 So, for the harmonic oscillator, we started with the assumption that the potential energy looked like quadratic, one-half kx squared. We were able to solve the quantum mechanical problem and find the energy levels of the molecule. Those are equally spaced energy levels for the harmonic oscillator. Equal gaps between everyone. Those energy levels we can use to obtain a partition function. That partition function, we call that the vibrational partition function because this is describing the vibrational motion of these diatomic molecules that we're trying to study. That had this particular form, and that's where we've gotten to so far. The next step, as in previous examples, like the rigid rotor and the particle in a box, will be to use that partition function to derive some thermodynamic properties. So we're about to learn the properties of the vibrational motion of harmonic oscillators, diatomic molecules that we can treat as harmonic oscillators, such as the energy or pressure and so on. So to do that, remembering that these harmonic oscillators are, we're using this model to describe diatomic gas molecules, molecules like carbon monoxide or H2 or N2. Those molecules, as gas molecules, we typically are interested in not just one of them, but a whole box full of many identical gas molecules. So we can write the partition function for a whole collection of identical and indistinguishable molecules as 1 over N factorial times Q to the N for a collection of N molecules, and so identical and indistinguishable harmonic oscillators. And that's enough to start using the thermodynamic connection formulas. For example, if we want to know the energy of a box of N gas molecules, the diatomic molecules, specifically the energy that's due to their vibrational motion, that's going to be kT squared times the temperature derivative of the log of this partition function for that collection of molecules. So we can take that step by step. That's going to be kT squared times the temperature derivative. The log of this quantity, so I want log of Q, so log of 1 over N factorial is minus log of N factorial. Log of Q to the N looks like N log Q. So that's the temperature derivative I want to take. In order to do that, we need to think about what the log of Q is going to be. So let's go ahead and write down. So I've got kT squared, temperature derivative. I'll hold off on taking the derivative for just a second. Temperature derivative of this minus log N factorial plus N times log of little Q. Little Q is this quantity here, so I need to think about this log. Log of this fraction is going to be the log of the numerator minus the log of the denominator. Log of the numerator is easy. Log of an exponential just pulls down the exponent. So I've got the log of e to the minus h nu over kT. Actually, let me pause here and say, when we first derive this partition function, we derived it in terms of these energy gaps being h times the fundamental vibrational frequency. We've since chosen to write h nu divided by k as the vibrational temperature, which is often more convenient. So in particular, in this case, I'm going to collect those constants h times nu divided by k and call those the vibrational temperature. And this equation gets a little bit simpler when I go to take derivatives of it. Notice carefully that there's a 2 in the denominator on the top, but there's not a 2 in the denominator on the bottom. So I want to take this derivative of the log of little Q, log of the numerator, log of this e just gives me this exponent. So I've got minus theta over 2T, minus theta vibrational divided by 2T. And I've also got n times log of the denominator. So there's a minus sign because it's log of the numerator minus log of the denominator. n from here and then log of that denominator e to the minus theta vibrational over T. All right, so if I haven't dropped a negative sign, that's the temperature derivative we're interested in taking. Let's see what that works out to. And it simplifies a little bit. So the temperature derivative of log n, that one's easy. There's no temperatures involved, so that one disappears. Temperature derivative of 1 over T gives me minus 1 over T squared. So the negative sign kills the one that was already there. And I can write that as, so I've got factors of n and theta and 2. I've got 1 half n theta and a 1 over T squared. So I've got the 1 half n theta, 1 over T became 1 over T squared, with a negative sign that killed that negative sign. Next I need to take the temperature derivative of this term. This one I'm going to have to be careful and make sure I don't make an algebra mistake. Derivative of log is 1 over. But chain rule says I also have to include the derivative of the argument of that log. So the derivative of 1 minus this exponential is just minus, derivative of an exponential is an exponential. So I've also got the derivative of minus e to the minus theta over T. But chain rule again, derivative of an exponential is the exponential multiplied by the derivative of the exponent. So I need to take the derivative of this one over T. So I've got minus theta vibrational. 