Lecture 22. Effusion Continued and Real Gasses
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Brindley, Amanda
UCI Chem 1A is the first quarter of General Chemistry and covers the following topics: Atomic structure; general properties of the elements; covalent, ionic, and metallic bonding; intermolecular forces; mass relationships. 
 So today we're going to finish up with a new material and we're going to finish up with a fusion. We still have the example problems to go over for that and then we're going to move on to real gases and we won't spend a huge amount of time on that just because there isn't a lot of material to cover. When we start getting into real gases things get really complicated really fast. So we'll show you one model of how we can deal with real gases and we'll also go ahead and then show you when you should be able to use the ideal gas and when you shouldn't be able to use the ideal gas. So we'll do that as well. So first we need to go over to the document scanner and finish up some work. So remember we were talking about a fusion. So we have a gas moving into an empty container and we want to see what the rates of those are both in terms of being able to calculate them and also in terms of being able to go through and figure out how likely one is to effuse X amount faster than the other. And that's really handy when we want to separate gases. We can use the fusion to separate a mixture of gases which is handy. So I showed you one example of where that would be useful in taking uranium and getting one isotope out of it and we can do that because different masses effuse at different rates. And so we'd done one of them and we'd calculated the differences in rates. So now let's go ahead and do another one. In this case I had actually kind of given you the fraction. I had said well we have rate A and rate B, we have our two different gases. And I already told you that fraction. I told you that it was a 1.55. So one effuses 1.55 times faster than the other one. So now let's move on to this one where we're going to kind of do the opposite. Now we have nitrogen and hydrogen and I want to know the two gases and I want to know the difference in the rates. Okay, so we have to go ahead and probably should write this equation to be on the safe side. Now it's N2 to H2. So we'll go ahead and just write out N2 and H2 now so that we can keep track of everything since we have real numbers. And we know that those have to be proportional to the square root of the molar masses. Now does the N2 molar mass go on the top or the bottom? Bottom, right? It's always opposite. They're inversely proportional. So we're going to take this and we'll put the mass of nitrogen down here and we'll put the mass of hydrogen up here. Okay, so when we go through and do this then, do I have to convert this to anything special as far as units go? Or can I just put in the molar masses like I'm doing right now? Should I put some conversion factors there? No, why? It's a ratio so they cancel out. So we don't have to put anything there because it's a ratio and they cancel out. Other times we have to worry about it. Mostly in this chapter when do we have to worry about it? When you see R. Yeah, basically in any form. If you see R in the form that has joules in it, what are all your masses have to be converted to? Kilograms. Okay, so at this point it's calculator pushing and you get 0.269. So our rate of fusion of nitrogen is 0.269 times the rate of that of hydrogen. So now we've done some calculations with them. Let's go ahead and see if we can just do this conceptually too because it's nice to have numbers but sometimes it's just good to be able to look at a bunch of compounds and say which one's going to fuse faster? Which one's going to have a higher speed and just be able to rank them? So we're going to do that now. Sorry, you said that 0.269, what did you say that unit was in there? You said that it was in the right? So you mean unit? Or if you just repeat what you said. Okay, yeah. I'll attempt to repeat what I said. I'll repeat something like that. So this is a rate over a rate. So those units as long as they're in the same units will always cancel. This is going to be in whatever unit we happen to have it in, it's usually molar mass, right? So we'll have it in grams per mole and grams per mole off the periodic table. Now when we used an equation that we derived this from which was the square root of 3RT over M, that's where we got this equation from, right? Well we had to worry about converting it to kilograms in that case because R had joules in it and joules has kilograms in it. In this case, if this is grams per mole, and this is grams per mole, what happened to these units? It can't fall out. And so this becomes a unitless quantity. And so since it's unitless, it's just a ratio. And so we don't have to worry about whether we have to convert these. They have to be in the same. If for some reason you had one molar mass in kilograms, you looked it up off the internet or something random, you would have to make sure you put it in grams. But as long as these are the same, it doesn't matter. So you're saying that hydrogen is 0.269 times faster than nitrogen? Or is it faster? Well so which one's going to be faster? Nitrogen or hydrogen? Hydrogen. So because this is a decimal, you would say that the speed of nitrogen is 0.269 times that the speed of hydrogen. So if we're just like, let's just throw out a number and say hydrogen's 10, whatever. You would multiply 10 by 0.269 to get your speed of nitrogen. Because you want nitrogen to be slower. Now the nice part of that is what happens if you forget that? Well, let's say I tell you the speed of, it's a good question. So let's, if I told you the speed of hydrogen at some given temperature. So if I said that H equals, let's make the numbers easy since I'm doing this kind of, you know, without notes, let's just make it 10, whatever. So at a given temperature, it's 10 units per second. And I wanted to know the speed of nitrogen. How would we do that? Well, we would take this 0.269 and multiply it by the 10. You can think of the ratio as being kind of like a percent too, if that helps. You can think of it as saying, well, nitrogen has 26.9% the speed of hydrogen. And so that would mean that nitrogen, whatever those units were, would be 2.69. So that's how you would go about determining the speed. I'd have to give you one of the speeds. The other way you can think about it is you can just fill it in too. That's the, you know, if this would be if I gave you the ratio, if I gave you this final answer and I asked for you to solve it. But in this case, we have all sorts of information. You could just fill in the rate of, in this case, we could take the nitrogen and just fill it into, excuse me, hydrogen and just fill it into here, which if we did that, we would multiply this up and it would be the same thing. And it just sounded up like that because nitrogen was slower than you had in the beginning. Yeah. So you, you, when, when we started at this ratio and we had to multiply this by the 10, then it was because we, I just had started from this ratio to show you how to do it. The other way is to fill into the problem, exactly as we did up here and fill in the 0.269 here and your masses here and do it the exact same way we did here. The thing with ratios is there's just, there's as many ways to do it as there are people. So yeah. No, it just depends on how you write it. I could have said write hydrogen to nitrogen and if you had written the hydrogen to nitrogen ratio, instead of being 0.269, it would have been one over that. So we could have gone through and we could have written this, the rate of hydrogen over the rate of nitrogen. Now where would the hydrogen mask go top or bottom? Bottom. And then filled in the same numbers. Now it would have been opposite. And then that number is definitely going to be over one. So we could have done it the reverse. It's just in this case I had asked for nitrogen to hydrogen. Okay. Oh, hang on. I see people writing that. Let me put it back. And then we'll do the next problem. Okay. So next one. Now I want you to rank in order of increasing speed of effusion and increasing time of effusion. I haven't talked about time of effusion, but we can figure that out. We can think about that. So let's start with speed just because we've kind of already discussed it. So I say increasing. So what makes something go very quickly when it effuses? A small mass or a large mass? Small. So when we go through and we do this, which one is going to be our smallest? So how do you figure it out? Yeah, you look at your periodic table and you would look at where all of these are and which one would be your smallest? Yeah. So your speed of it, oh, you're not sure which one I'm asking. That's the problem. All right. So speed of effusion. So we need to do increasing speed, so we need small speed here. And so we need the larger molecule. So which one's the largest molecule? The 205. Okay. So we have an increasing speed and we want to go all the way up to fast