 So the Van der Waals equation of state, it turns out, does sort of surprisingly well at predicting not just gas phase behavior but liquid and gas and liquid gas coexistence behavior as well. So what that means is we can use the Van der Waals equation to make predictions to understand where the critical point is for a particular gas. Remembering that the critical point is the point where the critical temperature is the temperature where the isotherm exhibits this special point where the function becomes horizontal for exactly one point and exhibits a critical point where the curvature changes from positive to negative at this point as well. So in other words, the slope of those isotherms becomes zero and the curvature of that isotherm is zero as well at this particular point. So when we're at the critical temperature and the critical volume on the critical isotherm at the critical volume, this is the point that we're describing. So if we remember what the Van der Waals equation of state tells us, so that's P described as a function of molar volume and temperature. I can take derivatives of this expression, set them equal to zero and therefore solve for where the critical point is going to be. So the first derivative of this function with respect to V bar, this looks like RT times V bar minus B to the minus one. So the derivative is going to be negative one times RT times V bar minus B to the negative two. The second term I have a minus A V bar to the minus two. So the derivative is going to be negative two times this negative sign times A and I decrease the power of V bar by one more to negative three. So that's the first derivative. Take one more derivative. The second derivative looks like V minus B to the negative two becomes V minus B to the negative three. And I pull down a minus two, so I've got positive two and RT. Here my V bar to the minus three becomes minus four. Pull down a negative three to have negative six A. All right. Those equations describe the slope and the curvature of each one of these isotherms at any temperature I want, at any molar volume I want. What's special about the critical point is this is equal to zero and this is equal to zero when I'm at the critical temperature and critical molar volume. So if I set this expression equal to zero, this negative term must equal this positive term. I'll put the positive term on the other side. So I'll say RT V minus B to the negative two is equal to twice A V to the negative three. That's rewriting this first equation. That's only true at the critical point, so I need to be careful and say that's true when I'm at the critical volume and critical temperature. Doing the same for the second expression. I've got twice RT V minus B to the negative three at critical conditions is equal to six A V to the minus four at critical conditions. We can rearrange these and solve for T and V bar. Let's first get rid of the negative power. So I'll just take this V to the minus three and move it to the other side. Same thing with this quantity to the negative two. Move it to the other side. So I'll end up with RT V bar cubed is going to equal to A V minus B squared. These are now positive powers. This one looks like twice RT V bar to the fourth is equal to six A and I bring V bar minus B over here and its power is now positive three. The easiest way probably to solve this is to break this six A V minus B cubed. Let me write that as two A times three V minus B cubed looks like V minus B squared which is this term right here times an extra V minus B. So all I've done here is I've rewritten this term. I've broken it apart so that one piece looks like the term up here and this is the pieces that are left over. After rewriting it so that this term matches this term I can replace that term with the left hand side so that's equal to RT V cubed times three V minus B. The left side is equal to RT V fourth. Now comes the simplification. There's an R on both sides. There's a T on both sides. There's a V cubed on both sides so that gets rid of the four and turns it into one. And what I'm left with is just two times a molar volume is equal to three times a molar volume minus three times B. Or if I rearrange that a little bit on the left side I can put the B's on the right side three V's minus two V's is equal to just one V and we've solved for the molar volume. In other words the unique molar volume at which the critical isotherm exhibits its critical point happens at three times the constant B for a Van der Waals gas. Three times the molecular volume of Van der Waals gas. What I can do with that now is I can go back to one of these equations let's say this one. If I know what V bar is I can solve for one of the other things I don't know like the temperature. So if I take this equation and solve it for the critical temperature that will look like T critical is equal to on the right hand side I have two A V bar minus B. So I now know what V bar is under critical conditions it's three B. Three B minus B is equal to two B. That I'm going to square. Now I need to divide by an R and a V bar cubed. V bar is three B and I need to cube it. So if I rearrange that equation two times two times two gives me eight. There's an A in the numerator there's a B squared in the numerator. There's an R in the denominator three cubed is twenty seven and there's a B cubed in the denominator. The B's cancel to leave me with none of them left over in the numerator one of them left over in the denominator. So a little bit of algebra has now told me that specifically what this temperature is the critical temperature at which the Van der Waals gas exhibits this critical point. If I know A and I know B I know the two constants for the Van der Waals gas I can combine them in this way and figure out what the critical temperature of the gas is. That gives me V bar gives me T but of course we also would like to know the critical pressure at what pressure the critical point takes place. To do that I just need to go back to this equation. If I know V bar and I know T this is how I solve for P. So I can say the critical pressure is R times the critical temperature over critical volume minus B and then subtract A over the critical volume squared. Now I just plug in what I know for V bar and for T under critical conditions. T looks like 8A over 27 Br. V bar minus B. 3B minus B leaves me just 2B. Now I need to subtract A in the numerator. The denominator looks like V bar squared so 3B quantity squared gives me 9B squared. I can simplify that a little bit. Bars cancel. I've got the 8 and the 2 cancel leaving me just 4 up top. So I can write this as 4A in the numerator. I've got a 27B and an extra B in the denominator. So 27B squared. I'm subtracting from that A over 9B squared. Let's put those over the same. If I make sure the denominator looks like 27B squared I've had to multiply by 3 in the denominator so I'll multiply by 3 in the numerator. Now they're over the same denominator so I can say 4A minus 3A is A and the denominator is 27B squared. So that's my critical pressure. So it was worthwhile to go through all that algebra partially because the approach to doing that in a relatively simple way is not necessarily obvious. But essentially we had two equations that we could use to solve for two of our unknowns, the temperature and the molar volume. This equation allowed us to solve for our third unknown, the critical pressure. And now we know in terms of the constants of the Van der Waals gas, the A and the B constants, we can calculate the critical pressure, critical molar volume, and critical temperature. So if you have the Van der Waals constants for gas you can calculate the critical point, all the thermodynamic properties at the critical point. And we'll do an example of that coming up next.