 So, before we get started, can I ask a question? How many people took the chemistry GRE? G.N. Mark raises his hand. Nobody took the chemistry GRE? How many people are going to graduate school? And you guys didn't take the chemistry GRE? Just the general GRE. Just, I'm just, okay. So, most of you know by now the midterm scores are posted and the key is posted. You can pick up your exam in your discussion section. Stephen and G.N. Mark are going to have those for you. Next time, I'll put the cover sheet on the exam properly and you'll get it back electronically. It takes longer actually. It's not the greatest system right now, but that's the way we'll do it. If you have questions about the grading of your exam, don't be afraid to talk to Stephen and G.N. Mark. Those are the two guys who did the grading. And so, ask them first if you have a question or a conflict about the way your exam was graded, then come and see me. We'll work it out. Quiz Force Friday, and in a moment, I'm going to show you that these quizzes are very important to us. I'm going to show you that in just a second. But let me just tell you I looked very carefully about what's going to be on the quiz and what's going to be on it is the stuff at the beginning of the chapter and chapter 15, sort of the first two-thirds of the chapter, which is all the way through page 496. And if I look at the exercises, the stuff that you need to be able to do is sort of in this range of exercises right here, 15.20 is already stuff for next week. Okay, so these first 19 problems in the exercises, any one of those would be a great quiz question. Okay? Now, this is what the histogram looked like for midterm one. I'm not any happier with it than you are. So if you did better than a 49, you actually did pretty well. I mean, in this class, all right, the mean should be a B. All right, of course, that's not the case when you have a straight scale. These are A's, these are B's, and these are lower grades over here. Okay? So I don't know why it didn't go better than this. One concern I have is that you get in the groove of taking the quizzes with the multiple guests and they're pretty straightforward problems, the sort of plug and chug type problems on the quizzes. I mean, let's face it, because they're pretty short, these quizzes, and then when you get to the exam and it's harder problems that have to be worked out, you're just not used to that, maybe, I don't know. Let me just say, if you think I think this material is easy, you're wrong. I mean, I'm not even a physical chemist by training. I'm an analytical chemist. I don't even know why I got assigned to teach this class. I mean, I have enormous difficulty with this material. If you think all the symbols look the same to you, look the same to me, too. U, S, Q, W, I mean, I get them wrapped around my neck constantly, all right? When I took that trip last week to Kansas, I was sitting on the airplane with my calculator, working out problems. When I got off the airplane, I had about 10 pages of problems. I'm trying to remember how to do this stuff, all right? So you guys, realistically, you know, if I've got 10 pages and I'm teaching the class, I've had this material before, mind you, I mean, it's just been a while, all right? You guys should have 100 pages of notes, 100 pages of problems worked out in your notebooks at a minimum. Wouldn't you agree? So I'm just saying that I don't know that you don't have 100 pages. But if you don't, right, midterm two is three weeks from Friday and it's going to be just like midterm one in terms of length and difficulty. I mean, I don't think midterm one should have been too difficult, all right? Maybe midterm two will be a little bit less difficult, but it's going to be pretty close to midterm one. And so I'm just telling you, in these three weeks, you need to really buckle down. If you haven't done this 100 pages of problems, I'm really suggesting that you focus on doing the hard problems, working them out. I mean, I think you realize that open book doesn't really buy you very much on this type of exam. I mean, if you don't know, if you're not really familiar with the book because you haven't been paging back and through it, working problems, I mean, if you own the book because you've been doing all these problems, then having the book there is pretty helpful, all right? But if you're opening the book and you're like, gosh, I'm trying to find a problem like this one that's been worked out here in the book, you just don't have time for that on this kind of test. I mean, time is really an issue. Okay, so I know this material is hard. It's really hard for me, too. And the only way around that, and we do need to really learn this stuff, the only way to learn it is to do a lot of problems. That's the only way to learn this stuff. I try to make these good lectures, but these lectures are not going to teach you this material. You've