 So, let me remind you, we're going to take a quiz on Friday during the six minutes of class. We'll be a stack of stantrons right at the front of the lecture hall there. Pick them up, pick up one, please, and find a seat. Turns out there's 116 of you, and there's 240 seats in this lecture hall. And so what that means is we don't need two forms for this exam. Everyone, we can sit every other seat in this room, and there shouldn't be a problem with that. Okay, so if you would, please make sure that you don't sit exactly next to somebody else. Leave an empty seat between you and then it's first and next to you. Okay, and we shouldn't have any problem. There's plenty of seats in here for us to do that. Okay, it's going to be five short problems. Key non-concepts covered in, lectures one and two, homework one, and our discussion study guide one, which was posted to the website on Tuesday morning. That's the discussion study guide that you're going to be talking about in discussion all week. I also posted three quizzes from last year. They're already posted to the announcements page. You can get an idea of what these problems are going to look like. One of the things you'll notice is that you have to be able to calculate a factorial using your calculator. This might be the first time you'd have to do that in any class. I don't know. But I know when I had to calculate factorial on my calculator, I had to pull up the owner's manual and figure out how to do it. Because there's no key that says factorial on it. Okay, so Friday morning would be the wrong time to try and figure that out. Look at it before that. Yes. I'm sorry? There's no calculator policy. Anything that you can carry in here is fine with me. All right, not only that, but this quiz is open book, open notes, but it would be a mistake to imagine that you're going to have the luxury of time to sort through your notes and pause and look through the book and think about what you might want to time is going to be an issue for you on this quiz. You're going to have to know what you're doing, find the right equation if you don't remember it. Boom, knock it out. There isn't going to be much time to do the problems on this quiz. You're going to have to know what you're doing. Okay, so write anything down that you want on a piece of paper if you want to have a list of constants or conversion factors or equations that have been involved in the stuff that you've been studying. This week, put that on a piece of paper. Anything you want. Yes. Since you emphasized not putting up a lecture notes, should we bring our laptops to have the lecture slides and be able to discuss them? You are not going to have time to click through this lecture slide. What was on slide 40, Dan? No. You are not going to have time to do that. That would be a waste of time for you, I think. All right, any questions I could have on the slide? Yes. What if I had a discussion on this Friday? Because as you said, you don't think we're going to be asked to do it? Yes. Well, that's why I posted that discussion study guide on the website. Well, it turns out you can go to any discussion section. Right? You know, you might be assigned to the Friday discussion section, but you can go to the one on Tuesday, Wednesday, Thursday. There are six of them. So if you want to go to one earlier or one because you're concerned that if you want to be, okay, other questions? Okay. We're talking about the Boltzmann Distribution Law today. It will be about microstates and configurations and W. We're going to derive the Boltzmann Distribution Law using a slightly different approach from the one that's taken in your book because you can read what's in your book already. You don't need to hear me regurgitate that to you. I'm going to introduce you to partition functions. We're going to do a few examples. Boo. This is a molecule. All right? It has evenly spaced vibrational states to zero order. We're going to assume it's a harmonic. Vibrations to this molecule are harmonic. Okay. Now, it's tedious for me to draw this monstrous thing. I'm going to draw something smaller that looks like this. This is three molecules. We're going to label them A, B, and C. I've condensed these three molecules into this little notation that I'm showing you here now. These are the quantum numbers for the vibrational states. V equals zero. That's the ground vibrational energy level. One, two, three. You remember all this from quantum mechanics. I'm calling this a three-dimensional array or a three-dimensional lattice because I want to emphasize that these three molecules are distinguishable in terms of their position. In other words, I can tell where A is. I'm not going to confuse it with B. It's going to, because my lattice has A next to B next to C, and I know where A, B, and C are, and they're not going to move around. In other words, these molecules can be labeled and kept track of. That's an important point because later on, we're going to relax that constraint. But for the time being, we know where these molecules are. We can call them A, B, and C. We're not going to lose them. So now imagine you have a three-dimensional array of these molecules. Each occupant identically gives it a lattice. To keep it simple, let's say that we've just got three molecules, A, B, and C. Now let's add three quant of energy to these three molecules. All right? There's three ways to do that. We can put all the energy into molecule A. We can put all the energy into molecule C, or we can put all the energy