 In this video we're going to find the critical point using Van der Waal's equation. We call that the critical point is PCVCTC, pressure, critical volume, critical temperature, and at this point the system is teetering between liquid and gaseous phases. Van der Waal showed that this is the saddle point of his equation, so that means it satisfies the two, so that means that it satisfies these two equations. The first partial of pressure with respect to volume evaluated at the point Tc and Vc must be zero, and the second partial derivative of P with respect to V evaluated at the same point must also be zero. So let's see if we can work this out mathematically and find the point like Van der Waal did. The first thing we need to do is to solve his equation for P and then take the partials with respect to V. So solving this equation for P we would get this is equal to RT over V minus P, so P plus A over V squared equals RT divided by V minus B, so P equals RT over V minus B minus A over V squared. Notice that V, which we're going to take the derivative with respect to, is here and here. So this is our equation that we're working on with respect to the partial derivatives. So let's do the first one, delta P delta V is equal to, and RT would be, in this case, in this place they're quite constant, so we have RT over delta V of 1 over V minus B minus A times, same thing it's constant, delta V of 1 over V squared. So this is V minus B to the minus 1, so its derivative is minus 1 V minus B to the minus 2. So we're going to put the minus up front and then we have 1 over V minus B squared. And now this is V to the minus 2, so its derivative is minus 2, so we will have plus 2A here, and then V to the minus 3, so 1 over V cubed. So this is the first partial with respect to V. Now what we're going to do is, because we're looking for Tc, Vc, and Pc, we're going to set this equal to 0, like it says here, and write T equal to Tc and V equal to Vc, and because it's easiest to solve this for Tc, we're going to solve it for Tc. So what we would have is at 0 equal to this, 0 equal to RTc times 1 over Vc minus B squared plus 2A times 1 over Vc cubed, and we can solve that fairly easily for, we'll solve it for RTc. So it would be this times this, so we would have 2A times Vc minus B squared over Vc cubed. Now we're going to find the second partial, which means we take the partial derivative of this with respect to V and set it equal to 0 and then use this equation to find the rest. Remember we have the first partial is this, and setting it equal to 0, we got this equation. So now we're looking for the second partial, so that's delta squared of P delta V squared, and so now we have plus 2 here, RT times 1 over V minus B cubed minus, and then there's three times this, 6A times 1 over V to the fourth. So now we set this equal to 0 with Tc and Vc, and we get 0 equals 2RTc times 1 over Vc minus B cubed minus 6A times 1 over Vc to the fourth. And we substitute this into that, we get 0 equals 4A Vc minus B squared over Vc cubed times 1 over Vc minus B cubed minus 6A times 1 over Vc to the fourth. Now we can cancel these things because they're not 0. Because A is different than 0, we can cancel A. Because Vc is different than 0, we can cancel Vc cubed, and because Vc minus B is different than 0, we can cancel, we can multiply by it, right? So we're going to cancel this part and then multiply. We're left with 4 over Vc minus B equals 6 over Vc. And now we can cross-multiply, we get 4Vc equals 6Vc minus 6B. So we get 2Vc equals 6B, or Vc equals 3B. So that's our first thing that we got, Vc equals 3B. We substitute that back in here to get Tc, and then we'll go to the original equation to get Pc. So RTc equals 2A times Vc, which is 3B, so 2B squared over 3B. Cute. Equals 2, 4, 8A over 27, and there's 1B left at the bottom. So Tc equals 8A over 27B, R. And that is our second equation. And now we need to go back to the original Van der Waal equation to get Pc. So here we have Van der Waal's original equation, here's Vc and Tc, and we're going to substitute these into this to get Pc. So we have Pc plus A over V squared is 3B squared, parentheses, parentheses, times 3B minus B equal to R times 8A over 27B, R. So we see the R cancels here. This is 2B. And so the first thing we have is Pc plus A over 9B squared is equal to 8A over 27B divided by 2B. So this would be 4 and B squared. So we have Pc equals 4A over 27B squared minus 8 over 9B squared. So that's equal to, this is going to be 3, so A left up at the top and 27B squared. So Pc equals A over 27B squared. So there's our critical point. And a final thing here is that Pc times Vc over Tc, we would find that P equals A over 27B squared times 3B over 8A over 27B, R and all the A's and B's cancel and we get 3 over 8. And the R comes up at the top, R.