hey everybody we're going to be going through the topic of gases working our way through chapter 11. we're going to start with kinetic molecular theory so kinetic molecular theory is a model so being a model it's different from what's actually occurring but it gives us a simpler way to look at the behavior of gases so think of a lego model versus an actual ferrari are there differences absolutely but we can see the basic structure of the ferrari we can see roughly how it operates it's got wheels it's limited to operating on a flat surface or relatively flat surface so we can get some of the behaviors by looking at a lego model now kinetic molecular theory think kinetic energy of motion and molecular we're looking at molecules and this is a theory describing behavior so we're looking at the behavior of moving molecules or gas behavior so under uh typical conditions it's going to work and we're going to focus where the model does work we're going to wrap up talking about chapter 11 where the behavior of this model goes a little bit wonky but this is a very solid model to go by especially getting a tangible visual feel for how gases behave okay so getting into the four basic tenets of kinetic molecular theory so number one we've got a collection of particles in constant motion so i'm going to insert a word here i would suggest you do the same in random so random constant motion i want you to think of blindfolded bumper cars write this here for you so blindfolded bumper cars now what do i mean by this these little particles that are in the picture they can't say oh no hit the brakes i'm gonna hit another molecule or let me swerve i'm about to hit a wall they have no idea what they're doing they don't have consciousness so we're behaving as blindfolded bumper car drivers or randomly moving pool balls just bouncing off of walls and bouncing into each other they are separate independent particles moving around independent of each other so number two we're saying there's no attraction there's no repulsion between these particles so this particle here if it approaches this particle it's not being drawn in it's just randomly moving if they happen to collide that just happens because they randomly happen to come together no attraction no repulsive forces these are like billiard ball collisions so energy is conserved so if this ball rams into this ball they're going to bounce off and conserve their energy and continue moving if this ball bounces into the wall energy is conserved and it bounces and continues moving all right tenet number three a lot of space between the particles compared to the size of the particles themselves so we can assume that these little gas particles they are super small so i want you to picture a walmart checkout line where we have an ant down here and we have an ant way over here so aunt number one is at checkout lane number one this ant over here at number two is all the way down at checkout line number 18. so walmarts are huge if we were to think of this huge distance between these two checkout lines i seriously doubt you could be standing at checkout line number one and even see ant number two off in the distance so do these ants take up space technically they don't take up a lot but they do take up space but we could say that this huge distance that separates them is so much bigger than the ants themselves that these ants they have negligible volume let me spell that out over here negligible so this tenet says a lot of space between the particles this huge space here between these ants which we could think of as gas particles so a lot of space between them relative to the size of the particles themselves means that these ants take up negligible volume or we could really say that essentially gas particles take up no volume now i want to remind you that when we look back here this is a model and we said models are simplified they're not perfect the gas particles actually take up no space no but this is important according to kinetic molecular theory gas particles take up a negligible or essentially zero volume that's a part of the model now let's think back on number two no attractions or repulsions we said that these particles are really far apart there's huge distances now if i'm going to shake somebody's hand if i'm going to interact with someone talk with someone typically the distance has to be close so no attractions or repulsions make sense because if my particles are so far apart that's also what's lending to this idea of no no attractions no repulsions so these four ideas really tie together to support one another now our fourth idea here the speed of the particles increases with increasing temperature so we're going to say this as an increase in the absolute temperature and increase in temperature is related to an increase in the average kinetic energy which is the energy of motion so we could say that as temperature increases the average speed or the average kinetic energy of these particles would increase too so just the speed that these bounce around this box and bounce around into one another is going to increase as we warm up this box that's going to be important when we start to think about the behavior so let's start predicting behaviors but first let's generally think about gases so when we think about the gas phase versus the solid or liquid phase the solid or liquid phases these are called condensed phases if you look at the space between the particles on the solids and the liquids very closely packed and that's important because when we look over at the gases over here we've got more space between particles space and i'll write this in here between particles so that means that gases are going to be compressible so this piece here in the picture this is called a piston so when i move this piston down what i'm doing is i'm pressing down on that gas sample so this whole arm is