okay now we're going to continue with the ideal gason get into the heart of the matter and how we're going to do all of our calculations um for pressures and temperatures and volumes um for this semester okay so it all revolves around the ideal gas equation um this is one equation that includes pressure and volume and temperature and number of moles um in relationship to to each other and it includes all of the uh relationships that we saw in the last lecture in chapter 5.3 that includes um Bo's law Charles law mattin's law um and uh uh avagadro's law all of that is included in this one equation and it's relatively simple equation that's it PV equals nrt you're going to use this until you're really sick of it um pressure times volume is equal to the number of moles times R times the temperature so the only thing we don't really know about this is what is R so let's get to that okay let's talk about what R is this is the ideal gas constant it comes in many forms um that is to say it has different values depending on which units you're using units units units um that's what ra matters entirely for this um this uh constant um it relates the dimensions of pressure and volume and uh moles and temperature so it has all of these values included in it um so um the one we're going to use most often is r equal 0.082 57 I usually use 0.08206 to four significant figures that's in units of lers time atmospheres divided by moles per Kelvin so those are the units that you have to use for your volume and pressure and moles and temperature moles and temperature are always going to be in these units but liters or rather volume can be in liters or cubic meters or gallons there's lots of different volume units and pressure as we saw in the first set of SL that can also be in lots of different units so we have to be careful that we're using the right units um for instance another value and I wouldn't bother learning any of the rest of these but uh because we'll do our calculations by converting everything into liters and atmospheres doing the calculations and then converting out of it rather than looking for all of these different um values of R I'm just trying to show you that it its value depends on which units you're using so if you're using liters um Millers millimeters of mercury in t then you'll have this value if you're using bar then you'll have this value and if you're using me cubed time pascals you'll have this value this value is going to be important later and we'll talk about what meters cubed pascals means um but for right now um I really just want you to focus on this r equal 0 0 8206 um to do our calculations um right is also equal to that that's the one we want to talk about okay um so let's do a couple of calculations um with this um let's just do one of the basic calculations this is a number that all chemists know um by heart uh it is called the molar volume this is the volume of an ideal gas at standard temperature in pressure it's a rule of thumb we all know that it's 22.4 L um if you ask anyone in the chemistry Department what's the molar volume of gas they'll tell you they don't have to calculate it they just know it um but we're going to do the calculation um so it is the volume occupied by an ideal gas of exactly one mole so that's why it's the molar volume uh under the conditions called standard temperature and pressure talked about this earlier that is defined as 0 degre Celsius and one atmosphere now I'm writing these to just um one significant figure but in the calculation I'm going to keep three significant figures um as we go forward okay so here's our balloon or container holding one mole at STP so uh 0° uh 1 atmosphere and I don't know what the volume is but I do know what the ideal gas equation is so I'm going to use that here it is PB equal n RT and I'll look in here and say okay what did this what here don't I know uh R is a constant I always know what R is um the rest are variables I know what n is that's one mole I know what T is I can calculate that from the temperature I mean and that is the temperature but I'll find the right units for that and and I know what V is so I need to find I'm sorry and I know what pressure is I need to find volume so let's rearrange this equ equation to find volume so rearranging it we just divide both sides by P and now we can say volume is equal to the number of moles times the gas constant times the temperature over pressure now let's make sure we do everything in the right units n is one mole right that's given and I said three significant figures that's why I'm including three significant figures R is 0.08206 lers atmospheres time inverse moles time inverse temperature it's the same as dividing by temperature is 0 deg C but those units aren't going to work I need to do all of my gas LW calculation in kelvin so I'm going to convert this to 273 technically 273.15 these are not significant so I put them in parenthesis we'll just keep three significant figures in this calculation now let's uh and pressure is in is given as one atmosphere and atmospheres is the unit in the gas constant so I'm good I don't have to convert that now let's just plug these values into the equation to solve for v v equal 1 mole time the gas constant times the temperature in Kelvin divided by the volume in atmospheres before we do the mathematical calculation let's look at the units atmospheres can cancel with atmospheres moles cancel with inverse moles and inverse Kelvin cancels with Kelvin the only unit left here is liters perfect I need this answer to be in volume and the volume is a volume unit is liters that's the one we're going to use for this chapter so when I go ahead and multiply all the numbers through and divide I get that the molar volume is 22.4 L at STP so that's a rule of thumb all chemists know there's another version of standard temperature and pressure this one's called standard ambient