got to work the problems. Okay. Now, fortunately, the quizzes are saving you to a large extent. This is how you're doing. All right, I calculated this yesterday morning. No quizzes were dropped in the generation of this score. I didn't drop any quizzes, right? So some of you missed quizzes. Some of you did poorly on one or more quizzes. Everybody, virtually everybody would move to the right on this diagram if we dropped two quizzes or even one. Okay, so actually this histogram is a worst case scenario. It would be further to the right if we had dropped quizzes. Okay, so what this means to us is that the quizzes are, in fact, important. All right, they are moving us from a mean in this range over here to a mean in the low B range. And hopefully, by the end of the quarter, this mean is going to move further to the right after we drop these quizzes and we do better on midterm, too. Right, that's the goal. Okay, so that is actually a low B. And we're hoping that this moves over to the mid B range over here somewhere. And I think that's a realistic objective for us. All right, but it's not going to happen on its own. I mean midterm, too, could be just as bad as midterm one. I hope not. All right, help me out here. Okay, so the way I calculated this, in case you're wondering, is I just added your three quizzes together, quizzes are out of five points. I multiplied by eight, because each quiz is in reality worth 40 points. So that would be 120 plus 100 is 220, right? That's where I got this, so this is scaled from zero to one, right? So that would be 100% of the course points. If you've got 5s in all the quizzes and 100 on the midterm, the midterm was actually worth 110 with the extra credit. Why? You'd be over here somewhere. Okay, in case you're wondering how I calculated it. So you, if you want to, you can drop your own quiz, recalculate your, how am I doing score and see where you compare on this diagram. All right, I understand most folks are going to move to the right a little bit. Okay, so we have this guy, Sadie Carnot. He's interested in steam engines and at the time that he's doing his research, steam engines are about 5% efficient. The Carnot limit for a steam engine turns out you can run a steam engine where the steam is at about 300 degrees C and the cold reservoir, it's not room temperature, it's actually the boiling point of water. That's 100 degrees C, okay? And so in principle, you could operate this thing at sort of 65% efficiency. That's the Carnot limit for a steam engine where the steam's at 300 degrees C and the cold reservoir is the boiling point of water, all right? So that really is quite a bit lower than the Carnot limit. So he was asking two questions, could he do better and would changing the fluid from water to something else, could that help? We haven't really addressed that question yet. Okay? So the question is, what limits does nature place on the amount of work that we can derive from this temperature gradient right here, right? Nature doesn't allow us to get an arbitrary amount of work out of this temperature gradient. There are well-defined limits, that's what we're talking about. That's why this is sort of an important subject. So Carnot derived this thing called the Carnot cycle. Turns out he didn't really talk about entropy in his writings, he never used the word entropy and it actually took a second guy to sort through everything that he had done to reformulate it in a way that's useful to us in terms of what we now call the second law of thermodynamics, all right? But he derived this thing called the Carnot cycle which is the most efficient existing cycle cable of converting any given amount of thermal energy to work, he did that and there's another way to look at this, you know, if you inject a certain amount of work into the process of transferring heat from a cold reservoir to a hotter reservoir, why the temperature difference that you're able to generate using a certain amount of work, that's also, it's the same thing, all right? And so in his cycle, we start here on this isotherm and we do an isothermal expansion and then an adiabatic expansion and isothermal compression and then an adiabatic compression. And the reason we care about this, this looks like a completely artificial thing to be focusing so much attention on, all right? I mean, when are we actually in real life going to want to go from here to here or from here to here or from here to here, all right? Why are we going to care about this cycle? Well, the answer turns out to be that any arbitrary cycle like this thing shown in red here, any arbitrary cycle that you care about can be decomposed into a large number of Carnot cycles, all right? This is a diagram from your book. Here's a Carnot cycle right here, all right? Here's another one right next to it right here, all right? I think if you look at these two Carnot cycles, you can see that this one is going