moving down this is our before this is our after so after we compress it you'll see that gas had space between particles so we were able to push down and compress that gas so gases are compressible because they have that space between particles because solids and liquids are condensed they're closely packed they are not compressible we'd also say gases assume the shape and volume of their container if i have a cup and that cup is filled only partially with water that water doesn't surprise us and suddenly fill the container that's because liquids don't assume the volume of their container but if we had a gas if this was just a cup of air then the air isn't going to be all at the bottom of the cup the air is going to be throughout the cup it's going to be outside the cup it's going to assume the shape of whatever container it's in we'd say gases have low densities compared to solids and liquids why because density is mass over volume let's look at this container here these particles are really far apart so i don't have as much mass that's in the cup so i would say density in this case would be lower mass and then i have a lot of empty space in here so bigger v so i have more volume and less mass in it that's why i tend to see low densities in gases compared to solids and liquids all right so let's think a little more but make it tangible on the density of gases so picture a typical soda can 12 ounces standard soda can the density of the water in the soda cam can is one gram for every one milliliter so to be able to picture what one milliliter looks like think about your fingernail and about the length of your fingernail is going to be about one centimeter so if i had a box that was one centimeter on all sides so one cm again about the length of your your fingernail that would give us one milliliter of volume that's in this box now one gram think about that as about a half of an m m so about a half of an m m's mass trapped in a box that's one centimeter on all sides that's about the density of water now this is a liquid and we said that liquids were condensed phases so i have more mass packed into a particular amount of space when we're in the liquid phase now steam is water once it's vaporized and turned into a gas so this one atm pressure that's just roughly room pressure the room that you're in right now and 100 degrees c well that's boiling for water it's about we would say one half of one thousandth of one gram that's huge i mean the difference there think about the difference for water for the liquid phase we said it's a condensed phase more matter packed into an amount of space so in that little one centimeter box i've got a half of an m m about one gram now i've got one half of one thousandth of one gram way less mass that's packed into that little box when i get to a gas so if i had that can of water that 12 ounce can of water and we vaporize it to turn it to a gas that single can of water would now take up 1700 cans of space so just the water the liquid water trapped in one can after we vaporize it and it turns to a gas we're looking at not one can but 1700 cans worth of steam so when we said the particles were further apart once we moved to the gas phase we're serious so that uh that characteristic uh four properties of gas that we saw in kinetic molecular theory of a lot of space between particles has a huge impact when we get to densities so let's think about pressures gas pressure is the result of constant collisions remember we said little blindfolded bumper cars so our atoms our molecules our particles are constantly moving constantly bumping around so a pressure is a force in this case in this picture we're looking at a force hitting a wall and if this is a gas particle it's about that big and it's moving towards the wall it's going to hit well gas particles are tiny so who cares what kind of pressure is that going to cause well we do care because room pressure air pressure is not just caused by one tiny little particle it's caused by more particles than we can fathom moles and moles of air moles of particles molecules are striking in random directions constantly now how do i know that the gas particles aren't just hitting me on this side because if all the gas particles only hit me on this side i would fly off in the other direction constant random motion means i'm being struck roughly evenly all across my body at any given moment that is that pressure of room pressure or air or atmosphere that i'm feeling right now now when you feel your wind blowing when you feel your ears pop really what you're feeling is the result of these tiny gas particles causing pressure and again they do that by striking the walls of their containers uh on and surfaces throughout the room okay so let's think of ways that we could affect pressure so remember just to simplify this picture when i have a gas particle and that gas particle hits a surface that's what causes pressure so we'll have a little phrase off to the side here more hits or harder hits equals increased pressure so i can increase the pressure by hitting more or by hitting harder so pause the video for just a minute and think about this picture i've got a bunch of our blinded bumper cars a bunch of our molecules and atoms just flying around this box independent of one another striking walls striking each other how can i increase the pressure pause the video in 30 seconds we'll keep going and work through this together okay so how can we hit more let's think about temperature think about kinetic molecular theory if i increase the temperature we said that i would increase the average energy of motion the average kinetic energy so they're moving faster so what does that mean if this box stays the same size and they're whizzing around faster that means i'm going to up the number of hits and i'm also going to increase harder hits so both i hit more and i hit harder