 All right, so we have a partition function now for what we've called the Van der Waals model for gas, which we hope is going to be more realistic than the ideal gas model because it accounts for finite molecular volumes as well as intermolecular attraction between the molecules of the gas. But just having the partition function is only halfway there. What we'd like to do with that partition function is use that to calculate some thermodynamic properties and two that we know how to calculate so far are the energy and the pressure. So that's what we'll do next. So we need to remember that we can calculate the internal energy directly from the partition function by taking kt squared times d log q dt for that partition function. So in this case for the Van der Waals model it's kt squared times the temperature derivative of the log of this partition function. So if I break down the log of this large product into the sum of several different log terms, I've got log of 1 over n factorial is minus log of n factorial log of this term in parentheses is going to be 3n over 2 the power times the log of 2 pi mk h squared. And once again rather than take the log of the entire term in parentheses, I'll look ahead and notice that I'm going to be taking the temperature derivative. So I'll split that up. Log of 2 pi mk t over h squared I'll write as log of 2 pi mk over h squared and log of t each of which gets multiplied by 3 halves. I've also got n log of v minus nb. Most of these terms are very similar or exactly like they were for the ideal gas. This one's a little different because now I have the term representing the molecular volume. This term is new. Luckily the log of an exponential is pretty easy to write down. Log of e to the an squared over vk t is just an squared over vk t. So if I simplify that expression, that's what will tell me the internal energy of this Van der Waals gas. So the derivatives I have to take here, most of them, if I take the temperature derivative of most of these terms, they don't contribute anything. There's no t's in a log of n factorial. So the temperature derivative goes away. There's no t's in this term. I am left with a temperature derivative of log t. So I've got altogether kt squared. Temperature derivative of this term looks like the 3n over 2 out front. And then derivative of log t is 1 over t. There's no temperatures in the volume term. And then the last term does have a t in it. So the temperature derivative is going to look like an squared over vk. And then the derivative of 1 over t is minus 1 over t squared. So that is simplified quite a bit. And if we clean it up further, kt squared, multiplying each of these terms in brackets, 3n over 2 times kt squared divided by t is just one factor of t. And in this last term, I have a kt squared out front, which is going to cancel the k and the t squared both in the denominator. And all I have left is the minus sign a n squared over v. So we see that the internal energy for a Van der Waals gas is 3 1⁄2 nkt. That's exactly what it would have been for an ideal gas. Minus a n squared over v. And that's the change in the internal energy that comes from the intermolecular attraction that we have included in the Van der Waals model. So the internal energy is lower than it would be for an ideal gas because we've allowed the molecules to attract each other, which lowers their energy. So we can do the same thing for the pressure. If we remember the thermodynamic connection formula for the pressure, pressure is kt dLnqdv taking the derivative of log q with respect to v. So I need to write down the volume derivative of log q. So the volume derivative of the same term in brackets, log q. So I've got n log v minus nb and then a n squared over vkt. So we're taking the volume derivative. We can see there's only two places that volume shows up in this expression once in this n log v term and once in this fraction that has a v in the denominator. So when I take the derivative, the first terms don't contribute anything. This third term, volume derivative of log v minus nb is 1 over v minus nb. Volume derivative of 1 over v is negative 1 over v squared. And I have these coefficients a, n squared, and kt to include. And now I can just simplify by distributing this kt across both of these terms. And we find that the pressure of the Van der Waals gas is nkt over v minus nb from the first term. And then minus kt cancels the kt in the denominator. And I have just a n squared over v squared. So that's the pressure of the Van der Waals gas. Notice again that that's different than the pressure of an ideal gas. This much of it looks like the pressure of an ideal gas, pv equals nrt or p equals nrt over v and kt over v. So the correction, this correction, accounts for the fact that the molecules themselves have some non-zero volume, some finite volume. And notice the effect that this has on the pressure. If I'm dividing not by the total volume, but the smaller volume after subtracting the size of the molecules, dividing by that smaller quantity makes the pressure go up. So the effect of the finite molecular volume is to increase the pressure of the Van der Waals gas when compared to the ideal gas. And then this last term, the one with the A in it, describes the effect of the intermolecular interactions, which we've assumed are generally attractions. So since the molecules are attracting themselves, the forces are mostly directed toward other molecules toward the inside of the box, and therefore the molecules collide with the outsides of the box a little less strongly. So that has the effect of decreasing the total pressure of the gas in the box. So molecules attract each other and that decreases the pressure. They take up some room and that increases the pressure for this Van der Waals model as compared to the ideal gas model. So that's showed us how to calculate two of these thermodynamic properties from the partition function for the Van der Waals model. And what we can do now, because mostly when we use the Van der Waals model, mostly what we're interested in is this relationship between the pressure of the gas, its volume, and its temperature. So we can actually spend a fair amount more time discussing this relationship, which we call the Van der Waals gas equation of state, that describes how P, V, and T are related to each other for this Van der Waals gas. And that's what we'll do next.