this direction, this one's going this direction. And so the two Carnot cycles that share that edge, all right? What's going on at that edge is just going to cancel, all right? And this edge cancels between these two Carnot cycles and this edge cancels between these two Carnot cycles because what that edge represents is the two cycles going along the same path in opposite directions, all right? And what you're left with after you cancel all of these paths here is you're left with the path on the outside of this thing that doesn't cancel, all right? Because there's nothing, there's no Carnot cycle next to it to cancel it out. And so I think you can see that here there's maybe 30 Carnot cycles, all right? And if you made this 3 million, you could map out this red circle pretty accurately with little tiny Carnot cycles, okay? So the point is, is if we understand how one of these cycles works, we can generalize what we learned from one cycle to this infinite number of cycles or a large number of cycles, okay? So any cyclical process like this is constrained by these Carnot considerations, these Carnot efficiencies that we're talking about. This applies to this, that's why we spend so much time studying it, all right? We constantly bump into this Carnot limit in chemistry, okay? So this is what we said, blah, blah, blah, blah, blah, how efficient is the heat engine, the efficiency is the work that we get out of it divided by the heat that's absorbed from the hot reservoir, that's how the efficiency is defined, okay? So in this case, it would be 5 over 20, we said this on Monday and we can derive then an expression for this efficiency which is 1 minus the temperature of the cold reservoir divided by the temperature of the hot reservoir. Where did we get that equation, all right? Can we derive it? So this is just going back to the stuff that we did on Monday, we can prove that this is right and the way that we do that is we write out the work for each of the cycles, cycle, this should not, I don't mean cycle, I mean path, step, step one is the isothermal expansion and step two is the adiabatic expansion, step three is the isothermal contraction, step four is the adiabatic contraction, okay? And we have equations for each one of these terms and it should be obvious to us that these two terms right here are going to cancel because my goodness they're identical except for the ordering of these limits. TC is here, TC is here, all right? And so if I reverse those limits, I'm just going to put a minus sign in front of one of these integrals and they're going to cancel, all right? So I'm only left with these two guys right here and these volume ratios here are related to one another and that's not obvious, all right? But the way that you can tell that they are is this equation right here which is on page 471 of your book, this describes the relationship between the volume at one point on an adiabat and V1 and T1, volume and temperature and the volume and temperature, this is supposed to be four, that's supposed to be four. So I'm talking about this point one and this point four so that's V4, that should be T4, V1, T1, those two points apply to this, okay? So these, because these two points are both on the same adiabat, this ratio has to be satisfied where gamma is Cp over Cv, okay? And the same thing is true here, did I mess this up? Gosh, oh, sorry, no, this is, actually the way I wrote this is fine, right? This is T2, that's this temperature isotherm right here, this is T1, that's this isotherm right here, and these are the two volumes right here, right? So this, actually this is, okay, sorry, so is this, okay? So if I write these two equations and rearrange them, I get this equality right here and this equality right here, one from each one of those equations. And then, if I divide this guy by this guy, the T1's go away, the T2's go away, right? And I'm left with these volume ratios and so I can simplify this into this, all right? And once I've done that, it's obvious that these two logarithmic terms are going to cancel one another, they're the same, okay? So rather than canceling, I can add them, all right? And so I can express them both in terms of V2 and V1 or I could express them both in terms of V3 or V4 if I wanted to, all right? And so this is the expression that I get for the work. Integrated over the whole cycle now, okay? This is the hot reservoir temperature, the cold reservoir temperature in these two volumes, okay? That's the work for the whole cycle, the heat that's transferred in the hot part of the cycle is given by this equation right here, this is just the expression for the isothermal expansion that's occurring in the first step of the cycle, okay? And so the efficiency is just that whole thing divided by that whole thing. According to this equation right here, I just take the total work, divide by the heat that's transferred in the hot part of the