by increasing the temperature which is going to increase the pressure all right let's think about another way that we can do this so let's say we took this box and instead of having the box that size we were able to compress the box to a smaller size i contain the same number of particles but now i have a smaller box well if i have the same particles in a smaller box they're going to hit the walls more often so by let's see decreasing the volume i've increased the number of hits now i didn't necessarily make these particles hit any harder but i have increased the number of particles that are going to hit at any given moment and because i've increased the number of hits i increase the pressure all right one last one here what if i took a little more gas and put it into this container so if i increase in a way that we've learned to express this is by saying moles so increased moles which in this chapter moles is going to be represented by the letter n so increasing moles is going to give us more particles which doesn't mean they'll move faster but it does mean we have more bumper cars more bumper cars more hits so we'd have increased pressure so table 11.1 in your book goes over some common units of pressure the ones that we're going to focus most on millimeters of mercury tor and atmospheres those will be what i tend to stick to most but i will give you any of these units that you need for conversions so just to reiterate will give for exams but you will need to have this table handy for homeworks just to have those conversion factors when you need them all right so we want to convert between different units of pressure in this case we want to go between atmospheres this is what i have and millimeters of mercury this is what i want so our little solution chart here this is where i start this is where i finish i want to cancel atmospheres so the conversion factor that i use is going to have to have atms on the bottom atmospheres it's going to have excuse me millimeters mercury on the top and if we go back to our chart we would say that one atmosphere is going to be equal to 760 millimeters mercury now this chart is set up in such a way that if i wanted to move between pounds per square inch and pascals i would just take these two units 14.7 psi is equivalent to 101 325 pascals in this case we were just interested in going between atmospheres and millimeters mercury though so let's use these two numbers we had one atm for every 760 millimeters mercury atmospheres cancel leaving us with millimeters mercury which is what we wanted so let's plug this in 0.311 atms are where we're starting one conversion factor will get us where we want to go one atm and 760 millimeters mercury again i'll give you these relationships know how to use them atms cancel leaving with millimeters mercury 0.311 times 760 divided by 1 gives me about 236 millimeters mercury all right let's move on to what we're going to call our changing gas laws now what do i mean by changing i mean we're going to have befores and afters so we're going to cover boyle's charles law as we move through you'll see some similarities in these equations in that ones and twos are going to show up the ones mean before the change and the twos mean after or initial and final however you want to word it now as we go through these changing gas laws we're going to have four variables p v n and t and p is pressure v is volume we said n would be moles and t is temperature now each of these changing gas laws we're going to have two of these that change and two that stay constant so for boyle's law it says pressure and volume are indirectly related if moles and temperature are constant so we have four variables that we use to describe gases it says moles n and temperature are constant so we're just looking at p and v so that's why in boyle's law since n and t are constant they don't show up and i'm just looking at my pressure and volume before the change and my pressure and volume after the change now this is a really important word here indirectly so pressure and volume are indirectly related and what that means is they move in opposite directions so we'd say that as pressure increases volume is going to decrease in specific terms we look at doing the opposite to the gas sample so if pressure doubles what's the opposite of doubling something having it or multiplying by a half so if pressure doubles we can press a gas or pressuring it the volume is going to shrink by a half when pressure doubles now looking at that the opposite way if v doubles so v times two we have to do the opposite to p we're gonna go in the opposite direction what's the opposite of doubling something having it so as one goes up the other goes down that's what makes it an indirect relationship so let's think about bike pumps so with my bike pump as i pull that bike pump handle up i'm going to increase the volume that's inside this cylinder right here and when i increase the volume i'm going to decrease the pressure that creates a vacuum that's inside this little tube so when that vacuum's created air is going to be sucked in or pulled through this valve at the bottom of the bike pump so increasing volume decreases pressure ends up drawing air to fill that vacuum into the bike pump now when i move my bike hand or bike pump handle down i'm going to decrease the volume of this little cylinder here and when i do that i increase the pressure of the air that's inside here and that increased pressure forces the air through the tube and into your bicycle so we have this opposite or indirect relationship between volume and pressure on the up stroke and the downstroke so let's think about what's happening here we're going to quantify it we're going to put numbers to it this is my before and this is my after so here i've got a sample of gas and it says that the volume inside this cylinder is about one liter so it takes