into molecule B. That's one way to configure the energy. Put all the energy into one of the three molecules. We could also put two quant into one of the molecules and one into the other, leaving the third with no energy in it at all, and there are six ways to do that. Finally, we could just put one quantum of energy into each of the three molecules, and there's only one way to do that. I think you can see that intuitively. Now notice I'm not labeling these molecules anymore. Here they're labeled A, B, and C, but I'm going to get rid of that because I'm lazy. Don't be confused. I'm always talking about three molecules here. Somebody sent me an email and asked me about that. It's a good question. A, B, and C are still here. Okay. We're going to call each one of these guys a microstate. And we've got this short form notation that we can use to identify not microstates, but families of microstates, configurations. These curly brackets are going to help us identify different configurations that these microstates can be in. This first digit inside the curly brackets is going to be the number of molecules in the ground state. The next digit is going to be the number of molecules in the first excited state, and so on and so forth. That's what this notation will mean. So you can see these guys are all in the same configuration. How do I know that? There's two that have a ground state occupied, and one that has the excited state, and the third, B equals 3, is occupied. Right? So that's also true here, 2 and 1. Also true here, 2 and 1. So they all have this configuration here, 2, 0, 0, 1. And these guys, analogously, 1, 1, 1, 0, boom. And of course there could be any number of 0s. Okay? This guy is 0, 3, 0, 0. So we have a notation that we can use to identify these different configurations. That's all this is. There's no interesting science here. Okay. Now, short of figuring out and drawing every single configuration, we need a formula that allows us to calculate how many microstates there are for each configuration. Let's see if we can derive one from the simple model that we've looked at already. Because instead of having three molecules, we're going to have 10 to the 23. Right? So it will not be practical for us to think about these molecules in the terms that we've been using so far, where we're just thinking about three. Okay? So as I build up, let's think about configuration 2. You know what we call configuration 2, that's 1 molecule that has 2 quanta, 1 molecule that has 1 quanta, 1 molecule that has 0 quanta. Alright? As I build up that configuration, I have to, I can put that first V equals 2 into any one of these three molecules, A, B, and C, which are no longer on here. Right? You can go into any one of the three molecules. So I put it in A, but it could have gone into B or it could have gone into C. In terms of where I put the second one, once I put that one there, I've got two possibilities of where I can put the second one. And the third one then, once I put that into A and that one into C, the third one can only go into B. The last parcel of 0 goes to the remaining one molecule. Okay? And so if I want to know how many versions of this are there, it's 3 times 2 times 1. 3 times 2 times 1, that's just 3 factorial. That means there should be six ways to do this and there are. What about configuration 1? Configuration 1 was 3 quanta and 0 and 0. Right? Starting with the first, I can put the first one in A, B, or C. This time I re-labeled these guys. Then the next one can get all going to B here. In this case, it could go into A because I already put the 3 quanta into B. In this case, it can go into A because I already put the 3 quanta into C. Okay? And then final increment, it's going to go into a place that's left over. I mean here it can only go into C. Here it can only go into C. Here it can only go into B. Now, in principle, the counting in this case works exactly the same as it worked in this case. Here we've got six possibilities, but here I think you can see that there's only the possibility of having three microstates that have this configuration. What changed? What's the difference between configuration 2 and configuration 3, or configuration 1 rather? Well, the difference is that this state is degenerate. V equals 0 has got two molecules. And in other words, there's two verbally distinguishable ways to create that microstate. I can put this guy here, and this guy here, or this guy here, and this guy here. It makes no difference. I'm going to get the same microstate when I do that. So when I count microstates, even though there's, in principle, six ways, just like there was in the first configuration, 3 times 2 times 1, we're going to need to reduce the total number of configurations by the degeneracy factor. And in this case, there are two molecules that have the same energy, and so it will turn out to be the case that we need to reduce by dividing by two factorials. So instead of three factorial or six, we're going to have three factorial divided by two factorial. That's six divided by two. That's three, there should only be three ways to make this state enter our heart. Now, it may not be intuitively obvious that it has to work this way, and it's rarely the case for me that anything is intuitively obvious. And so I work on an example, and it's like, well, that's not going to be true, but I'll work this way. In other words, you always need to divide by the number of states that have the same energy. And if there's multiple states that have