 So for the partition function for rigid rotor, we've managed to understand what the answer should be in half of the cases. So the partition function, which looked like an infinite sum, in the classical limit, we can simplify that quite a bit and it works out to be just temperature divided by the rotational temperature. But under conditions where we can't assume the classical limit, we don't necessarily know yet what the partition function should be. So it turns out we're left with no option, really, other than to compute the sum, which sounds like bad news because it looks like there's an infinite number of terms in that sum, but it doesn't actually turn out to be all that bad. So to show you what I mean by that, let's work an example. So now instead of doing carbon monoxide, let's take a molecule that's a little bit more quantum mechanical, HCl. We'll see in a moment why that's a more quantum mechanical molecule. And let's do it at a temperature that's quite a bit lower, so quantum effects become more important. 77 Kelvin is a temperature that sounds very cold, but it's not terribly difficult to reach. That's liquid nitrogen temperature. So if we take some HCl molecules and cool them down with liquid nitrogen, then we can ask ourselves what the partition function might be. To answer that question, we'll have to know things like the rotational temperature. So we would have to compute first the reduced mass of an HCl molecule, mass of H, mass of chlorine divided by the sum of the masses of hydrogen and chlorine. Now skip the arithmetic because we've seen examples of these types of calculations before, but 1 gram per mole times 35 grams per mole for chlorine divided by 36 for the sum. That works out to be about 0.97 grams per mole or some small number of kilograms. I'll just point out again that the mass of the reduced mass for HCl is smaller than either the atomic masses of either hydrogen or chlorine. It's smaller than either of the atoms. The fact that it's very close to the mass of hydrogen means it's this object with the reduced mass spinning at a bond length of, and I should tell you the bond length, the bond length for HCl, we would need to complete this problem. That's 1.27 angstrom. So an object with mass mu spinning about the origin is behaving an awful lot like a bare hydrogen atom spinning around the origin. Because in an HCl molecule, the chlorine is very heavy, the hydrogen is very small, the center of mass is in fact quite close to the chlorine, and the rotating HCl molecule is pretty much a hydrogen atom spinning around the chlorine. It's actually something a little bit lighter than a chlorine, sorry, a little bit lighter than a hydrogen spinning around at this distance from the origin. So with that information and the bond length, we could calculate the rotational constant. Now that we know mu, I'll skip the arithmetic and tell you that that works out to be 2.11 times 10 minus 22 joules. We're almost to being able to calculate the rotational constant, I'm sorry, rotational temperature, which is just this value divided by Boltzmann's constant. And again, skipping the arithmetic, that works out to 15.4 Kelvin. So the rotational temperature of this HCl molecule is 15.4 Kelvin. That's the number that we can stop and think what it means. What does it mean? Is that hot or cold? That sounds like a small number, 15.4 Kelvin. It's a little larger, higher temperature than we saw previously for carbon monoxide. So that means it's more quantum mechanical, largely because its reduced mass is much smaller than carbon monoxides was. But notice that 15.4 Kelvin, although it's less than 77 Kelvin, it's not much, much less than 77 Kelvin. It's no longer multiple orders of magnitude, less than 77 Kelvin. So it's in the same ballpark as 77 Kelvin. So maybe we're not comfortable using the classical approximation and saying that the system is very, very hot compared to the rotational temperature. So instead, we're left with the task of saying, let's go ahead and calculate our rotational partition function directly from this sum. In other words, first let's do the L equals 0 term, then the L equals 1 term, then the L equals 2 term, and so on until we get to the end of the sum. So let's go ahead and start doing that. When L equals 0, the degeneracy, 2 times 0 plus 1, that's just 1. And that's multiplying an exponential e to the sum stuff times 0s. So e to the 0 is just itself 1. So that first term is going to be pretty easy to calculate. The next term is where it gets more complicated. The L equals 1, so this is the L equals 0 term. Next we're going to do the L equals 1 term and so on. When L equals 1, 2 plus 1 is 3. The degeneracy is 3. The Boltzmann factor looks like e to the minus 1 times 2 times theta rotational over t. Next term, the degeneracy is 5 and then it's e to the minus 6. The next term will be e to the minus 12 and so on. So e to the minus 6 theta rotational over t and then 7 e to the minus 12 theta rotational over t and then the bad news is we have a plus dot dot dot that in principle goes on forever. But if we begin to stick numbers in, now that we know what theta rotational is, 15.4, we know the temperature. So this ratio, 15.4 over t, if we plug these numbers in, the first term 1 times 1 is 1. The second term I've got 3 times e to the sum, pretty small number that works out to be 0.67. The next term will be 5 times an even smaller number because of this e to the minus 6 has become more negative than e to the minus 2. So this even smaller number, that's 0.3 and then the next number, the Boltzmann factor is 0.9 and so on. So you can begin to see what's happening here. These Boltzmann factors are becoming smaller and smaller essentially because the probability of occupying each one of these states is getting smaller and smaller as we climb the energy ladder. Since we're not under classical conditions, those numbers are getting small pretty fast. So we don't have to include too many of these terms before they get small enough, we don't have to worry about them. In fact if I go ahead and multiply out these out, 3 times 0.67, 5 times 0.3, I think with rounding that actually comes out to 1.51 and then 0.63. The next few terms look like 0.16, 0.03 and now it does indeed look like we won't have to include too many more of these terms before they're so small that they're not actually contributing anything. So the good news is this infinite sum is infinite but most of the terms are so small that we don't need to worry about them and only the first few terms in the sum are important enough to include. So if we calculate the first 10 or so of these terms, that's all we need and it turns out that the partition function works out to be about 5.3 and we can get that without doing an infinite number of terms but just a handful of them. So that's the good news, in fact that turns out to be typical. When we're not under classical conditions, when we're under quantum mechanical conditions, by definition because we're under classical conditions and this temperature is not very large compared to the rotational temperature, these numbers will come small quite quickly and will always reach a point relatively quickly where we can terminate the sum. So what does that mean for the partition function? First of all, the fact that we got a partition function that was about 5, that confirms our expectation that we're under quantum mechanical conditions when temperature is not that large compared to the rotational temperature and we're quantum mechanical, not very many of these states are occupied. In fact, the partition function tells us that the effective number of states that are accessible is about 5.3. So not just, not much more than these first few states are occupied, essentially these numbers tell us that the L equals 1 states are only about two thirds as occupied, 0.67 as occupied as the ground state, these states are only about 30% as occupied, these states are only 9% as occupied and when I add up all those fractional occupations, I get an effective number of five or so states that are accessible, counting these fractional occupations. It also confirms our expectation that we couldn't have used this more simple formula. In fact, if we had, so we'd expect that the rotational temperature is not exactly T over theta rotational, not what we would have gotten if we used the classical approximation just to double check and see if that's true. If we had said 77 Kelvin divided by 15.4 for the rotational temperature, so that works out to be not 5.34, it actually works out to be 5.0. So that's actually not too far off, we'd have only made less than a 10% error in the partition function if we had been a little bit lazy and used this classical approximation even though we know we didn't deserve to be able to use this, but we're quantum mechanical enough under these conditions that the correct answer, the full sum of all these Boltzmann factors turns out to be something a little bit different than what the classical approximation predicts. So that tells us what we need to do in order to calculate the partition function If the temperature is quite large compared to the rotational temperature, we can use this rotational, this classical approximation, either the version for heteronuclear diatomics or more generally with a symmetry number if we want to be able to use homonuclear diatomics as well, or if we are afraid we can't use the classical approximation, then there's no harm in just calculating the sum directly when we're under quantum mechanical condition.