 So by thinking about the critical point for a Van der Waals gas, we've been able to figure out that the critical point, the volume, temperature, and pressure at the critical point can be expressed in terms of the Van der Waals coefficients in this way. So the critical point, of course, is relatively high temperatures, high pressures, conditions where we might not expect the gas to behave very ideally. So let's remember that we've got this concept of the compressibility factor that tells us exactly how ideally a gas behaves. If we calculate PV bar over RT, for an ideal gas, that would be one. For a non-ideal gas, it'll be not exactly equal to one. It might be greater than one if the finite volume of the molecules of the gas causes it to occupy or have higher pressure than ideal gas would predict. It'll be less than one if the pressure or the volume is less than the ideal gas law would predict. So what we can do relatively easily is say at the critical point, what would be the compressibility factor of a Van der Waals gas? How ideal or non-ideal is it at the critical point? So if we just plug in the values of the critical pressure, critical molar volume, critical temperature, using these terms we've obtained. So the critical pressure is A over 27B squared. Critical volume is 3B. Divide that by R and the critical temperature, which is 8A over 27BR. There's clearly going to be quite a bit of cancellation here. I've got an A in the top and an A down here that cancel. Once B squared, we'll cancel the B in the numerator and the B in the denominator of the denominator. So those two B's cancel these two B's. The 27s are both in the denominator on top and bottom, so those go away. R goes away, almost everything is going away. The only things that survive are the three and the eight. So what we've found is that the critical compressibility at the critical point for a Van der Waals gas is 3 eighths, about .375. So that's fairly remarkable, not for the numerical value, but because all of the A's and B's have canceled. So what that means is regardless of what the value of A is, what the value of B is, it doesn't matter whether I'm talking about the Van der Waals coefficients for water vapor or for CO2 or for oxygen or nitrogen, whatever gas we want to think about. Regardless of the constants A and B, the critical compressibility works out to be 3 eighths. So that's pretty surprising. Critical compressibility of every gas should be the same number, predicts the Van der Waals equation. So is that, it's true for Van der Waals gases, is it actually true for all real gases? We can answer that question. I'll bring up a table of some experimental data over here with the critical compressibility factor measured experimentally for a range of different gases. And what we can see here is rather than being 3 eighths, so Van der Waals equation predicts this to have a value of .375. Real gases don't have a critical compressibility of .375, they're all lower than that. But notice they're all very similar to one another, argon .29, benzene .27, carbon dioxide .27. They're mostly in the range of .27, .29, .28, .30. Some of them are a little bit lower, water is the lowest one on this table. It's the most non-ideal with a compressibility factor of .23 roughly. But notice how similar all these values are to one another. The compressibility factor, the non-ideality of these gases at the critical point is roughly the same for all these gases. So clearly they're all behaving similarly in some way when they're exactly at their critical point. So it turns out that similarity between different gases is true not just at the critical point, but also at other points as well if we think about them in the right way. So let's go back to the Van der Waals equation of state and I'll show you what I mean by that. So here's what the Van der Waals equation of state tells us a gas will behave like. So what I want to do is I want to rewrite this equation in terms not of the pressure, molar volume and temperature, and not just at the critical point, but in the conditions of the gas relative to that critical point. If I'm 10% over the critical point, then maybe all the gases will behave the same way. If I'm 10% below the critical point in some of the properties. So what I'm going to do in these next few steps is I'll define for each one of these properties, molar volume, temperature, pressure, I'm going to define the reduced volume, the reduced temperature, the reduced pressure as whatever the actual molar volume is divided by the critical volume. So likewise for reduced temperature, that will be the actual temperature over the critical temperature. Reduced pressure will be actual pressure divided by the critical pressure. So if we're at the critical point with this molar volume, this temperature, this pressure, then the reduced properties will be exactly one. If I'm 10% at a temperature 10% larger than the critical temperature, then the reduced temperature will be 1.1. The reduced properties just tell us how far above or below the critical point we are. So my goal now is to rewrite this equation so it has these ratios of volume to critical volume, temperature to critical temperature, and so on. So I