speed. So we go slow to fast. So we have increasing speed. So that means we need to go large mass to small mass. Does that help? Okay. So we know then that our biggest one is going to be our slowest. So that one goes first. And then which one? Good. And then which? Good. Okay. So that was speed. Now what about the one that we haven't talked about? Time of effusion. So what that's saying is how long does it take to effuse? How long does it take to get completely effused through the barrier? So if something is very fast, is it going to have a slow time or a large time? Yeah. Something's very fast. It's going to have a very small time, right? It's going to be zipping around all over the place and it's going to get to the other side very quickly. And so it's going to have a very small time. A fast molecule has a very small time. So what's our fastest molecule in this bunch? And two. So it's the smallest time. And then what? Yeah. So it's the exact opposite, right? Okay. So that's one where we didn't really have to calculate anything. We could just kind of look at it. I suppose we had to calculate the molar masses, but that's it. Okay. Let's zoom back out of it. Next one. Oh, yeah. What do you mean? Okay, yeah, I guess I should say something like, you know, till reaching equilibrium or sure total effusion. Yeah, good point. So what he had asked us is they're all going to start effusing instantly, right? So the question is a little bit ambiguous because they are all going to instantly start effusing. The question is when have they completely effused? Or if we're just going into one container. Okay. Next one. It takes 12.6 seconds for that much carbon monoxide to effuse through a penhole. Under the same conditions, how long would it take for CO2 to effuse? So same sort of idea, same sort of thing. We're just doing a ratio with the square root in. So we go through and rewrite it down. Okay. And now we have to figure out which one is which. Now notice one quick thing here. What did I ask for? How long will it take for the same amount to effuse? So to effuse completely, to undo any ambiguity. So did I ask for the rates necessarily? No, I asked for the time of effusion. Now what did we just decide up here? How are they related? If something's fast, is it going to take a long time or a little bit of time? A little bit of time. So they're going to be proportional or inversely proportional? Inversely proportional. So we can just sort of add another ratio on to here and do that. So let's be real clear about where all the 2s and the 1s are. So we have 1 over 2 here. We have 2 over 1 here, right, so that they're inversely. But now our time and our rates are inversely proportional. Which what does that mean about our times and our masses? They're proportional, which makes sense, right? Now let's do this in words. So let's go the more something weighs, the slower it effuses, or excuse me, the slower it moves, and so the longer time it takes to effuse. So once again, if we say this is a higher mass, just kind of whatever, that means that it's going to take a long, it's going to have a much slower rate, and it's going to take it a much longer time to effuse. So our rate and our masses are inversely proportional, because if something is big, it takes, it's moving slow. If something is big, it takes it a long time to get completely effused, okay? Because it's moving slow. Maybe one more time? So if something is large, if something weighs a lot, it's going to be moving slower, and so it's going to take it a longer time to effuse. So that's what this is saying in terms of the ratio. Okay, so now we can pick and choose what we need. So we need the time one, right? We need to relate the masses, which we know, because we know we can add up CO and CO2, and we can relate it to the times. So we go ahead and we write down each one. So we related our CO2 to the time that we're trying to figure out. We related our CO to the time that we know, so that we can figure out the new time. Now does this matter? That 1.73 moles, do we need to put that into the calculation somewhere? No, I say the same amount, right? So how long will it take for the same amount? So that's what you call a red heron. It's to give you more information than you need, because in real life, you have a lot of information and a lot of it you don't really need, and you have to pick and choose what you need. So you may have noticed that happens on exams too. Okay, so now we sell for time. We'll go ahead and fill in some numbers here for you guys though. There you go. I wasn't going to fill them in, but just so you have them so you know where I got the time from. Then you do the algebra, which I'll trust you to do, and you get 15.8 seconds. Now with ratios, it's really easy to flip-flop things, so you always want to look and see well does it make sense? And sure, the larger thing, this carbon dioxide, that's what we solved for, right? We can add a little subscript here if it helps. It's bigger, so should it take longer or shorter to diffuse? Longer, does it? Yeah. So we got a ratio right. So it's nice that if you have a general feel for how these work, you can double check yourself too and make sure that you set up the ratio correctly and you didn't accidentally fill in the wrong number, which is a nice double check on exams. Okay. Here we have one more. Okay. So you guys give this one a shot. I've done three or four of them today. So I'll give you a minute to get started and then I'll give you some hints. Ready for your first hint? We have temperature and we have mass, right? So can we use the equation we've been using? No. What equation do we need to use? We need, we're kind of stuck going back to this equation, right? Okay. That's your first hint. What do they use stand for here? In the ages? Yeah, you're going to need a hydrogen. It's not an exam. You can cheat off your neighbor. It's okay. All right. Next hint. Cancel everything that stays the same, right? The root, the three and the r. And now what are you down to? Fill everything in, right? We know all of this or everything except one. Now am I filling in 37 up here? No. What am I filling in? Good. Do this and we solve for t. It's a pretty large number, right? So we'd have to get uranium hexafluoride up until that temperature in order, or excuse me, to have the same. That's how we would get the same speed. We'd have to move uranium hexafluoride up to that. So pretty warm. Okay. So that's all I have for you for a fusion. I'll give you guys a few more seconds. Oh yeah, because hydrogen is not going to be hanging out on its own. So since it's in a gas form, yeah, that's actually a really good point, especially for the exam. I'm pretty sure having written the exam, it'll come up somewhere. If I just say hydrogen gas, what is that? F2? Or fluorine? Chlorine. Yeah, and that's something that at this point you're expected to know. So if I say fluorine gas, I'm not going to say F2. I'll just say fluorine gas. If someone just says oxygen or chlorine without actually necessarily even saying gas, if it's implied that it's a gas, you should be assuming it's diatomic at this point. Especially more for any of the diatomics, it's assumed diatomic unless stated otherwise. Or unless it's in some weird situation where it's find the number of oxygen atoms in a molecule, something like that. If you're ever just told, you know, find the molecular mass of oxygen, you shouldn't write down, you know, the 16 off the periodic table. You should know it's diatomic and you should put in the 32. So that's just sort of a good all around point. Okay. So moving on to real gases now. So a quick reminder, before we can talk about real gases, we have to talk about ones that aren't real. So a quick reminder on ones that aren't real. What is, or excuse me, that things that are ideal, in other words, the things that make a gas ideal, the things that we had to assume to be able to use the ideal gas law is that there's no attraction between molecules, right? So when molecules are bouncing around doing whatever it is that they're doing, they're not, when they come near each other, they're not going to attract each other and they're not going to repel each other. They're just going to kind of be. We also had to assume that they had no volume, right? That their electrons weren't really there, I guess in a sense, that they weren't taking up any volume. We call it a point mass. They have a mass but no volume. Now we've been doing a lot of calculations with this, so we can assume that we can use this sometimes, right? Otherwise it wouldn't really be a useful thing to teach. Well for a lot of conditions it works quite well, but not for others. So we need to think about when that's going to be true and when it isn't. So when molecules are moving really quickly, attraction is negligible. So what does that mean? So this kind of goes back to something that I've hinted at earlier in the chapter just because you almost need to talk about it at times. If things are racing by each other, they're not going to have as much time to attract. For people who