 Alright, so we have the Van der Waals equation of state which tells us how to calculate the pressure of a gas that obeys the Van der Waals model if we know of course the temperature and the molar volume but also these two constants A and B, the strength of the intermolecular interactions in the gas and the finite molecular volume of the molecules in the gas. And just as a quick reminder, when the molecules have some finite volume that's subtracted from the volume that the gas is occupying, that reduces this denominator which causes the pressure of the gas to increase compared to if we just had a ideal gas. And on the other hand, when the gas molecules have some attractive interactions, some positive value for A, then we subtract something from the pressure so this attractive intermolecular interaction term causes the pressure to decrease. So we can sometimes run into a situation where those two corrections from the attractive interactions and from the size of the molecules cancel each other out. So if the correction that reduces the pressure happens to exactly cancel the correction that increases the pressure, then it turns out the ideal gas law is doing a coincidentally extra good job under those conditions. So that's what we're interested in exploring now is finding out when that happens. And on this graph of compressibility factors, remember that compressibility factor of 1 is the case when a gas is ideal. So we've seen that at cold temperatures, the compressibility factor drops as we increase the pressure. At high temperatures, the compressibility factor increases. Again, the drop is from attractive intermolecular interactions. The increase is from finite molecular volume. At some intermediate temperature, there's bound to be a temperature where the compressibility factor doesn't drop or increase as the pressure increases but stays flat. So let's think about what that temperature would be. So what we're interested in doing is rearranging this equation. Let me go ahead and write this equation instead of the way I've got it written here. Let me multiply it by a V bar. So I'll multiply it by V bar on the left. And so on the right, I've got not RT over V minus B, but RT V over V minus B. And then in the second term, multiplying by V bar just removes what are the V bars from the denominator. So now that I've got it written as PV bar is equal to this quantity for a Van der Waals gas, if the gas is behaving ideally, then we know that PV equals NRT, PV bar is equal to RT for an ideal gas. So if the Van der Waals gas is coincidentally behaving just like an ideal gas, then we'd expect the Van der Waals value of PV bar to be equal to the ideal gas value of PV bar. So if I take these two expressions and set them equal to each other, we'll rearrange a little bit. We've got an RT on the left and an RT on the right. So if I collect those terms, I've got an RT times V bar over V bar minus B. If I subtract this RT when I bring it over to the left hand side, I can get this. And then that's all equal to, if I put the A over V bar on the other side, all right? So we've got this expression, which if we solve this for, actually let's first rearrange this quantity in parentheses to put that all in one fraction. So if I make the denominator look like V bar minus B, the first term is V bar in the numerator. The second term, if I put it over a denominator of V bar minus B, I need to subtract. So this minus one is minus V bar minus B over V bar minus B. And then I can see that I'm going to have some cancellation. The V bar minus V bar is going to cancel, and on the left things will be a little simpler. So I've got, what I'm left with is RT times positive, negative, negative makes a positive B over V bar minus B, and that's all equal to A over V bar when our Van der Waals model is behaving ideally. So that looks a little bit simpler. We can rearrange this now and solve for temperature. Remember what we're looking for is the temperature where the gas doesn't have its compressibility drop, compressibility factor drop as we increase the pressure or have it go up, but we want to find the temperature where it stays flat. So we're solving this expression for the temperature. So leaving the T on the left, all we have, if we move everything over to the right is I have an A, I'll bring the B and the R over to the right-hand side and put them in the denominator, and the volume terms look like this. I'll do two things here. First, let's go ahead and name this quantity. That quantity is called the boil temperature, just the name we give to the particular temperature where a gas behaves ideally because of a cancellation between the positive deviations from the finite molecular size and the negative deviations from the attractive intermolecular interactions. So the boil temperature would be this collection of constants for Van der Waals gas. If we know A and B and the gas constant, turns out that depends on the volume. The temperature where any of these curves cross this axis, this 200 Kelvin gas is coincidentally just for a moment behaving ideally at a pressure of, let's say, 225 bar because the effects of A and B both cancel each other. But we're not really interested in that, and in fact, if we say, under most conditions, the volume of the molecules is much, much smaller than the volume of the container they occupy. So if I, in this step, if I say B is much, much less than the molecular volume, if I'm letting the gas occupy a box that's much bigger than the molecules themselves, then V bar minus B is pretty close to V bar, and then this ratio is pretty close to one. So I'm left with just this expression, a boil temperature of A over B times the gas constant. So that's now a constant for any given gas. If I know A and B and R, I can calculate this quantity, the boil temperature. And what that's telling us is, in fact, let's go ahead and do an example before I decorate that graph. If we do that for a particular gas, one that we've talked about the interval's constants for before is nitrogen gas, where we know the value of A, we know the value of B, that should be a three. So if we were to calculate the boil temperature for nitrogen gas, treating it as a Van der Waals gas, that's just going to be this A over B, R. I'll write the numbers in just to make sure we choose numbers with the right units. Since I've got A in units of liters squared bar per mole squared, that's going to affect what value, what units we use for R. So there's A and B. If I divide by R, I want to use a value for R that's in units of liters bar per mole Kelvin. So that's the value 0.08314. And double checking the units, I've got liters twice in the denominator, twice in the numerator, bars in the numerator, bars in the denominator, moles twice on the bottom of the numerator, moles twice on the bottom of the denominator. And all I'm left with is one over one over Kelvin. So the units are going to end up being Kelvin. And the value if we plug those numbers into a calculator, the value that we get is 426 Kelvin. So what that tells us is if we were to plot the compressibility factor, not at 200 Kelvin, not at 300 Kelvin, but a little higher at 400 Kelvin, that's the temperature where it's going to be flat for an extended range of pressures, it's going to remain fairly ideal over a range of conditions, eventually it will still begin to deviate. And what that's representing is the point where the volume is in fact, we've increased the pressure to the point where the volume is in fact, no longer much, much greater than the finite molecular volume. So at those pressures, it does start to deviate from the ideal gas law. But the main point of this quantity, the boil temperature is to describe the temperature where because of this coincidental cancellation between attractive and repulsive effects, the pressure can be predicted pretty well by the ideal gas law. So nitrogen gas normally, we would say the ideal gas law applies best at high temperatures, but coincidentally it applies extremely well at this temperature of 426 Kelvin, specifically at the boil temperature.