 We take a step back and remind ourselves what we've done for the rigid rotor so far. We started out by saying our rotating rigid object has no potential energy. We'd only expect its potential energy to change as it rotates. That was enough plus use of spherical polar coordinates to derive some wave functions for the rigid rotor and the energy levels for the rigid rotor. We can use those energy levels to obtain a partition function. And then the next step, which we haven't done yet, is to use that partition function to calculate various thermodynamic properties. We can calculate the energy, the pressure, the entropy, anything for which we have a thermodynamic connection formula. We can calculate thermodynamic properties from the partition function. So we do in fact know what that partition function is for the rigid rotor, at least under classical conditions. It takes this simple form. When the temperature is large compared to the rotational temperature of the diatomic molecule, it takes this form for heteronuclear diatomics. Or remember, if we want to allow homonuclear diatomics, we need that symmetry factor in the expression as well. So that's our task right now is to use that partition function and some thermodynamic connection formulas to calculate properties. And we could do that with our small q partition function. That would tell us the thermodynamic properties for a single molecule of a gas. As usual, when we talk about gases, we're more interested in a collection of many molecules. We're talking about n molecules of a gas that are all indistinguishable from one another. Then the partition function for those n identical and indistinguishable molecules would be little q to the n divided by n factorial since they're indistinguishable. And we can, so I've got 1 over n factorial t over theta raised to the nth power. And I can either include the symmetry factor or not. If the symmetry factor, I won't include the symmetry factor. It won't end up changing our result. But if the symmetry factor is included, it would show up in here as 1 over sigma to the n as well. So what do we do with that? If we want to know a property like the internal energy, then we need to remember what the thermodynamic connection formula is that connects the partition function to the energy. And that connection formula tells us just take d log q dt multiplied by k d squared. That'll give us the internal energy. So I want to know what is the temperature derivative d dt of log q. So log q is a product of multiple terms raised to some power. So log of this partition function, log of 1 over n factorial is minus log of n factorial. Log of t to the n is n log t. Log of 1 over theta to the n is minus n log theta. So I need the temperature derivative of everything here in brackets. There are not many temperatures here. Only the log t involves the temperature. All the other terms don't involve t. Theta rotational is called a temperature, but that's just a constant. That's a number that describes an individual molecule. So the derivative of that constant is 0. The derivative of n with respect to t is 0. The derivative of everything but this log t is 0. So this derivative is, I've got a kt squared out front. And then the temperature derivative of log t is 1 over t. And I've got an n multiplying that term that I need to include as well. So this 1 over t cancels one of these factors in the t squared. And I've left with, let me put the n out front, n times k times t. So it turns out the internal energy for my box containing n, identical and indistinguishable diatomic molecules. This energy is n kt. The, if I want an intensive energy, divide by the number of molecules. And I get kt, or if I prefer, rt since k and r are the same quantity. Let me point out that the energy that we're talking about, this is the rotational energy of the molecules, because we're talking about the rotational partition function. What I mean by the rotational energy will be more clear in just a second. So our answer is that this energy per mole is kt or rt. We've already seen, let's say, we already know something about monatomic ideal gas molecules from when we studied the 3D particle in a box. We got a similar but not identical answer before. For monatomic ideal gases that we described with the 3D particle in a box, we found that the internal energy was 3 halves rt. Now we found rt. The reason for the difference between these two and the reason I'm saying monatomic ideal gases for what we described with the particle in a box, let me draw a small cartoon here of some gas molecules in a box. If I have monatomic gases like argon or neon or helium that just bounce around in this box, the only type of energy they have is kinetic energy. The only type of kinetic energy they have is this translational energy as their center of mass moves around the box, as the atom itself moves around the box. On the other hand, if I have a diatomic molecule like N2 or O2 HCl, those molecules can also translate and bounce around the box, but they can also rotate. So the energy that these molecules have is one portion that's translational energy that comes from the particle in a box, and another portion that's rotational. And that's the part we've just figured out the answer to. So the rotational energy in particular, the energy that we get by describing the molecule as stationary but spinning, is rt. So a diatomic molecule has an internal energy not of 3 halves rt, but of this 3 halves rt from bouncing around in the box, translational energy, and an extra factor of rt from its rotational energy, so a total energy of 5 halves rt. So now, finally, these expressions are beginning to agree with what you may have learned in a previous chemistry course, like general chemistry. The energy of a monatomic ideal gas is 3 halves rt, diatomic ideal gas is 5 halves rt. If we also calculate the heat capacities, constant volume heat capacity, remember, is du dt. Just take the temperature derivative of these expressions and the temperature goes away. We've already known that the heat capacity of a monatomic ideal gas with the 3D particle in a box tells us that's 3 halves rt. Now we find that for a diatomic ideal gas, it's 5 halves rt, 3 halves of which comes from the translational energy, particle in a box energy, one factor of r of which comes from the rotational contribution to the energy. We also know about a monatomic ideal gas that pressure times volume equals rt. So the energy and the heat capacity both changed when we started including rotational motion. So let's see what happens to the pressure of an ideal gas when we include this rotational motion. And to do that, we need to know, we need to back up to this portion, this step, and begin with another thermodynamic connection formula. This was the thermodynamic connection formula for the energy. The thermodynamic connection formula that lets us calculate the pressure is similar. Not kt squared, it's kt. Not d dt, but d dv, the volume derivative of log q. So I need to take the volume derivative of the same log q, the same quantity in brackets that we have discussed before. So that's minus log n factorial, maybe a log of a 1 over sigma to the n, which I'm not including, a plus n log t minus n log theta. Theta, OK. So I need to take the volume derivative of this quantity in parentheses. And that's even easier than the last derivative. There's no volumes that appear in here anywhere. The number of molecules in the gas doesn't depend on the volume. The rotational constant of the gas doesn't depend on the volume. None of these quantities depend on what the volume is. So the volume derivative of this quantity is 0. So the pressure of a diatomic gas is 0. Does that make sense? It does if you remember that this is the rotational contribution to the pressure. This is the extra contribution to the pressure that comes from the fact that this molecule is rotating. The pressure, the force per unit area we feel on the wall when the gas collides with the wall, that all comes from the translational energy. The fact that it's rotating as it collides with that wall doesn't affect the pressure. It may. Here's a diatomic molecule about to collide with the wall. If it happens to be spinning in this direction, it might have a little bit extra kick as it hits the wall. But equally often, it's spinning the other way, and its kick is a little bit reduced as it hits the wall. So on average, the contribution to the pressure from the rotational motion of the molecule is 0 as the thermodynamic connection formula has told us. So for a diatomic molecule, it's also true that PV equals nRT, pressure times molar volume is equal to RT for a diatomic gas, which has this rotational contribution as well. So the rotational contribution has changed the energy and the heat capacity. It has not changed the ideal gas law, the ideal gas equation of state. So this tells us about the rotational contributions to the energy, heat capacity, and pressure. We haven't yet talked about the vibrational contribution. So in addition to rotating, this molecule, a diatomic molecule, also has a vibrational degree of freedom. The bond length can change. We've ignored that bond length when we talked about a rigid rotor. But the next step will be to continue this analysis and do some more quantum mechanics, feeding eventually into thermodynamic properties and see what affect the vibrational properties of the molecule has. So that's coming up next.