 Okay, how are you guys doing? How'd the exam go? How many people thought the exam was too easy? How many people thought the exam was too hard? Hmm. So, we're going to post the scores later on today. You don't want me to post the scores? Steven, G, and Matt, Mark, Mac, graded all of these exams by themselves. So, I'm very grateful to them for doing that. And I don't honestly have any idea how you did right now. So, I'll find out later today just like you. But don't worry, we'll deal with the outcome no matter what it is. I don't think it'll be that bad. Okay. Unfortunately, I did make one major mistake which was, you know, I'm supposed to put one of those exam return forms on the front of the exam and I always forget to do that for some reason on the first exam of the quarter, quarter after quarter. So, we're going to have to return these exams to you in discussion this week. And so, in your discussion section, your TA is going to return your exam to you. Okay. You can look at it. I'll post the key later on today as well. So, you can look at the key, compare that with how your exam was graded, talk to me. Okay. We do have a quiz Friday. We have quiz four and it's going to be about chapter 15 which is all about the thermodynamic definition of entropy. We'll be talking about that today and Wednesday. Okay. And I'll tell you more about what's going to be on the quiz on Wednesday. Now, where are we? Well, we're right here, right? We've been going at this for four weeks now. We had your midterm on Friday. We're coming to the end of this topic of statistical mechanics and thermodynamics. We're starting chapter 15 today and we're going to just sort of zoom through 16 and 17. We're going to, at like 20,000 feet, we're going to look at these chapters. All right. We don't have time to devote all the time that we ought to to these two chapters because we have to talk about kinetics and reaction dynamics and so we're going to sort of skim these two chapters, 16 and 17. All right. So believe it or not, we are near the end of where we're going to be talking about this topic right here and I'm personally happy about that because I find this topic difficult myself. If you think it's difficult, I think it's difficult too. Okay. So we're going to transition to chemical kinetics probably next week. Okay. Probably maybe some place in the middle of next week. And we are pretty much on track, which is unusual for us. All right. By now, we've usually trashed the schedule and we're behind and we're starting to throw stuff out but we're pretty much hanging in there and pretty happy with where we are. Okay. So just so you have an idea. All right. You're going to need to read 16 and 17. You're going to have to have some knowledge of what's in those chapters but probably not everything and I'll tell you what's important. Okay. So in chapter 14, chapter 14 was all about the first law, conservation of energy. Energy is conserved for an isolated system. Change in the internal energy is equal to zero for any process. Okay. The central concept is internal energy. We're constantly talking about that in chapter 14 and in connection with the first law. All right. And what we learn from the first law is to understand whether a transformation is allowed, a transformation that doesn't conserve energy is not allowed. In other words. All right. And so we can tell immediately whether a particular process is even possible or not but that doesn't tell us whether it will actually happen. It only tells us it's an allowed process. In the second law, the second law is the entropy of an isolated system increases in the course of spontaneous change. We find out whether the process that's allowed will also happen. All right. Not all allowed processes will happen, all right. And we need to be able to sort allowed processes into processes that will not happen and ones that will. All right. The central concept here is entropy, not energy and the answer to the question is a transformation of spontaneous, will it happen? If it's allowed, will it happen? Pretty important question to be able to answer in chemistry, I think you'll agree. Remember this? We talked about entropy weeks ago, all right. We talked about statistical entropy. We put nickels in a Kenny shoe box, 100 of them, all right. And we shook the box and then we looked inside, made sure we had one layer of nickels, okay. And we saw this evolution in the distribution of heads and tails as we shook the box, shook the box, shook the box. And we concluded that what was changing here is not the energy because the energy of this state is just virtually identical to the energy of this state, right. The energy doesn't really depend on whether the coins are face up or face down. Those are energetically degenerate states for the coin. And yet we see the system evolve consistently in this direction and what we concluded is that this is the direction of increasing W, right. The system is optimizing on increasing W. It wants a configuration that maximizes W. So we concluded for any isolated assembly, we can always predict the direction of spontaneous change as that in which W increases and remember W is not energy. It's something else. So Boltzmann postulated that this