the same energy, you take the product, two factorial times three factorial times four factorial, for example, we'll do an example like that in just one second. In this case, there's three guys, right, in configuration three, and so that's three factorial, and three factorial divided by three factorials, one. Okay, now you can do more examples like this and convince yourself that this is true as I have to when I first looked at this formula. So, in other words, we do have a general formula that we can use, this is it, that's the number, total number of microstates, that big W there, that is the number of molecules, that's the number of molecules that have no energy, one quantum of energy, two quantum of energy, and so on. So, consider a system of eight molecules considering four energy quanta, eight molecules, yep, one, two, three, four, five, there's eight molecules there, four energy quanta, really, one, two, three, four, and these are all zeros, all right, what's the shorthand notation for that configuration? Well, there's five guys that have zero, two guys that have one, one guy that has three, and all the rest of these are going to be zeros. Problem one on quiz one, what's W? Eight factorial, one, two, three, four, five, here's another one that's got two and one, what's zero factorial, okay? Now, if you can't, if your calculator will not do that, one times two times three times four times five, you know, so eight is sort of doable, but when you get to like 27, you know, you really need to figure out how your calculator does a factorial, right, you will need to calculate that formula on Friday through the quiz, okay? Boom, 168 weights to generate that configuration. Pretty useful formula for us, it turns out. Now, let's look at this a little bit more closely because there's something hidden here that is very, very important, right, that we haven't noticed yet, right, this is a fundamental fact of nature that you have never seen before, almost certainly, right? Really, related to this, I think so, here it is, right? Ten molecules now, one, two, three, four, five, six, seven, eight, nine, 10, put all five quant of energy, now let's make it five in this case, put all five into that guy, all right, how many microsates are there, 10 molecules, nine of them are all in this energy state down here, so I'm gonna divide by nine factorial there, there's 10 ways to generate this configuration and you can see immediately that that has to be true, right, 10 ways to do this, 10 microsates. What about if I put four in one and one in another, all right, what's the W gonna be for that? Still 10, I'm just calculating factorial, 10 times nine times eight factorial is 10 factorial, right? There's eight guys that have this ground state energy assignment that have eight factorial there, one factorial, one factorial, bunch of zero factorial, so that's 90, or how do I know that, eight factorial cancels with eight factorial and I'm left with nine times 10 is 90, so there's 90 ways to generate that micro state right there, or 90 microsates that have that configuration. You with me so far? Now, let's work out all of the other possibilities, which we can readily do, we understand how to do that now. All right, and here are the number of microsates, that's the one we calculated first, then we calculated 90, there's another 90 and this guy right here is 360 and this guy right here, he's 360, this guy is 840, all right, 840 has two quanta in one, one quanta in three and zero in six. All right, and then there's a 252, that's this guy right here, right, those are the only possibilities that exist. What's profound about this? Well, that number is bigger than all the rest of these numbers. Configuration six, it's just randomly labeled six, six doesn't mean anything. It's favored by more than two to one compared with any of these other configurations. And two to one, it's special, not super special, it means only special by a factor of two. All right, now I'm just plotting w versus the configuration number, which remember is just arbitrary, don't worry about configuration number, it doesn't mean anything, but I'm just pointing out, here's that configuration, 840, all right, it's special. Now, how special is it? Well, we've only got 10 molecules here. What if we had 10 to the 23rd? We've only got 10 here, but if we have what we're dealing with in chemistry is not 10 molecules very often, once in a while, most of the time we're dealing with 10 to the 20, 10 to the 23rd, 10 to the 24th. 10 to the 23rd is a mole. We're dealing with fractions of a mole or a mole. All right, so it's that level that we want to be able to understand here. We want to be able to understand the statistics of large numbers of molecules. So let's make the system a little simpler. Let's make it a two-state system, heads and tails, right? Only two states for the molecule now. Excited or not excited, let's say. Now our w, here's the number of coin flips. That's just a way to think about molecules, okay? Now I think it's changed. I'm just gonna evaluate the system multiple times. That's the number of heads, that's the number of tails, that's the number of microstates. For example, if I flip four times, I could get heads, heads, tails, tails. I could get heads, tails, heads, tails. I could get tails, heads, heads, tails. I think this notation is now obvious too. All right, if I wanna now, I'm gonna get two of each, all right? That's a configuration. How many microstates are there for that configuration? Boom, six. How