 So at this point, we've seen what the heat capacities look like, the constant volume molar heat capacities due to translational motion, which we obtained from the 3D particle in a box model. That worked out to be 3 halves r. We've seen the constant volume molar heat capacity for rotation of a diatomic molecule that we examined with the rigid rotor molecule. That worked out to be r. And now, most recently, we've seen that the vibrational contribution to the constant volume heat capacity is either 0 at temperatures that are cold compared to the vibrational temperature of the molecule, or perhaps r if we're at temperatures that are considerably larger than the vibrational temperature of the molecule. And in between, it makes a gradual transition from 0 to r. So looking at these three heat capacities next to each other points out that something strange is going on here. Each of these heat capacities looks like a multiple of r, or a multiple of some ratio of small integers times r, some number of 1 halves r. So that's a little bit mysterious. And it seems like there must be something going on that causes that to be the case. And in fact, that's true. And the reason for this behavior, the fact that all these heat capacities work out in multiple of r, or 1 half r, is because each of these three models turns out to have had something in common that we didn't really notice when we were talking about them the first time through. To illustrate what that is, let's just remind ourselves some details about those models. The 3D particle in a box, so we were modeling diatomic molecules, let's say, the translational motion. So here's a diatomic molecule. If it's translating, then all we're modeling is just that molecule moving through the air in some direction, maybe the x or the y or z direction. It has only kinetic energy, no potential energy. And that kinetic energy, classically, we think of as 1 half mv squared. The rotational energy of the molecule on the other hand, if I redraw the diatomic molecule and think about what it looks like when it's rotating, that also has kinetic energy, but no potential energy. So the rigid rotor has kinetic energy. We could think of that in Cartesian coordinates. We could think of the velocity of this atom and the velocity of this atom each moving in different directions to make the analogy with this equation more clear. If you remember what you may have learned in a calculus, I'm sorry, a physics 1 class, you can describe the kinetic energy of a rotating molecule, specifically a rigid rotating molecule or object as 1 half its moment of inertia times the angular momentum squared. And you'll notice that these two equations look very similar. Don't worry too much if that's not a familiar equation from physics. The point is that the kinetic energy has the same form of some constants times something squared, whether it's a velocity or an angular velocity. For the harmonic oscillator, we also have a kinetic energy as well as a potential energy. So remember, we started the harmonic oscillator by saying the potential energy looked like 1 half kx squared, a quadratic potential energy. And I'll draw a cartoon of that as well. So here's the diatomic molecule, the vibrational motion of the molecule. The two atoms are either moving apart from each other or they're moving towards each other. The kinetic energy of that motion, when we represent the diatomic molecule in reduced coordinates, remember we have a reduced mass and the bond displacement. So the kinetic energy again has the same form of some constants times the variable squared. The potential energy, actually this should be an x dot squared. It's the velocity of the bond. We can call that 1 half times mu times v squared if we like. So these are all types of velocities. The potential energy also looks like some constants times something squared. So it turns out all of the energies of all these three different models are quadratic. They all have the form of a variable squared. And that's going to turn out to be the key behind why the heat capacities all are a multiple of r. So let's ask ourselves the general question. What if we had, well, first let's remind ourselves how we got heat capacities from statements of the energies of the system. We started out by doing some quantum mechanics. If we define what the potential energy of a system is, we can do some quantum mechanics to get wave functions and energies, do some statistical mechanics to get partition functions, use some thermodynamic connection formulas to get variables like internal energy, and then heat capacity is the temperature derivative of the internal energy. So that's the general process we went through. We went through a fair amount of work for the particle on a box, and then the rigid rotor, and then the harmonic oscillator, ending up with a heat capacity at the end in every case. What if we skip those steps and just skip the initial steps and say the energies are all quadratic. Let me just think about what the partition function would look like for a system with a quadratic energy. Partition function, remember, is the sum of the Boltzmann factors, e to the minus energy over kT, summed up over all the possible states of the system. If our energy has the form some constants times some variable squared, so maybe that variable is a velocity or an angular velocity, maybe that variable is a position, but these all appear to have the form of one-half times something that is a little bit like a mass or a force constant, some other constant, multiplying a variable squared. So this lambda is not a variable you've seen before. I don't expect you to know what lambda is. Lambda is just standing in for maybe a velocity, maybe a position, maybe something else. And if we use that for the energies, and if we additionally say in the classical limit, I don't