In this video, I'd like to employ the tools of statistical thermodynamics that we've developed thus far to try better to understand the concept of the internal energy of an ideal gas. To do that, let me begin with a fundamental postulate of statistical thermodynamics. And that postulate is the following. The observed energy of a given system corresponds to the average energy over all possible states of the ensemble, each weighted according to its respective probability. And so if I were to express that mathematically, I would write it in the following way. Expectation value of the energy, that is an observation, is equal to sum over all possible energies times the probability that energy would be encountered. So it is the average of all the energies probabilistically weighted. And recall that we have a way to express that probability. So I'll just keep the energy here. Here's the energy, emphasizing that it depends on the number of particles in the volume. And here's the probability. E to the minus beta times that energy divided by the partition function. So condensing that and keeping it up on the upper right hand side so we can keep track of it, here is our expectation value of the energy expressed as involving the energy and the partition function. So remember what the partition function is. It is the sum over all possible energy states, e to the minus beta e j, energy depending on number and volume. So, let me do a little bit of elementary differential calculus for a moment. Let me remind you, that if you take the partial derivative of log Q with respect to Q. In this case I guess it's a normal derivative cause I haven't specified that Q depends on anything. But, I wrote it with a partial differentiation sign. But in any case, the derivative of log Q with respect to Q, you recall, is 1 over Q. And now let me take a different derivative, let me take the derivative of Q, the partial derivative of Q, with respect to beta and in thermodynamics we're usually pretty careful to go even beyond just saying partial derivative. We really explicitly specify, by indicating at the lower right hand side, that we're holding the other coordinates fixed, that is the other things on which Q depends, N and V. And so if I go look at the expression for Q and I ask about its dependence on beta, it's e to the minus something times beta, so that something would come down. In this case, it's the energy and you get again, the exponential, that's how the derivative of an exponential works. So there's a minus sign as well, so a minus sign comes out front here. And the reason I wanted to do that was, if I now take the negative of the partial derivative of the log of the partition function, with respect to beta, holding N and V fixed. Well, by the chain rule, partial log Q, partial beta, is equal to partial log Q, partial Q, times partial Q, partial beta. And I just worked out what those two things are. Partial log Q partial Q is 1 over Q. And here's partial Q, partial beta written again. And notice that's that. All right, Here we have 1 over Q. Here's our 1 over Q. Here' s energy times e to the minus beta E sub j. Here' s energy e times the minus beta E sub j. So the expectation value of the energy is minus the partial derivative of the log of the partition function with respect to beta. Right? And that's a key formula of statistical thermodynamics that we'll want to keep in mind and explore it's utility. And then note that on occasion, it's a little more convenient to work with temperature directly than it is to work with beta, which you'll recall is 1 over kT. And in that instance, given that beta is 1 over kT, I can again use the chain rule. So I can establish that the partial derivative of the log of Q, now with respect to temperature, is equal to partial log Q partial beta times partial beta partial T. I already know partial log Q partial beta, that's minus the expectation value of the energy, that's right here, with the negative sign moved to the other side. I know partial beta partial T, because it's 1 over kT, so T to the minus 1. The derivative with respect to T is minus p to the minus 2, and so it's minus 1 over kT squared, and I can then rearrange that. The two negative symbols cancel out, and I get that the expectation value of the energy is equal to kT squared partial log Q, partial T. Again, N and V held constant. So, what's the practical utility of Q then? So, we've already seen that the partition function is a measure of accessible states, but here, it seems to be an enormous utility. Namely, that given a large collection of particles with an associated partition Q, we should be able to compute the energy of the system. However, how do we go about getting that partition function? So, that is a brobdingnagian task, which is to say very, very, very, very large. Let's do a simple example though. A monatomic ideal gas, maybe helium. So in that case, one can write the partition function as Q depending on N, V, and beta is equal to a quantity little Q that depends on volume and beta. All taken to the nth power, divided by n factorial, where little Q is, a big series of constants here the three halves power of 2 times pi, times the mass, in this case of helium, divided by Planck's constant squared and theta in the denominator. All multiplied times the volume. So, Q is the partition function for one atom of Helium, m is the mass, h is Planck's constant. And so, I'm actually going to derive these not too long, but for now just take my word for it. And I want to see what we can do with them given this form for the partition function. So we're going to do a little calculus again. here are the two forms for the atomic partition function and the ensemble partition function capital Q. And remember, what I want to do is I want to work with log of Q. So when I take the log of this expression, I get the log of something raised to a power, so that's N times the log of the thing itself. So I'll get N log q. And I also have a log of a ratio. So that's equal to a difference in log. So I get minus log of N factorial. And so now let me expand this log of q, because q is a whole bunch of things taken to powers and divided by one another. So the power comes out in front, three halves there is still this factor of N that was multiplying log q before. So three halves N and then log on 2 pi m minus log of h squared minus log of beta, because there in the denominator. Plus N log times the volume, and I still have this leftover minus log N factorial term. Okay, but what's my goal here? Well, I want to take the negative of the partial derivative of log Q, with respect to beta. And so I can just wander through these terms. Does N factorial depend on beta? No. So that derivative would be 0. Does log of the volume depend on beta? No. That derivative is 0. This one is 0. This one is 0. Here's the only thing that has a beta dependence, minus three halves N log of beta. So, what do you get when you differentiate the logarithm, with respect to its argument? You get 1 over the argument. So I'll get 1 over beta. And there's still a factor at three halves N. And there was a negative sign, which multiplies this negative sign, and result, three halves N, beta to the minus 1. And beta is one over kT. So beta to the minus 1 is just kT. So I get three halves NkT for the expectation value of the energy. So, that's delightfully simple, of course. So just to recapitulate that, we've got the expectation value of the energy, three halves NkT. And from experiment in the early days of measuring gases, from the kinetic theory of gases, we know that the molar internal energy of a monatomic ideal gas is three halves RT. Many of you may, may remember that from beginning physical chemistry, that there's a half an RT associated with translational motion in each of the three directions, so three halves RT. And that's a molar quantity. And this result then demonstrates the power of statistical thermodynamics, because I used microscopic properties to compute a macroscopic property. evidently this expectation value of the energy is equal to the internal energy. So if it's the molar internal energy, I must have a mole, this capital N is Avogadro's number of atoms. And remember I told you that Boltzmann's constant times Avogadro's number is the universal gas constant. In a way, you can think of this as being how you determine Boltzmann's constant. We didn't have to determine it previously. We could have just gone through this to predict the energy, compared it to what's known from other other kinds of measurements, and assess what Boltzmann's constant must be. But in any case then, three halves RT, derived from first principles from a partition function. A very impressive result. I hope you're impressed. And so that establishes then, that the internal energy, which is a macroscopic classical thermodynamical concept is related to the statistical mechanical concept, the expectation value of the energy. All right. Well, that's enough working with the partition function for the purposes of energy. Next, we will take a deeper look at how the partition function allows us to derive the ideal gas equation of state. See you then.