do I know that? Four factorial, four flips, two of each. Six possible ways to do it, boom, there they are. Here are all the different ways that you could do that. All heads, all tails, what's the probability of that? One, two, three, four, five, six, seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, one part and 16. You got one chance in 16 of getting all heads flip four times. You wish me on this? All right, here's the profound thing. On this axis, I'm plotting the number of microstates divided by the total maximum number of microstates. In other words, the maximum number of microstates for different numbers of coin flips or different numbers of molecules. Here's six coin flips, 10, 20, 100, 1,000. Okay, of course, the number of microstates goes way up as the number of molecules goes up, but let's normalize by the maximum number of microstates that can exist for each one of these systems as it gets larger. Because we want to see what happens to the shape of this distribution. All right, what happens is very interesting. Here's six coin flips, here are the data points that I just showed you, plotted on here, all right? And now, if you increase that to 10, 20, 100, 1,000, look what happens to this distribution, right? It becomes needle sharp, all right? It becomes almost infinitely narrow. It's not infinitely narrow, but it is darn narrow. That's only 1,000 molecules. We're not dealing with 1,000 molecules very often. We're dealing with 10 to the 23rd. What is that distribution going to look like for 10 to the 23rd? It's just going to be a needle. It's just going to be a delta function at this preferred configuration. What is on this axis here? These are just different configurations for the system, all right? This axis is just different configurations. This axis is the normalized number of microstates. What am I saying here? I'm saying there is one configuration or a small ensemble of configurations that dominate as the system gets bigger. That is a very surprising conclusion, all right? Are you telling me that I can't have four heads? No, you can't have four heads when you've got 10 to the 23rd coin flips. In other words, you can't have 10 to the 23rd heads. No, that's not going to happen. Well, it's sort of intuitively obvious, isn't it? All right, but it's less obvious that you can't have any other possibility either. In other words, there's really only one possibility that's going to be exactly, probably turns out to be half heads and half tails in the case of coin flips, all right? But more subtle if there's multiple states available to the molecule, more subtle. So the main point here is that we really need to understand this configuration right here. What is that? What are its properties? If we can understand that configuration, that is sufficient for us to understand how this chemical system is going to function, what its properties are. We only have to understand that guy. We don't have to understand this one or this one or this one or this one or any of the other 10 configurations that may exist here. We only have to understand one or a small ensemble of them. The properties of these other systems are going to be so rare that they're not going to contribute to the behavior that we observe in chemistry for macroscopic amounts of stuff. Imagine if that's 10 to the 23rd and not 1,000, boo. What does the N stand for in Cornhusker Stadium? Who knows the answer? Knowledge is the correct answer. Knowledge of this highly preferred configuration equates to knowledge of the system as a whole. We only have to know about that guy. We only have to know about him. All the rest of these guys we don't need to know about. It gets sharper. That's the whole point. Your book derives the Boltzmann Distribution Law by finding the W where the maximum value. Right now I think you can see that at the maximum W by Dolly, that's the one we want. That's the preferred configuration right there. So if you find the maximum of any one of these curves, you're going to locate that point and that's what you want to do, right? But there's another approach that one can take and that is that the derivative here is zero, isn't it? The derivative with respect to configuration of W over W mass at this point is zero. In other words, if I draw a tangent there, I'm just going to get a horizontal line. That's another way to derive the Boltzmann Distribution Law. That's what I'm about to do. Today I find the configuration for which W does not change when the configuration of the system is infinitesimally altered. In other words, if I shift the configuration left or right of that peak point slightly, alright, and W doesn't change because the derivative there is zero why I've located the peak of that distribution. So I've located that magic configuration. That's the whole point of the Boltzmann Distribution Law. Consider, okay, so we're going to do this derivation. You ready? Consider it isolated. In other words, the number of molecules is constant. Q, what's Q? Q is the total number of energy quanta in the system so when I say isolated, I mean we're not going to change the number of particles. We're not going to change the amount of energy we're going to leave that constant. Metroscopic, large number of molecules. Assembly of N, and oscillators that share a large number of energy quanta. Large number of oscillators, large amount of energy. Either one of those things is going to change. Which configuration is preferred? How do you, in other words, where is the peak in that curve