represent compression okay now place your thumb at the common Point curl your fingers in the clockwise direction that curve which touch your fingers first that's your isothermal so this is your isothermal and this your adab batic I hope you got to know how do we check whether the curve is adamatic or isothermal am I clear am I clear people am I clear am I clear one more see guys I'm not teaching you all the stuff in detail right now in detail I'll be doing it after some time I'll just I'm just giving you the overview of the things I'm just giving you the overview of the things okay I'm giving you overview of the things there is one more process which we have to discuss in detail after some time that's what you call a cyclic process what happens in cyclic process what happens in cyclic process in cyclic process I would say in cyclic process I would say system returns system returns to its initial State system returns to its initial state after going through a lot of steps system returns to its initial state after going through a lot of steps system returns to its initial state after going through a lot of steps for example for example let's say this pressure this volume let's say we are starting from point A from point A we are going towards point B from point B we are going towards Point C C and then from C again we are coming to a right we started from a we reached we started from a and we reached a can I say the initial State and the final state of the system is same right can I say the initial State and the final state of the system is same can I say system is returning to its initial state after going through a lot of steps whenever you see such kind of the process in which system returns to its initial state after going through a lot of steps you call that particular process cyclic process so this is a cyclic process this is a cyclic process if it is a cyclic process in which systems initial State and final state are same can I say change in the value of a state function change in the value of a state function for a cyclic process will be zero right change in the value of a state function for a cyclic process that has to be zero because initial State final state are same and what are State functions they only depend on initial and final state if initial final state is same that means the value of State function will be also same initial value of State function and final value of State function will be the same if initial value of the state function final value of the state function is same that means change in the value of the state function that has to be zero so do remember in case of cyclic process change in the value of State function that is always zero so your Delta U your Delta H your Delta s your Delta G all these terms for the overall cyclic proces they're always what they're always they're always zero am I clear with this am I clear with this people the change in the value of a state function for a cyclic process that's always taken as zero yes perfect now let's move on let's move on to few more things let's move on to few more things let's move on to few more things okay my dear students if you ask me what is heat how do you define the term heat how do you define the term heat how do you define the term heat let me tell you first of all heat is defined as it is defined as the mode of heat is defined as the mode of transfer of energy heat is defined as the mode of transfer of energy between the system and surrounding it is defined as the mode of transfer of energy between system and surroundings what does it mean my dear students whenever energy has to be exchanged between system and surroundings whenever energy has to be exchanged whenever the energy has to be transferred either from system to surrounding or from surrounding to system whenever energy has to be transferred either from system to surroundings or from surrounding to system this transfer of energy can happen via heat so heat is nothing but the mode of transfer of energy heat is nothing but it is the mode of transfer of energy it is the mode of transfer of ener energy and you know heat always flows due to heat always flows due to heat always flows due to temperature difference heat always flows due to temperature difference right Whenever there will be difference in the temperature between system and surroundings then only heat will flow and heat always flows from higher temperature to lower temperature it flows from higher temperature to lower temperature it flows from higher temperature to lower temperature now people hair two things you have to remember two things you have to remember that is the sign convention that is the sign convention okay remember whenever heat is absorbed by the system when heat is absorbed by the system when heat is absorbed by the system at that point of time Q value is taken as positive Q value is taken as positive and if heat is released when heat is released by the system Q value is taken as negative Q value is taken as negative Q value is taken as negative for example for example let's say this is the container which I have and in this container I got the ideal gas which is my system perfect let's say this system is at temperature T1 let's say the system is at temperature T1 outside surrounding is at temperature T2 surrounding is at temperature T2 my dear students if for example if T2 is greater than T1 if T2 is greater than T1 can I say there is temperature difference between system surroundings so heat will automatically flow heat will flow from higher temperature to lower temperature right now surrounding is at higher temperature and system is at lower temperature so heat will flow from surrounding to system if heat is Flowing from surrounding to system what does that mean that means system is absorbing heat that means system is absorbing heat and when heat is absorbed by the system Q value is taken to be positive right Q value is taken to be positive perfect similarly if for example T1 is greater than T2 if T1 one if system is at higher temperature surrounding at lower temperature heat will flow from system to surroundings at that time can I say heat is released by the system and when heat is released by the system Q value is taken to be negative these are just few things which you need to remember about what about heat am I clear am I clear people am I clear yeah say it in the chats quickly am I clear perfect now comes one more term what is that that is work that is work and already we know what is heat is it a state function or path function you know that heat is a path function yeah heat is a path function now if I talk about work if I talk about work which is denoted by W if I talk about work which is denoted by W how do we Define work work is again it is again the mode of it is again the mode of transfer of energy it is again the mode of transfer of energy between the system and surroundings if if energy has to be transferred either from system to surrounding or from surrounding to system the transference of energy can happen via work as well so work is again the mode of transfer of energy between system and surroundings mode of transfer of energy between system and surroundings and already we know work what is it it is a path function work it is a path function it is a path function now guys try to understand try to understand what exactly I'm going to say for example this is a container this is a container okay over here it is the Piston which is frictionless massless whatever okay and here I kept an ideal gas this is the ideal gas in the contain let's say the Piston is right now at Point a when the Piston is at Point a let's say volume of this gas right now is V1 volume of the gas right now is V1 okay my dear students if I ask you if I ask you if I ask you on this piston how how many pressures are exerted on this piston this internal gas these internal gas molecules will be colliding with the Piston so in this direction there'll be pressure of gas in this direction there will be pressure of gas similarly outside there is atmosphere those atmospheric gas molecules will be also colliding from outside so this your P external so on the Piston two pressures are acting pressure of the gas P external right now case one case one my dear students tell me if P gas is greater than P external if P gas is greater than P external if P gas is greater than P external will the Piston go inside or piston go outside piston will go outside let's say the Piston went outside and stopped somewhere here stopped somewhere here right where the volume is V2 let's say Pistons stopped here when piston is going out outwards what is happening to the volume of the container volume of the container is increasing if volume of the container is increasing that means volume of the gas is increasing if volume of the container is increasing that means volume of the gas is increasing so if P gas is greater than P external can you say at that point of time V2 is greater than V1 what is happening to the volume of the gas volume of the gas is increasing increase in the volume of the gas is what we call as expansion increase in the volume of the system is what we call as expansion Whenever there expansion we always say work is done by the system we always say work is done by the system Whenever there is expansion we always say work is done by the system and let me tell you whenever work is done by the system by the system W value as per sign convention in chemistry is taken to be negative W value is taken to be negative in what in chemistry not in physics okay remember these points these are super important okay I hope I'm clear I hope I'm clear this was Point number one point number two here Point number two if P external is greater than P gas if P external was greater than P gas if P exal was greater can I say at that point of time final volume will be less than initial volume right right so what is happening decrease in the volume of the gas is happening decrease in the volume of the system is happening decrease in the volume of the system is what you call as compression and whenever there is compression we always say work is done we always say work is done on the system and whenever work is done on the system whenever work is done on the system W value as per sign convention is taken as positive right is taken as positive this was case number two this was case number two case number three I'm writing here tell me if P gas was great if P gas was equal to P external if P gas and P external were equal what would have happened will there be any moment of the Piston no if there is no zero moment of the Piston that means that means transfer of energy between system and surroundings is not happening in the form of work that means transfer of energy is not happening in the form of work right if P gas is equal to P external work at that point of time is equal to zero am I clear am I clear people okay guys so basically basically remember one thing whenever you see piston moving going outwards or inwards remember at that point of time between system and surroundings energy is transferred in the form of work right except in Cas of free expansion that I'll teach you separately whenever you see piston moving either going outwards or inwards what is happening at that point of time exchange of energy between system and surroundings is happening in the form of work except in case of free expansion free expansion will we discussed later on okay now my dear students how do we calculate how do we calculate this work done how do we calculate this work done I'm not deriving the result I'll just give you the result w w is equal to minus integral P external DV from V1 to V2 from V1 to V2 this is how this is the formula by means of which we calculate work this is the formula by means of which we calculate work okay minus integral V1 tov2 P external DV my dear students here only we have to Define two processes which I'll be defining in some time but right now I want you just to remember them I just want you to remember them there are two processes one is called as a reversible process and one is called as irreversible process what are reversible irreversible process you'll get the idea in some time right now you just remember things in case of reversible processes in case of reversible processes what you'll observe you'll observe pressure of the gas and P external they'll be almost equal and if P gas and P external be always equal that means w reversible is equal minus integral V1 to V2 instead of P external we are going to WR P gas multipli by DV this is the result which is used to calculate work done in case of reversible processes this is the result which is used to calculate work done in case of reversible processes now similarly in case of irreversible processes what you'll observe you'll find external pressure as constant why is that you'll get the idea in some time okay just remember things first of all external pressure will be constant if external pressure is constant P external will come out integral of DV is V upper limit lower limit so at that point of time w irreversible will be minus P external vs2 minus V1 that means minus P external Delta V so these are the two general results which you need to remember in order to calculate work in order to calculate work one for reversible process one for irreversible process one for reversible process and one for irreversible process one for reversible process one for irreversible process yeah perfect guys am I clear with this am I clear with this am I clear with this just remember the units of energy this is something which we shall be frequently using you know the