cycle, all right, I should get the efficiency. And so when I do that, look, NR is going to cancel when I look at this ratio, right? Log V2 over V1, that's going to cancel, all right? So I'm going to be left with TH minus T over TH and that simplifies right away to that. So we can prove this, we know enough to prove this. Remembering these equations is the hard part but you don't have to remember them, all right? You can look them up, okay? Now, what's the entropy change for each of the four steps of the reversible Carnot cycle? We need a thermodynamic expression for S. Let's say it's this. We'll come back to this later. In other words, the change in the entropy is the change in the heat provided the process is reversible divided by the T that the process is occurring at, all right? Now this reversibility issue, it turns out to be important, that's the reason this is an equal sign, turns out. We'll come to that later, okay? And so if I want to know the delta S for this whole cycle, I can just add up the delta S for these steps. The reason I can do that is because delta S is a state function, okay? And so if I want to know what it is for the whole cycle, I can just add up the delta S for these individual steps and I will get delta S for the whole cycle, all right? Steps 2 and 3, steps 2 and 4, rather, sorry, are adiabatic, right? Step 3 is reversible, isothermal compression, all right? So Q is zero for those two steps, all right? That's why that's zero, all right? The entropy is zero for those adiabatic steps because Q is zero, all right? For the isothermal steps, which are the first one and the third one, all right, these are the entropies, right? Q is not zero, otherwise they'd be adiabatic. And so what can we say about steps 1 and step 3? Well, we can look up Q, all right, for step 1, all right? It's going to be equal to, according to equations that we talked about on Monday and last Wednesday, a week ago, all right? This is the equation that applies to an isothermal change in volume. Hopefully that looks familiar. That was one that you needed for the midterm, all right? What's not totally obvious to me is that Q is equal to minus W, all right, how do we know that this is true? Well, we know that D U equals D Q plus D W, right? We know U doesn't change during any isothermal process, all right? U doesn't change. In other words, the change in U is equal to the constant volume heat capacity. In fact, the constant volume heat capacity is defined as the change in U divided by change in T. That's how the heat capacity is defined. And so the change in U is just the heat capacity times delta T, all right, and it's an isothermal process. Delta T is zero, all right? So we know U's not changing. That means Q's got to equal W. Rather, D Q has got to equal minus D W, right, if U's not changing. So that's why that has to be true. Don't take anything for granted when you're looking at this stuff. I mean, try to prove it to yourself that stuff like this, as simple as that seems, is true. I'm constantly having to do that myself. Okay, so what do we know? Well, we know for step one, V2 is bigger than V1. It's an expansion, right? We're going down that to bigger V. We know W's minus P DV, that's true always, all right? So we know that's going to be negative because V2 is bigger than V1 and there's a minus sign in front of this thing. So delta V is positive, but there's a minus sign, all right? That means Q's got to be positive. If work is negative, Q's got to be positive. In fact, it's got to be the opposite of work. And remember, we're talking about QH. This is the Q at high temperature because we're talking about step one along the top of that isotherm. Okay, and then if you look at this guy, this guy is exactly like this guy except he's happening now at TC, isn't he? And it's a compression, not an expansion. And so, okay, so we've got this equation right here. We've got this equation right here. We're trying to conclude something about this delta S for the whole cycle, the sum of all the four steps. And what we have to remember is that we showed back like 10 slides ago that V2 over V1 equals V3 over V4. And if you make that substitution, these two terms are going to be equal and opposite because check this out. TC is in the numerator. It's going to cancel with TC in the denominator. The same thing with TH, that cancels with that. And so these terms, if that equals that, these two terms are going to be equal to one another, right? Well, that wasn't obvious to me that that had to be true so that's why I went through and added these slides. So that means that this, I mean that's why I had to prove this, that this really is a square box. In other words, the entropy is not changing here or here and the temperature obviously is not changing here and here and so if you make a temperature entropy diagram out of a Carnot cycle, it's a square box. Now, everything that we've said