parameter is the entropy. He defined it according to this equation here. This is really a postulate, right. And this is the statistical definition of entropy. This is the statistical definition. We're going to talk in a moment about the thermodynamic definition. Now we can already take the statistical definition and apply it to the expansion of a gas. We're going to try and tie together statistical entropy and thermodynamic entropy. What is the probability of finding gas molecule in the entire volume of a closed vessel? I've got one molecule in a vessel, right. The probability that it's in the vessel somewhere, this vessel is airtight, all right, that molecule can't get out, the probability is one, right. There's a 100% chance that the molecule is in the entire vessel. I put it in there, right, it can't be anywhere else. Now what's the probability that it's in half of the vessel? Well, we know intuitively the answers, one-half. How did we get that answer? Well, if we're talking about half the vessel, V over 2. Now, let's say there are two molecules. The probability that both are in V prime is the product of these two one molecule probabilities. In other words, we can take the probability for molecule one, multiply it by the probability for molecule two, that's the same as the probability for molecule one, we just multiply these together, that's the same as squaring that, okay. And so for N molecules, this isn't going to be squared anymore, it's going to be this ratio to the Nth power, big N, okay. So we're just using the statistical definition of entropy to arrive at this conclusion. This is going to tell us something about volumes. Let's see if we can apply it, right. Here's an experiment, I've got gas A in this half of this vessel and gas B in this half of this vessel. A and B are located in the two halves of a container. Now, I take away the barrier between the two halves. What's going to happen? We know with 100% probability these two gases are going to mix, we don't have to wonder whether that's going to happen or not, and we also understand that that's a manifestation of the entropy of the system increasing, right. That's a direct manifestation of the second law. Here they've noticed how I switched from blue to red to purple. Isn't that beautiful? Show this process of all this increase in entropy. We know it does, can we show that it does? Here's the Boltzmann law, S equals K log W. The change in the entropy is the final entropy of the system minus the initial entropy of the system because entropy is a state function. So that's the entropy associated with the final volume. That's the entropy associated with the initial volume and we can just use Boltzmann's law and plug in Boltzmann's law to this equation right here. So we've got the final state, final number of states and the initial number of states. And so this is a minus, this is a subtraction. I can just write this log as WF over WI, right. That's the change in entropy right there. And we have an equation that relates these number of states with this volume. We just showed that W sub N for V prime is given by this equation right here. So we can just substitute this guy for that and a term just like him for that, right. VF and VI play the role of V prime, okay. So I'm going to make V prime VF over V in the numerator and I'm going to make V prime VI over V in the denominator. All right, and these two V's are just going to cancel for us and so we're just going to end up with VF over VI and I can move that N out front in the logarithm. Okay, so the change in the entropy is going to be K and the number of molecules log VF over VI and I've got a term like that for delta S for gas A and delta S for gas B because there's two gases involved here, right, and there's a VF and a VI for both gases, right. There's, if I took only gas A, all right, and I allowed it to expand into a larger volume, I think you'd agree it would do that spontaneously, okay. So if there was no gas B, we could arrive at the same conclusion. There just would be no second term here. This would be 1, NK log 2, right. Both of those are going to be spontaneous processes. A expanding into a vacuum if we take the barrier away and A and B spontaneously mixing if we take the barrier away. This is a positive number. We just need to know N to calculate it, right. We can calculate this delta S is, but it's positive. That means this is a spontaneous process. Now, going all the way back to lecture 4, all right, what if instead of a change in entropy we wish to calculate the absolute entropy of a monotomic gas? We actually talked about this briefly in lecture 4. We derived an equation for this purpose. We started off with this equation right here, excuse me. And we did a bunch of extra steps, but I'm not going to talk to you about, you can see it's lecture 4 for the rest of this derivation. And we derived this thing called the Sacher-Tetrode equation. All right, and the thing is this is not a delta S now, it's an absolute entropy. Usually we're calculating a change in the entropy, but here for a monotomic gas we've got an equation that tells us the absolute entropy. So it's a pretty important equation, right. It can anchor our entropy calculation. And so this