up one liter of space now it says the measure of pressure remember we said pressure was caused by these little particles hitting the walls is one atmosphere now after and let's let's look and see what happens we had a little weight that we placed on top of this little moving piston we replaced that with a much heavier weight and what happened to that little moving piston will it pushed it down and what happened to this space here well the space in the cylinder used to be one liter of space well now it's got cut down to 0.5 liters of space so how do we get from 1 liter to 0.5 liters we would say that volume went down by a half or volume times one half well let's look and see what happened to pressure pressure went from one atmosphere when we started then we compressed that gas and the pressure went up to two atmospheres so if it went from one atmosphere to two atmospheres pressure was times two so volume went down and pressure went up that's our indirect relationship we're not going to see that with all of our changing gas laws but we do see that here now if you were to work that out according to our equation p1v1 equals p2 v2 you'd see that pressure for where we started was 1 atm and volume was 1 liter and pressure for where we ended was two atms and our volume was 0.5 liters now when you plug all this in you'd see that everything works out right because both sides are equal once you multiply them out so this does follow our trend for boyle's law now let's think of another scenario we've got a diver and that diver inhales a lung full of air so he's all the way down here we'll have to draw his little dive back on there's his dive back and he's got little flippers okay and a little mask so you can breathe okay there's my doodle so here's our scuba diver he's down here he takes a breath of air and the pressure is higher down here so it's three atmospheres he comes up towards the surface and the pressure at sea level is about one atms so when he gets up here what happened to the pressure well the pressure was multiplied by one third we went from a higher pressure at three it dropped to one as he came to the surface and going from three to one that's the same as uh cutting our pressure by a factor of three so what's going to happen to the volume of that air in his lungs now remember your lungs are basically bags made of flesh so you drop the pressure of the air in your lungs that rapidly the gas that's in your lungs would do the opposite pressure goes down volume goes up this is going to go up by a factor of 3. we're looking at rapid expansion so he comes up too quickly because it says he swam quickly to the surface he is going to suffer some lung damage that's probably a vast understatement because the volume of air that's in his lungs is going to triple so there's a reason why these divers surface slowly they're going to breathe slowly and continuously on the way up they may spend some time in decompression before surfacing because we're looking at the relationship between breathing at that high pressure environment versus breathing at the lower pressure environment at the surface so let's work through using our equation p1v1 equals p2v2 and let's label what the problem gives us it says a cylinder equipped with a movable piston so that's that picture we've been looking at so far has an applied pressure of four atmospheres so this is our original pressure p1 remember this is where we start the twos are where we finish and it says it also has a volume of 6 liters liters is a volume so that's going to be my original volume v1 and it says what is the volume of the cylinder if the applied pressure is decreased to 1 atm so if it wants what is the volume after a change then we want v2 and it says the applied pressure is decreased to 1 atm well this is what it goes to this is where the pressure finishes so that's my p2 so let's pull these all down into one spot we'll say p1 equals 4 atms we've got v1 equal to 6 liters and we've got p2 equal to atm and we're solving for v2 that's what we want the way we're going to get from what we have to what we want is by using the equation for boyle's law p1 v1 equals p2 v2 now i will give you these equations i'll give you all the equations on the equations page but you have to decide which one to use and you have to know how to use it okay so let's plug everything in so p1 four atms v1 6 liters p2 1 atm and v2 is what we're solving for okay let's make a prediction here when we went from four atms to one atm we were seeing pressure atmospheres were dropping so what should volume do if pressure decreases volume should do the opposite this is an indirect relationship so volume should increase this is going to give us a way to figure out what kind of answer to expect so if volume should increase it should be an answer higher than where we started so we would expect a number of liters higher than six so to get v2 by itself we need to divide both sides by one atm we're going to divide by one atm atmospheres cancel and our v2 is going to be equal to 4 times 6 divided by 1. the unit that's left over is liters and that's going to equal our v2 now we predicted that volume should increase and volume went from 6 to 24 liters so this relationship checks off um we should always look for that kind of trend and make sure that our answers make sense decreased pressure increased volume all right now let's think a little more on that trend we went from four atms originally to one atm afterwards so we would say that pressure went down by a factor of four so volume should have done the opposite volume should have gone a factor of 4 higher so 6 would go up to 6 times 4 or 24 liters exactly what we saw okay moving on to our second relationship here we've got charles law charles law is going to be represented by v1 over t1 equals v2 over t2 now wait a minute this is a little different the way the equation's