applies to only reversible processes. We've been talking about reversible processes so far. And real life processes are virtually never reversible because to be reversible, it's got to be infinitely slow. So what if one or more of these steps are irreversible? The efficiency of irreversible is not, in other words, if the process is irreversible, it's going to be less efficient than a reversible process or at best equal in efficiency, right? Limited, the efficiency of the irreversible process is bounded by, if you will, the efficiency of the reversible process, okay? So what does that mean? Well, it means that if I write the sufficiency in terms of heats, here's the efficiency of the irreversible process here, here's the efficiency of the reversible process here, if it's reversible, now I'm going to write rev on it to make sure that we understand and remember that it's reversible. Okay? So these ones are going to cancel. So we've got this guy right here, all right? And now what you need to remember is that these QCs here, those guys are negative, right? QC and QC reversible are both negative. We showed that back on slide 36. I have no idea what slide this is, 45. How do you know that? Oh, it's down here. Okay, so if I write these Qs as negative, I've got to flip that equality sign around. Here it's greater than, sorry, here it's less than and here it's greater than, right? Greater than or equal to. All right, so if I make these negative, I've got to flip that equality sign around, all right? And these reversible heats can be equated with the temperature of the cold and the hot reservoir, okay? And so if I set this equal to this, I've got to use this greater than or equal to sign. And in other words, if I just cross multiply here, I've got the heat transferred from the cold reservoir divided by its temperature and the heat transferred from the hot reservoir over its temperature and there's a minus sign in front of this guy because it's negative. So for everything that, so this applies to an irreversible process. If I move both of these guys over to the same side, I can put a zero over here and say in general, this has to be true. In other words, if I evaluate these heats from the cold reservoir and the hot reservoir, it's got to be less than or equal to zero. For any irreversible process now or any process, all right? This applies to reversible or irreversible processes because there's an equal sign here. This is called the Clausius inequality. Who's Clausius? All right, so in other words, if I integrate Q over T for a series, you'll recall this is the Q over T that applies for step one. This is Q over T that applies for step three. All right, we've integrated for the whole cycle, all right? So if I integrate over a whole cycle Q over T, that's got to be less than or equal to zero. This is really the most general statement of the Clausius inequality. He's the first German that we've talked about, all right? And what he did is he took Carnot's ideas and translated them into what we now call the second law of thermodynamics. He named entropy and he tried to give it the symbol, you know, good one, but no, sorry, we're going to call it S. I don't know where S, where do you get S out of entropy? Santa? Oh, I don't think so. Now, you can memorialize Clausius with a T-shirt of your own, all right? And you can order this T-shirt online at a website that I will show you, but look at this right here, all right? No, I'm sorry, that was Juul, all right? The energy of the universe is constant, sorry, that wasn't Clausius. So this is totally wrong and I would ask that with me you join in the silent protest by not purchasing one of these $24 T-shirts, yeah, inaccurate and expensive. Okay, so this is the Clausius inequality but the second law of thermodynamics is this, all right? Now, is it obvious to you that this comes from this? Well, it wasn't obvious to me, all right? So let's see if we can figure that out. Now, batteries are dying. Here's how we can show that this is right, all right? State one, state two. We're going to go by some irreversible path between state one and state two and then we're going to go back on a reversible path, all right? Irreversible, reversible, very simple. Now, according to Clausius inequality, these two steps, I can write two terms like this for these two steps. I'm going to go from state one to state two and I'm going to have a DQ over T for that and I'm going to go back from state two to state one except it's going to be reversible and what the Clausius inequality says is, hey, if there are any irreversible steps, that's not the equal sign goes away, all right? See how that's less than or equals to? Equals only applies if everything's reversible, all right? So if this is just this, sorry, this is just going to be less than zero now because of this guy, okay? So since the second step is, so since this is reversible, we know it's equal to