is just the residual end of this derivation. And so we can use the Sacher-Tetrode equation to calculate the standard molar entropy of something like neon gas. This is actually right out of lecture 4 as well. That's the thermal wavelength. What is this dismissive units of meters? Yes, yes, yes, yes, show that that was true. Then we can plug everything in. That mass has to be in units of kilograms, never forget, right. The mass of the neon molecule has to be in units of kilograms. And so that's, neon has a mass of 20.18 grams per mole, but that's 20.18 times 10 to the minus 3 kilograms per mole. Okay, Avogadro's number, blah, blah, blah. So that's the mass of a single neon atom in units of kilograms. Yes, yes, okay. And so we can calculate the thermal wavelength. It's always very, very short. We expect to get absurdly small number like this. So never be surprised by that when you're calculating the thermal wavelength. And then we can plug everything into the equation. And S is equal to 138 joules per Kelvin per mole, those of our units of molar entropy. Joules per Kelvin per mole, okay. So this is just straight out a lecture four. I'm only putting it in here to remind you that we can calculate the absolute molar entropy of a monowatomic, ideal monowatomic gas. Because that's something you might need to do hypothetically on a quiz. Okay, now, this is Sadie Carnot. He's the first French guy that we've talked about this quarter. And I could say something bad about the French, but I won't. All right, he was interested in steam engines. He was a mechanical engineer. And he wanted to understand how to make them more efficient. All right, you know, as we go through and we look at these people who contributed to the thermodynamics and statistical mechanics, they all had a practical motivation for what they were doing. All right, they weren't just interested in pure science. I mean, they were interested in pure science, but they needed to get an answer to make something happen better. Make better beer if you're a jewel. Make it more efficiently. Make a better steam engine if you're Carnot. All right, he wanted to know two things. Was the energy available from a heat engine and steam engine is a type of heat engine? Is that unbounded or are there some fundamental limits involved? How do I know when my steam engine is operating as efficiently as I could possibly expect it to? All right, or is there no limit to how efficient it could be? He was doing this stuff actually in the 1820s. And this is before Joule had figured out the first law. All right, so conservation of energy wasn't a defined concept at this point in time. He was thinking about entropy before conservation of energy was even worked out. All right, and the second thing is can a steam engine be made more efficient by changing the working fluid using something besides water as the working fluid in the steam engine? All right, he was thinking about that. So I told you, Carnot, he's French. How about Maxwell? Scotland. Yes, Maxwell's a Scott. Joule, you guys know that one. England. Gibbs. American. Yes. We're in there. We're going to be saying more about him. Boltzmann. Austrin, yes, there they are. This would be a good quiz question. Okay, so there's a statistical definition of entropy. That's the Boltzmann equation. And there's a thermodynamic definition, which is this. Maybe we'll derive this equation on Wednesday. I don't know. It takes about six slides to do that. All right, the change in the entropy is the change in the heat for a reversible process divided by the temperature. We're going to talk about what happens when the process is not reversible on Wednesday. All right, but today we're always going to be talking about a reversible process. All right, the change in the entropy is the change in the heat that flows at a defined temperature. That's what the entropy is. That's a thermodynamic definition. Okay, so what did Carnot do? He worked out how much work you can extract from a temperature gradient. All right, a steam engine is a temperature gradient. You've got steam at maybe 150 degrees C. You've got the ambient temperature, which is roughly 20 degrees C. All right, so you've got a huge temperature gradient and you're extracting work from that. You're pulling the train with that temperature gradient for all practical purposes. Okay, and so he devised this thing called the Carnot cycle. We talk about the Carnot cycle because it's extremely important. It places an upper limit on the amount of work that we can extract from a temperature gradient. Alternatively, it tells us how much temperature, how much heat we can pump with a given amount of work. All right, that's a very practical problem that we need to be able to solve. All right, Carnot cycle is the most efficient existing cycle capable of converting a given amount of thermal energy into work, or conversely, it's the largest temperature difference we can establish using a particular amount of work. All right, it's both things. So if you've heard of a heat pump, all right, a heat pump is, its efficiency is bounded by the efficiency of a Carnot cycle. We have to be able to understand this. Now, a heat engine extracts work from a temperature gradient. Here's