written before it was p1 times v1 equals p2 times v2 and instead of multiplying these two are divided probably means there's a different kind of relationship and that's because instead of being indirectly related temperature and volume are directly related and remember two of the variables stay constant so where we have pressure volume moles and temperature to distract gases we're focusing on volume and temperature that means that moles and pressure and sorry this should say moles in the notes and pressure are going to be constant that's why they're crossed out here so let's focus on volume and temperature for a moment now in indirect relationships we said we had opposite directions in their relationship being directly related means they move in the same direction so as t increases v also increases we could also say the flip of this relationship as t decreases volume would also decrease so we're looking at thermal expansion thermal contraction and it has the same kind of mathematical relationship that we saw in boyle's law where if t doubles instead of going the opposite direction where we saw in boils we're going to see a direct relationship where they go in the same direction so if t doubles b is going to double we say if t have v halves because they are a direct relationship what happens to one is going to happen to the other so let's think about hot air balloons if you've ever been to a fair or a festival and you've seen a hot air balloon it's not all that exciting when you first get there when you first get there it's just some strings kind of not coming out of a basket they've got a balloon laying flat on the ground and it's not doing a whole lot of exciting floating now as they heat up the air that's inside the balloon we would say that temperature is increasing so the volume inside that balloon the air sample is going to increase also we said that was a direct relationship and we see that balloon start to expand and we can look at the relative density of the air outside the balloon and the air inside the balloon we'd see a lower density inside we'd see a higher density outside and we know that lower density materials tend to float in higher density materials so we would see this balloon start to rise but it all starts with this idea of increasing temperature increasing volume thermal expansion and filling the balloon with hotter air so let's try out a sample problem we said our equation was v1 over t1 equals v2 over t2 and let's start digging out what they give us says a sample of gas has a volume of 2.80 liters so this looks like where we start to me so this is going to be our initial volume and it says at an unknown temperature so where we start we know the volume but i don't know where my temperature started it says when the sample is submerged in ice water at 0 degrees celsius so this here after it submerged because that's an after this would be a t2 and it says its volume is decreased to 2.57 liters so this is after the change after it's decreased so this is going to be a volume that's after so it's going to be v2 it says what was its initial temperature in kelvin and in celsius so i'm looking for this value this unknown temperature from the start that i don't know so let's write down what we know v1 equals 2.80 liters we know that t1 is what we want our v2 was 2.57 liters and our t2 equals zero degrees celsius now here's a little note uh we've got to be in kelvin so we're going to work all gas problems that will wire require calculations in kelvin why because kelvin is never negative it's always a positive value so to get to kelvin we have to take celsius and add 273 to that temperature 0.15 so 0 degrees celsius that makes our t2 273 0.15 kelvin okay so let's plug in what we know v1 is 2.80 liters my t1 is what we're looking for that's going to equal my v2 2.57 liters divided by my t2 which was 273.15 k okay now we have to get t1 by itself but also on top so my first step i'm going to move it from the bottom to the top how do we do that by multi multiplying both sides by t1 on this side it cancels so i'm going to be left with 2.80 liters is going to equal 2.57 liters times t1 whoops see if i can erase that extra set of parentheses there and boom we are going to divide that by 273.15 now our goal here we got t1 on top now we have to get it by itself so let's multiply both sides by 273.15 all right and now we're left with 273 0.15 k times 2.8 liters is going to equal 2.57 liters times t1 now the last step we have here to get t1 by itself is to divide both sides by 2.57 liters oops go back there that crosses out liters cancel i'm going to be left with kelvin which is what i wanted a temperature so i'm going to say t1 is going to equal 273.15 times point eight divided by two point five seven so i'm going to get 297 point six roughly for my t1 and this unit was in kelvin now this is going to be the temperature in kelvin it also wanted the temperature at celsius remember to get to kelvin we added 273.15 so to get back to celsius we have to subtract that so my temperature in kelvin was 297.6 so this is going to be about 24 degrees celsius for my t1 after i convert back to celsius so you will have the equation for converting between kelvin and celsius on the exam it'll be on your equations page all right so we're moving on to the combined gas law so our combined gas law we're going to see similarities to the laws that we've already covered if you focus on the top part of the equation here we've got the relationship built in for boyle's law p1 v1 equals p2 v2 if you look on the right hand of each side of the equation we've got v1 over t1 equals v2 over t2 now if instead of looking just at two variables at a time remember we had pressure volume moles and temperature if we just look at the relationships between p v and t changing and n as a constant so we have a constant gas sample size then we can get out the