want to get these types of expressions in here. So I think the easiest way to do that is P a. Let's multiply each of these expressions to start. Let's multiply by 27 b over a. And I'm doing that because there's a 27 b and an a in each of these two expressions that's going to come in handy. So right now I'm just taking the Van der Waals equation of state and on both sides I'm multiplying by 27 b over a. So the left side is still equal to the right side. I want to make this pressure eventually look like a pressure divided by a over 27 b squared. So I don't have yet enough b's here. If I add one more b in the numerator, if I add a b in the denominator somewhere else, that'll help me get towards this v bar looking like v bar divided by its critical volume of 3b. In fact, if I want to divide it by 3b, I'll throw in a 3 on top and bottom here, that v bar divided by 3b is going to help get me closer to the reduced volume. Let's see, over here the temperature looks pretty close to temperature divided by an a over 27 br. I've got the 27b and the r and the a. What I'm missing is an 8. So I want to multiply this whole thing by an 8 over an 8. So all I've done is either multiply by 1, something over something, 3 over 3, b over b, or I've included something on the left side that I've also included on the right side. And now I think I've got things in a good enough shape that I can rewrite. Let's bring this a over 27 b squared into the parentheses. So I'll write pressure divided by a over 27 b squared. When I do that to this term, the a's cancel and I've got a v bar squared in the denominator. 27 b squared I'll write as 3 and 9. So the 3 times 9 is my 27. The 3, this 3 I'll leave outside the parentheses. The 3b I'll bring inside the parentheses, so v divided by 3b and minus a b divided by 3b. I'll go ahead and cancel those b, so it's just 1 divided by 3. All that is equal to, now let's see, this temperature on the top. I want to divide by 8a over 27 br. So divide by 8a over 27 br. And that takes care of this 8 and the a and the 27 br. And I've got an 8 left over. All right, so now all I've done is I've just rearranged all the numbers in this term. So they're in a position that what makes me more easily write things in terms of these reduced quantities. So now I can say pressure divided by a over 27 b squared. That's pressure divided by critical pressure. So that is the reduced pressure. Likewise, 9b, I think I've lost my fraction line there. v bar squared under 9b squared. That's like v bar under 3b quantity squared, which is the same as v over the critical volume. So that I can write as 3 times 1 over reduced. Critical reduced molar volume squared. I've still got a 3. v bar over 3b is again the reduced molar volume. And I've still got a 1 third. All that is equal to 8 times temperature over this collection of constants. Temperature over 8a over 27 br, that's temperature over tc. So that's my reduced temperature. I'll just rewrite all this one more time. And what I've got is reduced pressure plus 3 over reduced molar volume squared. Let me go ahead and take the 3 inside these parentheses. So I'll write that as 3 times the reduced molar volume minus 3 times a third gives me 1. On the right side, all that's equal to 8 times the reduced temperature. All right, so that's all the algebra work finished. Let me go ahead and put this equation in a box. Notice what's significant about this. I've succeeded in rewriting my pressure as a reduced pressure, my temperature as a reduced temperature, and so on. But notice what happened when I did that. When I rewrite all of my thermodynamic variables in terms of the reduced quantities, what happened is that the a and b constants all disappeared. There's no a's or b's left in this expression whatsoever. So what that means is this Van der Waals equation of state written in terms of these reduced variables does not depend on the identity of the gas. So again, H2, N2, O2, H2O, they all have different values of these Van der Waals coefficients. So the Van der Waals equation of state will look different for each individual gas. But if I write it in terms of reduced quantities, the Van der Waals equation of state has this one universal form for all gases. So not only is it true that all gases behave somewhat alike at the critical point, but in fact all gases behave the same at all points according to the Van der Waals equation if we think about them in terms of reduced properties rather than actual pressure, volume, and temperature. So that's what we mean by the principle of corresponding states. What that means is any gas that's at its critical point behaves the same as another gas at its critical point, at least according to the Van der Waals equation. Any gas that's at half of its critical temperature and twice its critical molar volume will behave the same as any other gas that's at half its critical temperature and twice its molar volume, for example. So if we think about properties in terms of reduced properties, all gases behave the same. So that may seem a little surprising, maybe even hard to believe. So that'll get a little easier to understand if we think about a few examples and show some data. So that's what we'll do next.