are into astronomy and things of that sort, you can think about that like the planets or the sun and comets and things like that, right? If a comet is whipping by a long ways away past the sun and it's moving really, really fast, it'll be able to escape the gravity of the sun and it won't become part of our little solar system. If it doesn't, it's going to grab onto it and that has to do with the speed that it's going at and how that's being done. So if molecules are moving really fast, they're not going to have very much attraction. They're just going to whip by each other and not really attract each other. Now if they're moving very slowly, then they have some time on their hands and they can actually have some of those intermolecular interactions that we've talked about. So they have some time to attract or repel each other. So we have to think about when this isn't going to happen. Because okay, great, we can say, okay, well, I can assume something's ideal if something's moving quickly. But am I always going to be giving you the rates of the molecules? No. But what will I give you most of the time? What is it that we can measure that tells us about the rate of molecules? What decides the speed of molecules? Temp- okay, math, good. What else? Okay, let's say for one given molecule, right? It's not only dependent on the math, it's dependent on something else too. Temperature, good. So I can heat it up, right? So does that mean then that gases are going to be ideal at high temperatures where things are moving really fast or low temperatures where things are moving really slow? Yeah, they're going to be ideal when the temperature is high and they're moving very fast because they don't have time to attract each other. When you cool the gases down and now they have time to attract each other, then they lose their ideal- ideality. Okay, now, next thing, how is mass related to speed? We touched on that one or someone had mentioned that one too, right? So mass is related to speed. So how- what can we say about heavy gas as opposed to a light gas? Which one's going to be more ideal? Something like helium or something like nitrogen? Which one's more ideal? Helium, right? Because it's small. So something that's small is going to be moving very quickly at the same temperature. And so those are more likely to be ideal or to fit into the ideal gas law. Now there's something else to do with the smallness. What do we know is true about really large molecules? How does our interactions compare to something with really small molecules? It's bigger attraction. Why? What does it have more of? London dispersion forces and why does it have more London dispersion forces? The more electrons and what's the buzzword that goes along with that? More polarizable. So let's condense that little mini review session down. So something that's big is going to be less ideal both because of its speed, which we just talked about, right? We know that something that has a large mass has a much smaller speed and so it has more time to be attracted. We also know that something that's very large has more London dispersion forces, which of course means more electrons. Is London dispersion forces a mass? Is that a causation thing? No, right? It's all about the molecules and the shape of the molecular cloud. So the more electron clouds. So the more electrons you have, the more polarizable something is. And so the larger something is, the more electrons it has and the more attraction it has. So something has a lot of attraction. It's not ideal. Everything on this slide is relating to how ideal something is. So low temperatures, it's not going to be very ideal. High intermolecular forces, it's not going to be very ideal. High mass, it's not going to be very ideal. And there's one other that we'll talk about in a minute, but this is all the reasons relating to just the attraction part. Because ideal gases aren't allowed to be attracted to each other. Actually, you guys are still ready. I'll give you a sec. So that's the first part of being an ideal gas. They can't be attracted to each other. So if they are, and they are to a very high level, they're not going to be ideal. Or you're not going to be able to treat them as ideal. Your answer will be a little bit more off. The other part then was what else does an ideal gas have that real gases, of course, do? No attraction and no volume, right? OK. So now we need to talk about what changes the volume aspect. If something has a high volume and it's very apparent that it has a high volume, then it's not going to be as much of an ideal gas. So there's one that kind of works quickly there, right? What one is that? Volume is equal to size, so the bigger the molecule is, the little bit less ideal it's going to be. That comes into play with volume as well. Now there's another one. And this one relates to pressure. So ideal gases, we don't have any volume. So that means that the space between the molecules is very large. So this kind of works well to think about people. So let's say I empty out this room and I put five people in this room. I blindfold them and I tell them to walk around. Is what percentage of volume of this room do they make up? Pretty small, right? Five people, that's not very much volume. The room is huge. I give them jet packs too so they can fly around. So they're not making up a very large percentage of the volume. Most of the volume is what? Empty space. OK. So I give all of you jet packs and I put you all in this room. How much does your volume make up for the room now? A lot more. Now let's say I bring over the lecture hall next door and I give them all jet packs and we send you all flying around. First of all, there's going to be a lot of broken bones, but setting that aside, it's going to, now you're making up even more of the volume. OK. I bring in another class. Now you're like 98% of the volume. You can barely move too, but that's besides the point. Now your volume, the volume of people, make up a lot more of the room, right? So suddenly the volume of people really makes a difference. It's not mostly empty space. To be an ideal gas, you have to assume that most of the volume doesn't matter. So what is that related to when it comes to pressure? Yeah. So having like five different lecture halls worth of people all in here moving around, that's a high pressure, right? Having five people in here moving around in jet packs, that's a low pressure. So when is a gas more ideal? Low pressure or high pressure? Low pressure. All right. Well, I ask the opposite one on the slide. When does this become false? High pressures, right? At high pressures, there's not a negligible volume. At high pressures, if we start packing molecules in there, suddenly almost all the space is made up of the volume of the molecules. If there's only five molecules in that whole empty space, it doesn't really matter. Their volume is very small in comparison to what's around them. Most of the room is empty space. Okay. Now, let's put this in a little bit of graph form. And this next graph comes from your book, and it's one of the main reasons I want to talk about it a little bit, because it can be a little bit misleading if you aren't careful about what you're looking at. And so I want to spend some time talking about it so that you're not kind of misled. So keeping in mind what I just said, there's a measure for this, and that measure is called a compression factor. So let's talk about what that means. So the idea behind a compression factor is how close something is to ideal. If something is an ideal gas, if we have everything set up perfectly, so something is an ideal gas, then it's going to be one. So that's what we have here for all conditions is just one. Now, this kind of reiterates what we just talked about. Molecules are going to attract when they're near each other. They're going to repel when they come close enough that their electron clouds interact. So everything to do with this graph decides, well, are they going to attract or are they going to repel? If it deviates from one, so up or down, it doesn't matter. If it's away from one, then it's not very similar to an ideal gas. Now where it gets a little bit misleading and what I want you to be careful of is this little section right here. What do you notice right here? This is almost ideal, right? The other is at 400 atmospheres. That's just a fluke of a transition thing. The part of this graph that we spend most of our time in is where? It's in this little inset, right? If we move right to here, we're already at 200 atmospheres. That's already at a really high pressure. So we've hit high pressure a long time ago by the time we get to this point on the graph. So what we've been talking on the previous two slides about high pressures and things like that, that's really just talking about this inset, this little section where it took this little piece of the graph and blew it up. Because right here is 10 atmospheres, which is already pretty high. So I just wanted to talk about the graph a little bit so that you aren't misled into