that I just showed you? Where is it? Let's consider just three states. There could be thousands. Let's consider just three. Let's label them L, M, and N randomly. If we want to know how many microstates there are, we use our standard formula. That's the total number of molecules. And if this is L, M, and N, then there must be A, B, C, B, E, and so on all the way to L, M, and N, and then O, B, Q, R, S, Z, and so forth. Okay, now what we're going to do is, this is the configuration that we're starting with. Let's shift it slightly and see if we can find the configuration that doesn't change W. That's the strategy here in this derivation. We're going to shift it infinitesimally and we're going to maintain the energy constant while we do that. How are you going to possibly do, well, here's what you do. Energy on this axis here. Here's our state, N, M, L. Here's the energy difference between M and L, the energy of M minus the energy of L. That's the difference between M and L. Here's the energy difference between N and M. It's just that difference right there. Graphically, this is our situation. There's a bunch of states below this and a bunch of states above this, okay? So now what we're going to do is we're going to shift. We've got some number in state N, some number in state M, some number in state L. Now we're going to shift. We're going to move Q guys from M into N and P guys from M into L. We're going to choose P and Q so that the total energy here doesn't change. We're going to choose P and Q so that we maintain the same total energy. We're going to move some guys from here to here and some guys from here to here. I think you can see because we're going, we're moving some to lower energy and we're moving some to higher energy. It's reasonable that you could maintain the same total energy, right? Yes? No, they're not supposed to be evenly spaced but thank you for asking that question. Okay, so what concerns energy? Well, Q times that difference N minus L, N minus M rather. Q times that difference. Minus P times that difference has to be equal to zero so we're going to choose P and Q so that they satisfy this equality right here. Now, so before, here's our number of microstates. We already said they're given by this equation before the shift, all right? This equation is true. After the shift, I've added some number of molecules to L, I've removed molecules from M, moved them to P, moved P to L and Q to M, all right? And N got larger by N amount Q and so this is the shift here. It's described by these three equations right here. All right, and what I'm saying is that we want to consider the case where the before equals the after. But we want to find the microstate where after we do this infinitesimal shift, W doesn't change. That's the whole point. All right, so if the rate of change of W with configuration zero, that means that these guys have to be equal to one another. In other words, write the peak of this curve, W before equals W after, if and only if we are right there. If we're here and we shift in either direction, it's not gonna be true. W changes with configuration, right? It only doesn't change with configuration up there. So that has to equal that. And if you work out the math, we're gonna skip some steps. Canceling and rearranging terms. You don't have to do this on quiz one. Take my word for it, lots of cancellations occur here, and when you're done, this equality holds. But a more useful version of this is if you just take the natural logarithm of both sides here, you get this, right? P times log of NL over NM equals Q times log NM over NN. Does it look super useful? Okay, and substituting from the expression that you defined earlier, we have, right? I plug this in for Q, I plug this in for P, and I get these guys right here after actually I cross multiply. And then you've got this equation right here that looks completely useless, but what you have to realize is that M, M, and L, M, and N are just random letters. I could have chosen any three states and done this manipulation, okay? And so it turns out to be the case that these two things can only be equal to one another if I permutate these states, right? If it's the case that I could have used A, B, and C, or X, Y, and Z and gotten the same result, that implies that this has to be a constant, right? Or this has to be a constant. And that may not be intuitively obvious because it's really not intuitively obvious to me, but it turns out to be the case. This can only be true if the left and right sides of this equality are a constant. They do not depend on exactly which states are involved, which states we chose. We chose L, M, and N. Let's call this constant beta. In other words, I'm gonna take one of these guys and just set them equal to a constant. We're gonna call that constant beta. Okay, so now I can rearrange this. I can multiply by that energy difference. Boom, moves over to the right-hand side. So I've got log of N of N over N of I. I chose just N and I now to label these states. Completely general expression. It doesn't matter which two states you choose, right? Subject to the constraint that that's the energy difference between the two states right there. That's the only constraint. Okay, now if we choose one of these guys to be the ground state, let's say that guy, N sub I, right? So instead of N sub I, we'll have N sub zero. Let's assume that the energy of the ground state is zero. In other words, that there's no zero point energy. Okay, and so instead of having E sub I