 So, we have an expression for the energy levels of a rigid rotor model that's been simplified by the definition of this rotational constant, which is a collection of fundamental constants and properties of the individual molecule that we're talking about. And that has given rise to this energy ladder with its different degenerate levels and different spacing between these energy levels. So, an interesting question is whether these energy levels are large or small compared to kT. So, we would like to know how does delta E between these energy levels compare to kT? And the reason that's interesting is because it tells us whether the energy levels are likely to be populated, whether there's many of the energy levels that are populated or whether only the lowest few of the energy levels are populated. And the reason that's true is because Boltzmann tells us the probability of occupying some state, the quantum number LM state, that's just 1 over a partition function, e to the minus energy of that state over kT. So it's these energies and kT that the ratio between these two that are important. In fact, since we know these states are degenerate, so the formula I've written down here is the probability of occupying a particular state, the LM state, maybe this state where the L equals 2, M equals negative 1 state, or this state that equals 2, M equals positive 1 state, any one of these individual states. More interesting is the probability not of occupying an individual state, but an entire energy level. What's the probability that I'm somewhere on this L equals 2 level without really caring which M quantum number I have, because they all have the same energy as each other, because they're degenerate. And that's going to be the probability of each one of these states is exactly the same. So the probability of the entire energy level is going to be the same as the probability for an individual state multiplied by the degeneracy of that energy level. And that makes sense because we don't usually think about different energies for the LM state. We think about an energy of an entire energy level. All we care about is the value of L, not the value of M, when we're computing these energies. We know what the degeneracy is, that's 2L plus 1. In fact, let me also write the energy, not as E sub L, but as the rotational constant times L times L plus 1, and then I still divide all that by kT. So now we can see that the quantity that tells us whether these energy levels are populated or not is this ratio of the rotational constant relative to kT, because that's what's up here with a small integer multiplying it in the exponent. So if that ratio, rotational constant divided by kT is a large number, then this E to the negative large number is going to be very small, and the energy levels won't be very populated. But if BE over kT is a small number, then E to the negative small number is not that small, and the energy levels might be well populated. So we're particularly interested in this quantity BE over kT. We know what Bayt E is, so the rotational constant is H squared over 8 pi squared mu R squared. If we divide that by kT, that's the ratio that we're interested in. And what we often do, we've already lumped a bunch of these constants and renamed them the rotational constant. We're going to do that yet one more time. What I'm going to do now is now that Boltzmann's constant has entered the chat, we're going to say let's define a whole new constant, and let's say let's let H squared over 8 pi squared mu R squared with the Boltzmann's constant. So I'm going to take all of these constants that I could know ahead of time before knowing what temperature I'm asking a question about, lump all those constants into one, give that thing a new name. We're going to call that capital theta with a sub rotational, and we're going to call that the rotational temperature. We'll see in just a second why that's a reasonable name for that quantity. So notice that most of those constants, H squared over 8 pi squared mu R squared, most of those were already present in the definition of B sub B. I've just divided by an extra k. So if you want, you can think of the rotational temperature as the rotational constant that we've already defined divided by one more factor of k. And once we do that, B E over kT just becomes theta rotational over T, because I've lumped everything in this circle into one constant. I just have theta rotational divided by T. And I can rewrite the probability 1 over q times the degeneracy times E to the minus theta rotational over T times L times L plus 1 that I haven't lumped into the constants. So I've simplified this expression up in the numerator a little further. Now it just looks like theta rotational over T. So to get an understanding of what this rotational temperature means and how it's different than the rotational constant, that is the number that we multiply some integers by to get the levels of these energy levels, let's calculate a value for a rotational temperature. We've already asked a few questions about carbon monoxide. So let's continue with carbon monoxide. If I say carbon monoxide has a rotational constant of what is the value for carbon monoxide? It's 3.82 times 10 to the minus 23 in units of joules. Then what is its rotational constant? That's relatively simple math. The rotational constant from the rotational temperature from the rotational constant I just divide by Boltzmann's constant. So take this quantity in units of joules divided by Boltzmann's constant, which is also about 10 to the minus 23 joules per Kelvin. That ratio, so 3.8 over 1.4, that works out to be a little smaller than 3. Specifically, it works out to be 2.77. And the units, joules will cancel one over one over Kelvin, leads me with units of Kelvin. So now you can see why we call it a rotational temperature, because the units on this quantity work out to be units of temperature. Anytime I divide an energy by Boltzmann's constant, I'm going to get units of temperature. It also illustrates the reason we prefer to talk about the rotational constant many times. I'm sorry, prefer to talk about the rotational temperature many times. When I give you the value of the rotational constant, 3.82 times 10 to the minus 23 joules, other than saying 10 to the minus 23 is a very small number, you don't get much of a sense of the scale of this, whether it's big or small compared to kT, whether it represents a lot of energy or a little bit of energy for something as small as a molecule. But when I tell you 2.77 Kelvin, you immediately have a sense of what that number means. Room temperatures, temperatures that we're used to dealing with are up in the hundreds of Kelvin, 300 Kelvin or so. So 2.77 Kelvin immediately sounds very cold. You have an immediate intuitive understanding that this number is quite cold compared to room temperature. So you know, without thinking about it too hard, that this rotational temperature is quite small compared to temperatures that you're likely to be studying in a chemistry class. So what does that mean if the rotational temperature is very small? That's the same thing as saying the gaps between these energy levels are very small compared to kT. If I just multiply both of these by a k, k times theta is equal to the rotational constant, which is similar to these differences in energies. And if I multiply by k on the right side, I get kT. So anytime the rotational temperature is small compared to room temperature, that means the gaps between these energy levels are equally small, comparably small, compared to the thermal energy that's available at room temperature. What that means is for molecules that are at this temperature and have this much thermal energy, they can access many, many of these energy levels. So that means many of these states are occupied. On the other hand, what if the opposite is true? What if theta rotational isn't much, much smaller than the temperature? Maybe it's bigger than the temperature or in the vicinity of the temperature. That means, likewise, the gaps between these energy levels are gonna be big or comparable to kT. So if kT represents the size of these gaps between the energy levels, then I can't climb many, many of these energy levels higher. Boltzmann tells me that's gonna be progressively less and less likely as I climb the ladder. So that means relatively few states are gonna be occupied. So what it's telling you when we see that the rotational temperature is a cold number, really what that means is room temperature is quite hot compared to the properties of this molecule. And therefore, because room temperature is hot compared to that number, we can access many, many of the states. So quite a few of these states are occupied. The other molecules under other conditions might work out a different way. So now that we've started talking about how many states are occupied, of how many of these states are effectively occupied under a given set of conditions, that's beginning to sound like partition function. Remember one of the meanings of the partition function is telling us how many states are occupied. So we've begun to relate the properties of this rigid rotor molecule to something that sounds close to a partition function. So a natural thing to do next would be to calculate the partition function for the rigid rotor. So that's what we'll do.