temperature and pressure I don't really use this one very much I'm just going to go through this exercise to show you why we have to be careful of units um so standard ambient temperature and pressure is defined as 25° C and one bar of pressure so the temperature is different and the pressure is different um it's one but the units are different so it's a different pressure remember a bar is slightly less than an atmosphere so let's set up our numbers n and R stay the same but uh um T and P are different temperature this time is the uh pressure is one bar it's a little bit it's right right there there one bar okay um next uh let's plug that into our equation I got to get uh uh um got to try and get my picture out of here um I'll come back and show you this slide without my face um in a minute so let's do that sorry about that I'll go up here let's get back to where we were um okay so we plug in these values into our equation we plug them in down here so we have the volume now is going to equal the number of moles times the gas constant times the temperature this is a new temperature and divided by the pressure so let's look at our units moles cancel with inverse moles Kelvin cancels with inverse Kelvin but bar and atmosphere are not the same unit they're close but they're not the same so we need to use the conversion factor from um our pressur uh uh units uh lecture to make sure that these will cancel with each other so we're going to look that up and we'll see that 1.01325 bar is equal to 1 at atmosphere so I'm going to multiply by this fraction with the bar on top and the atmosphere on the bottom and that's because now I can cancel the bar on the top here with the bar on the bottom and the atmosphere on the bottom with the atmosphere on the top then just like last time the only unit left here is liters so now with this adjustment because there's a difference of about 1.3% well 1 .325 per to be exact um we will be able to um find our molar volume um so it's a little bit bigger because the temperature is higher and it's a little bit bigger because uh bar is a smaller pressure um so it needs a bigger volume so when we multiply all of this out we get 24.8 um lers at standard ambient temperature andess pressure this is a unit that if you ask most chemists my age what's the um molar volume of a gas at standard ambient temperature and pressure they might ask you what's standard ambient temperature and pressure so we don't use this very often um but it's important to know that some some uh disciplines do use it engineering disciplines might work in bar because it's a unit that uses SI units um so you got to know which unit you're working and you can't just assume the volume is uh you can't just assume the pressure is in atmospheres um okay so that's our first uh PV equal nrt calculation let do a couple more hi there um this is the only place on the slide I can be without getting in the way um it's fine I'm comfortable in here nothing bad can happen I'm in a jar okay here we go um so we're going to ask the question at 31.5 de C how many moles of gas are there in each container um so in each case I'm giving you some of the values um but well I'm giving you the I'm giving you the volume and the pressure um and the temperature is over here so you got to calculate in from these equations so let's start with pbal an RT rearrange that this time to solve for n n is equal to PV over RT and then we just have to do that um for each side so for this side in the 2.50 atmosphere um uh bubble um we have that's the pressure the volume is 1.50 L uh the r is always for these calculations in units of liters atmospheres 0.826 uh per mole Kelvin and the temperature has to be converted to Kelvin otherwise we'll get the wrong answer and I'll show you that um a little bit too much I'm going to go over and over why we get the wrong answer if we don't convert and we'll get answers that look like they should work but they'll just be wrong um okay so 304.5 and again I'm keeping an extra significant figure for the calculations but I'm only going to keep four significant figures um in the answer actually I'm only going to keep three in the answer plug those in 2. 5 * 1.50 over 0.286 uh * 30465 uh again units cancel atmosphere with atmospheres this time liters are going to cancel um and Kelvin are going to cancel and we'll be left with uh inverse moles in the denominator which is the same as moles in the numerator right this is to the negative 1 move it upstairs and then it's to the positive one um and then we just multiply uh that out and the answer is to three significant figures .15 moles just that easy okay so over here in the uh jar that I'm in uh the we set up the equation I've got the pressure also in atmospheres liters uh volume also in liters over the same R and temperature and same thing units cancel got moles and there are 0.191 moles in this jar and me hi everybody um okay so that's um that's how we do these calculations all right I'm back down here so um so let's do a uh calculation where we're going to start changing variables so we'll have an initial set of conditions and a final set of of conditions we want to know what's going to happen uh to our um system after this change so um in this question we're going to ask what is the new pressure in millimeters of mercury so we got to change the pressure unit when the temperature is lowered to uh from the initial temperature to minus 20.9 degrees C okay so let's look at what we have oh I'm sorry let's not worry about what we have here we're just going to use the information from the last slide where we've already calculated n so n doesn't change because it's still in the same container the volume doesn't change still in the same container R never changes because it's a constant but the temperature went from 31.5 de C to - 20.9 - 29.9 de C C is 500 wow let me read that again is 25225 Kelvin so I'm using a different