SI unit of energy that is Jewel TGs unit of energy that's urg one Jew is equal 10 7 SGS other units of energy other units of energy which we shall be using in this chapter one is ATM lit one is bar liter one is calories remember 1 atm lit is0 1.3 Jew one bar liter is 100 jewles 1 calorie is 4.2 Jew 1 atm lit is 24.2 calories these conversions are important you have to remember them you have to remember them right and I'll be using them in some time I'll be using them in some time okay these conversions are really important guys you have to you need to remember you need to remember them okay perfect so can you solve one simple question here this question I gave you yesterday in that small video I gave you one video yesterday I I don't know whether you have watched that video or not one small video I gave you yesterday yeah look at this particular equation my dear students students if you look at the first reaction see first of all all these process are happening at STP STP means pressure temperature constant so all these proc are happening while pressure and temperature is kept constant pressure temperature is kept constant now look at the first process 4 * NH3 gas plus 7 * O2 gas it gives 4 * NO2 gas plus 6 * H2O gas this is the process which is carried out at constant pressure temperature okay which carried out at constant pressure temperature my dear students have you heard about this equation PV isal n RT right since pressure temperature is constant R is already constant can I say volume is directly proportional to moles absolutely volume is directly proportional to moles so if moles increase volume will increase if moles decrease volume will decrease right you know this you know this now if I ask you on the reactant side how many gaseous moles do we have 7 + 4 is 11 on reactant side we have got 11 gaseous mole on product side 6 + 4 10 gaseous mole so when this process is happening when this process is happening are moles increasing or decreasing moles are decreasing moles are decreasing and you know if moles decrease volume will automatically decrease decrease in the volume is what you call as compression decrease in the volume is what you call as compression Whenever there is compression work is done on the system work is done on the system and W value is taken to be positive that's what we had to check work is done on the system and W value is positive look at the second case gas gas liquid on reactant side you have got gases on product side you have got liquid on reactant side you have got gases on product side you have got liquid so gases are getting converted into liquid you know gas molecules are far liquid molecules are closed right so gas is getting converted into liquid so that means molecules are coming close what is that what does that mean expansion or compression compr Whenever there is compr work is done on the system whenever work is done on the system W value is again taken to be positive right W value is taken to be positive look at the next one or leave the third one aside look at the fourth one solid is getting converted to gas solid is getting converted to gas that means molecules are going far expansion is happening whenever there's expansion work is done by the system W value is taking to be negative look at the third one solid is getting converted into liquid but it is ice basically it is H2O solid ice is getting converted into liquid water ice is getting converted into liquid water it's a special case guys ice is getting converted into liquid water normally if you think solid is getting converted in liquid so it has to be the case of expansion right it has to be the case of expansion but in case of ice due to the special structure due to the special structure of ice volume of ice is more than that of volume of liquid water right volume of ice is more due to the special cage like structure volume of ice is more than that of volume of liquid water so when this process is happening decrease in the volume is happening decrease in the volume is what you call as compression Whenever there is compression work isn't on the system W value is taken to be positive W is positive ha also am I clear am I clear with all these questions people am I clear with all these questions am I clear with all these questions okay one simple and basic question let me give you let me see if you can solve this or not one simple and basic question the question is calculate the work done calculate the work done in Jewels when an ideal guas when an ideal guas when in ideal gas under goes an expansion under goes an expansion under goes an expansion from from 10 ra power -3 decim Cub to 10 ra^ -1 DM Cub against against a constant external pressure off against against a constant external pressure of 10^ minus 5 Newton per M squ can you solve this question it's a very simple and basic question it's a formula based one it is a Formula based question can you solve this can you solve this people quickly e done done guys is it done first of all tell me we have to calculate work how many Expressions I gave you till now two expressions one was for reversible another one for irreversible right in reversible I told you P gas and P ex they're almost equal in in irreversible I told you P external is constant here it is mentioned that external pressure is constant external pressure is constant so I'll be considering this process to be irreversible so which formula I'll be using W irreversible W irreversible is nothing but minus P external V2 minus V1 right so minus what is p external 10^ minus 5 Newton per M Square what is V2 what is V2 final volume final volume is 10^ minus 1 what is ini volume 10^ minus 3 right I taken pressure as for SI system right sorry this is not decimeter Cube let me just change the units Meer Cube okay write it directly as Meer Cube 10^ minus 3 m Cube to 10^ minus 1 M Cub okay pressure I have taken as as per SI system volume also I took as per SI system that means the value of w which we'll be getting at the end that will be also as per s system right so it's minus 10 - 5 asset okay if I take 10^ 3 as common here if I take 10^ 3 as common it becomes 10^ -1 / 10^ - 3 -1 correct so since I've taken 10^ minus 3 common so this becomes - 10^ minus 8 okay this is 10 ra this 10us 3 goes up so it is 100 - 1 99 so 99 into 10^ - 8 Jew this is the value of w is that clear is that clear people is that clear quickly let me know in the chats only bit of calculation here nothing else did you get the same answer did you get the same answer say perfect guys perfect now before talking about before talking about before talking about internal energy and first SL thermodynamics etc etc before talking about that stuff there's there are few more things which you need to remember I'm not going into the details of those things but I want you guys to remember I want you guys to remember them okay I want you guys to remember that my dear students there's a term called as heat capacity heat capacity heat capacity is represented by what it is represented by C heat capacity is represent by C how do you define heat capacity how do you define heat capacity heat capacity is simply defined as the amount of heat the amount of heat which is required which is required to raise the temperature of a given mass of the substance by 1° right it is basically the amount of heat which is required which is required to raise to raise the temperature of a substance of a substance by 1° the amount of heat which is required to raise the temperature of the substance by 1° for example understand for example this is a object it's a block it's an object right it's an object let's say I have to increase the temperature of this object by 1° I'll have to supply heat the amount of heat which I have to supply the amount of heat which is required to raise the temperature of this object by 1° that is something which I'll be calling is heat capacity that is something which I'll be calling as heat capacity now people try to understand try to understand for example for example in order to increase the temperature of this object by DT units heat required was DQ let's say in order to raise the temperature of this particular object by DT units right heat required was DQ I'll say something like this in order to raise the temperature by DT units heat required heat required is how much heat required is DQ heat required is DQ this something which I'm assuming in order to raise the temperature of this object by DT it requires DQ if I use the unitary method if I use the unitary method in order to raise the temperature by one unit in order to raise the temperature by 1 unit how much heat is required heat required is equal DQ upon DT so what is this DQ upon DT this is the amount of heat which is required to raise the temperature of that substance by one de and the amount of heat which is required to raise the temperature of the substance by 1° that's called as heat capacity the heat capacity C is equal DQ upon DT right heat capacity C is equal DQ upon DT perfect now guys you can make its units as well over here in the numerator it's heat right so Jews jewles per Kelvin Jews per deg Centigrade calories per Kelvin calories per degre Centigrade whatever correct okay lot of units you can make away one thing which I already told you heat capacity is it the example of extens or intensive property it is the example of extensive property it is the example of extensive property now people one thing which we have to realize over here we got to know C is equal DQ upon DT so from this I can make a result DQ is equal CDT if I integrate under the limits T1 to T2 over here I'll get a result Q is equal C delta T which is T2 minus T1 this result we'll use in the equ C Q is equal C delta T where do I use where do I use this particular equation where do I use this particular equation for example my dear students I'll tell you where do we use this particular equation for example this is the object that's given to me its heat capacity is given its heat capacity is given something right 10 Jew per Kelvin whatever 50 Jew per k what heat capacity is given now people I want to increase the temperature of this object from 10° to 100° I want to increase the temperature of this object whose heat capacity is given from 10° to 100° I want to check how much total heat is required how much total heat heat is required to raise the temperature of this object from 10° to 100° if it heat capacity is given so this formula I use C delta T C T2 minus T1 right per right now similarly there is one more terminology that is molar heat capacity have you heard or leave the M heat capacity aside specific heat capacity specific heat capacity which is represented by S which is represented by S specific heat capacity which is represented by S how do you define the specific heat capacity specific heat capacity is defined as heat capacity per unit mass of the substance and C you already know that is DQ upon DT so it is DQ upon MDT DQ upon MDT perfect DQ upon MDT or you can make the result over here DQ is equal msdt or if you integrate this equation under the limits T1 to T2 integral of DQ is Q it becomes Ms delta T T2 minus T1 this is one more result to calculate the heat this is one more result to calculate the heat this is one more result to calculate the heat let me tell you the specific heat capacity it the in property I hope you know I've already told you whenever an extensive property is expressed per unit Mass extensive property Express per unit Mass becomes intensive it becomes intensive yeah right people I hope I'm clear to here I hope I'm clear till here so this is one more way of calculating Q already we we already know one formula to calculate q q is equal C delta T if heat capacity of the object is given then C delta T if specific heat capacity is given then m is delt T then this is the formula I'll use okay now we have got M heat capacity now we have got M heat capacity which is represented by S sorry which is represented by CM molar heat capacity how do you define M heat capacity it is defined as it is defined as heat capacity per unit mole heat capacity per unit mole of the gas now people you are done C is nothing but DQ upon DT so DQ upon n DT right DQ upon n DT or from this particular equation you can calculate DQ DQ is equal n CM DT or if you can integrate under the limits T1 to T2 so integral of DQ becomes Q so Q you can calculate from this result as well NCM delta T NCM delta T again this m heat capacity it's also which property it's also an intensive property it is also an intensive property so my point is basically my point is there are different ways of calculating heat whenever you are supposed to calculate calculate the heat absorbed by the system calculate the heat released by the system or calculate the heat exchanged between system and surroundings basically you're supposed to calculate Q right whenever you get a question on heat exchange whenever you get a question on heat exchange between system and surroundings Q how many ways do we have to calculate Q how many ways do we have to calculate Q how many ways do we have number one if C is given if heat capacity is given how do I calculate q q is equal C delta T right number two if if cm is given if M read capacity is given then how do we calculate q q is equal n CM delt T right similarly if specific heat capacity of the system is given then Q is equal Ms delta T Q is equal m s delta T these are three ways by means of which we can calculate Q now guys here you can further classify into two parts at constant volume if volume of the system is constant over here I'm writing add constant pressure add constant pressure If