combined gas laws now we can't do the same kind of basic predictions of increasing and decreasing that we did before because now we have a three variable system that's a little hairier to work with but we can still look at using our equation so we're going to use p1v1 equals p2v2 we're going to add t into the mix on the bottom here on either side to get out our combined gas law we're looking at a sample of gas having an initial volume of 158 milliliters because that says initial that means my initial volume v1 is going to be 158 ml i have a pressure of 735 millimeters mercury so pressure initially was 735 millimeters mercury and it says a temperature of 34 degrees celsius so when i started my temperature was 34 degrees c but remember i've got to convert this over to kelvin so i need to add 273.15 that'll give me 307. whoops 0.15 kelvin now let's look at the rest of what it tells us it says the gas is compressed to a volume of 108 milliliters so that is a volume so v and it says after it's compressed so v2 is going to equal 108 milliliters and it says it's heated up so temperature is going to increase to 85 degrees celsius so my new temperature after heating is going to equal 85 degrees c which when we convert to kelvin is going to give us 358 0.15 kelvin and it says what is the final pressure in millimeters of mercury so that means my p2 is what i'm solving for so let's plug everything in i've got v1 and p1 going on top so we've got 735 millimeters mercury multiplied by 158 mils whoops is going to equal our p2 which is what we're solving for multiplied by v2 which is 108 milliliters all divided by on the left side t1 which is 307.15 k and on the right we're going to divide by the final temperature which was 358.15 k okay so step number one we want to get p2 by itself what we're trying to solve for so i need to get it by itself and on top it's already on top so we're going to start peeling off the layers so 358.15 k to get rid of it on this side need to multiply by 358.15 k i'm going to multiply by the same value on the other side 358 0.15 k my kelvin cancels so now i'm going to have 358.15 times 735 millimeters mercury times 158 mils divided by 307.15 and that's going to be equal to p2 times 108 mils remember our goal here is to solve for p2 get it by itself so now i need to peel off that 108 mils and i can do that by dividing both sides by 108 milliliters so i can cancel out my milliliters what i'm left with on this side is millimeters mercury which makes sense because i'm trying to solve for a pressure so we end up with p2 being equal to 358.15 times 735 times 158 divided by the product of 108 and 307.15 and that takes us to approximately 1200 millimeters mercury okay now avogadro's law this is a little more intuitive moles remember we said moles are n so we're looking at a relationship between n and v or moles and volume moles we said represented the number of items so really we're looking at the number of gas particles being related to the volume and remember we have four variables to describe gases we've got pv and t sorry that should be a capital v and in this case n and v are changing and we're assuming that pressure and temperature are going to be constant so when we look at the relationship it looks more similar to boils or more similar to charles because we've got that division going on v1 over n1 equals v2 over n2 and this looks a lot more similar to charles law and why because they have a similar relationship they're directly related to one another meaning they move in the same direction so as volume increases moles increase whereas moles increase volume increases so n goes up v goes up so we see the same type relationship that we've seen before because this is a direct relationship what happens to one if moles double volume doubles so same thing happens to both because they're direct n cuts in half v cuts in half so let's think about blowing up a balloon so do a little thought experiment with me play along grab your balloon hold it up to your mouth and start blowing it up what are you doing when you blow up the balloon well when you blow it up you are increasing the moles or number of particles of gas that show up in that sample by increasing the number of gas particles we are increasing the volume inside the balloon is a direct relationship and that's exactly what we see in avogadro's law now something that falls out of this also good to bring up here and by the way i'm going to write this different this go around so be prepared n1 over v1 equals n2 over v2 uh-oh over here we had v1 over n1 over here we had n1 over v1 well it doesn't matter as long as both ends are on top or both are both v's are on bottom as long as they're in the same place relative to one another we're fine you'll see it in both representations depending on what resources you gather on the web but as long as you're consistent with n's on top or v's on top either way is fine to represent it now we said n's represent moles and we said that it represented moles of gas so think about this for a minute we didn't say moles of a particular gas we just said moles of gas so this is an important point here n n doesn't care what type of gas it is if it's oxygen gas carbon dioxide nitrogen gas n just cares about moles of gas how many gas particles are in the sample now could we see slightly different behavior if we change the gas or change the conditions absolutely but remember this is all based on the idea of kinetic molecular theory so we're just looking at moles of gas we're looking at those blindfolded bumper cars not specified at this point what kind of gas we're looking at okay so let's look at an avogadro's law sample problem we're going to try to make a prediction before we do any of the equation in math work just based on what we've done so far so it says a 4.8 liter sample of helium gas so that sounds to me like the volume originally or