 So, since the Van der Waals equation can predict a critical point for a gas, we can determine the critical properties of the gas, its critical molar volume, temperature, and pressure directly from the Van der Waals constants for the gas. So to see how that works, let's take an actual gas, let's talk about water for whom I can give you the Van der Waals coefficients. The Van der Waals constant A is equal to 5.464 liter squared atmosphere per mole squared. And the molecular volume given by the Van der Waals equation is .03049 liters per mole of water. So those are the Van der Waals coefficients. We can use those to plot isotherms. We can now use those also to calculate these critical properties. So let's say the most interesting one perhaps is the critical temperature. We just plug values into this expression, 5.464 for A. In the denominator we have a 27 and a B and an R. So let's think carefully about what units we want to use for R. Leaders and moles are only going to partially cancel leaving with a liter and an atmosphere up top. So we'd use .08206 liter atmospheres per mole kelvin as the gas constant to save myself a little work converting units. And then leaders and leaders cancel liter squared, moles and moles cancel mole squared, atmospheres cancel atmospheres. And 1 over 1 over kelvin gives me a critical temperature in units of kelvin. And that quotient works out to 647 kelvin. So Van der Waals constants predict a critical temperature of 647 degrees kelvin for water. And in fact that's very close to what the actual critical temperature is for water. What that means physically, remember what the critical temperature means, that's the temperature above which we can no longer see phase transitions between the liquid and the gas. So at room temperature water can evaporate and become a gas. At 100 degrees Celsius we can boil water and turn it into a gas. At elevated pressures we can boil water at higher temperatures at let's say 400, 500, 600 kelvin. If the pressure is high enough we can boil water and form a gas. But at 650 kelvin we can heat the water up and it won't boil it will just become it will smoothly transition to a very dense fluid, a high temperature dense fluid that we call a supercritical fluid. The pressure of that critical point we can calculate as well. A over 27 B squared. So plugging in our values for A and for B, 27 times .03, 049 liters per mole and that one is squared. At arithmetic if our units work so I have a liter squared on both sides I have a mole squared on both sides that leaves me with just atmospheres. So in atmospheres this quotient works out to be 218. So again stopping to think about what that means, water boils at one atmosphere 373 kelvin. If I increase the pressure the boiling point increases, Clausius-Cleperon tells us roughly how that works. At the time the boiling point gets up to 647 kelvin the pressure is up to 218 atmospheres but at temperatures above this and or pressures above this we can't liquefy the gas, we can't boil the water anymore. So water above 218 atmospheres is a dense fluid, we don't call it a gas or a liquid. Just a supercritical fluid. The last one is the one with the easiest arithmetic, the molar volume and I should say that's also again very close to the actual critical pressure. The molar volume at the critical point 3 times B, 3 times that .0349 in liters per mole, that is .09 in units of liters. That one probably easier to understand if I move the decimal place over by 3 to give a value in milliliters per mole. So 91 milliliters, a mole of water, liquid water at room temperature remember takes up about 18 milliliters so at the critical point at this very elevated temperature and this particular pressure water will take up according to the Van der Waals equation should take up 90 milliliters rather than 18 for the liquid or several liters for the gas. Actually this one is not a terribly good estimate. Experimentally we find that the critical volume of water is about 57 milliliters per mole. So now it seems like the Van der Waals equation has done a much worse job at predicting the critical volume and it turns out there's a good reason for that. In fact if we go back to these equations we have two Van der Waals coefficients A and B and we have three critical constants, molar volume, temperature and pressure. We can do that conversion in the direction we've got here. If we know A and B we can calculate the critical constants. The other way we can think about that problem is to do it the other way around. If I know the experimental critical temperature, critical pressure, critical volume I can use those to solve for the Van der Waals coefficients that best predict that critical point. So for example these equations look quite similar they've both got an A's in the numerator B's in the denominator. So if I take the critical temperature divided by the critical pressure, so if I take that quantity divided by that quantity, 8A over 27Br, if I divide by this one so I'll write it as multiplying by the reciprocal A in the denominator, 27B squared in the numerator when I divide there's going to be a fair amount of cancellation between the A and one of the B's and the 27's all cancel, leaving me with 8B in the numerator. This B survives and an R in the denominator. What I can do with that, remember what I'm trying to do is given the critical temperature and pressure solve for some Van der Waals coefficients. So if I rearrange this equation to solve for B, I'll find B is equal to, if I bring the R to the other side it shows up on top, that 8 shows up in the denominator. So if I know the critical temperature and pressure I can solve for B. Likewise I can solve for A, let's go back to this expression which has an A and a B in it. I know now what B is equal to, so if I say critical temperature is equal to 8A over 27B and for B I'll use that expression we've just derived, RTC over 8PC and an R. Then let's see what happens here. We've got 8 divided by 1 over 8 that gives me a 64 in the numerator, an A in the numerator, a PC in the numerator. I've got an R and an R in the denominator, 27 in the denominator. So if I rearrange that expression solving for A, I'll get A is equal to 27 and an R squared and a T, I'll bring those over to the other side divided by a 64 and a pressure. So there's an expression for how to calculate the Van der Waals constant A if I know the critical temperature and critical pressure. So if I know these critical constants I can solve for A and B, but notice what's just happened. I only needed two equations to solve for the two unknowns A and B. I didn't use this expression at all. So in fact if I have a critical temperature, critical pressure, critical volume, I've just used these two to solve for A and B. It's actually no surprise at all that when I used A and B to solve for the critical temperature and critical pressure I got exactly the right answer. You can perhaps guess now where these Van der Waals constants came from. We use these Van der Waals parameters because they perfectly predict the critical temperature and critical pressure. These Van der Waals coefficients were obtained using these expressions. What that means is we didn't guarantee that the volume was going to come out because we didn't end up using the volume to solve for these. I could if I wish, if I'd like to do a good job at predicting the critical temperature and critical volume, I can rearrange these equations in a different way to solve for slightly different Van der Waals parameters that will get the critical temperature and volume perfect, but mispredict the critical pressure. Or of course I could use the critical volume and pressure to predict a pair of Van der Waals constants that would get those right but mispredict the critical temperature. I can't because the Van der Waals model is not a perfect model of a fluid or a gas or a liquid. I can't perfectly predict all three of them. I can perfectly predict any two I want but one of them is going to be wrong and the fact that it's wrong by this much tells you how approximate the Van der Waals equation gets when you get near the critical point.