volume of the system is kept constant I'll write QV is equal n CV delta T here I'll write QP is equal n CP delta T these are two very very very important results what is the meaning of these results have a look this QV is something which I call as QV is what I call as heat exchanged between system and surroundings heat exchanged between system and surroundings at constant pressure if your system is kept at constant pressure at that point of Time how much heat is absorbed or released by the system that is given by sorry constant volume at constant volume if your system is kept at constant volume if your system is kept at constant volume at that point of Time how much heat is absorbed or released by the system that is given by this equation similarly this is what I call as heat exchanged this is what they call is heat exchanged between system and surroundings at constant pressure if your system is kept at constant pressure if your system is kept at constant pressure at that point of time the amount of heat absorbed or released by the system is given by this formula Q is equal NCP delta T so basically these two results are very important these two results are very important what is CV CV is what I call as molar heat capacity at constant volume this CP I'll be calling as M heat capacity at constant pressure right mol heat capacity at constant volume molar heat capacity at constant pressure right people so these are the ways by means of which you can calculate Q sometimes they'll ask you calculate the heat absorbed at constant volume calculate the heat released at constant volume that means QV you have to calculate calculate the heat absorbed or heat released at constant pressure that means QP you have to calculate accordingly you'll have to use the result I'll show you its application in some time okay now there are few more things which you need to know there are few more things which you need to know okay you must be knowing CP it's always greater than CV what is CP molar heat capacity at constant pressure what is CV M heat capacity at constant volume CP is always greater than CV number one CP minus CV is equal r or how many moles of ideal gas for one mole of ideal gas remember these results directly okay there is one term which I'm defining Poison's ratio gamma gamma is CP / CB since CP is greater than CB so gamma value will be greater than one right perfect there is one more result which you need to know CV is equal R / gamma minus one this is one more result which you need to remember CP minus CV is equal R that's for one mole of ideal gas if you'll be having two moles of ideal gas it becomes 2 R 3 moles of Ideal 3 r n moles of IDE NR so remember CP minus CV is equal n * R for what for n moles of ideal gas or n moles of ideal gas yes for n moles of ideal gas right people for n moles of ideal gas now here one more thing we can do over here we got the relation between CV and Gamma here we can get the relation between between CP and Gamma as well we can get the relation between CP and GMA as well right we can get the relation between CP and Gamma as well right people we can do that you know your gamma is equal CB by CV so I can write it as gamma is equal I can write as gamma is equal CP divided by instead of CV I'll write R / gamma minus one R ID gamma minus one right so you got the result like this the result will be like this it will be CP is equal gamma R TP is equal gamma R / gamma minus 1 this is the result which you'll be remembering not this one CP is equal to gamma R Gus 1 this is one more result which connects CP with gamma and this is something which you have to remember as well okay now guys apart from all these things there are few more things which we have to remember what are those things let me give you the results let me give you the results for example over here you have got two chemically non-reacting gases 1 and two let's say you have got two chemically non-reacting gases 1 and two two chemically non-reacting gases 1 and two okay let's say number of moles of one is N1 this N2 CV for gas one is cv1 this is CV2 this is cp1 this is cp2 the mixture of gas which we got over here the mixture of gas which we got over here the mixture of gas which we got over here let's say you have to calculate CV for the mixture how do you calculate CV for the mixture N1 cv1 Plus N2 CV2 divide by what divide by N1 + N2 how do we calculate CP for the mixture how do we calculate CP for the mixture N1 cp1 plus N2 cp2 divide by what divide by N1 plus N2 how do we calculate GMA for the mixture gamma for the mixture is calculated as CP mixture divide by CV mixture these are few results which are very important which you have to remember I hope all these are clear okay remember the directly okay now guys one more table which I'll be giving you directly then we'll go into the actual stuff right now I'm just giving you the basic basic things which you have to remember first of all before solving the equation yeah before solving the questions see guys in case of a monoatomic gas in case of the monoatomic gas like like helium helium is monoatomic in case of monoatomic gas what is the value of CV what is the value of CP and what is the value of gamma let me tell you CV is nothing but 3id 2 R CP is 5id 2 R gamma is CP by CV so this divid by this which makes it 5 by3 which makes it 5 by3 number one these values you have to remember these value you values you have to remember these values you have to remember okay now if you'll be having a diatomic gas if you'll be having a diatomic gas like what like O2 diatomic or if you have got triatomic linear gas if you have got triatomic linear gas like what like carbon dioxide which is triatomic linear it is CV is 5/ 2 R it is CP is 7/ 2 R and it gamma is this by this so 7 by 5 7 by five and by chance if you have got triatomic nonlinear triatomic nonlinear gas for example you have got S2 it is triatomic as well as nonlinear triatomic as well as nonlinear it CV is 3 R CP is 4 R right and Gamma is 4x3 4x3 right now people can you let me know one thing 5 by3 is what 1.66 this value is 1.4 this value is 1.33 perfect this value is 1.33 now tell me initially start with monatomic then diatomic then triatomic on M from top to bottom what is happening to the atomicity of the gas on M from top to the bottom atomicity is increasing first diatomic triatomic triatomic Etc atomicity is increasing what is happening to the value of gamma gamma is decreasing so can I generalize a statement over here as the value of as the atomicity as the atomicity atomicity means number of atoms in a molecule as the atomicity of the gas increases the value of gamma decreases and these these these things guys CP CV and Gamma values for these different types of gases will be frequently using everywhere in the chapter thermodynamics am I clear with this am I clear with this guys yes am I clear with this right people I hope you remembered all the things which I gave you till now now comes the actual stuff so basically your thermodynamics is starting from here you'll know whatever I was telling you right that's all complete you know Basics and stuff which you need to remember degrees of freedom you need not to remember here I mean we have got nothing to do with degrees of freedom do that in physics why do you need that in chemistry from here the show begins yeah from here the show begins please do have some fun sir yeah sure sure sure sure no all the chapters cannot be completed today I told you already there will be two marathons right for class 11 physical chemistry half we'll do today half we'll be do in the next marathon yeah chemistry is fun let's wait for 2 minutes then we'll start okay let's take a short break for 2 3 minutes hair's secret like what is good in this hair do you want a break for 10 minutes do you need a break for 10 minutes okay let's take a break for 10 minutes and get back quickly but get back okay from here the chapter is starting huh so the time now is 550 so break till 6 p.m. I would want everyone to be back on 6 p.m. at 6 p.m. huh let's take a break for 10 minutes or let's let's keep it a 615 wait not 615 sorry just a second let's get back at 610 I'll see you all at 610 right 610 but everyone has to be back because this is something which is super important in this chapter see you e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e e are you guys back are you guys back with the positive energy yes are you all back are you all back guys positive energy in the chats is everyone back what's up how are you all call up call up everyone call up everyone let's get going perfect perfect perfect perfect can not see properly perfect then let's start with one more important topic that's internal energy right internal energy what this internal energy is exactly all about my dear students internal energy which is represented either by U or E right in some books they represent internal energy by U or or by E okay what is internal energy how do we Define it my dear students internal energy is basically defined as the sum of all the kinetic sum of all the kinetic and potential energy present in the system internal energy is basically defined as the sum of all the kinetic and potential energy present in the system now what does it mean imagine that you got a container and in this container container just assume you have got real gas molecule these are for example your real gas molecule okay now if I ask you a question whether these real gas molecules will be in rest or motion these real gas molecules definitely will be in motion right translational motion vibrational motion rotational motion right so can I say basically these real gas molecules they have got different types of kinetic energies translational energy vibrational ktic energy rotational ktic energy etc etc so there are different types of kinetic energies in the system number one number two between the real gas molecules there'll be interactions as well between the real gas molecules there'll be interactions as well and which energy comes into play due to interactions potential energy potential energy arises due to the interactions so these real gas molecules which are there in the container in short can I say this real gas has got kinetic as well as potential energy if I take the sum of all the kinetic and potential energy present in the system the sum of all the kinetic and potential energy present in the system that is something which I'll be calling as internal energy so internal energy is basically the total energy present in the system internal energy is basically the total energy present in the system right which is the sum of kinetic and potential perfect now my dear students for a real for an ideal gas for an ideal gas if you talk about ideal gas in case of ideal gas there is no interaction between the molecules in case of ideal gas there is no interaction between the molecules therefore potential energy in case of ideal gas will be zero so in case of ideal gas your internal energy is basically the total kinetic energy present in the system system in case of ideal gas your internal energy is basically the total kinetic energy present in the system right now people if I ask you this internal energy is it a state function or path function this internal energy is absolutely a state function if it is a state function we shall be calculating change in the value of internal energy which has to be U final minus U initial right if I talk about internal energy what do you think is it going to be extensive or intensive it is absolutely going to be an extensive property perfect in case of the cyclic process change in the value of internal energy change in the value of internal energy change in the value of a state function in case of a cyclic process that's always that's always zero remember this thing as well okay now people two things which you need to remember internal energy for a real gas internal energy for a real gas depends on two parameters volume and temperature if the moles are fixed okay for a real gas internal energy depends on two parameters volume and temperature or a fixed amount of moles so if you change the volume of the real gas internal energy will change if you change the temperature of the real gas internal energy will change but in case of ideal gas internal energy only and only depends on temperature for fixed amount of gas right this is valid for ideal gas in case of the ideal gas your internal energy it only only and only depends on temperature internal energy only only depends upon temperature in case of what in case of ideal gas right I'm writing it again internal energy it is a function of temperature only for what for an ideal gas if the moles if the amount of gas in the container is fixed now guys what does it mean what does it mean it means something like this let's say you got a container which contains fixed amount of ideal gas which contains fixed amount of ideal gas if you change the temperature of the gas if you change the temperature of the gas its internal energy will change if you change the temperature of the gas its internal energy will change how do we calculate change in internal energy due to change in temperature for an ideal gas I'll give you the direct result Delta U or small change du is equal n cvdt du is equal ncv DT it is valid only and only for the ideal gas this is the this is the ideal guas in the container if you are changing its temperature by DT units its internal energy will change by du units that du is equal n CV DT number one right total change in internal energy for an ideal gas Delta U that has to be ncv delta T that has to be ncv delta T and both these