DS only because it's reversible, all right? And so I can flip these limits here. If I flip these limits, I think you'll agree I've got to make this plus sign a minus sign, right? And now I can move this guy over to the other side and when I've done that, I just derived this equation right here, all right? So this is the second law of thermodynamics, all right? It says that, well, okay, so what does it say? Look, it makes, it says there's three kinds of processes. First of all, if DS is greater than DQ over T, integrate it over the whole cycle, all right? That cycle can be spontaneous and if it's greater than DQ over T, it's also irreversible, all right? If these two things are equal to one another, it's reversible and if DS is less than DQ over T, it doesn't happen spontaneously. You can make it happen, all right, but it doesn't happen spontaneously. Did it just get yellow again? Oh, what's going on? Okay, if we're talking about an isolated system, an isolated system is one where there's no heat flow and in and out, no mass flow either, okay? And so if I want to understand what these inequalities translate into in an isolated system, I just substitute zero for Q, all right, because there's no heat that can go in and out of the system if it's isolated, okay? And so all of these inequalities just turn into I just substitute zero for this, boom, I get this, okay? So the second law of thermodynamics allows us to classify a process as one that occurs spontaneously or one that doesn't, very important to us as chemists. Okay, what are some examples? Here's some simple examples and this is the kind of problem that's right at the beginning of the exercises from exercise one to exercise 19, all right, if you're going to be looking at the reversible phase transition, vaporization, sublimation, melting, all occur at a particular temperature and with a defined enthalpy, all right, delta H. In this case, the phase transition occurs, if the phase transition occurs reversibly, delta S is equal to delta H over T. So for example, here we have a plot of entropy on this axis and temperature on this axis. We are heating a frozen solid, all right, entropy increases until we get to a phase transition where melting occurs and there's a stepwise increase in the entropy at the melting point, all right, what is that delta S? It's that, all right, if we know the delta H, we just take this temperature, all right, we can calculate what that stepwise increase in entropy is and now we're heating the liquid. The entropy continues to go up, now we get to the boiling point and by golly there's going to be a different delta H that applies there, delta H vaporization, all right, we divide by that temperature, we get that change in the entropy right there, okay, so these stepwise increases in entropy happen as we're heating a solid through these phase transitions, all right, melting and boiling. There could also be reversible heating or cooling of a gas, right, DS is DQ reversible over T, that means the heat capacity can be used to calculate the DQ, multiply that by DT or integrate, divide that by T and we're going to get DS, all right, we use the constant pressure heat capacity if it's a constant pressure process or constant volume heat capacity, if it's a constant volume process, that part's pretty easy, okay, so if I evaluate this integral, all right, it's an integral of 1 over T so I'm going to get log T final over T initial, all right, that's what the delta S is going to be equal to and so S goes up with the logarithm of this ratio, so here's what that looks like, here's delta S plotted against that ratio right there, Tf over Ti, all right, this is a logarithmic increase, these are just different heat capacities, okay, entropy goes up, all right, that's why when you look at this diagram right here, here we're talking about a gas, all right, in principle that's what this is right here, all right, it's the entropy going up for this gas as we heat it, all right, it's happening in accordance with this equation right here, all right, which would predict a logarithmic increase in the entropy with temperature, all right, so we can calculate that very easily, we've got these, we've got this equation right here, we can also talk about the expansion or compression of a gas, right, dQ is minus dw is pdV, so the change in the entropy is just Q over T, again dQ over T rather and in this case it's dQ is pdV and so if the gas is ideal why we can recast that in this form right here, we can just talk about the final and the initial volume and we can use that equation to calculate the change in the volume, so at constant Tv is inversely proportional to p, all right, so I can express this in terms of volume or in terms of pressure, either one, all right, so I can calculate the delta S for that process too, that's another example from the beginning of chapter 15, those exercises, okay, so because S like uses