have to sum over each individual energy state if there's a whole lot of different states. If there's a lot of positions, the molecule can exist in a lot of different velocities. If I want to integrate over all of those because the system is behaving classically, instead of quantum mechanically, I'll replace that sum with an integral. I can replace that energy with what we've just decided it's going to take the form of a quadratic, so minus one-half C times some variable squared over kT. And I guess I should point out that this C is not any constant in particular, it might be a mass, it might be a moment of inertia, it might be a force constant. It's not the speed of light, it's just some arbitrary constant that multiplies some variable. So if I want to compute this partition function, what am I integrating over? I'm integrating over all the possible values that this variable can take. The energy is one-half C lambda squared, and I need the value of that energy for any possible value of x, or any possible value of v, or whatever the variable is. That's an integral we can do, that variable, whether it's a position or a velocity, can have any value, any possible value from negative infinity to infinity. And we know how to do integrals of e to the minus variable squared over all values of the variables. That's a Gaussian integral. So that's the square root of pi over the coefficient that multiplies the variable. So the coefficient is a one-half C, and we can't forget that divided by kT. So that's the value of this integral, not the first time we've seen Gaussian integrals in physical chemistry. If I clean that up a little bit, put the two and the k and the T on the top, that looks like square root of two pi kT over this constant C, whatever that is. So this tells me the partition function, anytime I have a quadratic energy, an energy that depends quadratically on some variable, the partition function is going to be square root of some constants and a temperature. What I can do with that partition function is use it to find the internal energy. That's kT squared d log q dt, or kT squared temperature derivative of the log of this q that I've just calculated. Log of q, so q is some constants and a temperature raised to the one-half power, that's what the square root is. The log of it is going to be one-half times the log of the stuff inside the square root. I will go ahead and break that up. Log of two pi kT over C, I'll write that as log of two pi k over C and log of T. Looking ahead to the fact that in a minute, I'm going to be taking the temperature derivative, so I've just split the temperature apart from the other pieces. So I've rewritten log of this square root in this particular form, which makes it relatively easy to take the derivative. That leaves me with a kT squared. Temperature derivative of this first term is zero by design. That's where I put all the terms that don't contain a temperature. Temperature derivative of one-half log T is just one-half one over T. And after some cancellation, I get one-half and a k and one T survives. Or if we prefer to use r instead of k, that's one-half times r times T. So I'll pause here and point out something fairly remarkable. Regardless of what our particular energy is that we're interested in, whether it's kinetic energy or potential energy, what the variable is that it depends on, what the constant is that it depends on, the partition function did depend on that quantity. If this constant C is a mass or a reduced mass or a force constant, that affects the partition function, but somehow that constant has completely disappeared by the time we get to the internal energy. The internal energy is gonna be one-half rT regardless of whether we're talking about kinetic energy, potential energy, rotation, vibration, translation, and so on. Likewise, the heat capacity, if I take the temperature derivative of the internal energy, du dt, the T goes away and I just have one-half r. So that's a pretty significant result that clearly begins to explain why our heat capacity has always looked like either three-halves r or two-halves r or two-halves r or zero-halves r. The heat capacity is always a multiple of one-half r because the energies that we're talking about are always a collection of some quadratic energies of some sort and each of those quadratic energies gives rise to a contribution of one-half r. So that fact is known as the Equipartition Theorem. And if I phrase that Equipartition Theorem in words, what it says is that quadratic degrees of freedom, quadratic variables, like the velocity when we're talking about translation or like the position when we're talking about covalent bond vibration, each one of those quadratic degrees of freedom that shows up in the energy, actually let's say it slightly differently, each degree of freedom, what I mean by a quadratic degree of freedom is a degree of freedom that contributes quadratically to the energy, like any one of these terms, that degree of freedom will also end up contributing one-half r to the heat capacity. I'll just write that as CV. One-half r is what we get from each one of these quadratic degrees of freedom, as long as this step was okay, as long as we can convert the sum to the integral, so we'll say in the classical limit. So that's a statement of the Equipartition Theorem. If we have a degree of freedom that contributes quadratically, so that's important to the energy, it contributes exactly one-half r to the heat capacity, and so all we have to do to be able to predict what the heat capacities are is to count up how many of these quadratic degrees of freedom there are. Before we do that, for a few examples, we'll pause and make sure that we understand what we're talking about when I talk about degrees of freedom and how to identify those degrees of freedom in a molecule.