results they're only only and only valid for the ideal gas they only valid for the ideal gases right du is equal ncv DT Delta U isal ncv delta T perfect so if you are changing the temperature of this ideal gas from 10° to 20° for example what will be the change in internal energy Delta U you can calculate from here you can calculate from here number one number two number two for an ideal gas for an ideal gas undergoing an isothermal process for an ideal gas undergoing an isothermal process isothermal process means temperature is constant if temperature is constant delta T has to be zero if delta T is z that means Delta U will be also zero so remember for an ideal gas which under goes isothermal process its Delta U has to be zero its Delta U has to be zero right this is the next point this is the next point now guys look at this particular equation du is equal n cvdt if I take one mole of ideal gas for one mole of ideal gas for 1 mole of ideal gas n is equal 1 I can say CV is equal du upon DT what is Du upon DT what is Du upon DT D upon DT gives you the slope of the tangent SL slope of the tangent at a point at a point on internal energy versus temperature curve on internal energy versus temperature curve internal energy versus temperature curve okay internal energy versus temperature curve for example for example I'm giving you equation like this let's say on this side I'm plotting internal energy let's say this is a hypothetical question on this side I'm plotting temperature okay and the curve which I'm giving you let's say that's valid for one Mo of ideal gas one Mo of ideal gas for example right hypothetical situation this is the graph which we have I'm taking two points here this is point a this is point B now we have to check at which point A or B where is the CV value more where will be the CV value more what I'll be doing I'll be extending this line after the extension is done I'll be drawing the tangent here I'll be drawing one more tangent here perfect after drawing the tangents I'll measure this angle let's call this as Theta a let's call this as Theta B now tell me which angle is more I'll say Theta a is more than Theta B if Theta is more than Theta B that means slope of this tangent is more wherever the slope is more CV value is more so CV value will be more at a than that of B so this sort of a question can be asked yeah this sort of a question can be asked I believe it's clear I believe it's clear yeah perfectly done now my dear students let's have a look on few more things let's have a look on few more things there is something called as first law of thermodynamics there is something called as first law of thermodynamics okay if you look at the statement first law of thermodynamics it is an energy conservation principle it is an energy conservation Principle as per first law of thermodynamics when the state of the system chains there is the change in the internal energy of the system when the state of the system changes there will be change in the internal energy of the system and that change in the internal energy it can be brought about in two ways number one when heat is absorbed or released by the system when work is done by the ststem or on the ststem let's have a let's let's try to understand what it means let's try to understand what it means first of all I want you guys to remember this it is the energy cons principle and you know when the state of the system Changs there will be definitely change in its internal energy there'll be change in the internal energy of the system and that change in the internal energy of the system can be brought about in two ways number one when heat is absorbed or released by the system number two when work is done on the system or by the system what does it mean let's try to check it let's try to check it see guys for example this is the container which I have and over here you have got a pistol this is a pistol this is the ideal gas here in the container okay this is the ideal guas this ideal guas the system is right now present at its initial State the system is right now present at its initial State let's assume its total energy total energy present in the system I mean the internal energy of the system that's U1 that is U for example what am I doing I'm supplying some Q amount of heat to this system I'm supplying some Q amount of heat to this system now if I ask you what will be the total energy of the system now I have supplied some Q amount of heat to the system so the total energy of the system now will be U1 + Q it'll be U1 + Q right my dear students let's say I'm bringing this piston down let's say I'm bringing this piston down when I'm bringing this piston down what am I doing compression Whenever there is compression work is done on the system let's say I have done some W amount of work on the system let's say I have done some W amount of work on the system and for example whatever amount of work I'm doing on the system that is stored in the gas so what will be the energy now U1 + Q + W I'm assuming this is the final state of the system if this is the final state of the system then this has to be the final value of intern energy of the system if you take U1 on that side it's U2 - U1 U2 - U1 is Delta U and Delta U is nothing but Q + W this is the result which you have to remember this is the result which you need to remember Delta U is equal Q + W Delta U is equal Q + W okay Delta U is equal Q + W perfect this is the result which we have to remember Delta U equal Q + W first law of thermodynamics first law of themodynamics now people try to understand try to understand for example one simple question which is just a standard question right by means of which you'll understand this first law the question is a system expands system means your ideal guess expands from 5 L to 10 L so initial volume of the gas is 5 L final volume of the gas is 10 L expansion is happening against a constant external pressure expansion of the gas is happening against a constant external pressure of 28m if it absorbs 800 Jew of heat during the expansion of the gas system is absorbing heat when heat is absorbed Q value is positive when heat is absorbed Q value is positive right during the process calculate the value of delta U calculate the value of delta we have to calculate Delta U okay right now few minutes back only I gave you the first law and first law says that your Delta U is nothing but but Q + W okay Q + W you just tell me one thing is the process reversible or irreversible is the process reversible or irreversible just tell me that is the process reversible or irreversible just tell me that expansion is happening against constant external pressure constant external pressure means whenever it mentioned that external pressure is constant remember the process irreversible okay it's irreversible if it is irreversible how do we calculate work done in case of irreversible processes it is simply minus P external Delta V which is vs2 minus V1 so minus what is p external P external is 28m V2 minus V1 V2 minus V1 that comes out be 5 so the value is - 10 ATM L but I'll convert this in jewles minus 10 1 atm L I told you already 10 1.3 Jew so the value comes out be -113 jew this is the value of w right what is the value of Q what is the value of q q is plus 800 Jew right so Delta U is nothing but plus 800 Jew uh W is -113 Jew right the value will be minus 213 Jew this is the value of delta U which I was supposed to calculate is it clear is it clear people is it clear so this sort of equation I believe you can easily solve okay similarly one more type of question can you do it on your own check it out check it out quick will you be able to do it on your own check it out carefully again you have to use Delta = Q + W Delta U is given Q is given P external is given Delta V you have to calculate change in volume right yes harini you can watch it it's a simple question guys as far as I remember it answer will be I think 2 L yeah you you just check it by using the same first law of thermodynamics nothing else okay now comes one more thing that is enthalpy enthalpy yeah when is the second part of the marathon next week next week okay I'll let you know I'll let you in the telegram my dear students now comes your enthalpy what is enthalpy it is represented first of all by H how this enthalpy function was introduced how this enthalpy function was introduced let's get to know see guys see how this enthalpy function was introduced how this enthalpy term was introduced you know your Delta U is equal Q + W you know that right Delta U I can write as U2 - U1 is equal Q + W is equal Q + w w is minus P Delta V V2 - V1 okay if I write the same equation at constant pressure it becomes QP at constant pressure if I write the same equation so here I can write U2 - U1 is equal QP minus pv2 + pv1 okay if I take it on this side it's going to be U2 + pv2 U2 + pv2 minus U1 + pv1 U1 + pv1 is equal QP my dear students do you see two similar terms here u+ PV as well as U + PV it is just this term U plus PV it's written at final State this is written at initial State otherwise the terms are same U plus PV U plus PV right U plus PV U plus PV both the sides yeah this U plus PV only we call as H in mathematics what do we do we say let Y is equal 5x² + 4x plus something something right perfect just to reduce the calculations etc etc over here U plus PV U plus PV this U plus P only I'm calling as H so how do I Define enthalpy enthalpy is the sum of internal energy and the pressure volume work right if you need to Define if you need to Define enthalpy it is defined as the sum of internal energy and the PV work okay so people this U2 plus pv2 can I write it simply now as H2 this U1 plus pv1 H1 is equal QP H2 minus H1 is nothing but Delta H and Delta H is equal QP now this is something which is very important what is meant by this Delta equal QP Delta is equal QP it means that enthalpy change it means that enthalpy change is defined as is defined as the the total heat absorbed or released or released by the system absorbed or released by the system at constant pressure at constant pressure whatever ever will be the at constant pressure whatever will be the amount of heat absorbed or released by the system if you have go to the system if you got the ideal gas right which is kept at constant pressure whatever heat will be absorbed or released by this ideal gas which is kept at constant pressure that actually gives you what that gives you the enthalpy change that gives you the enthalpy change of the system so remember enthalpy change of the system is nothing but the heat absorbed or released by the system when the system is kept at constant pressure yeah when the system is kept at constant pressure am I clear am I clear people again if I talk about enthalpy the way we talked about internal energy again if I talk about enthalpy enthalpy let me tell you it is a state function enthalpy is a state function if it a state function change in enthalpy has to be equal Edge final - H initial right again enthalpy is an extensive property it is an extensive property why is it extensive property because you know H is equal U plus PV U is extensive p is extensive V is extensive right so H automatically becomes extensive okay oh I told you the reverse thing I had to say it is a state function just a second I gave you the wrong logic here sorry I don't know where was I I'm telling you enthalpy is a state function why is this why is it a state function because because because enthalpy is the state function because enthalpy is U plus PV U is a state function p is State function as well as V is State function okay perfect right people so enthalpy is a state function now enthalpy is extensive as well as well enthalpy is an extensive property as well because its value chains when you change the amount of substance present in the system its value change when you change the amount of substance present in the system for a cyclic process for a cyclic process delt has to be for a cyclic process Delta H has to be zero okay perfect now similarly enthalpy for fixed amount of gas is the function of pressure and temperature for a real gas for a real gas enthalpy depends on two parameters pressure and temperature if you change the pressure of the gas enthalpy of the gas will change if you change the temperature of the gas enthalpy of the gas will change but only valid for real gas in case of the ideal gas enthalpy is only the function of temperature right for a fixed amount of gas or a fixed amount of gas so basically let me tell you one thing internal energy gives you the total energy present in the system and enthalpy gives us the idea about the heat content present in the system okay these are two things one is internal energy one is enthalpy internal energy gives us the idea about total energy present in the system enthalpy gives me the idea about the heat content present in the system it gives me the idea about the heat content present in the system okay heat content present in the system so for example imagine that you have got an ideal gas in the container enthalpy of the ideal gas only depends on temperature so if you change the temperature of this ideal gas it's enal will change its enthalpy will change right its enthalpy will change now if I'm changing the temperature of this ideal gas from T1 to T2 what will be the total change its enthalpy what will be the total change in the enthalpy remember guys remember one thing your DH Small Change in enthalpy is nothing but n ppdt or an ideal guess for an ideal guas and total change in enthalpy Delta H is nothing but n CP delta T for an ideal gas these are the