1 over T becomes 1 over T squared when I take its derivative. And I think it becomes minus 1 over T squared. So I need an extra negative sign thrown in there as well. And I think that takes care of all of my chain rule. So now let's see what we've got after putting all that together. So I've got this negative sign, cancels this negative sign. I've got a T squared out front that kills this 1 over T squared in the first term. There's also a 1 over T squared in the second term that gets killed by that T squared. So if I rewrite all this in the simplified form, I've got 1 half times n times k times theta vibrational. That's all that survives out of this first term, 1 half n k theta. The second term with this plus sign looks like n times k times, let me go ahead and use this theta from the end so that my k's and theta's appear next to each other. And then I've got this fraction of an exponential e to the minus theta over T divided by 1 minus e to the minus theta over T. All right, that's simplified a little, but we can make it a little bit cleaner still. Make it make more physical sense by doing two different things. Let's say, so this is the vibrational energy that we're talking about. First of all, both these terms involve include an n, and that makes sense. The energy of these n molecules should be proportional to the number of molecules we have. We expect the energy to be an extensive property. If I go ahead and divide by, so this is gonna be my vibrational energy, divided by the number of molecules, calling that an intensive molar energy. Molar vibrational internal energy, so I'm just gonna divide away the n's. I'm also gonna recognize that k times theta, this quantity k times theta is equal to. So h nu is equal to k times theta. So sometimes as when we're doing the math it's more easier to think of things in terms of this vibrational temperature. Sometimes when we want to understand the physical meaning of this equation, it's gonna be easier to think of it in terms of h times nu. So I'm gonna replace some of these kthetas by h nu's. So one half k theta becomes one half h nu, and now maybe you see why I replaced it. And I'm gonna add to that after this n disappears k theta is equal to h nu. Multiply by this exponential, e to the minus theta over t, over one minus e to the minus theta over t. So there's an equation worth putting in a box. It'll take a minute of explanation to convince ourselves we understand what that equation means. What this is telling me is the energy per molecule, the intensive energy due to the vibration of these molecules, is equal to one half h nu. That's actually not a surprise. We know that the molecules have to have at least one half h nu. That's the ground state energy. One half h nu itself is equal to the ground state energy, the zero point energy of the molecules. So they have to have at least that much energy, plus some. And the amount of additional energy they have, at least when we've written it in this form, is equal to h nu. You'll recognize that as the increment of energy it takes to get up to the next level, or to the level beyond that, or again to the level beyond that. So we've got h nu, the zero point energy, plus some number of increments of energy. So this is the increment of energy, and this quantity that I'm multiplying it by must therefore be the number of increments of energy, the number of excitations, the number of levels up the ladder that this molecule has climbed. So the average thermodynamic energy is the ground state energy, plus some number of excitations. Each excitation is this quantity h nu. So that's, now that we know how to look at it, a reasonably intuitive way to think about what the internal energy is. That has given us the energy as one of our thermodynamic properties. We can consider other thermodynamic properties like the pressure. And I will do the pressure because that one is quite easy mathematically. So remember the thermodynamic connection formula for the pressure is not kT squared, but kT, and derivative of log Q not with respect to temperature, not with respect to temperature, but with respect to volume. So luckily we've already done the work of writing down what log Q is. Log Q was this whole thing in brackets. So before I took the temperature derivative, then I took the T derivative. Now I want to take the V derivative. What's the volume derivative of this entire section in brackets right here? We can look through, there's no volumes that appear anywhere inside these brackets. None of these quantities depend on the volume of the system. So that derivative just works out to be zero. So the pressure of this diatomic gas, in particular the pressure due to the vibrations of the diatomic molecules is equal to zero. So if I have a gas of molecules bouncing around in a box, all the pressure is still coming from the translational energy when they move through the box. The fact that they're vibrating as they move doesn't contribute anything additional to the pressure. So that emphasizes that what we know so far on this board anyway is the properties due to the vibrational energy of the molecules. What we can do next is ask ourselves how that combines with what we already know about the translational energies and the rotational energies. And then we'll be in a position to talk about the full thermodynamic properties of diatomic molecules.