minus E sub N, all right, we're just gonna have minus beta times E sub N because E sub I is equal to zero, all right? So we get this expression here and if we just take E to the both sides, take the exponential of both sides, rather, we get this expression right here. This is the famous Boltzmann distribution law. What it says is the ratio of the populations of these two states, state N, which is elevated by energy E above the ground state, right? The energy of the state is E higher than this ground state. All right, we can calculate the population of that excited energy level using this equation right here where beta turns out to be equal to one over KT, where K is Boltzmann's constant. Is that on the next slide? Yeah. Beta is one over Boltzmann's constant times temperature. One over KT. Leonard Nash, who is my hero, says, for any isolated microscopic assembly of species with any kind of spacing between the states, any kind of state spacing, predominant configuration has been fully defined by an explicit functional relation between the energy and the population of each quantum level. In other words, this equation holds true not for every configuration, right? For the only one that matters, right? The one that dominates the number of configurations that are out there. Remember, when you have a large system, it's dominated by one configuration or a very small number of them, right? What we're saying is that this equation here characterizes that preferred configuration. That preferred configuration is very important. Any chemical system is dominated by it. But there are all these other ones that we're really gonna ignore. And we ignore them in all thermodynamics. At least all of it that I know about. Okay. Now, we also need a different form of the Boltzmann distribution law. And you can look at these slides later and there's a little derivation of that here. The total number of molecules, the molecules in the ground are stateless. The number of molecules in the first is excited stateless. Still the molecules in the second is excited state. We can write a summation formula in terms of that. There it is. These are summations over all of the states that are occupied. And so we can write one of these Boltzmann distribution laws for each one of these terms. Look at that. Isn't that nice? All right, N2. If that's N2, then that's gotta be E2. Yep, sure enough, there it is. All right, N2, E2, N0, yes. Let me just click through this. So solving for N0, we have this equation right here. And I can then plug this N0 into this N0 right here. Nope, let me do that backwards. Let's put solve for N0 here, plug that into this guy. I get this equation right here. This is your equation, 13.68. You don't actually have to do this derivation yourself. I'm just showing you where this equation came from. It's not totally obvious where this thing comes from. You can just derive it from the simplest version of the Boltzmann distribution law. All right, what matters is how we use this equation. This thing in the denominator looks innocuous, but it turns out to be super important. All right, it's something called the partition function. And we're gonna keep coming back to this. As innocuous as this seems, this is the central idea that we keep returning to in statistical mechanics. What is that thing? It doesn't look like anything in particular, does it? It's the summation of this exponential where we're gonna put a bunch of different energies in here and sum them up. I mean, how important could it possibly be? Okay, so this is the molecular partition function for that particular case that we're talking about where we've got evenly spaced states in a ladder, just the partition function for that case. The molecular partition function, we usually call it Q. There's two versions that turns out. There's a version that contains the degeneracy G. The degeneracy is the number of molecules that have the same energy in a particular state. So if the ground state happens to have four molecules in it, no, no. Let me say that a different way. If there are four ground state energies that all have the same energy of zero, the degeneracy would be four. We'll do an example. Okay, this is your equation 13.9. Two versions of this equation. I never learned anything until we started doing exams. So let's see some examples. So if any of this starts to make sense. What are the relative populations of the states of a two level system when the temperature is infinite? We've got two states that are gonna be very high. What's the relative population of those two states? In this system it's got a lot of molecules in it. This happens to be a case that we have all the time. Here's our equation for the population of some state I. Two states. So let's call one state two and one state one. Let's write one of these equations for state two and one of these equations for state one. What's the difference between them? I've got E2 here for state two. I've got E1 here for state one. Otherwise, the denominator is the same summation. It's identical. Okay, so if I cancel terms that I'm asking about two states in the same system, so big N is the same. There's only one system here. It has N molecules in it. This thing in the denominator, this partition function, that's the same between these two terms. The only thing that's different is this exponential and so I can just pull that out. Everything else cancels. So I've got this exponential for energy two corresponding to N2 and this exponential for energy one corresponding