results by means of which you can calculate the enthalpy change of an ideal gas when you change the temperature so when you change the temperature of the ideal gas its enthalpy will change how do we calculate that change in enthalpy with the help of this result I'll show you the questions in some time right perfect now people one more thing one more thing for an ideal gas for an ideal gas undergoing isothermal process for an ideal gas undergoing isothermal process in isothermal process temperature is constant if temperature is constant delta T has to be zero if delta T is zero your Delta H automatically becomes zero so whenever the ideal gas under goes isothermal process it's Delta H has to be zero its delt has to be zero its delt has to be zero its Delta H has to be zero I hope this is clear oh similarly people one more thing one more thing one more thing DH is equal ncpd if I'll be having one mole of an ideal gas if I'll be having one mole of an ideal gas n is equal to 1 if n is equal to 1 CP is equal to what CP is equal DH upon DT what is DH upon DT graphically it is the slope of the tangent at a point slope of tangent at a point on enthalpy versus temperature curve on enthalpy versus temperature curve now for example for example one hypothetical question if I'm plotting a graph between enthalpy versus temperature and this graph is for example for one mole of ideal gas it is for one mole of ideal gas I'm marking two points there A and B can you let me know where will be the an CP value more can you let me know where will be the CP value more quickly can you let me know where will be the CP value more CP value where will be the CP value more A or B how you how you guys are going to do it is this color visible no use a different color today I think this is visible what I'll be doing at this point I'll be drawing a tangent and at this point I'll be drawing a tangent I I have drawn two tangents let's say this is Theta a this is Theta B now which tangent has more slope at a tangent has more slope right at a tangent has more slope more the slope more the CP value so I will say CP at a will be definitely greater than that of CP at B okay am I clear people yes Wasim are showing off different colors all right is it clear guys is it clear is it clear okay one thing one thing one thing I don't want go into the details but one thing I would want to tell you if you remember long back like like 1 hour back when I taught you the heat capacities when I taught you the heat capacities I gave you two results there QV is equal one was n CV delta T and one was QP QP is equal NCP Delta perfect I gave you these two results long back when I taught you heat capacities but now you got to know your ncv delta T is basically Delta U your NCP Delta is basically Delta H for the ideal gas right so basically the questions can be asked in different formats the questions can be asked in different formats either they'll ask the question like this calculate the heat absorbed or released at constant volume that means you have to calculate Delta U for the ideal gas which is ncv delt T they can ask you they can ask you calculate the heat absorbed or released at constant pressure at constant pressure that means you are they're asking you to calculate Delta H for the gas and Delta H is what for the ideal gas is NCP delta T nothing else either they'll be asking the questions directly calculate Delta U or Delta H or they can ask you the question in terms of heat exchange as well calculate the heat absorbed released at constant pressure or at constant volume if it is constant pressure that means Delta H if it constant volume that means Delta U okay perfect let's try to do one question by means of which you'll get the idea of what kind of question can be asked okay one simple question calculate the Delta U and Delta H for example in Jewels let's say you have to calculate both in Jewels look at it look at it people look at this question manika I did not teach entropy yet this fate these are two different things entropy andal can you give it a try can you give it a try it's a simple question guys as per the question you have got 10 decim cube of helium 1 decim cube is 1 lit so basically you have got 10 L of helium 10 L of helium helium is monoatomic and this gas is present at STP this gas is present at STP right that means temperature of this gas is right now 0° C right 273 Cal perfect now as per the question the gas which is there in the container you are changing its temperature you're heating it up till till how much till 100° Cen till 373 kin that means initial temperature of the gas was 273 final temperature of the gas is 373 yes so basically as for the question you had you add 10 L of helium you are heating it you're increasing its temperature you're increasing its temperature from 273 kin to 373 kin since you know in case of ideal gas internal energy as well as enthalpy internal energy as well as enthalpy they are temperature dependent so if you are changing the temperature there will be change in the internal energy as well as enthalpy of this ideal gas now that's what I'm supposed to calculate calculate the change in internal energy and calculate the change in enthalpy of the ideal gas these two terms I'm supposed to calculate right these two terms I'm supposed to calculate first of all how do we calculate Delta U for the ideal guess n CV delta T which is T2 minus T1 how do we calculate Delta h n CP delta T T2 minus T1 now guys helium since helium is monoatomic if it is monoatomic it CV is nothing but 3x2 R it is CP is nothing but 5x2 R it is monatomic correct it is monoatomic now at the same time we had 10 L of helium and STP can you tell tell me how many moles of helium were there given volume divid 22.4 so these many moles of gas we have in the container n value we know n value is 10id by 22.4 what is CV value P / 2 and R is nothing but 8.314 Jew per Kelvin per mole because we have to calculate Delta U Delta H and jewles T2 T1 T2 minus T1 will come out to be 100 so solve this answer will be in Jewels similarly n is 10 ID 22.4 CP is 5 by 2 R value is 8.314 T2 - T1 is 100 right solve this and this will come in Jewels so this is how you are going to get Delta U as well as Delta H am I clear okay perfect I hope this sort of question you can easily solve from now on now guys there is one more important topic from which questions have been frequently Asked frequently I would say that is the relation between Delta H and Delta U relation between Delta H and Delta U this one very very very important top relation between Delta H and Delta U I won't be deriving things I'll be just giving you the results and I'll show you how to apply them that's all okay that's all that's all see guys first of all you know as you know your H is equal U + PV H is equal U + PV you know it correct If I multiply throw by D it becomes DH is equal du + D of PV if I integrate throughout it becomes Delta H is equal Delta U plus Delta of PV this is the general result that relates Delta H with Delta U this is the general this is the general result that relates Delta H with Delta U Delta H is equal Delta U plus Delta of PV okay now my dear students if in the question pressure of the gas pressure of the system is given as constant if the pressure of the system is constant then you'll write Delta H is equal Delta u+ if pressure is constant P comes to the front then it becomes P Delta V Delta V is V2 minus V1 this is the equation which you'll be using IF pressure of the system is given constant if volume of the system is kept constant as per the question if volume of the system is kept constant then take this V out your formula becomes Delta H is equal Delta U + V Delta p V Delta P this is used when volume is constant similarly my dear students if if pressure and volume both are changing as for the question if pressure and volume of the system is changing which formula do I use I'll be using Delta H is equal Delta u+ I'll be simply writing P2 V2 minus P1 V1 P2 V2 minus P1 V okay P2 V2 minus P1 V right for example you have got let's say you have got a homogeneous gaseous phase reaction let's say you have got a homogeneous gaseous phase reaction which is carried out at constant temperature carried out at constant temperature let's say the reaction is like this N1 a gas plus n2b gas it gives n N3 C gas plus n4d gas this your homogeneous gaseous phase reaction okay my dear students for this particular reaction remember Delta H will be equal Delta U plus Delta ngrt Delta ngrt what is Delta NG here what is Delta NG for this reaction Delta NG means number of moles of gaseous products minus number of moles of gaseous reactants so Delta NG is equal number of moles of gaseous products number of moles of gaseous products this is gaseous product this is gous product N3 plus N4 gives you the number of moles of gaseous product minus number of moles of gaseous reactants this is gaseous reactant this gaseous reactant their moles are N1 plus N2 minus N1 + N2 these are certain results which you need to remember and with the help of these particular results with the help of these particular results you can solve the questions wherein you have to relate Delta H with Delta U am I clear am I clear am I clear people am I clear people with this dark mode is going to happen in some time okay let's try to utilize all these formulas let's see what kind of different questions can be asked okay uh let's start with this question it's an as it's asked question look at the question carefully look at the question carefully one mole of ideal gas under goes a change in state one mole of ideal gas under goes a change in state 2 ATM 3 l so you have got the ideal gas whose initial pressure was 2 ATM and its volume was 3 l let's say the ideal gas was present at its initial State okay pressure of the gas was 2 ATM its volume was 3 l now as for the equation we are changing the state of the gas the final State parameters of the gas are 2 ATM 7 L now the gas for example is present at its final state so look at the things which are given volume of the gas initially was 3 l now the volume of the gas is 7 L that means the state parameter is changing right and when the value of State parameter Chang the state of the system automatically Chang so basically the gas is undergoing a change in state but if you look carefully during the change in state is the pressure of the gas changing no pressure is remaining constant pressure is remaining constant right so we identifi during the change in the state p is constant if p p is constant then which formula do we need to use Delta H is equal Delta U + P Delta V plus P Delta V right as for the question your Delta is given as 30 ATM lit I mean Delta U is given as 30 ATM ler pressure is 2 ATM V2 - V1 V2 - 7 - 3 is 4 so this is ATM this is liter so this also comes in ATM liter so the value comes out be 38 ATM L this is the value of delta h which you are supposed to calculate perfect now you can convert this into jewels as well multiply this 38 with 10 1.3 the answer will come in Jewels I hope I'm clear with this yeah I hope I'm clear with this this is the first type of equ question that can be asked number one number two number two number two look at this particular question look at this particular equation which of the following is correct for the following exothermic exothermic reaction carried out at constant pressure you know your reactions are mainly carried out at constant temperature and with constant temperature you have kept pressure constant as well so first of all this is the reaction that's given to me as per the question the reaction is exothermic now you tell me if the reaction is exothermic if the reaction is exothermic heat is absorbed or released released whenever heat is released by the system whenever heat is released by the system Q value is taken to be positive or negative negative right over here Q is positive this cannot be the answer this cannot be the answer done number one number two the reaction is carried out at constant pressure temperature the reaction is carried out at constant pressure temper you know your PV is equal to what nrt so pressure temperature constant R is constant volume is directly proportion of moles if moles are increasing volume will increase if moles are decreasing volume will decrease now try to understand on the reactant side how many gaseous moles do we have 3 + 1 4 on the product side how many gaseous moles do we have 1 + 1 2 so when this process is happening are the gaseous moles increasing or decreasing when the process is happening are the gaseous moles increasing or decreasing gaseous moles are decreasing therefore volume will decrease decrease in the volume of the system decrease in the volume of the system is what you call as compression Whenever there is compression work is done on the system and whenever work is done on the system W value is taken to be positive W positive W has to be positive over here they mentioned W negative so this cannot be the answer okay now can you let me know the Delta for the reaction number of moles of gaseous products which is 1 + 1 2 minus number of moles of gaseous reactants which is 2 + 1 I mean 2 + 3 sorry 3 + 1 4 the value comes out be min-2 so Delta is-2 if Delta n is min-2 you know Delta H is equal Delta U + Delta NG RT it becomes like this so your Delta U becomes equal Delta H + 2 r so from this particular equation is Delta U equal Delta h no I'll say Delta U is greater than Delta Delta U from this particular equation you can see Delta U is greater than Delta by this much of amount by this much amount Delta U is greater than Delta so this cannot be the answer only B is the answer right am I clear am I clear people quick quick quick okay I believe it's clear moving ahead moving ahead okay this is one more common question which is asked look at the question carefully look at the question carefully guys calculate Delta U when 2 moles of liquid water vaporizes at 100° C so you vaporizing liquid water you're vaporizing liquid water what does that mean so you converting H2O liquid into what you're converting H2O liquid into H2O gas you converting H2O liquid into H2O gas right Delta H vaporization is given to us as 4066 KJ per mole what do we have to calculate we have to calculate Delta U during the process Delta is given Delta U we are supposed to calculate yeah first thing guys can you let me know Delta n for this reaction number of moles of gas gous product which is one minus number of moles of gaseous reactant this not gaseous reactant so zero so Delta n is 1 number one number two you know Delta H will be equal Delta U plus Delta ngrt right I need to calculate Delta U so Delta U is equal Delta H minus Delta ngrt okay Delta U is equal what is Delta H 40. 