thinking, oh, well, right around 400, all of these become ideal. Well, that's just a fluke of it transitions from being attracted and repelled. So this part of the graph is where we spend most of our time in life at, right? From 0 to 10, that's our atmospheric pressure is what? What is it about? It's about 1, right? You ever take a little bit depending on the weather. So most of the time you're here. So now let's spend some more time looking right in this section then. As you increase pressure, what happens to all of these actual gases? Do they stay ideal or do they get more away from ideal? They get more away from ideal, which is what you would expect. That's what we talked about, right? Impressing pressure makes it become less ideal. And why is that? Why does that matter? Because the volume isn't negligible, right? Because suddenly the volume matters. You're packing them in, you're pushing them in closer to each other, and now their volume matters more. And their attraction and repulsion matter more too. Okay. Now a little bit of thinking about why these are different. What's different from hydrogen and then all of the rest of them? Smaller so it has less intermolecular forces, right? Dispersion forces work too. And then why is N3 so strange here? What does ammonia have? What type of intermolecular force? Hydrogen bonding. So what happens if you put it under pressure? Like this right here. That is a line just stop. Turns into a liquid. So it has enough intermolecular forces that it's just going to turn into a liquid at this pressure. So that hopefully clears up that graph a little bit. And remember this little inset is kind of where we're talking about in those previous slides to some extent. I mean, further out goes to, it's just there's this weird flip here where for this little while it's not true. And that's just the difference between these attractions and repulsions, which one's winning. Okay. So now we've talked about it kind of in theory and we know that at what kind of temperature is it at Ax ideal? High or low? Okay, maybe we need a review. All right. When something is moving fast, is it more ideal or less ideal? More ideal. And when do you, when is something moving fast at a high temperature or low temperature? High temperature. At a high temperature, is it more ideal or less ideal? More ideal. Okay, good. So we've talked about that. And then we've also talked about pressure. Where is it more ideal? High pressure or low pressure? Low pressure, right? Because everything's all spread out. So my mildly lame analogy for the interaction thing, my jet pack one is much better. But my mildly lame one for the interaction is if you're walking down the ring road and you come across a friend and you're both running late to class and you're running past each other, how much interactions can you have? Not very much. Not very much because you're moving too fast, right? Now on the other hand, if you're both just kind of, you know, waiting in between classes and you're walking really slowly down the ring road and you come across to each other, how much time do you have to interact then? Quite a bit. So that's kind of a good analogy to help you remember it. Since that one seems to be the tougher one. Okay. So now Van der Waals equation. So this takes what we just learned and what we basically just learned is that no gases are really ideal, right? We can put them under conditions to make them more ideal. We can put them under conditions to make it so that we can probably use the ideal gas law. But they're still going to be, it's not going to be perfect. So we need ways to be able to make it at least a little bit better. If you're trying to do real calculations with these things, you want to at least be able to get closer. And that's where the Van der Waals equation comes in. What happens if you ignore this A and V in this NB? What does it become? PV equals RT, right? It's the ideal gas law. So what this does is it just takes and it puts a couple of different adjustments on it. And so this is one of the ways that you can adjust for deviations from being ideal. You can adjust for the fact that real gases aren't quite ideal. They do have attractions and they do have volumes. There are other ways too. There's other equations. Some of them a little bit more accurate. Some of them just different, maybe more accurate in certain situations. A and B are experimentally determined. So what that means is that in order to get A and B, you have to have a table. Should you go to the back of the book and memorize all the A's and B's? No, you have been going to the wrong class this whole time. You most certainly should not. He's just looking for a good curve. Don't listen to him. Don't memorize the A's and the B's. Those are just numbers. They're just numbers you look up. Even people who work with these sorts of things all the time are not going to go to the back of the book and memorize the A's and the B's. So I will give you them. They will either be in a table, added on to that front page where all of your constants and stuff are, or it will be put inside the question itself. And I'll just list the A and B values. You'll notice that this equation is on your equation sheet. It's been on your equation sheet since the beginning. What equation isn't on your equation sheet, by the way, that you maybe want to notice? PV equals NRT. Do you want me to give you that on the equation sheet? No. I can ask you to memorize things occasionally. PV equals NRT is one of them. This one you'll get. You'll get this on the equation sheet, and you'll get any A and B that you need. OK. So they're experimentally determined. So the way that they do this is they take a real gas, and they measure it under a bunch of different circumstances, and they fit it to this equation, and they figure out the A and B values by fitting it to the equation. They do this in other forms, too. So we can take this, and we can use this to actually do some calculations then. And it's not really at its heart to any more or less difficult than normal ideal gas law. It's just a little bit different, a little bit more algebra. OK. So some examples. First we'll do some conceptual ones, and then we'll fill into some equations. A sample of methane gas is 50 degrees C in 20 atmospheres. Would you expect it to behave more or less ideally if it was reduced to 1 atm? So we're at 20, we reduce it to 1. So is the pressure lessened or increased? Lessened. So is it going to be more ideal or less ideal? More. OK. Good. Now we've reduced the temperature, and we've made it negative 50 degrees C. So if we've reduced the temperature, what have we done to the speed of the molecules? Slowed them down. Is it going to be more ideal or less ideal? Less ideal. Less ideal. Good. So in each of those circumstances, maybe at that point, or excuse me, in this circumstance, maybe you should be thinking about, well, maybe I should use the Van der Waals equation to start the ideal gas law. OK. So an example now using ideal gas law, or not ideal gas law, Van der Waals equation. So we have 37 degrees, 3 moles of helium, in a container of 2.0 liters. And you would have to be told something along the lines of, this isn't very ideal. I don't expect you to guess that. I would tell you. OK. I think sapling just said something like use the Van der Waals equation to do it. So first we write down our equation. Now we fill everything in. So we're solving for pressure, so we leave pressure as is. Now on a homework question, where are you going to go find A? Some table. If you're doing it on sapling, I think they usually have a link to it. If you're doing it in your book, there's a table in the back. So if you do that, you get this. And then what is n? 3 squared. 3 squared. Or n is 3 and we square it, so it's going to be 9. And then over v. And so we put 4, or v squared, so we put 4. So this is 3 squared. This is 2 squared. We put in our v. How do we, or n is 3 again? How do we get our v? Go to a table and look it up. On an exam, you will be given it. Say it one more time. On an exam, you will be given it. Don't listen to him. He didn't write the exam. OK, I'm running out of room, so I'm going to have to just write it down here. That's equal to 3.0 times r times t. So we filled everything in. Now we go through and calculate things out. I'm going to do one step of algebra in between to show you, I think, probably the easiest way to do the algebra. Solve this, and then you solve this, and you distribute it in. You should play around with the algebra on this a little bit. I'm not going to, because it's the last five minutes of the last day of class, and we need to move on. But you need to go home and play with the algebra a bit, and make sure that you can do it. So you subtract that over and divide, and you get 39.5 atmospheres. So go home and convince yourself that you can do that.