state function you can add da, da, da, multi-brunch, yes, so any random process that you're thinking about, you can break up into steps yourself, you can decompose any process you want into steps, calculate the delta S for those steps, you will always get the right final delta S because it's a state function, it doesn't matter what path you take, right, you will get the same right delta S if you calculate the delta S of each path correctly and add them, calculate the entropy change when argon gas is 25 degrees C in one ATM in a container in volume 100 cubic centimeters is compressed of five cubic centimeters will simultaneously be heated to 45, so we compress the gas, we compress argon and we heat it, okay, we don't have any equation for heating and compressing but we have equations for each one by itself and so we can just heat and then compress, or compress and then heat, add them up, so the total S for this process is just going to be the delta S for compressing plus the delta S for heating, this is process one, we'll call it isothermal compression, we'll do that first, we'll compress first and then we'll heat without changing the volume. Isothermal compression, we expect delta S to be negative for that based on the equations that I just showed you and delta S for the isofolumetric heating, that's going to be positive, remember that's that logarithmic increase, right, entropy goes up logarithmically with temperature, all right, so in this case we don't know if this is going to be a spontaneous transformation or not, we don't know if delta S is going to be positive or negative because we've got one term that's positive, one term that's negative, we don't know which one's bigger, okay, so it's useful to actually think that through and just say, you know, if these were both positive or both negative, you could say, all right, delta S is definitely going to be overall positive or overall negative, okay, so process one is compression to constant temperature, that's this equation right here and all I've done is substituted for the number of moles, I've just substituted PV over RT because I don't know, it doesn't tell me how many moles I've got but I can calculate that from the information I've got here, all right, I can just cancel these Rs, oops, better not, all right, they don't cancel for gosh sakes because I need this to be moles and that is Joules per Kelvin per mole, right, it's different Rs, this one's 0.08206, this one is 8.31451, okay, and if I do the dimensional analysis by Golly, yeah, that's the way I need it to be because I want the units to be Joules per Kelvin, right, be a little careful, okay, so if I plug the numbers in 1 atm, 0.1 liters, 0.08206, 298.15, 8.31451, volumes, boom, boom, boom, minus 0.102 Joules per Kelvin, good, we expected it to be minus, all right, do I have any intuition about the size of this number, zero, I have absolutely no clue if that makes sense, I mean, the fact that it's not 10 of the 20 reassures me a little, all right, so I don't know, now we're going to do step two, all right, that's the equation that we need for step two because we're only changing the temperature at constant volume, that's the heat capacity right there, all right, for an ideal gas which we can expect argon to behave pretty darn ideally, I think you'd agree, we know what that heat capacity is going to be for constant volume, it's 3R over 2, that's the constant volume heat capacity of any ideal gas, that's the number of moles, all right, 3R over 2, yes, units will be yes, we need to multiply by mole, all right, so always think about the units, all right, the units of this heat capacity if they're 3R over 2 are joules per Kelvin per mole, so I've got to multiply by moles to get rid of that because I want the units to be joules per Kelvin, okay, and so that's the same number of moles that we used before, that's the constant volume heat capacity, we know the initial and final temperatures are, yes, yes, yes, yes, yes, yes, yes, yes, yes, Kelvin temperatures and this is the delta S that we calculated, it's a lot smaller, all right, it's positive, we expected it to be positive, okay, and so then the total delta S is just the sum, that's the, what did we do first? Heated, and then we change the volume, right? No, we can, what did we, oh, change the volume for, then we, sorry, yes, all right, okay, and so that's the sum of those two things, minus 0.0987 joules per Kelvin, do I have any clue whether that number makes sense? No, I really don't have any intuition about these entropy changes, all right, but I can say, because that's a minus sign that this is not a spontaneous process, all right, that's what the minus sign tells me, it's not a spontaneous process, and we care about the magnitude too, but the most important thing is to understand that, okay, oh, my goodness, it's exhausting, isn't it, okay, so study for the quiz on Friday, we'll see you then.