a hot temperature. This would be like the steam in a steam engine. Here's a cold temperature. This would be like the ambient temperature outside the steam engine. All right, this delta T is the temperature gradient that we're talking about, th minus tc, okay, and some of that can be extracted as work, right? Some of that temperature gradient can be extracted as work, what the Carnot cycle tells us is how much? How much work is it possible to extract from this temperature gradient? So, here it is. All right, here's a pressure volume trace. This is an isotherm at T1, and here's an isotherm at T2. All right, and pressure volume space. All right, we're starting here at this point in the cycle. Okay, and so this is step one, step two, step three, and step four, I haven't labeled them that way. All right, but we're starting here, so this is step one. All right, step one is an isothermal compression. Step two is an adiabatic compression. Step three is an isothermal expansion. Step four is an adiabatic expansion. Isothermal, obviously it's on the isotherm. Adiabatic, obviously it's not on the isotherm. All right, so that helps you remember it's adiabatic. That's an isotherm also, okay, so that's an isothermal, sorry, did I say expansion? Well, expansion, expansion, compression, compression. Isothermal adiabatic, isothermal adiabatic, okay? So the first thing we have to know is what this is. Now, there's an infinite number of these, all right, but we're going to be able to derive some conclusions about them that's general and the same for all of them. All right, I think you can, in principle, I could stop this here and then do an adiabatic expansion here, then move here, and then do an adiabatic compression here. I could concoct any number of these Carnot cycles. What do we know for sure? Well, step one, Q's greater than zero. Because I'm doing an expansion, it's an isothermal expansion, work is less than zero, that means Q's got to be greater than zero. This is an isothermal compression, work is going to be greater than zero for that, that means Q's got to be less than zero, and I'll show you why that's true in just a second. Q, of course, is zero for these two steps because they're adiabatic. These are the things that we know for sure. Okay, so how efficient is a heat engine? Any heat engine, not just a Carnot cycle, right? The efficiency is defined as the work that's performed divided by the heat that's absorbed from the hot reservoir. All right, in a heat engine, there's a hot part and a cold part, okay? In your car, the hot part part is inside the cylinder. The cold part is outside the engine. That's the delta T that matters in an internal combustion engine. In a steam engine, there's this temperature of the steam and there's the temperature outside the steam engine. What we care about is the heat that's transferred from the hot reservoir, the inside of the cylinder, the steam, this guy. Okay, so there's some numbers here and the units don't matter. It could be kilojoules, okay? But if we wanted to calculate the efficiency of this particular heat engine right here, it's easy. All right, I'm transferring five units from the hot reservoir, I'm sorry, I'm extracting five units of work from the hot reservoir and dividing by 20, that's the total number of units that were transferred out of the hot reservoir, all right? So that's the work right there. That 20 is the number coming out of the hot reservoir. The efficiency is 25%, all right, you with me? Okay, now, how efficient is a Carnot cycle? Well, it turns out that a Carnot cycle has an efficiency that's given by this equation right here where that's the cold, temperature of the cold reservoir, that's the temperature of this guy and that's the temperature of the hot reservoir, that's the temperature of the hot reservoir, all right? The efficiency depends on what these two temperatures are and it can't be higher than this. So let's see if we can prove that this is true. Let's prove this because that's the central thing that Sadie Carnot was able to do. All right, we want to calculate the work for each of our four cycles, okay? This is step one, two, three, and four, all right? This is isothermal expansion. This is adiabatic expansion. This is isothermal compression. This is adiabatic compression. Everybody recognize those equations? Yes, and that's also, this is problem two on your exam. Now, that guy is the opposite of that guy. All right, because I look at the limits of the integration. That's Tc to Th, that's Th to Tc, so that's the negative of that, so those two terms are going to cancel. Exactly, all right? What we're left with are these two guys, okay? And to further simplify them, we have to notice that these two data points here lie on an adiabat. So that means that these temperatures and volumes are related to one another through this equation right here, where gamma is just the ratio of the constant pressure and the constant volume heat capacities. These two guys right here, these two data points are also related through an analogous equation because they're also located on an adiabat, all right? And so this equation holds true for any two temperature volume data points on any adiabat, okay? So using these