v1 is going to be 4.8 liters and it says it contains 0.22 moles of helium so n1 is going to equal 0.22 moles it says how many additional moles would you add to obtain a volume of 6.4 liters okay so we've got 4.8 liters going to a v2 of 6.4 liters and it's saying this was the moles i started with but what moles would i need to even get to 6.4 liters so i'm solving it first for an n2 but then we're going to have to dig a little bit further so let's go with v1 over n1 equals v2 over n2 and let's start plugging in so 4.8 liters divided by our n1 0.22 moles is going to equal our new volume 6.4 liters divided by n2 which that's what we're solving for so i'm going to start out by multiplying both sides by n2 to get it on top and that's going to give me n2 times 4.8 liters divided by 0.22 moles is going to equal 6.4 liters okay our next step here is going to be peel off the layers on our n2 to get it by itself so here i'm going to multiply both sides by 0.22 moles so that cancels out on the left and then i'm going to divide both sides by 4.8 liters cancels out on the left that cancels out the unit of liters on the right so then i'm left with n2 is going to equal 6.4 times 0.22 moles now we're going to divide all of that by 4.8 i'm left with a unit of mole which makes sense because i'm looking for an n2 so going back here we said that volume is going to increase so moles thinking about the type of relationship here if volume increases mole should increase and the problem also says how many additional moles so it makes the assumption that moles should be increasing or going up so in other words i should have a number when i solve this out greater than .22 6.4 times 0.22 divided by 4.8 so n2 should equal about 0.2933 moles now we got a problem here this is not what the problem is asking for the problem says how many additional moles would need to be added so before we started we had 0.22 moles so i'm going to need to take 0.22 away from this and that gives me about .07 moles of gas that i would have had to add so that .07 is really my final answer because that's how many additional moles i had to add to get to that new volume of 6.4 liters so just be careful what the problem is asking for if we had stopped at 0.2933 that's the total number of moles of the gas sample after we got to the volume of 6.4 liters it wasn't the additional gas we would have had to add okay so so far we've been looking at relationships between pressure volume moles and temperature under changing conditions meaning we had befores and afters well what if i have unchanging conditions for unchanging conditions this is really where i'm looking at one set of conditions pressure's not dropping temperature isn't increasing i have static or unchanging conditions again i'll give you all the equations that you need but we can tell a big difference between this and what we've been looking for there are no ones and twos because there are no initial and final or before and after conditions we just have pressure volume moles temperature and this new introduction here of the universal gas constant r so this is a constant i will give it to you on your equations page it's equal to 0.0821 and the units are a little funny but they're useful it tells me the unit is liter atmosphere per mole kelvin so remember we had pressure volume moles and temperature that we were working with well that tells me if i look at these units if i use this value of r pressure needs to be in units of atmospheres volume is going to be in liters moles really another option there is going to be in units of moles and temperature should be in kelvin so good reminder if we want everything to cancel out when we're using this equation make sure we're using the units that show up in r okay so let's try this out we want to calculate the volume occupied so we're going to find v that's what we want it gives us 0.845 moles that's n that's the number of moles it gives us a pressure of 1.37 atm so that's my p and a temperature of 315k so that's my t so let's start writing things down my p pressure equals atms my v is what i'm solving for my n was 0.845 my r is a constant and my t is equal to it's already in kelvin that's good 315k so let's plug it into the equation we've got p times v which is what i'm solving for is going to equal n which is 0.845 moles times r which is a constant we said that was .0821 liters atmospheres per mole kelvin multiplied by t which is 315k so that gives us our pv equals n times r times t now our goal here is to get v by itself so i want to divide both sides by the pressure 1.37 atms cancels it out here and we would take 0.845 times 0.0821 times 315 divided by 1.37 and that gives us volume equal to about 16 liters all right so we said we've been dealing with gases behaving according to kmt or kinetic molecular theory they're behaving as ideal gases so do they always follow that model remember we were looking at the lego versus the actual ferrari the model is great at predicting behavior but certain conditions things start to break down so just to recap ideal behavior we were assuming that these gas particles are far apart and because they were far apart they didn't interact they didn't repel they didn't attract and part of that was due to the spacing so what happens to make this gas behavior ideal how do we get that far distance between our particles we're going to have low pressure we're going to have fewer particles that allows for that increased space between the particles so it's not compressed we're looking at high temperature so lower pressure higher temperature that's where we start to see that ideal behavior with increased temperature we see these particles zipping past one another they don't spend an appreciable amount of time near one another even when they pass close so reduces even further that ability to interact now when we see non-ideal behavior see these little blue