temperature um and because of that I'm going to get a different pressure so let's do that multiplication oh so use the value that was supposed to pop up first so here's how I calculated the uh the temperature uh these are just well these are equivalent to each other that works out that units are all going to cancel I'm not going to go through that we'll be left with just the unit of atmosphere and it will be 2.07 atmospheres so that is the new pressure so we went from 2.5 at 31° which is hot you know but not super hot it's like a hot day outside um and then we cool this to minus 20 which is cold like it's it's like a super cold cold Chicago day this is like um 30° below 0 fah so hot to cold and the pressure of the gas went from 2.5 and it decreased a little bit well it decreased quite a bit almost by 25% um so that's what we expect and so this is a good answer um the question is asking for this to be in millimeters of mercury so let's convert our 2.0 seven atmospheres using the conversion factor of millimeters of mercury to atmospheres and we get 1,570 that's three significant figures millimeters of mercury don't truncate the the um significant figures because of the conversion factor um okay so that's what we get um when we solve this problem notice that I didn't convert the pressure to millimeters of mercury first because then I would need to use a gas constant that has millimeters of mercury um built into it uh rather than atmospheres it wouldn't work um it just wouldn't work I have the wrong number uh so um I'm I do my conversions in atmosphere and then at the end of the problem I convert from my answer into whatever units the question is asking for um when I do this pval nrt okay hi I'm up here now um so the next uh thing we're going to do is solve this exact same problem we're going to do it a different way let's prend let's pretend that this is the question I didn't give you a question to calculate how many moles there were before so you don't know how many moles are in here you can calculate that from this information from 31.5 de in this but we're going to do this a different way we're going to look at how things change and it's a faster way to calculate a new value a new pressure a new temperature a new volume um by just looking at the things that change so let's do that so what are the two variables that are changing in this problem and the answer is pressure and temperature the volume of this is not variable it's going to stay at 1.5 L and uh the number of moles that are in here even though I you don't know what they are now uh or we're going to pretend that we don't know what they are those aren't going to change it's a sealed container so the only two things that'll change are pressure and temperature so I want to rearrange the equation PV equal nrt so that everything that changes I'm going to put on the left side and everything that stays the same I'll put those on the right side so I'm going to get pressure divided by temperature right if I pressure is Chang changes and temperature does so I have to move that over by dividing both sides by temperature and that's going to equal R * T and now I got to get the volume because that doesn't change in this problem over to the other side so I divide both sides by V so I get pressure divided by temperature equals NR over v um the reason I want to do that is because NR and V for this problem are constant so they don't change so so all I have to do is consider that the relationship of pressure divided by temperature is going to be constant um and therefore I can say my initial pressure divided by my initial temperature is equal to some constant so that's going to also equal my final pressure divided by my final temperature which means I don't even have to think about what happens here in the middle I don't need to know the volume I don't need to know the number of moles and I don't have to think about what the gas constant is this question is asking for us to calculate a new pressure from an initial temperature a final temperature and an initial pressure so let's rearrange that equation um but this time since I don't have to worry about my pressure units getting um getting interfered with by R's pressure units I'm going to convert my pressure one into Mill of mercury first I could do it at the end of the problem but I already did that so let's do it a slightly different way on this problem so it is also true to say that the pressure initial pressure is 1.90 * 10 3 millim of mercury okay so that's my initial pressure now let's rearrange the equation to solve for P2 P2 is equal to P1 that's this P1 and then T1 is in the denominator and then this T2 has to move up to the numerator so rearranging this equation I get P2 = P1 * T2 over T1 now let's substitute in our values 1.9 * 10us 3 mm Mercury not minus 1.90 * 10 3 millim Mercury times 22.25 Kelvin that's the um final temperature over the initial temperature which was 34.657214 um in this method you can use any pressure unit which is why I put it in millimeters of mercury first because without r i don't have to worry about what the pressure units are of my starting and ending numbers in red down here is the wrong answer in red down here is the wrong answer in red down here is the wrong answer why is it the wrong answer well because it didn't take my temperature units and convert them to Kelvin so these are on a relative scale not the absolute scale it's not going to work pval nrt doesn't work in degrees celsius doesn't work at all so if I use these values what happens is um I would get a number it looks like Celsius cancels with Celsius and it looks like I get a legitimate answer in the correct units dimensional analysis seems like it worked here but but it didn't because I'm in the wrong units um and in fact I get a nonsensical answer here I get a value for a pressure for pressure of a negative pressure there is no such thing as