two equations, I can, this is just a statement of the one on the left, this is a statement of the one on the right, and now I can make substitutions. In fact, I can divide this guy by this guy, and I get this very simple expression that I can use to substitute now into this equation right here. I'm substituting that into this to obtain that, all right? The total work is just minus NR, temperature difference, multiply by log V2 over V1. And the transfer of heat in the first compression is. So that's the total work for the whole cycle, okay? And the transfer of heat in the first compression is this, that was just the first term that we wrote. And since we already agreed that the efficiency of the heat engine is just that divided by that, why it turns out if you make that substitution and you cancel terms, you get this equation right here. Try it yourself. Okay, so heat pump is used to maintain the temperature of a building at 18 degrees C when the outside temperature is at minus 5. For a frictionless heat pump, how much work must be expended to obtain a joule of heat? Is a joule a lot of heat or a little heat? Pretty big unit of heat, pretty big heat unit. Answer, here's the efficiency, all right? How much work must be expended to obtain a joule of heat, all right? Here's our Carnot expression for the Carnot efficiency. And so solving for work, all right? Work is just equal to QH times this guy, right? And so a joule, all right? So Q is a joule of heat, all right? One minus these temperatures are, of course, converted to the Kelvin scale from those two guys, all right? And what I calculate is 0.079 joules, pretty small number of joules, all right? But we agreed joules are pretty big energy unit, okay? And so the work necessary to pump one joule of heat from minus 5C to 18 degrees C, I think you'll agree that's going to be thermodynamically uphill to do that because that's colder than that, all right? That's going to be 79 millijoules. One, two, three, 79 millijoules, very simple. Okay, what is the entropy change for each of the four steps of the reversible Carnot cycle? There's another question concerning the Carnot cycle. We need the thermodynamic expression for S, yes, all right? Obviously, if we can convert this differential into a delta, then we just can pay attention to these delta, the change in entropy during step one and step two and step three and step four. The sum of those is going to be equal to the change in the entropy for the whole cycle. And that's what we want to know, all right? Because S is a state function. So steps two and three are adiabatic, so Q is zero for those and that means S is zero because S is Q over T, all right? And so if Q is zero because that's an adiabatic step, step two is adiabatic and step four is adiabatic, those two change in entropies are also going to be zero. You can tell that right away, all right? But that's not true for steps one and three, all right? Steps one and three are going to be Q1 divided by TH. All right, we can make these total derivatives because this process is occurring entirely at TH. It's occurring in an isothermal way. And process three is going to be Q2 over TC because that process is occurring entirely on the cold adiabat, a cold isotherm. And so if we know what the pressure volume carno cycle looks like but if we rewrite the carno cycle in terms of temperature entropy, it's a box, right? There's no change in temperature but there's a change in entropy for process one. Process two occurs with no change in entropy but a change in temperature between the hot and the cold, right, you see what I'm saying? So the carno cycle represented as a temperature entropy process is a box. It tends to be a helpful thing, at least for me to remember, helps me remember that the entropy is not changing in process two or in process four, all right? It's constant because there's no heat flow. Okay, so steps two and three are adiabatics, so Q1 goes zero, that means this, so let me emphasize one thing. So these two guys cancel. Did I prove that? Did I leave a slide out? I don't think I proved that. All right, that guy and that guy are equal and opposite. What's that? Yes, but I don't think I adequately prove that that has to be the same distance as, I mean, in other words, that this, sorry, this has to be the same distance as this. I think I still need to prove that to you, I guess, on Wednesday. Now, everything that we've said has so far pertained to a reversible process, but real processes are not reversible, right? Reversible processes happen infinitely slow, real processes happen at some finite rate. So we have to address what that means for us in terms of the thermodynamics of real processes versus reversible processes. We're going to get to that on Wednesday. We're going to talk, first of all, about this equation right here, and then we're going to talk about what happens when the process is not reversible. Okay? And I think, actually, this is all I've got, even though I've got 20 minutes left, Sean, somehow I managed to go through these 75 slides faster than I thought I would. Does anybody have questions? Yes? So steps two and four are adiabatic. That should be Q3, sorry. Okay, so we'll see you on Wednesday. Thank you.