 So today we're going to finish up with a new material and we're going to finish up with  a fusion.  We still have the example problems to go over for that and then we're going to move on to  real gases and we won't spend a huge amount of time on that just because there isn't a  lot of material to cover.  When we start getting into real gases things get really complicated really fast.  So we'll show you one model of how we can deal with real gases and we'll also go ahead  and then show you when you should be able to use the ideal gas and when you shouldn't  be able to use the ideal gas.  So we'll do that as well.  So first we need to go over to the document scanner and finish up some work.  So remember we were talking about a fusion.  So we have a gas moving into an empty container and we want to see what the rates of those  are both in terms of being able to calculate them and also in terms of being able to go  through and figure out how likely one is to effuse X amount faster than the other.  And that's really handy when we want to separate gases.  We can use the fusion to separate a mixture of gases which is handy.  So I showed you one example of where that would be useful in taking uranium and getting  one isotope out of it and we can do that because different masses effuse at different  rates.  And so we'd done one of them and we'd calculated the differences in rates.  So now let's go ahead and do another one.  In this case I had actually kind of given you the fraction.  I had said well we have rate A and rate B, we have our two different gases.  And I already told you that fraction.  I told you that it was a 1.55.  So one effuses 1.55 times faster than the other one.  So now let's move on to this one where we're going to kind of do the opposite.  Now we have nitrogen and hydrogen and I want to know the two gases and I want to know the  difference in the rates.  Okay, so we have to go ahead and probably should write this equation to be on the safe  side.  Now it's N2 to H2.  So we'll go ahead and just write out N2 and H2 now so that we can keep track of everything  since we have real numbers.  And we know that those have to be proportional to the square root of the molar masses.  Now does the N2 molar mass go on the top or the bottom?  Bottom, right?  It's always opposite.  They're inversely proportional.  So we're going to take this and we'll put the mass of nitrogen down here and we'll put  the mass of hydrogen up here.  Okay, so when we go through and do this then, do I have to convert this to anything special  as far as units go?  Or can I just put in the molar masses like I'm doing right now?  Should I put some conversion factors there?  No, why?  It's a ratio so they cancel out.  So we don't have to put anything there because it's a ratio and they cancel out.  Other times we have to worry about it.  Mostly in this chapter when do we have to worry about it?  When you see R.  Yeah, basically in any form.  If you see R in the form that has joules in it, what are all your masses have to be converted  to?  Kilograms.  Okay, so at this point it's calculator pushing and you get 0.269.  So our rate of fusion of nitrogen is 0.269 times the rate of that of hydrogen.  So now we've done some calculations with them.  Let's go ahead and see if we can just do this conceptually too because it's nice to  have numbers but sometimes it's just good to be able to look at a bunch of compounds  and say which one's going to fuse faster?  Which one's going to have a higher speed and just be able to rank them?  So we're going to do that now.  Sorry, you said that 0.269, what did you say that unit was in there?  You said that it was in the right?  So you mean unit?  Or if you just repeat what you said.  Okay, yeah.  I'll attempt to repeat what I said.  I'll repeat something like that.  So this is a rate over a rate.  So those units as long as they're in the same units will always cancel.  This is going to be in whatever unit we happen to have it in, it's usually molar mass, right?  So we'll have it in grams per mole and grams per mole off the periodic table.  Now when we used an equation that we derived this from which was the square root of 3RT  over M, that's where we got this equation from, right?  Well we had to worry about converting it to kilograms in that case because R had joules  in it and joules has kilograms in it.  In this case, if this is grams per mole, and this is grams per mole, what happened to these  units?  It can't fall out.  And so this becomes a unitless quantity.  And so since it's unitless, it's just a ratio.  And so we don't have to worry about whether we have to convert these.  They have to be in the same.  If for some reason you had one molar mass in kilograms, you looked it up off the internet  or something random, you would have to make sure you put it in grams.  But as long as these are the same, it doesn't matter.  So you're saying that hydrogen is 0.269 times faster than nitrogen?  Or is it faster?  Well so which one's going to be faster?  Nitrogen or hydrogen?  Hydrogen.  So because this is a decimal, you would say that the speed of nitrogen is 0.269 times  that the speed of hydrogen.  So if we're just like, let's just throw out a number and say hydrogen's 10, whatever.  You would multiply 10 by 0.269 to get your speed of nitrogen.  Because you want nitrogen to be slower.  Now the nice part of that is what happens if you forget that?  Well, let's say I tell you the speed of, it's a good question.  So let's, if I told you the speed of hydrogen at some given temperature.  So if I said that H equals, let's make the numbers easy since I'm doing this kind of,  you know, without notes, let's just make it 10, whatever.  So at a given temperature, it's 10 units per second.  And I wanted to know the speed of nitrogen.  How would we do that?  Well, we would take this 0.269 and multiply it by the 10.  You can think of the ratio as being kind of like a percent too, if that helps.  You can think of it as saying, well, nitrogen has 26.9% the speed of hydrogen.  And so that would mean that nitrogen, whatever those units were, would be 2.69.  So that's how you would go about determining the speed.  I'd have to give you one of the speeds.  The other way you can think about it is you can just fill it in too.  That's the, you know, if this would be if I gave you the ratio, if I gave you this final  answer and I asked for you to solve it.  But in this case, we have all sorts of information.  You could just fill in the rate of, in this case, we could take the nitrogen and just fill  it into, excuse me, hydrogen and just fill it into here, which if we did that, we would  multiply this up and it would be the same thing.  And it just sounded up like that because nitrogen was slower than you had in the beginning.  Yeah.  So you, you, when, when we started at this ratio and we had to multiply this by the  10, then it was because we, I just had started from this ratio to show you how to do it.  The other way is to fill into the problem, exactly as we did up here and fill in the  0.269 here and your masses here and do it the exact same way we did here.  The thing with ratios is there's just, there's as many ways to do it as there are people.  So yeah.  No, it just depends on how you write it.  I could have said write hydrogen to nitrogen and if you had written the hydrogen to nitrogen  ratio, instead of being 0.269, it would have been one over that.  So we could have gone through and we could have written this, the rate of hydrogen over  the rate of nitrogen.  Now where would the hydrogen mask go top or bottom?  Bottom.  And then filled in the same numbers.  Now it would have been opposite.  And then that number is definitely going to be over one.  So we could have done it the reverse.  It's just in this case I had asked for nitrogen to hydrogen.  Okay.  Oh, hang on.  I see people writing that.  Let me put it back.  And then we'll do the next problem.  Okay.  So next one.  Now I want you to rank in order of increasing speed of effusion and increasing time of effusion.  I haven't talked about time of effusion, but we can figure that out.  We can think about that.  So let's start with speed just because we've kind of already discussed it.  So I say increasing.  So what makes something go very quickly when it effuses?  A small mass or a large mass?  Small.  So when we go through and we do this, which one is going to be our smallest?  So how do you figure it out?  Yeah, you look at your periodic table and you would look at where all of these are and  which one would be your smallest?  Yeah.  So your speed of it, oh, you're not sure which one I'm asking.  That's the problem.  All right.  So speed of effusion.  So we need to do increasing speed, so we need small speed here.  And so we need the larger molecule.  So which one's the largest molecule?  The 205.  Okay.  So we have an increasing speed and we want to go all the way up to fast speed.  So we go slow to fast.  So we have increasing speed.  So that means we need to go large mass to small mass.  Does that help?  Okay.  So we know then that our biggest one is going to be our slowest.  So that one goes first.  And then which one?  Good.  And then which?  Good.  Okay.  So that was speed.  Now what about the one that we haven't talked about?  Time of effusion.  