66 k 4.66 KJ minus Delta NG is 1 R values 8.314 jewles per Kelvin per mole jewles temperature is how much 100° 373 Kelvin since I used the value of r as 8.314 jewles but this term is in KJ so this term has to be in kilj as well I'll be dividing this term with th then only it will come in KJ perfect right so my dear students after solving this particular equation what you'll be getting you'll be getting the value of uh Delta U as 37.5 37.5 is there any option as 30. 37.5 yes but that's not going to be the answer this is not going to be the answer why this is the value of delta U this is the value of delta U for the vaporization of one mole of liquid water we are vaporizing one mole of liquid water here we are vaporizing one mole of liquid water but as for the question we have to vaporize 2 moles of liquid water this is the value of delta U for the vaporization of one mole of liquid water therefore for the vaporization of two moles of liquid water you have to multiply this by two when you multiply this by two the answer comes out to be 75.1 K am I clear with this am I clear with this people am I clear with this perfect perfectly done guys all the things are clear till here say in the chats once are all the things clear perfect now guys let's have a look on different types of processes first of all you know I have given you the idea about the isocoric process what is an isocoric process in which volume remains constant if volume is constant Delta V is zero so work done in case of isocoric process will be zero perfect so if I use the first law of thermodynamics here Delta U equal Q Plus W but since volume is constant isocoric so w is zero so Delta U is equal q and I'll be just writing it QV what is the meaning of this equation is this equation valid everywhere no this equation is only valid for isocoric processes when volume is kept constant so at constant volume basically at constant volume whatever amount of heat will be absorbed by the gas at constant volume whatever amount of heat will be absorbed by the gas that entire amount of heat is utilized increase the internal energy of the gas guys it's simple see if Q value is positive that means Delta U is positive Delta U positive means inter energy increasing so at constant volume remember at constant volume whatever amount of heat is absorbed by the gas that entire amount of heat is utilized to increase the internal energy of the gas okay perfect right I hope this clear W value in case of isocoric process always remember this is zero because volume of the system does not change if volume does not change Delta V 0 if Delta V 0 W 0 okay perfect if w z right I hope this is clear number one number two here that you all must be knowing pressure volume C how it looks like pressure volume C how it looks like it's a straight line like this and this straight line is what you call us isocore it's what you call as isocore okay now when you talk about isobaric processes in case of isobaric process what do we have in case of isobaric processes pressure is constant you know it pressure of the system is kept constant Delta p is zero right perfect Delta U is nothing but Q + W here I'll write QP add constant pressure add constant pressure Delta U is equal Q + W how do we calculate work done in case of the isobaric processes how do we calculate work the first result you already know that is minus P Delta V perfect or you can use this as well for the reactions minus Delta RT and particularly this particular result is used when a reaction is carried out in an open vessel when a reaction is carried out in an open vessel because when the reaction is carried out in the open vessel we consider it as isobaric in nature and if it is isobaric then this is the result which we'll be using okay perfect right guys now let me show you certain or before showing you the questions tell me what will be the PV C here if you talk about pvco it will be like this isobar it'll be isobar for example I'm giving you certain questions let me see if you can solve them or not for example this is the question for example this is the question one mole of aluminium is dropped in an aquous solution in an open container open container means isobaric containing HCL if 95 calories of heat is released heat is released means Q is - 95 calories heat is released right calculate W Delta Del U Delta h calate w Delta U Delta H first of all it's isobaric so w has to be equal to minus Delta nrt minus Delta RT Delta NG is equal Delta NG is equal number of moles of gaseous product which is 3x2 this not gaseous minus number of moles of gaseous reactant there is no gaseous reactant so your Delta n is just 3x2 so w if I calculate from here if I calculate W from here minus Delta n g is 3x2 r value I'll be taking in calories for example which is two temperature is given to me as how much 300 Calin perfect so this two this two got canceled so w comes out be minus 900 calories this is the value of w number one so w you got okay W you got what is the amount of heat released at constant pressure right it is 95 calories now since W you got uh and you you got this QP what is QP QP is basically heat absorbed or released at constant pressure you know already QP is nothing but Delta H I've already told you what is QP QP is basically Delta H QP is basically Delta H so Delta H also you calculated QP is Delta H QV is Delta V sorry QP is Delta H QV is Delta U okay so Guys these two terms we got Now Delta U we have to calculate what is Delta U if I use first law Delta U is equal Q + W right since the process is isobaric so I'll write QP so Delta U has to be equal what is QP QP is - 95 W value is -900 the value comes out to be - 995 calories this is the value of what this is the value of delta U over here I believe I'm clear I believe I'm clear yes perfect let's try to solve one more question from the same concept let's try to solve one more question let me see if you can solve this or not this is the question just add one thing at 300 Kelvin in an open container in an open container can you solve this one my dear students whenever it's mentioned that reaction is happening in an open vessel take pressure constant always okay take pressure constant always work done during the decompos of five moles of this PCL 5 so first of all what is the Delta NG for this reaction Delta NG is equal to 1 + 1 2 2 - 1 1 so Delta NG is 1 Delta NG is one how do we calculate W minus Delta n RT right Delta NG is 1 r value in calories is 2 temperature is 300 the value comes out to be 600 calories but this is the value of w for the decomposition of one mole of PCL this is the value of w for the decomposition of one mole of pcl5 right this is the value of w for the decompos of one mole of pcl5 but as per the equation we have to decompose five moles of pcl5 so the value of w for the decomposition of 5 moles of pcl5 is equal 600 mtip with 5 the value will be 3,000 calories this is the final answer of the question this is the value of w for the decomposition of 1 mole of pcl5 and this is the value of w for the decompostion of 5 moles of pcl5 I hope I'm clear I hope I'm clear right yeah perfectly done now let's talk about the isothermal processes let's talk about ismal process proc es my dear students what do you have to remember in case of isothermal processes first of all we have got the ideal gas which is undergoing isothermal process ismal proc means temperature constant temperature constant means delta T is equal Zer delta T is z means Delta U is equal Q + W or wait Delta t z you know your Delta U for the ideal gas is nothing but ncv delta T if delta T is z that means Delta U is zero as well you know it already you know it already now as for the first law Delta U is equal Q + W for the isothermal process Delta U is zero so Q is nothing but minus W this particular equation Q is equal to minus W is it valid everywhere no it is only valid for the ideal gas undergoing isothermal process right Q is equal minus W now my de students from this particular equation Q is equal minus w we can generalize a lot of things from this particular equation which is valid for isothermal process we can we can predict a lot of things we can depict a lot of things now what all things can we depict from here yeah uh HMA what are you saying exactly the Phoenix batch complete in 7 months is there other batch like what okay one more thing guys we have got the ideal guess undergoing ismal process it's Delta H will be also zero you know because delta T is zero Delta H is also zero because Delta H for the ideal guas is n CP delta T already that is discussed I've told you that so look at this particular expression look at this particular expression understand W value can be positive or negative right W can be positive or negative these are the two things which can happen W can be either positive or negative W negative means expansion W positive means compression okay perfect if W is negative W is negative negative negative makes it positive therefore Q value is positive therefore Q value is positive if W is positive positive negative makes it negative so Q is negative Q negative means heat is released by the system Q positive means heat is absorbed by the system basically here I can generalize two statements if you look carefully I can generalize two statements I can say I can say during isothermal expansion of an ideal gas I can say during isothermal expansion of an ideal gas during isothermal expansion of an ideal gas heat is heat is absorbed by the gas heat is absorbed by the gas and similarly during during isothermal compression of an ideal gas he is released by the gas it is released by the gas these are the two statements which are important guys okay during ismal expansion of an ideal gas heat is absorbed by the gas during isothermal compression of an ideal gas heat is released by the gas two important statements okay perfect perfect guys now the main point how do we calculate work done in case of isothermal processes right how do we calculate work done in case of isothermal processes my dear students you know we have got two types of process basically one is reversible one is irreversible so on one side I'm writing a reversible isal process reversible ismal process on another side I'm writing irreversible ismal process irreversible ismal process the point is how do we calculate work done in reversible ismal and irreversible ISO how do we calculate work done in both first thing if you remember if if you remember if you remember in case of reversible process how do we calculate work done W reversible is equal minus integral V1 to V2 this is p gas multiplied by DV do you remember that perfect so w reversible will be equal minus integral V1 to V2 what is p p is nothing but nrt / by V multipli by DV if you do the integration etc etc what you'll be getting minus nrt will come out it is going to be lwn of vs2 / V1 lwn of V2 V1 or you can write it like this minus nrt lwn of instead of V2 by V1 I can write P1 / P2 I can write P1 divid by P2 this is how you calculate work done in case of a reversible isal process or my dear students you can write it like this W reversible is equal uh convert this in logarithmic form it'll be - 2.303 nrt it's going to be log of V2 by V1 or - 2.303 nrt is going to be log of P1 by P2 these are the Expressions which are to be used to calculate work done in case of a reversible isothermal process reversible isothermal process now what about irreversible what about irreversible isal process how do we calculate work done in irreversible isal process W irreversible you already know it is nothing but minus P external V2 minus V1 this is the first result which you'll be using or some times you'll be using this one as well W irreversible is equal minus P external instead of V2 I can write nrt / P2 minus instead of V1 I can write nrt / P1 these are the two expressions which can be used to calculate work done in case of irreversible in case of irreversible isothermal process reversible ismal work done irreversible ismal work done remember them remember these Expressions now if you remember these Expressions you can easily e easily and easily solve the questions for example for example for example look at this question look at this question guys can you solve this can you solve this question can you solve it see guys as per the equation calculate q w Delta U Delta H when two moles of ideal gas when 2 moles of ideal gas are expanded reversibly and isothermally ideal gas undergoing isothermal process ideal gas undergoing isothermal process so Delta U as well as