an absolute negative pressure there's no way for um a gas to exert negative pressure the gas on the inside of something can have less pressure than the gas on the outside so that's a relative negative pressure but this is not a relative pressure this is the absolute pressure of the gas inside this container um is according to the incorrect calculation would be negative that cannot happen this is wrong and it is incorrect because I used the wrong temperature units you cannot do these calculations using degrees Celsius this is not like our thermochemistry chapter where we were looking at changes in temperature these are absolute values of temperature um absolute measurements from no temperature up to 252 it's different you can't do ppal nrt calculations using um units of Celsius or Fahrenheit or any other temperature unit you might want to find okay good good let's do another example hey I'm back in here okay I'm fine nothing nothing bad can happen to me in here I said that before okay so now even though these questions will use words like when a sample of nitrogen dioxide is in the cylinder is compressed wait that doesn't sound good um when it says nitrogen dioxide all it means is an ideal gas for the sake of these calculations of new pressures we're going to treat the gases as ideal so right now the identity of this gas being nitrogen dioxide okay it's missing a two down here I'll fix that later nitrogen dioxide or nitrogen monoxide the reason I didn't this doesn't match this is because it doesn't matter and I wasn't paying attention sorry about that it just means there's some gas in here that starts off at 6 L uh three sorry 34.6 de Kelvin and 796 atmospheres so this is from the first problem that we did did we calculated the number of moles in here so we can do this calculation in two different ways you'll see what's coming up all right so when the sample is compressed from the initial conditions oh I'm okay um from the initial conditions to the final conditions of 3 l and 2.10 atmospheres what's the new temperature right so I'm not right so for this one we have some variables that are changing some that are staying the same so we can do this calculation again two different ways um the volume changes from 6 to 3 and the pressure increases from 796 to 2.10 um the question is what's the temperature of the gas so if we do this just by looking at the variables that change um we can solve this problem like we did the last problem problem where P1 V1 over T1 equal P2 V2 over T2 right cuz all three of these variables change the only variable that's not changing is the number of moles right that amount of gas didn't change we didn't lose any we didn't gain any gain any so the amount of gas that was in there is still in there okay so rearrange this equation and solve for the final temperature so t will equal T1 so I'm going to move the that up to that side and that up to that side so that leaves the P2 and V2 in the numerator and then I'm going to move these two guys by dividing both sides and that's P1 time B1 so here is the way to calculate the new temperature just plug in those values and then the units here atmospheres cancel with atmospheres liters cancel with liters and again we get well and we get to three significant figures a temperature of 42 Kelvin so the uh temperature um increases greatly right goes up by almost 100° during this compression okay what if you did this wrong because it looks again it looks like it's just temperature why do I have to do this in units of Kelvin I why bother changing it why can't I just leave it in Celsius like it was given in the problem and the answer to that is because you'll get the wrong answer so here's what happens if you do it that way if you leave the initial temperature which was given in Celsius a few slides ago of 31.5 de C and then you multiply by this which and in here all the units do cancel um and you just get a uh a a value um a value that's about 4/3 um it will show that your temperature goes from 31.5 to 41.6 right it's kind of the same percentage increase but if I convert 41.6 de C to Kelvin that's only 314. 7 so by doing this calculation in the wrong unit of degrees celsus I am about 90 degrees or what 88 degrees too small um because it's the wrong answer this is the right answer in the one up here that I'm pointing at in Black the one in red at the bottom is the wrong answer it's incorrect because of the wrong temperature units watch the temperature units please okay um good still in here um okay or we could have done it because uh we know all the variables we calculated in the first step what the number of moles was um that was a few slides ago when I introduced this jar um and we saw that um so we can rearrange pval nrt to solve for T where the volume is now 3 l the uh pressure is now 2.10 atmospheres the number of moles this is from a few slides ago that we calculated under the initial conditions um was 1.91 moles and now we can just plug those numbers in and um using now we got to use the gas constant and make sure the units cancel um liters cancel with liters atmospheres and atmospheres moles and moles we're left with one over one over Kelvin which is the same as Kelvin and we get the right answer of 402 Kelvin okay let's do another example um where uh we take uh850 moles of an ideal gas that has a volume of 2046 L at 123 C and 1.35 atmospheres so we're getting plenty of information at the beginning what is the pressure when the temperature is raised to 468 de C with the volume staying the same okay what do we know beginning conditions pressure volume number of moles and temperature so we have all of the information right which is weird usually we would be missing at least one of these maybe two so this is an overdetermined system uh the way this question is being phrased but I just wanted to do it this way before we get to more complicated questions the ending conditions