So what that's saying is how long does it take to effuse?  How long does it take to get completely effused through the barrier?  So if something is very fast, is it going to have a slow time or a large time?  Yeah.  Something's very fast.  It's going to have a very small time, right?  It's going to be zipping around all over the place and it's going to get to the other  side very quickly.  And so it's going to have a very small time.  A fast molecule has a very small time.  So what's our fastest molecule in this bunch?  And two.  So it's the smallest time.  And then what?  Yeah.  So it's the exact opposite, right?  Okay.  So that's one where we didn't really have to calculate anything.  We could just kind of look at it.  I suppose we had to calculate the molar masses, but that's it.  Okay.  Let's zoom back out of it.  Next one.  Oh, yeah.  What do you mean?  Okay, yeah, I guess I should say something like, you know, till reaching equilibrium  or sure total effusion.  Yeah, good point.  So what he had asked us is they're all going to start effusing instantly, right?  So the question is a little bit ambiguous because they are all going to instantly start  effusing.  The question is when have they completely effused?  Or if we're just going into one container.  Okay.  Next one.  It takes 12.6 seconds for that much carbon monoxide to effuse through a penhole.  Under the same conditions, how long would it take for CO2 to effuse?  So same sort of idea, same sort of thing.  We're just doing a ratio with the square root in.  So we go through and rewrite it down.  Okay.  And now we have to figure out which one is which.  Now notice one quick thing here.  What did I ask for?  How long will it take for the same amount to effuse?  So to effuse completely, to undo any ambiguity.  So did I ask for the rates necessarily?  No, I asked for the time of effusion.  Now what did we just decide up here?  How are they related?  If something's fast, is it going to take a long time or a little bit of time?  A little bit of time.  So they're going to be proportional or inversely proportional?  Inversely proportional.  So we can just sort of add another ratio on to here and do that.  So let's be real clear about where all the 2s and the 1s are.  So we have 1 over 2 here.  We have 2 over 1 here, right, so that they're inversely.  But now our time and our rates are inversely proportional.  Which what does that mean about our times and our masses?  They're proportional, which makes sense, right?  Now let's do this in words.  So let's go the more something weighs, the slower it effuses, or excuse me, the slower  it moves, and so the longer time it takes to effuse.  So once again, if we say this is a higher mass, just kind of whatever, that means that  it's going to take a long, it's going to have a much slower rate, and it's going to take  it a much longer time to effuse.  So our rate and our masses are inversely proportional, because if something is big,  it takes, it's moving slow.  If something is big, it takes it a long time to get completely effused, okay?  Because it's moving slow.  Maybe one more time?  So if something is large, if something weighs a lot, it's going to be moving slower, and  so it's going to take it a longer time to effuse.  So that's what this is saying in terms of the ratio.  Okay, so now we can pick and choose what we need.  So we need the time one, right?  We need to relate the masses, which we know, because we know we can add up CO and CO2, and  we can relate it to the times.  So we go ahead and we write down each one.  So we related our CO2 to the time that we're trying to figure out.  We related our CO to the time that we know, so that we can figure out the new time.  Now does this matter?  That 1.73 moles, do we need to put that into the calculation somewhere?  No, I say the same amount, right?  So how long will it take for the same amount?  So that's what you call a red heron.  It's to give you more information than you need, because in real life, you have a lot  of information and a lot of it you don't really need, and you have to pick and choose  what you need.  So you may have noticed that happens on exams too.  Okay, so now we sell for time.  We'll go ahead and fill in some numbers here for you guys though.  There you go.  I wasn't going to fill them in, but just so you have them so you know where I got the  time from.  Then you do the algebra, which I'll trust you to do, and you get 15.8 seconds.  Now with ratios, it's really easy to flip-flop things, so you always want to look and see  well does it make sense?  And sure, the larger thing, this carbon dioxide, that's what we solved for, right?  We can add a little subscript here if it helps.  It's bigger, so should it take longer or shorter to diffuse?  Longer, does it?  Yeah.  So we got a ratio right.  So it's nice that if you have a general feel for how these work, you can double check yourself  too and make sure that you set up the ratio correctly and you didn't accidentally fill  in the wrong number, which is a nice double check on exams.  Okay.  Here we have one more.  Okay.  So you guys give this one a shot.  I've done three or four of them today.  So I'll give you a minute to get started and then I'll give you some hints.  Ready for your first hint?  We have temperature and we have mass, right?  So can we use the equation we've been using?  No.  What equation do we need to use?  We need, we're kind of stuck going back to this equation, right?  Okay.  That's your first hint.  What do they use stand for here?  In the ages?  Yeah, you're going to need a hydrogen.  It's not an exam.  You can cheat off your neighbor.  It's okay.  All right.  Next hint.  Cancel everything that stays the same, right?  The root, the three and the r.  And now what are you down to?  Fill everything in, right?  We know all of this or everything except one.  Now am I filling in 37 up here?  No.  What am I filling in?  Good.  Do this and we solve for t.  It's a pretty large number, right?  So we'd have to get uranium hexafluoride up until that temperature in order, or excuse  me, to have the same.  That's how we would get the same speed.  We'd have to move uranium hexafluoride up to that.  So pretty warm.  Okay.  So that's all I have for you for a fusion.  I'll give you guys a few more seconds.  Oh yeah, because hydrogen is not going to be hanging out on its own.  So since it's in a gas form, yeah, that's actually a really good point, especially for  the exam.  I'm pretty sure having written the exam, it'll come up somewhere.  If I just say hydrogen gas, what is that?  F2?  Or fluorine?  Chlorine.  Yeah, and that's something that at this point you're expected to know.  So if I say fluorine gas, I'm not going to say F2.  I'll just say fluorine gas.  If someone just says oxygen or chlorine without actually necessarily even saying gas, if it's  implied that it's a gas, you should be assuming it's diatomic at this point.  Especially more for any of the diatomics, it's assumed diatomic unless stated otherwise.  Or unless it's in some weird situation where it's find the number of oxygen atoms in a  molecule, something like that.  If you're ever just told, you know, find the molecular mass of oxygen, you shouldn't  write down, you know, the 16 off the periodic table.  You should know it's diatomic and you should put in the 32.  So that's just sort of a good all around point.  Okay.  So moving on to real gases now.  So a quick reminder, before we can talk about real gases, we have to talk about ones that  aren't real.  So a quick reminder on ones that aren't real.  What is, or excuse me, that things that are ideal, in other words, the things that make  a gas ideal, the things that we had to assume to be able to use the ideal gas law is that  there's no attraction between molecules, right?  So when molecules are bouncing around doing whatever it is that they're doing, they're  not, when they come near each other, they're not going to attract each other and they're  not going to repel each other.  They're just going to kind of be.  We also had to assume that they had no volume, right?  That their electrons weren't really there, I guess in a sense, that they weren't taking  up any volume.  We call it a point mass.  They have a mass but no volume.  Now we've been doing a lot of calculations with this, so we can assume that we can use  this sometimes, right?  Otherwise it wouldn't really be a useful thing to teach.  Well for a lot of conditions it works quite well, but not for others.  So we need to think about when that's going to be true and when it isn't.  So when molecules are moving really quickly, attraction is negligible.  So what does that mean?  So this kind of goes back to something that I've hinted at earlier in the chapter just  because you almost need to talk about it at times.  If things are racing by each other, they're not going to have as much time to attract.  For people who are into astronomy and things of that sort, you can think about that like  the planets or the sun and comets and things like that, right?  