dashed lines that's where we start to get interactions so these aren't full on chemical bonds we're just seeing interactions between the gas particles so that was not allowed to happen according to kmt that's where we have higher pressure so we have more particles per unit volume we have them moving slower so lower temperatures so anything that we can do to bring these closer to one another and have them moving slower is going to contribute to that non-ideal behavior so at the top here ideal conditions we would have lower pressures higher temperatures for non-ideal conditions we want high pressures we want those gases compressed particles in closer to one another lower temperature lower gas speed for these particles now thinking back to kmt we said gas molecules are far apart from each other they don't interact we said blind bumper cars so really the left hand doesn't know what the right hand is doing these gas particles are acting independent of one another so we said when a gas particle causes pressure what it's really doing is hitting the wall of a container so because these gases if they're in a mixture they don't know what each other gas is doing each gas is going to contribute its own pressure to the total pressure now each gas in a mixture contributes to p total or total pressure by taking the pressure caused by gas 1 and all of its particles hitting the walls of the container plus the pressure of gas 2 and all of its particles hitting the walls of the container if we think about the air we're breathing we could think about atmospheric pressure pressure of the atmosphere patm that's going to be the sum of all the gases that show up in the mixture of what we're breathing so we could say that the biggest component of that would be the pressure of nitrogen gas then we're going to see the pressure of oxygen gases plus we're going to see the pressure of argon and other trace gases that show up in the mixture but when i look at the total pressure it's the sum of the pressures caused by each one of the gases that show up in the mix so let's apply this we've got a box and inside these this box we've got a total pressure of one atm it says that the gray spheres are helium and the brown spheres are neon so this here is neon this is helium and if we count them i have one two three four five six seven eight eight helium atoms and if we count my neons i've got one two so two neons now the number of particles are related to their partial pressures we could say that the pressure caused by the helium and the pressure caused by the neon can both contribute to the total pressure in this box but when we look at the relative numbers i have more helium so the pressure caused by the helium is going to be greater than the pressure caused by the neon now how much greater let's start to think about how we looked at mass fractions so it's what we're interested in divided by the total so if i look at the number of helium atoms so i have eight helium atoms and how many total atoms do i have in this gas sample 10 so eight tenths of the particles that show up in that box are going to be helium so if eight tenths of the particles are helium then eight tenths of that one atmosphere eight tenths times one means i'm going to have point eight atmospheres being the partial pressure of helium i can also look at the partial pressure of neon the same way the partial pressure of neon would be two neon atoms divided by the total which is 10 total so 2 out of every 10 are neon the total pressure 1 atmosphere so this is going to work out to 2 divided by 10 times 1 or this is going to be 0.2 of those atmospheres are going to represent the partial pressure of neon now we said for dalton's partial pressures we said the total pressure is going to be equal to the sum of the gases that are in the mix so the pressure of helium 0.8 atmospheres the pressure of neon 0.2 atmospheres what do you get when you put them together you get one atmosphere so let's work a similar problem where we're working through partial pressures we've got a mixture so let's draw a little gas box represent our gas sample we've got a mixture of helium and we've got neon and we've got argon and it says it has a total pressure of 558 millimeters mercury so here i'm going to say the total pressure is 558 millimeters mercury i've got helium neon and argon particles bouncing around inside this gas sample it says the partial pressure of helium is 341 millimeters mercury so i'm going to write in here 341 millimeters mercury the partial pressure of neon is 112 millimeters mercury and it says what is the partial pressure of argon well applying dalton's law we should say the total pressure is going to equal the pressure of helium plus the pressure of neon plus the pressure of argon and we don't know the pressure of argon so we can solve this by plugging it all in and solving by subtraction the total pressure is 558 millimeters mercury that's going to equal the pressure of helium which is 341 millimeters mercury plus the pressure of neon gave us that at 112 millimeters mercury plus our unknown our pressure of argon so if i want to solve for the pressure of argon i'm going to subtract 341 from both sides i'm going to subtract 112 from both sides oops so my final answer here is going to be p of argon is going to equal 558 minus 341 minus 112 which is going to work out to 105 millimeters mercury now how do you check yourself to make sure that this answer is right we could put 105 millimeters mercury in for my pressure of argon and say the sum of all three of these partial pressures should equal our total pressure so go back and check yourself when you're done just because you can't you don't want to be off by a little bit there at the end all right so that is it for uh this chapter chapter 11. uh thank you for watching let me know if you have any questions by teams or email and have a great week