Delta h both have to be zero right initial pressure of the gas is given as 10 ATM final pressure of the gas is given us 5 ATM right 5 ATM temperature is kept 27° C right we have got two moles of ideal gas which is undergoing isothermal which undergoing isothermal process is it isothermal reversible yes it's isothermal reversible ismal reversible right it is ismal reversible how do we calculate work done in ismal reversible processes it is minus nrt lwn of P1 by P2 correct minus what is n value what is n value n value is two r value taken calories for example that is two temperature how much temperature 27° means 300 Calin lwn of P1 by P2 lwn of P1 by P2 10 by 5 is 2 lwn of two lwn of two is how much 0.7 right 0.7 how much this value comes out be uh do I have to calculate in calories yes n is 2 R is two temperature is 27° Cen 300 kin lawn of P1 / P2 so this is 2 2 4 4 3 12 12 into 0.7 comes out be - 840 - 840 calories so you got the W as well you got Delta U you got Delta H right you got W as well and since the process isothermal in case of isothermal process Q is equal minus W right W already calculated so Q value comes out be plus 840 calories so these were the things which we were supposed to calculate this Q This is w and Delta U Delta these were the four par MERS which were supposed to be calculated I hope this is clear yes am I clear people am I clear look at one more question look at this particular question let's see what the question is all about you read the question first of all read the question quickly 2 L of N2 gas at 0° and 5 PM are exploded isothermally against a constant exeler pressure it is mentioned that external pressure is constant so process irreversible process irreversible process irreversible isotherm process irreversible ISO right processes irreversible ISO calculate how much heat will be absorbed or released calculate how much heat will be absorbed or released first of all you have got 2 L of N2 gas 2 L of N2 gas kept at a pressure of how much 5 ATM now the gases exploded let's say the final volume of the gas is V2 as for the equation the final pressure of the gas is how much 1 18m expansion is happening against constant exteral pressure how much is the constant exal pressure that is also 1 atm P external is 1M P external is 1M the process is irreversible and at the same time it's mentioned that the process isal the process irreversible isothermal the process is irreversible isothermal okay irreversible isothermal since the process isothermal so it has to follow Bo's law it has to follow boils law boils law says that P1 V1 is equal P2 V2 correct what is P1 P1 is 5 what is V1 2 what is P2 1 what is V2 you can calculate from here so V2 is nothing but 10 L so you got to know the V2 that is 10 L correct since the process is irreversible isothermal if the process irreversible isothermal so w irreversible has to be equal simply minus P external vs2 minus V1 right which will be minus what is p external 1 atm what is V2 10 - 2 that's 8 the value comes out to be - 8 ATM lit which can be written as approximately - 808 Jew this is the value of w but am I supposed to calculate W or Q I'm supposed to calculate q and you know my dear students in case of isothermal processes Q is equal to minus W so it has to be plus 808 Jew plus 808 Jew right so these sort of questions can be asked from the isothermal processes be it reversible or irreversible okay okay guys yeah all right now mov ahead there is something called as adamatic process as well adamatic process what are adab batic processes you know it already in case of adaba process Q value is zero there's no heat exchange between system surroundings Q value is zero if Q value is Zer you know Delta U is equal Q + W so therefore Delta U will be equal W only this equation is valid for adaba processes Delta u w atic okay atic now here I can generalize few statements see guys the value of w can be either positive or the value of w can be here negative now you tell me if the value of w is negative that means the gas is undergoing expansion over here the gas is undergoing compression okay if W is negative if W is negative that means Delta is negative if W is negative that means Delta U is negative Delta U negative means internal energy is decreasing Delta U NE means internal energy of the gas is decreasing and you know internal energy is directly proportional temperature right if internal energy decreases that means temperature of the system increases sorry decreases they're directly proportional similarly similarly W is positive which means that Delta U is positive if Delta U is positive it means that intern energy increases if intern energy increases that means temperature of the system increases guys here you can generalize two statements during during during during adiabatic expansion of an ideal gas during adiabatic expansion of an ideal gas temperature of the gas decreases during adaba expansion of an ideal gas temperature of the gas decreases during adaptic compression of an ideal gas temperature of the gas increases two statements we need to remember here during adiabatic expansion of an ideal gas of an ideal gas during adamatic expansion of an ideal gas temperature temperature of the gas decreases similarly my dear students during during edotic compression during bring adabo compression of an ideal gas of an ideal gas temperature of the system I mean temperature of the gas increases temperature of the gas increases okay are these two statements clear guys whenever the ideal gas under goes adabo expansion its temperature will be decreasing with time whenever the ideal gas undergo under goes ad diotic compression its temperature will be increasing with time its temperature will be increasing with time right perfect now guys here also there are two things how do we calculate work done in reversible Tic and how do we calculate work done in irreversible tic that's a point here how do we calculate work done in reversible Tic and how do we calculate work done in irreversible tic see those are the two important things work done in reversible adic process and work done in irreversible adic process work done in reversible dat IC work done in irreversible tic how do we do it how do we do it see guys first of all in case of your adic process Q is zero in case of your adic process Q is zero so as per first law wal Delta U so w is equal to Delta U and Delta U for the ideal gas is nothing that is n CV T2 minus T1 is the first result which can be used to calculate work done in reversible tic processes or there can be few more results generated work done in case of a reversible tic will be equal n TV I'll write as rid gamma minus 1 and T2 minus T1 can be written as delta T this can be one more result by means of which you can calculate work done in case of reversible tic or W reversible is equal NR this is nrt2 minus nrt1 nrt2 is p2v2 nrt1 is P1 V1 divide by what divided by gamma minus these are the three Expressions by means of which we can calculate work done in case of reversible adamatic processes in case of reversible tic processes three Expressions you have to remember by means of which we can calculate work done in reversible adaba processes now what about irreversible irreversible already you know W irreversible is equal minus P external V2 minus V1 this can be the first expression the second one W IR ible is equal minus P external what is V2 V2 can be written as nrt2 / P2 V1 can be written as nrt1 / P1 out of these two expressions you can use anyone and get the work done in case of irreversible adamatic processes in case of irreversible tic processes okay these are the formulas by means of it we can calculate work done right right people now before showing you the question there are there are two more things asked as well from adiabatic part from adiabatic part you'll be asked see guys one thing you already know whenever the ideal gas under goes adab batic expansion its temperature decreases when the ideal guas under goes adamatic compression its temperature increases right so during expansion and compression during expansion during ad diabatic expansion or atic compression temperature of the system system temperature of the gas changes in short right now guys how do we calculate the final temperature how do we calculate the final temperature of the gas in case of a reversible anabatic process similarly how do we calculate the final temperature how do we calculate final temperature of the gas in case of irreversible anabatic proc in case of irreversible tic proc calculation of final temperature nature of the gas in case of reversible Tic and in case of irreversible tic just a second guys all right in case of reversible tic process how do we calculate the final temperature see General equation I'll be using T1 V1 ra gamma minus 1 is equal T2 V2 ra^ gamma minus 1 this is one equation by means of which we can calculate final temperature T2 of the ideal gas which will be undergoing tic process it will be undergoing reversible tic process the second one P1 ra^ 1us gamma P1 ra^ gamma is equal P2 ra^ 1us gamma T2 ra^ gamma this is one more expression with the help of which you can calculate final temperature T2 here so when the ideal gas will be undergoing a reversible tic process these are the Expressions by means of which you can calculate final temperature of the ideal gas okay now in case of irreversible tic how do we calculate final temperature let me give you the expression then I'll show you the directly first of all atic Q is0 q0 means Delta U is equal W and over here I'm talking about irreversible Delta U is nothing but n CV T2 minus T1 right W irreversible is What minus P external V2 minus V1 so this can be the first equation used to calculate final temperature T2 of the ideal gas right or you can convert it into different form as well you can say n CV is nothing but rid gamma minus 1 and this is T2 - T1 is equal minus P external V2 can be written as NR T2 / P2 V1 can be written as NR T1 / P1 so NR NR NR canceled so I'll be getting the final result as T2 - T1 / gamma minus 1 isal minus P external it's going to be T2 / P2 minus T1 / P1 right so 2 Expressions we have here to calculate the final temperature T2 of the ideal gas which under goes irreversible batic process okay so first of all you have to identify whether the process is reversible or irreversible accordingly you'll you'll be using the results I hope I'm clear to everyone I hope I'm clear to everyone let's try to solve certain questions so that it becomes clear to you see guys the first question it's an asked question too can you give it a try can you give it a try quickly quickly people done yes VMA you can do that you can do that this series will be very much helpful as per the question one mole of a monatomic gas so how many moles of the gas do we have one mole the gas is monatomic if it is monatomic it CV has to be 3/ 2 R perfect expands tically initial temperature is capital T initial temperature is capital T against a constant Exel pressure constant Exel pressure of 18m if external pressure is constant that means the process is irreversible irreversible from 1 L to 2 L initial volume of the gas was 1 lit final volume of the gas is 2 lit what am I supposed to calculate final temperature of the gas final temperature of the gas since guys the process is irreversible if the process is irreversible I'll be using the expression which I gave you a few minutes back n CV T2 minus T1 is equal minus P external vs2 minus V1 right now n value is one CV value is 3id 2 R 3id 2 R T2 we have to calculate T1 is capital t is equal minus P external is 1 V2 is 2 and this is 1 correct now from here you can calculate T2 and I believe T2 will come as T minus 2 / 3 R this is the final temperature of the gas which I was supposed to calculate this is the final temperature of the gas which I was supposed to calculate see uh Han gave the answer T ided 2 2x3 see this is the mistake that will be one of the options there that'll be one of the options there if it was mentioned that the process is reversible adamatic then your answer was correct then your answer was correct then you would have used then at that point of time you should have used this particular formula T1 V1 I mean this one right T1 is capital T V1 is 1 1 rais anything is 1 this is T2 V2 is how much 2 rais gamma minus one gamma for monom gas will be how much 5x 3 5 3 - 1 is 2x 3 so at that point your T2 would have been t by 2^ 2x3 but this is the final temperature of the gas if the gas was undergoing a reversible adamatic but in the question it's not reversible it's irreversible tic so your answer should be this one okay correct I hope it's clear guys just if in the question they have mentioned that external pressure is constant constant external pressure that process irreversible that's it okay that process irreversible let's try to solve one more question give this question a try give this question a try people a sample of fluorocarbon a sample of fluorocarbon was allowed to expand reversibly and AD diabatically to twice of it volume uh I think it'll take us 1 hour more okay maximum 1 hour more okay A sample of fluorocarbon was allowed to expand reversely and tically the process reversible adaba to twice of its volume if the initial volume is V so the final volume has to be 2 * V during the expansion the temperature dropped from initial temperature of the gas is given as 400 Kelvin final temperature of the gas is given as 100 Kelvin assuming ideal gas conditions calculate CV for the gas we are supposed to calculate CV CV is nothing but rid gamma minus one right CV is nothing but rid gamma minus one so basically I need