If a comet is whipping by a long ways away past the sun and it's moving really, really  fast, it'll be able to escape the gravity of the sun and it won't become part of our  little solar system.  If it doesn't, it's going to grab onto it and that has to do with the speed that it's  going at and how that's being done.  So if molecules are moving really fast, they're not going to have very much attraction.  They're just going to whip by each other and not really attract each other.  Now if they're moving very slowly, then they have some time on their hands and they can  actually have some of those intermolecular interactions that we've talked about.  So they have some time to attract or repel each other.  So we have to think about when this isn't going to happen.  Because okay, great, we can say, okay, well, I can assume something's ideal if something's  moving quickly.  But am I always going to be giving you the rates of the molecules?  No.  But what will I give you most of the time?  What is it that we can measure that tells us about the rate of molecules?  What decides the speed of molecules?  Temp- okay, math, good.  What else?  Okay, let's say for one given molecule, right?  It's not only dependent on the math, it's dependent on something else too.  Temperature, good.  So I can heat it up, right?  So does that mean then that gases are going to be ideal at high temperatures where things  are moving really fast or low temperatures where things are moving really slow?  Yeah, they're going to be ideal when the temperature is high and they're moving very fast because  they don't have time to attract each other.  When you cool the gases down and now they have time to attract each other, then they  lose their ideal- ideality.  Okay, now, next thing, how is mass related to speed?  We touched on that one or someone had mentioned that one too, right?  So mass is related to speed.  So how- what can we say about heavy gas as opposed to a light gas?  Which one's going to be more ideal?  Something like helium or something like nitrogen?  Which one's more ideal?  Helium, right?  Because it's small.  So something that's small is going to be moving very quickly at the same temperature.  And so those are more likely to be ideal or to fit into the ideal gas law.  Now there's something else to do with the smallness.  What do we know is true about really large molecules?  How does our interactions compare to something with really small molecules?  It's bigger attraction.  Why?  What does it have more of?  London dispersion forces and why does it have more London dispersion forces?  The more electrons and what's the buzzword that goes along with that?  More polarizable.  So let's condense that little mini review session down.  So something that's big is going to be less ideal both because of its speed, which we  just talked about, right?  We know that something that has a large mass has a much smaller speed and so it has more  time to be attracted.  We also know that something that's very large has more London dispersion forces, which of  course means more electrons.  Is London dispersion forces a mass?  Is that a causation thing?  No, right?  It's all about the molecules and the shape of the molecular cloud.  So the more electron clouds.  So the more electrons you have, the more polarizable something is.  And so the larger something is, the more electrons it has and the more attraction it has.  So something has a lot of attraction.  It's not ideal.  Everything on this slide is relating to how ideal something is.  So low temperatures, it's not going to be very ideal.  High intermolecular forces, it's not going to be very ideal.  High mass, it's not going to be very ideal.  And there's one other that we'll talk about in a minute, but this is all the reasons relating  to just the attraction part.  Because ideal gases aren't allowed to be attracted to each other.  Actually, you guys are still ready.  I'll give you a sec.  So that's the first part of being an ideal gas.  They can't be attracted to each other.  So if they are, and they are to a very high level, they're not going to be ideal.  Or you're not going to be able to treat them as ideal.  Your answer will be a little bit more off.  The other part then was what else does an ideal gas have that real gases, of course,  do?  No attraction and no volume, right?  OK.  So now we need to talk about what changes the volume aspect.  If something has a high volume and it's very apparent that it has a high volume, then it's  not going to be as much of an ideal gas.  So there's one that kind of works quickly there, right?  What one is that?  Volume is equal to size, so the bigger the molecule is, the little bit less ideal it's  going to be.  That comes into play with volume as well.  Now there's another one.  And this one relates to pressure.  So ideal gases, we don't have any volume.  So that means that the space between the molecules is very large.  So this kind of works well to think about people.  So let's say I empty out this room and I put five people in this room.  I blindfold them and I tell them to walk around.  Is what percentage of volume of this room do they make up?  Pretty small, right?  Five people, that's not very much volume.  The room is huge.  I give them jet packs too so they can fly around.  So they're not making up a very large percentage of the volume.  Most of the volume is what?  Empty space.  OK.  So I give all of you jet packs and I put you all in this room.  How much does your volume make up for the room now?  A lot more.  Now let's say I bring over the lecture hall next door and I give them all jet packs and  we send you all flying around.  First of all, there's going to be a lot of broken bones, but setting that aside, it's  going to, now you're making up even more of the volume.  OK.  I bring in another class.  Now you're like 98% of the volume.  You can barely move too, but that's besides the point.  Now your volume, the volume of people, make up a lot more of the room, right?  So suddenly the volume of people really makes a difference.  It's not mostly empty space.  To be an ideal gas, you have to assume that most of the volume doesn't matter.  So what is that related to when it comes to pressure?  Yeah.  So having like five different lecture halls worth of people all in here moving around,  that's a high pressure, right?  Having five people in here moving around in jet packs, that's a low pressure.  So when is a gas more ideal?  Low pressure or high pressure?  Low pressure.  All right.  Well, I ask the opposite one on the slide.  When does this become false?  High pressures, right?  At high pressures, there's not a negligible volume.  At high pressures, if we start packing molecules in there, suddenly almost all the space is  made up of the volume of the molecules.  If there's only five molecules in that whole empty space, it doesn't really matter.  Their volume is very small in comparison to what's around them.  Most of the room is empty space.  Okay.  Now, let's put this in a little bit of graph form.  And this next graph comes from your book, and it's one of the main reasons I want to  talk about it a little bit, because it can be a little bit misleading if you aren't careful  about what you're looking at.  And so I want to spend some time talking about it so that you're not kind of misled.  So keeping in mind what I just said, there's a measure for this, and that measure is called  a compression factor.  So let's talk about what that means.  So the idea behind a compression factor is how close something is to ideal.  If something is an ideal gas, if we have everything set up perfectly, so something is an ideal  gas, then it's going to be one.  So that's what we have here for all conditions is just one.  Now, this kind of reiterates what we just talked about.  Molecules are going to attract when they're near each other.  They're going to repel when they come close enough that their electron clouds interact.  So everything to do with this graph decides, well, are they going to attract or are they  going to repel?  If it deviates from one, so up or down, it doesn't matter.  If it's away from one, then it's not very similar to an ideal gas.  Now where it gets a little bit misleading and what I want you to be careful of is this  little section right here.  What do you notice right here?  This is almost ideal, right?  The other is at 400 atmospheres.  That's just a fluke of a transition thing.  The part of this graph that we spend most of our time in is where?  It's in this little inset, right?  If we move right to here, we're already at 200 atmospheres.  That's already at a really high pressure.  So we've hit high pressure a long time ago by the time we get to this point on the graph.  So what we've been talking on the previous two slides about high pressures and things  like that, that's really just talking about this inset, this little section where it took  this little piece of the graph and blew it up.  Because right here is 10 atmospheres, which is already pretty high.  So I just wanted to talk about the graph a little bit so that you aren't misled into  thinking, oh, well, right around 400, all of these become ideal.  