to state one. And so I can simplify that and here's what the resulting exponential looks like. Okay, and remember, beta is one over KT where K is Boltzmann's constant. All right, and so if I want to calculate this, I just plug in one over KT but I said T is infinite. That symbol is supposed to represent. Okay, and so if the denominator here becomes very large, I'm going to have E to the zero. E to the zero is one. So what we conclude is that the population in these two states will be equal. If the thermal energy of the system is way higher than either the energy spacing between them, I think that's already counterintuitive. Wouldn't you expect the excited state to be preferentially occupied at high temperature? I mean, to my simple way of thinking about it, that's what makes sense to me but that's wrong. All right, if the energy, the thermal energy of the system is dictated by the temperature, if that is high compared to the energy difference between these two states, all right, they will be equally occupied. That's an important thing to understand, it turns out, in lots of different kinds of physical chemistry that we're going to be getting to later on. Let's do another example. Certain atom is a three-fold degenerate ground level, the non-degenerate electronically excited level at 3,500 wave numbers and the three-fold degenerate level at 4,700 wave numbers. Calculate the partition function for these electronic states at 1,900 degrees Kelvin. Here's the situation. What do these words mean? A three-fold degenerate ground level, boom, energy zero, an electronically excited energy level at 3,500 wave numbers, boom, 3,500 wave numbers. A three-fold degenerate level at 4,700 wave numbers, boom, right? Calculate the partition function for these electronic states at 1,900 degrees Kelvin. Partition function. What is the partition function? It's this guy. This is the denominator of that expression that we were looking at earlier. I'm using the version, there's two partition functions. I'm using the version that has the degeneracy in it because we've been told about the degeneracy. Here's the partition function. Here's the version with the degeneracy. Here's the version without it. We need this one because we have degenerate states in this system. Okay. So, what does the summation mean? It means that we're going to have to add up, we've got three states, we've got three terms in our summation, ground state, first excited state, second excited state. Let's figure out what those terms are. The degeneracy of the ground state is three, boom, okay? So that's three times the exponential minus beta times zero because the energy of the ground state is zero. Plus, non-degenerate, e to the minus beta times 3,500 wave numbers plus triply degenerate e to the minus beta 4,700 wave numbers, three terms because there's three states. Okay. Now, one of the things that we're going to have to do constantly is move between different energy units in this class, wave numbers, joules, EV. The way that I do this is clumsy but reliable. All right. I remember two constants, the conversion from wave numbers to EV, that's 8065.5, wave numbers per EV and the conversion from joules to EV. There's 1.602 times 10 to the minus 19 joules per EV. If you remember those two conversion factors, I submit to you that you can obtain any unit that happens to be a joule EV or a wave number. For example, 3,500 wave numbers. What is that in joules? Why do I care? Because Boltzmann's constant is almost always 1.381 times 10 to the minus 23 joules per Kelvin. Boltzmann's constant is always in joules. So other energies, if they're in joules, it's very helpful. Because then you don't have to know Boltzmann's constant and different units. Always use 1.381 times 10 to the minus 23 for Boltzmann's constant and change the units of the other stuff. So 3,500 wave numbers, what is that? 3,500 wave numbers divided by 80. You know, go back to your camp one training and make sure that you cancel units and get the right ones. All right? Because we're going to be moving between energy units all the time. And this is the way I do it, right? Divide by 80, 65.5, wave numbers cancel. So that product gives me energy in EV. Then I multiply by 1.602 times 10 to the minus 19 joules per EV. Now I get energy in joules. And then I divide by kT. And I get 2.649. 2.649 is that number right there. There will be a minus sign in front of that right there. All right? And so these three terms then are going to be 3 because that's just e to the 0 times 3. So that's 3 times 1. That's 3. That guy turns out to be .07069. That guy turns about to be .0855. Now I add them up. I get 3.156. Does that make sense? What is the partition function? It's the number of thermally accessible states. How much thermal energy is there in this system? What is kT? kT turns out to be 1,300 wave numbers. If you convert that number, 1,900 kelvin, convert it into units of wave numbers. It's 1,321 wave numbers. All right? And so what this calculation tells us is that there should be a little more than three thermally accessible states in this system. Well, there are. These three states are thermally accessible. They're the ground states we're going to state. All right? And there's a tiny probability that these other states will be occupied. That's what this number indicates. 3.1 thermally accessible states. Yeah, because there's not enough thermal energy to significantly populate these excited states. Thermal energy is too low. Okay, so we'll review this on Friday.