Let's continue to work with the partition function for an ideal gas and see what else we can learn from it. Last time, we discussed the internal energy of an ideal gas. I'd like to begin this time by looking at the ideal gas heat capacity for a monatomic ideal gas, like helium, we were working with helium last time. And so the molar heat capacity of a substance expresses the energy that's required to raise the temperature of 1 mole of that substance by 1 Kelvin. And another way to think about that is, multiplication of the molar heat capacity by a measured temperature change, so I have a thermometer and a mole of my substance and I look at how much the temperature goes up. If I multiply the molar heat capacity by that temperature change, I will know how much energy had to be put in to get that change. Because the molar heat capacity is how much is needed for one degree. I multiply by how many degrees it is and there I have it. So let's pause for a moment and make sure thats clear and then we'll come back. [SOUND] So, now that you're comfortable that molar heat capacity is in units of energy per Kelvin, let's consider that we can express that as a derivative. So the molar heat capacity at constant volume in particular, so molar because there's a bar over the top, that's what a bar means, a molar quantity just as we did molar volume earlier in the course, v bar. And constant volume here because that's a variable we are holding constant within our ensemble. Is equal to the partial derivative of the internal energy, so here's the energy units in the numerator. With respect to temperature. So here's the temperature units in the denominator, holding volume constant. So just the same v subscript on each of these two terms. So, from our prior work, on the other hand, we know that the partial derivative of the molar, internal energy, excuse me, is equal to the partial derivative of the expectation value of the energy, right? That statistical, mechanical concept, is equivalent to the classical thermodynamic concept. So I can replace it. And we computed in the last video that for a mole worth of substance, that value of the expectation value is three halves RT. So this is now become an extremely simple partial differentiation. So we're going to differentiate three halves RT, with respect to T. And you end up with 3 halves R. And so statistical thermodynamics, then, has given us a route to understanding where does the energy go when a system is heated, and why the capacity to store energy is different for different substances. Because it's the form of the partition function, which we can work with in order to get energies and heat capacities by a differentiation of that energy. And we're going to see in subsequent examples to come, why heat capacities maybe different for a diatomic gas compared to a monotomic gas compared to a polyotomic gas. I want to continue with this manipulation of the partition function in order to derive important properties for the ideal gas. And this is just reminding you of some of the differential calculus we've already done. But now, I'm going to consider the partial derivative of the partition function with respect to the volume. So up till now we've already seen this one, partial log Q, partial Q is just 1 over Q. Partial Q, partial V. Well, we don't have an explicit form for v in the partition function itself, we just know that the energy depends on it, and so. Given that I have the exponential of a function of the variable I'm differentiating with respect to. I will pull down the multiplier of that function so thats minus beta and the minus beta goes there. And then, by the chain rule, I have to include the partial derivative of the function, with respect to the variable. So here it is. So that's the differential of this expression, with respect to volume. So, putting that all together then, if I take partial log Q, partial volume, that'll be this times this because of chain rule, this multiplying this. So I get minus beta over Q, that's the preceding factor. Here's this partial derivative that I don't yet know exactly how to expand, and here's the exponential of the energy. But this Q out front, this 1 over Q. Again, can be moved inside the sum and make this a probability. Because it'll be an exponential divided by the partition function. So I'll call this the probability of being in state j. And as always it depends on N, V, and beta. Or, of course I don't have to write it in terms of of beta. I can also include a T here. So I've got beta here, I move it over to another side and make it KT. I have some over J minus this partial derivative, probability. I've taken beta and I've moved it over here and multiplied this quantity. So I get KT. Partial log Q, partial V. And I want to continue to work with that equation for just a moment. So I've just reproduced it at the top of this slide. So, again, mostly, I'm interested in understanding the utility of the partition function this week. Something we will derive in the future. But for right now, I ask you to just accept on faith. Is that pressure is minus the partial derivative of energy with respect to volume for fixed number of particles. So if pressure is like energy, that is it's an average over weighted ensembles, recall that was a fundamental postulative, statistical thermodynamics. That an expectation value is an average over possible values. If that's true, that the observed pressure is probability weighted possible pressures, well I can replace this pressure with this expression minus partial E, partial V. Here's my probability and that's what I have up here. Alright, so I'll just equate it to what it also is. It says that the pressure that I aught to observe is kT partial log Q, partial V. So, something we've derived by manipulating the equations. And now let me go back to my partition function for the ideal monotomic gas, helium, that we were talking about. Just the equations we've seen before for the atomic partition function and for the ensemble partition function. And so once again I'm going to need log Q so I'll just expand it. I've shown you this equation before so I won't spend any time on it. And now I'll evaluate for the pressure, I get kT, so here's my kT. What is the partial of the logarithm of Q with respect to volume? This term doesn't depend on volume, this term doesn't depend on volume, this term doesn't depend on volume, Here's a term that depends on volume. Well, partial log V with respect to V will be 1 over V. It's multiplied times N, so I'll get N over V, and that doesn't depend on volume. So, I'm all done with that, so I have pressure equals kTN over V. And if N is Avogadro's number, if I'm using a mole of particles, then N over V becomes the molar volume. And N time, the Avogadro's number time Boltzman constant is the universal gas constant, so I get R. And I have P equals RT over molar volume. Alright and so I've just I've dropped my little expectation value brackets to imply observation. So the observed pressure is equal to RT divided by molar volume. Well that's the ideal gas equation of state. We've actually derived the ideal gas equation of state. From thermodynamic relationships of pressure, energy volume and partition function given certain partition functions for a monatomic ideal gas. So a, a tremendously powerful result. So what's the big picture so far. The partition function encompasses all possible states of an ensemble. Thermodynamic functions can be computed from the partition function. A central one being the internal energy and because the internal energy, it's dependence on other state variables gives rise to other functions. Its a very useful one to have. For simple partition functions, that is, like the ones I've so far told you were valid for helium and that we will derive in the future, the relevant calculations are actually terribly straightforward. That was not difficult calculus we did, of that logarithm, and that give results that agree with, and more importantly, rationalize classical thermodynamics as observed through experiment. Now we haven't yet derived that ideal monatomic gas partition function, but at least we've shown that it's consistent with the ideal gas equation of states. Which certainly looks like it's probably a good partition function. Alright. Well, that was the ideal gas. But, we've already seen in last week's series of lectures, most gases are not ideal only at very high dilution. So, is this partition function useful for other things. Well, next actually, lets take a look at the van der Waals Equation of State and see if there is a way to relate it to a certain partition function.