we don't know the pressure the volume didn't change so we know that the number of moles didn't change we know that and the temperature did change so we know that so I want to convert my temperatures first thing convert those to Kelvin so now I know what my temperatures are in units of Kelvin we can solve this two different ways since we know three out of the four variables at the end we can just do that directly use these three values with pbal nrt and calculate for p so we do that by saying P equal nrt over V plus plug in our numbers make sure the units liters cancel with liters um and we'll get our answer in atmospheres which we did so we I'm sorry I flicked through to the next slide too fast so we saw that we could get our answer in atmospheres just by doing that this problem also has a a change where only two things change pressure and temperature we did this before um so we can use this equation again and then solve for pressure again and if we do that P1 * T2 over T1 which is what we get when we rearrange this equation um that gives us the initial pressure of 1.35 atmospheres final temperature over the initial temperature gives us the same 2.53 atmospheres as our um uh ending pressure for for this example okay um again if you do it in Celsius you get the wrong answer um these numbers have a different ratio than these numbers um these numbers they don't work because these are not absolute um numbers these do not start measuring temperature from zero these do that's why um this doesn't work you'll get the incorrect answer if you do your calculation in units of Celsius must be in kelvin okay all right here we go another problem uh 750 milliliters of hydrogen which means ideal gas in this problem at 822 T and 325 Kelvin is heated to 475 Kelvin at constant pressure what's the ending volume what do we know okay beginning conditions pressure we know volume we know I'm going to keep three significant figures in this moles don't get it we don't get that in this problem uh temperature we do get ending conditions pressure stays the same volume we don't know number of moles we don't know but we do know that it stays the same and finally temperature we are given okay again two ways to do this problem we can let me see which goes first oh first um I'm going to convert my uh tour to atmospheres um uh 822 stays the same so this is at a constant pressure of 1.08 atmospheres um my volume I want to change that to liters and using uh method one where we calculate everything it's going to be harder because now we got to do this in two steps um so first I got to find out how many moles we're working with so rearrange picles nrt to find n Nal PV over RT plug in all the numbers and we got to use the values that we converted to so this is what I was saying we don't need to use a different value for R every time that's just cumbersome what we're going to do instead is make sure that when we report a pressure it's always in units of atmospheres make sure that when we report a volume it's always in units of liters that way the atmospheres will cancel with atmospheres and the liters will cancel with liters so you don't have to keep looking for different conversion factors just convert your temperatures uh just convert your pressures and volumes into liters and at atmospheres um so that they work with this value of R we do that that we find that we're working with a sample of 0.3 0.304 moles okay so now we have a value for that I wonder if I put it up here I didn't um but we'll take this value and we know that it doesn't change so now Part B we're going to look for the volume where um we're going to rearrange pbal nrt to get V equal nrt over p and then we'll plug in the new numbers right N2 well that's just the same as uh that's just the same as N1 which we just calculated so that number goes there times R times the new temperature T2 divided by uh the pressure which stayed constant so the pressure was the same here and here because it's constant pressure so we'll divide that out and we get um that the new vol volume is 1.10 L okay so that's the way to do it if we want to calculate everything so we're going to use these three values to find the fourth one the fourth one we bring over here and then we use these three values to find the fourth one we don't need to do that see me on the next slide oh hi there you guys are um all right so um what's what we're going to do same thing but this time we're only going to look at what changes so what changes here is temperature and volume so this is Charles law again you don't need to know the volume the the Law's name and you don't even need to know what the relationship is uh what the equation is you don't have to memorize any of that because you got PB equal nrt so we're just going to do that same thing rearrange it so that the things that change which are volume and temperature are are written on the outsides of this equation so V1 / T1 equals NR / P all of these are staying constant in this problem so that equals V2 over T2 and we're solving for v2 um and since it's constant we don't care what it is we just need to work with the relationship of what's on the outside rearrange that and we find that V2 = V1 * T2 T1 and of course we've got to do this calculation in kelvin but we don't have to change the volume units we can leave that in milliliters if we wanted to if our answer was going to be in milliliters um and we find when we do that the answer is 1,00 96 milliliters um but since we can only keep three significant figures we have to rewrite that as 1.10 * 10 3 milliliters in order to get the right number number of significant figures or we can rewrite this in liters um as 1.10 L exactly what we had on the last slide if you do it in Celsius you get the wrong answer you get um nearly three times the volume that is incorrect because of the wrong temperature units um do the problems in units of Kelvin not in Celsius