Well, that's just a fluke of it transitions from being attracted and repelled.  So this part of the graph is where we spend most of our time in life at, right?  From 0 to 10, that's our atmospheric pressure is what?  What is it about?  It's about 1, right?  You ever take a little bit depending on the weather.  So most of the time you're here.  So now let's spend some more time looking right in this section then.  As you increase pressure, what happens to all of these actual gases?  Do they stay ideal or do they get more away from ideal?  They get more away from ideal, which is what you would expect.  That's what we talked about, right?  Impressing pressure makes it become less ideal.  And why is that?  Why does that matter?  Because the volume isn't negligible, right?  Because suddenly the volume matters.  You're packing them in, you're pushing them in closer to each other, and now their volume  matters more.  And their attraction and repulsion matter more too.  Okay.  Now a little bit of thinking about why these are different.  What's different from hydrogen and then all of the rest of them?  Smaller so it has less intermolecular forces, right?  Dispersion forces work too.  And then why is N3 so strange here?  What does ammonia have?  What type of intermolecular force?  Hydrogen bonding.  So what happens if you put it under pressure?  Like this right here.  That is a line just stop.  Turns into a liquid.  So it has enough intermolecular forces that it's just going to turn into a liquid at this  pressure.  So that hopefully clears up that graph a little bit.  And remember this little inset is kind of where we're talking about in those previous  slides to some extent.  I mean, further out goes to, it's just there's this weird flip here where for this little  while it's not true.  And that's just the difference between these attractions and repulsions, which one's winning.  Okay.  So now we've talked about it kind of in theory and we know that at what kind of temperature  is it at Ax ideal?  High or low?  Okay, maybe we need a review.  All right.  When something is moving fast, is it more ideal or less ideal?  More ideal.  And when do you, when is something moving fast at a high temperature or low temperature?  High temperature.  At a high temperature, is it more ideal or less ideal?  More ideal.  Okay, good.  So we've talked about that.  And then we've also talked about pressure.  Where is it more ideal?  High pressure or low pressure?  Low pressure, right?  Because everything's all spread out.  So my mildly lame analogy for the interaction thing, my jet pack one is much better.  But my mildly lame one for the interaction is if you're walking down the ring road and  you come across a friend and you're both running late to class and you're running past each  other, how much interactions can you have?  Not very much.  Not very much because you're moving too fast, right?  Now on the other hand, if you're both just kind of, you know, waiting in between classes  and you're walking really slowly down the ring road and you come across to each other,  how much time do you have to interact then?  Quite a bit.  So that's kind of a good analogy to help you remember it.  Since that one seems to be the tougher one.  Okay.  So now Van der Waals equation.  So this takes what we just learned and what we basically just learned is that no gases  are really ideal, right?  We can put them under conditions to make them more ideal.  We can put them under conditions to make it so that we can probably use the ideal gas  law.  But they're still going to be, it's not going to be perfect.  So we need ways to be able to make it at least a little bit better.  If you're trying to do real calculations with these things, you want to at least be able  to get closer.  And that's where the Van der Waals equation comes in.  What happens if you ignore this A and V in this NB?  What does it become?  PV equals RT, right?  It's the ideal gas law.  So what this does is it just takes and it puts a couple of different adjustments on it.  And so this is one of the ways that you can adjust for deviations from being ideal.  You can adjust for the fact that real gases aren't quite ideal.  They do have attractions and they do have volumes.  There are other ways too.  There's other equations.  Some of them a little bit more accurate.  Some of them just different, maybe more accurate in certain situations.  A and B are experimentally determined.  So what that means is that in order to get A and B, you have to have a table.  Should you go to the back of the book and memorize all the A's and B's?  No, you have been going to the wrong class this whole time.  You most certainly should not.  He's just looking for a good curve.  Don't listen to him.  Don't memorize the A's and the B's.  Those are just numbers.  They're just numbers you look up.  Even people who work with these sorts of things all the time are not going to go to the back  of the book and memorize the A's and the B's.  So I will give you them.  They will either be in a table, added on to that front page where all of your constants  and stuff are, or it will be put inside the question itself.  And I'll just list the A and B values.  You'll notice that this equation is on your equation sheet.  It's been on your equation sheet since the beginning.  What equation isn't on your equation sheet, by the way, that you maybe want to notice?  PV equals NRT.  Do you want me to give you that on the equation sheet?  No.  I can ask you to memorize things occasionally.  PV equals NRT is one of them.  This one you'll get.  You'll get this on the equation sheet, and you'll get any A and B that you need.  OK.  So they're experimentally determined.  So the way that they do this is they take a real gas, and they measure it under a bunch  of different circumstances, and they fit it to this equation, and they figure out the  A and B values by fitting it to the equation.  They do this in other forms, too.  So we can take this, and we can use this to actually do some calculations then.  And it's not really at its heart to any more or less difficult than normal ideal gas law.  It's just a little bit different, a little bit more algebra.  OK.  So some examples.  First we'll do some conceptual ones, and then we'll fill into some equations.  A sample of methane gas is 50 degrees C in 20 atmospheres.  Would you expect it to behave more or less ideally if it was reduced to 1 atm?  So we're at 20, we reduce it to 1.  So is the pressure lessened or increased?  Lessened.  So is it going to be more ideal or less ideal?  More.  OK.  Good.  Now we've reduced the temperature, and we've made it negative 50 degrees C. So if we've  reduced the temperature, what have we done to the speed of the molecules?  Slowed them down.  Is it going to be more ideal or less ideal?  Less ideal.  Less ideal.  Good.  So in each of those circumstances, maybe at that point, or excuse me, in this circumstance,  maybe you should be thinking about, well, maybe I should use the Van der Waals equation  to start the ideal gas law.  OK.  So an example now using ideal gas law, or not ideal gas law, Van der Waals equation.  So we have 37 degrees, 3 moles of helium, in a container of 2.0 liters.  And you would have to be told something along the lines of, this isn't very ideal.  I don't expect you to guess that.  I would tell you.  OK.  I think sapling just said something like use the Van der Waals equation to do it.  So first we write down our equation.  Now we fill everything in.  So we're solving for pressure, so we leave pressure as is.  Now on a homework question, where are you going to go find A?  Some table.  If you're doing it on sapling, I think they usually have a link to it.  If you're doing it in your book, there's a table in the back.  So if you do that, you get this.  And then what is n?  3 squared.  3 squared.  Or n is 3 and we square it, so it's going to be 9.  And then over v. And so we put 4, or v squared, so we put 4.  So this is 3 squared.  This is 2 squared.  We put in our v.  How do we, or n is 3 again?  How do we get our v?  Go to a table and look it up.  On an exam, you will be given it.  Say it one more time.  On an exam, you will be given it.  Don't listen to him.  He didn't write the exam.  OK, I'm running out of room, so I'm going to have to just write it down here.  That's equal to 3.0 times r times t.  So we filled everything in.  Now we go through and calculate things out.  I'm going to do one step of algebra in between to show you, I think, probably the easiest  way to do the algebra.  Solve this, and then you solve this, and you distribute it in.  You should play around with the algebra on this a little bit.  I'm not going to, because it's the last five minutes of the last day of class, and we need  to move on.  But you need to go home and play with the algebra a bit, and make sure that you can  do it.  So you subtract that over and divide, and you get 39.5 atmospheres.  So go home and convince yourself that you can do that.
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