 So, the ladder of energy levels for a rigid rotor looks something like this with the states getting further and farther apart as we climb the ladder. Energy levels themselves being written either in terms of the rotational constant or if we prefer the rotational temperature and the degeneracy also increasing as we climb the ladder with this expression giving us the degeneracy of the state of the energy level with quantum number L. So, it's an interesting question to ask what is the partition function for this rigid rotor. Now that we know the energy levels we can calculate the partition function and then later go on and use that to calculate some quantum mechanics, I'm sorry some thermodynamic properties. So, as always the partition function is the sum of the Boltzmann factors e to the minus energy over kT summed over all the possible states of the system. So, in our case the states are all the possible, so the energies of all the possible states of the system, that's every possible pair of quantum numbers L and M. So, I need to sum over not only all values of L, L can go from 0 to infinity, but also all values of M and M can go from negative L to positive L. So, that's the work we have to do in order to calculate the partition function. Sum up all those partitions, all those Boltzmann factors. We can simplify that in one step right away because several of these at each level, I have five different states with all the same energy when L equals two or seven states with the same energy when L equals three. The degeneracy tells me how many of them there are. So, instead of summing over all the different states, I can instead sum only over the energy levels e sub L and just remember to include the degeneracy of that energy level. So, summing over energy levels rather than summing over states. So, let's go ahead and write that out as the degeneracy of two L plus one and the energy, e sub L looks like K times theta times L times L plus one. K times theta, when I divide it by K times t, the K's cancel and I've just got a theta rotational in the numerator, a temperature in the denominator. And I've still got this L and L plus one. So, that's the sum I want to perform over, unfortunately, an infinite number of L's, 0, 1, 2, 3, and so on as we climb this ladder all the way up. So, how do we go about doing that? There's a couple ways we can consider tackling that infinite sum. First, it may be true that the rotational temperature is quite small compared to the temperature. Which means that the energy levels are spaced quite closely compared to Kt. Delta E's are small compared to Kt. That also means that many of these states are occupied and that means we can consider there to be a whole lot of levels that are accessible and we don't have to worry about quantum mechanics too much. Like we're used to in classical mechanics, any energy's okay. There's anywhere we want to be, there's going to be an energy level there. So, that would be the classical approximation. On the other hand, if that's not true, if the rotational temperature is not much smaller than the temperature, then we're going to have a quantum mechanical system and we're going to have to consider only the bottom few levels rather than lots and lots of levels. The good news is the classical approximation, this condition's where theta rotational is much less than the temperature, makes this problem relatively easy because as we've in fact seen for the particle in a box, when we take the classical limit, we can approximate the sum over state 0, 1, 2, 3, and so on with an integral over all possible values of L. So, for case number one, if we're in the classical limit, I can say the partition function isn't this sum from zero to infinity, but it's going to be close to the integral from zero to infinity of the same quantity, 2L plus 1, e to the minus theta over T, L times L plus 1, and I'm integrating over all possible values of L. So that's an integral, and we can in fact do that integral relatively simply. Maybe it doesn't look super simple, but it yields quite easily to U substitution. The inconvenient thing is this L squared up in the denominator, L times L plus 1, that's the same as L squared plus L. So let's let U be L times L plus 1 to see if we can get rid of it. That's the same as saying it's L squared plus L. When I make that definition, dU will be 2L plus 1. The differential of L squared plus L is 2L plus 1, multiplied by dL. So notice 2L plus 1 is exactly the factor we have out in front of the exponential in this integral. This means the universe is watching out for us and wants our calculus to be relatively effortless today. So our partition function with that definition, 2L plus 1 times dL has become dU, e to the minus theta over T. Instead of multiplying L times L plus 1, now it's just multiplying U. So that's an integral we definitely know how to do. The limits on this integral, L used to go from 0 to infinity, U is just take that number times L plus 1. So 0 times 1 is still 0, infinity times infinity plus 1 is definitely still infinite. So the limits are actually unchanged. And when I do this integral, integral of e to the U with some constants out front is 1 over those constants times the same exponential. And I had to evaluate that between 0 and infinity. So I've got minus T over theta rotational. And then when I plug U in first at the upper limit, I've got e to the minus infinity. And then I subtract lower limit e to the minus 0. Minus 0 is the same as positive 0. So this is the same as saying e to the minus infinity is 0, e to the 0 is 1. So the quantity and parentheses is negative 1. That'll cancel the negative 1 out front. And I can say that the partition function is T over theta rotational. This is in particular the partition function for the rigid rotor for rotating molecule that we've treated with the rigid rotor. So I'll go ahead and put that in a box. Let's make it a dashed box for right now because I have some caveats to make about that statement. But first let's say we want to calculate the value of the partition function for a particular molecule. Let's say we want to do what's been our favorite molecule recently, carbon monoxide. If we want to know what is the value of the rotational constant at let's say 298 Kelvin, if the rotational temperature of that molecule is 2.77 Kelvin, which it is, then it's a fairly simple calculation to determine what's the rotational constant temperature divided by theta, 298 Kelvin divided by 2.77 Kelvin. Notice first thing that the units cancel, Kelvin cancels Kelvin and our answer is going to be unit less. And that makes sense because a partition function is just a sum of unitless exponentials or a partition function is a way of counting how many states are occupied and it's just a count, it's not a number with units. So that number when we divide 298 by 2.77, what we get is 108. So what does that mean? Partition function for carbon monoxide treating its rotational motion, its rotational partition function is 108. That means that effectively there's 108 or so states that are accessible to the molecule. Remember that when the temperature is large compared to the rotational temperature, many of the states are occupied. So not just this state and these states and these states, but many states, roughly 108 or so states are occupied at 298 Kelvin. So indeed quite a few states are occupied and our classical approximation was at least somewhat justified. Now back to the reason I put this equation in a dashed box, a few caveats about that molecule. Number one, this equation is only true under the conditions where we've derived it. We assume that the temperature was large compared to the rotational temperature. So we can only use this approximation in the classical approximation in the classical limit where the temperature is large compared to theta rotational, where this partition function comes out to be a pretty large number. Additionally, it turns out this equation is only true for heteronuclear diatomic molecules. Diatomic molecules where the two atoms are different from one another, like carbon monoxide, CNO, or different atoms. If I were to calculate the rotational partition function for O2 or N2 or any homonuclear diatomic molecule where the two atoms are the same as each other, I would find not that the partition function was T over theta rotational, but for those molecules it would end up being temperature divided by twice the rotational temperature. In this case, for the heteronuclear diatomic molecules, it's temperature divided by one times the rotational temperature. So in fact, in general, the more correct formula would be temperature divided by some other number, which we call the symmetry number or symmetry factor, that is one for a heteronuclear diatomic and two for a homonuclear diatomic. For now, I'll just put that equation in parentheses. We haven't seen enough information to see why that's true. That turns out to be true for some quantum mechanical reasons that we haven't covered yet, and that will be coming later on in the course. But for now, I'll just point out that the equation that we've derived so far is only going to turn out to be true for diatomic molecules with both atoms different. If we want to calculate the partition function for homonuclear diatomics, just remember to stick that factor of two in the denominator as well.