to calculate gamma first of all since the process is reversible and adamatic the process is reversible and adab batic so in case of reversible adab batic I can use this equation T1 V1 gamma minus 1 is equal T2 V2 ra gamma okay what is T1 check it out T1 is 400 what is V1 V1 is V raise bar gamma minus one is equal what is T2 100 what is V2 2 * V gamma- so 0 0 cancel V GMA 1 V cancel so 2^ gamma - 1 is 4 4 is nothing but 2 square so gamma - 1 is 2 you got to know gamma - 1 2 if gamma - 1 is 2 so CV is R / 2 right this is something which I was supposed to calculate isn't it simple people is it it's simple yes so you should be able to solve this sort of question question as well if it comes all right okay have you heard about something called as free expansion free expansion pre expansion yeah what is free expansion you know expansion of the gas in vacuum or you will say expansion against zero external pressure expansion against zero external pressure expansion against zero external pressure what it means for example there's a container which we have and in this container you have kept an ideal gas let's say okay so over here on the Piston there'll be two pressures one is pressure of gas and one is p external right assume right now P gas and P external they are equal if P gas and P external are equal the Piston won't be moving at all it be it'll be stick at one position if suddenly PX becomes zero if sudden suddenly PX become zero the Piston will suddenly go out if the Piston suddenly goes out the volume of the container increases volume of the gas increases increase in the volume is something which you call as expansion so over here I would say at that point of time expansion of the gas is happening against zero external pressure and expansion of the gas against zero external pressure expansion of the gas against zero external pressure or you can say expansion in vacuum you can say expansion in vacuum is something which is what you call as free expansion which is what you call as free expansion right perfect my dear students free expansion right expansion against zero exal pressure so suddenly if you make p exal 0 the Piston will go out fast right so this process is irreversible in nature it is irreversible in nature number one number two since external pressure is zero how do we work done in irreversible processes W irreversible is equal minus P external Delta V if P external is z so w value has to be zero W value has to be zero perfect so in case of free expansion W value is zero perfect during free expansion W value is zero now the same free expansion you can I mean further divide it into two categories let's see if you know it on or Not the Same free expansion you can divide into two categories the first one I'll be calling as atic free expansion ad diotic free expansion of an ideal gas second one is isothermal pre expansion of an ideal gas isothermal free expansion of an ideal gas atic free expansion of an ideal gas and second one is isothermal free expansion of an ideal gas the gas has to be ideal okay now guys understand first of all I'm mentioning that free expansion in free expansion W is zero in adaba Q is0 if I use the first law Delta U is equal Q + W so Delta U is z Delta U is z means ncv delta T has to be zero if ncv delta T has to be zero that means delta T has to be zero delta T will be zero only if temperature of the gas is constant temperature of the gas is constant is in case of what isothermal so can I say can I say an adiabatic free expansion of an ideal gas is isothermal as well an adiabatic pre-expansion of what anabatic free expansion of an ideal gas of an ideal gas is is isothermal as well is isothermal as well okay this is the statement which you need to remember now my dear students if you look at the next one isothermal pre-expansion first of all pre-expansion means w is equal 0 isothermal means isothermal means Delta U is equal Z right so as per first law Delta U is equal Q + W that means Q value has to be zero Q is equal to0 means adamatic so again I can generalize a statement I can say an isothermal an isothermal free expansion an isothermal free expansion of an ideal gas of an ideal gas is adiabatic as well is adiabatic as well am I clear with this so these can be the two statements which can be asked right two St statements can be asked okay guys there is one last topic of the today's session see guys one thing I'll tell you uh there will be the second marathon of physical chemistry class 11th as well in which we shall be doing thermochemistry thermochemistry atomic structure and equilibrium these three will be done in the second marathon of physical chemistry okay these three will be done in the second marathon of physical chemistry there is one last topic left that is cyclic process okay let's complete the cyclic process and we are done for the day in thermochemistry only I'll be doing Gibs free energy entropy in thermochemistry part I'll be doing Heats of reactions entropy entropy entropy gives free energy then atomic structure and equilibrium that will be done in the second marath okay so here let's talk about something called as cyclic process let's see what kind of question can be asked from the cyclic process okay let's see cyclic process so first of all As for the name is concerned you already know in case of cyclic process the initial State and the final state of the system is same initial State and final state of the system is same right change in the value of any state function for a cyclic process it has to be zero it has to be zero right so can I say Delta U net Delta U net as well as Delta H net in case of cyclic process has to be zero correct Now Delta U net will be equal as for first law Q Net plus W net so Q Net plus W net has to be zero so in case of cyclic process I would say Q Net is equal to minus * W net this is valid for the cyclic process Q Net is equal Min times W the first thing which you need to know the first thing which you need to know okay number one for example I'm making a cylic cross over here let's say this is pressure this is volume pressure versus volume we are starting let's say from point a point a from A to B we are going then from B to C we are going then from C to a we are going perfect so can I see the system is returning to its initial state after going through a lot of steps yes how many steps 1 2 3 1 2 3 right 1 2 3 now now my dear students understand one thing if I write Q Net what is qet it is the net heat absorbed or released during the cyclic Process net heat absorbed or released during the cyclic process Q Net has to be nothing but it'll be heat absorbed or released during the process AB heat absorbed or released during the process BC heat absorbed or released during the process CA similarly W net W net is going to be work done during the proc AB Plus work done during the proc BC Plus work done during the proc process CA work done during the process CA okay now one more thing if they ask you calculate the magnitude of net work done calculate the magnitude of net work done magnitude of net work done how you are going to do that it's always equal area enclosed by the Loop area enclosed by the loop this area you have to calculate right it will give you the magnitude of net work done and at the same time if the cycle is clockwise if the cycle is clockwise W net sign will be negative if the cycle is anticlockwise if the cycle is anticlockwise W net will be positive these are few things which you have to remember that's all that's all few things which you need to remember that's it okay in case of cyclic processes for example let's try to solve question so that you understand it properly see guys one basic question we have we are given with the cyclic process we have to calculate net work done net work done so first of all I'll calculate magnitude of network which will be area inclosed by the loop area enclosed by the loop means area this area area of the triangle which is half of Base into height base is V1 to 3 V1 that means 2 V1 and from here to here it's going to be 5 P1 so the value is going to be 5 * P1 V1 this is the magnitude of network done now the cycle is clockwise if the cycle is clockwise clockwise so right the W with the sign it has to be minus * 5 P1 V1 right these are of basic basic questions can be asked done understood right these are of basic basic questions can be asked for example what is this question all about let's see one mole of ideal gas one mole of ideal monatomic gas so n is one it is monoatomic as well so CB is equal 3x2 R right CP will be equal 5x2 R because the gas is monoatomic is taken through the cycle given calculate qab and qca qab we have to calculate qca we have to calculate so qab during the process A to B calculate the amount of heat absorbed or released and during the process C to a calculate the heat absorbed or released first of all if you analyze the ab process if you analyze the ab process AB process volume is constant it's isocoric it is isocoric so amount of heat absorbed Orly during the proc AB what does that mean it is basically amount of heat absorbed or released at constant volume right that will be ncv T2 minus T1 n CV T2 minus T1 what is n 1 CV is 3x2 r 3x2 r let's say I am taking everything in calories 3x2 R T2 minus T1 A to B this is initial this is final so final temperature minus initial temperature final is 600 initial is 300 600- 300 comes out be 300 right so 2 two cancel the value comes out to be 900 calories so qab is equal 900 calories now similarly qca heat absorbed or released during the process CA this CA process as you can see it's isobaric isobaric that means it absorbed or relased at constant pressure we have to calculate it has to be equal n CP P2 minus T1 and CP T2 mins T1 right what is n n is one what is CP CP is 5x2 R T2 minus T1 CA this was initial this is final so final temperature minus initial so 300 - 450 means - 150 cor so 2 two gets cancelled the value will be - 750 calories perfect these were the two things which I was supposed to calculate these were the two things which I was supposed to calculate yes perfect perfect guys I hope all these things are clear I hope all these things are clear to you right guys okay there is one more small topic which is left that is efficiency of the cyclic process that is efficiency of a cyclic process just a second efficiency of the cyclic process how do we calculate efficiency of the Cy process it's a simple 5 minute topic efficiency of the cyclic process area enclosed by the loop area enclosed by the loop means this area this the area enclosed by the loop divided by area under the curve divided by area under the curve what is the area under the curve area under the curve is this whole area right so first you have to calculate area enclosed by the loop then you have to calculate area under the complete Cur then multiply 100 you'll be getting the efficiency of the cyclic process for example see let me give the proper naming to everything this a b c let's call this point as D let's call this point as e okay so percentage efficiency percentage efficiency is equal area enclosed by the loop means area means area ABCA right perfect means this area area ABCA area under the curve means a d e c area a d e c multiplied by what multiplied 100 the percentage efficiency is equal area ABCA ABC means area of this triangle which is half of base base is from here to here 4 L height from here to here one perfect divided by this is a trapezium this is a trapezium trapezium means it is half of some of parall sides sum of parallel sides which is going to be 2 + 1 into height from here to here it's 4 L multiplied by 100 now we can solve this it's done right this term this ter cancel 2 + 1 3 3 4 12 4id 12 is 1x 3 1 by 3 * 100 is 33% so 33% is the efficiency of the cyclic process right I mean you just need to know the area enclosed by the loop divide by area under the curve divide them up multiply 100 that's all okay perfect guys I hope all these things are super clear okay perfect perfect guys all right in the next marathon what exactly are we going to do in the next marathon of your physical chemistry we are going to do thermochemistry including entropy and gives free energy number one number two atomic structure and number three we shall be doing the equilibrium we shall be doing the equilibrium these are the three chapters which I have planned in the upcoming week right so basically what I had planned in the today's session I had planned I'll be doing atomic structure instead of thermodynamics so that three chapters will get completed properly right uh wki you can I mean I don't know about that but I would rather tell you to write the test on priority okay perfect guys I think this much is it enough we are done we are done for the today we are done for the day yeah we are done for the day pdf also I'll send I hope all the things are clear yeah uh i y manzur is here yeah manour uh do attend the tomorrow session guys this is that is that is uh class 12th organic you used to tell me all the time Class 2 complete class 12th organic organic right HSP said is going to take a marathon on that so do join that session as well which topics come under thermochemistry Heats of reactions Heats of reactions enthalpy of reaction enthalpy of formation enthalpy of neutralization etc etc right perfect I'll send you the PDF for sure okay hello guys with this then I'll be a leave okay do let me know in the comment section of this particular video whether you like the session and I would request every one of you to comment on this particular session right those comments which will make me happy yeah tell take care guys see you by