 Hi, I'm Professor Stephen Nesheva, and I want to tell you a little bit about how to integrate the Clapeyron equation along the liquid vapor phase boundary. So here we have a phase diagram here, and the regions of the phase diagram are solid, liquid, and and a gas over here, and these lines are the phase boundary lines of lines of equilibrium, and in this case we're focusing on the liquid gas equilibrium. Sort of our anchor point in this is the triple point, which has a temperature we're calling T3, and a pressure we're calling P3, and what we have is the Clapeyron equation, which is this right here, and the Clapeyron equation relates the, it tells us about the slope of a phase boundary, and so the idea is, along this phase boundary, there's a certain slope, dP dt, and and here's what it's given by, a small change in the temperature as I move off to the right from the triple point, will result in a change in pressure multiplied by, by that factor. So because we're dealing with a liquid vapor phase diagram, that delta H down upstairs there would be the enthalpy of vaporization, which we're going to consider constant for this, for this derivation, but the change in volume, which would be delta V of vaporization, well by definition that's the volume of the gas minus the volume of the liquid, but at room temperature and pressure the volume of the gas might be 22, you know, one mole of the gas is 20, 20 some liters, and the volume of the liquid, the condensed phase, is much, much smaller, just on the order of, you know, 20 millimeters or so. So what we can say is that the volume of the gas is, to minus the liquid, is just about the same as the volume of the gas, and because when we go to integrate this, we are going to need to separate variables, we need to have this in terms of pressure and temperature, we're going to write that volume of the gas using the ideal gas law, which says for one mole, the volume of the gas is RT over, over the pressure. So, when we substitute those in this, in this Clapperon equation here, there's the delta H still, the delta V, which is the volume of the gas downstairs, I've turned it around so that we have the pressure upstairs now, and divided by RT 1 over T. So this will be the equation that we want to integrate, but a problem as it stands is that there's some pressure dependence on the right-hand side, and we need to get that to the left-hand side so that we can actually carry out the integration. Well, that's easy. We're just going to divide both sides by the pressure, and on the right-hand side that leaves delta H divided by R and the temperature squared in the denominator. So taking these, a piece at a time, if we integrate Dp over T, that's going to turn into a log of the pressure, and when we evaluate at the limits, which I'm going to say the upper limit is going to, we're just going to call it p star of the liquid vapor curve. The lower limit is just p3, so that leads to this term here, after we integrate the left-hand side, p star of the liquid vapor over p3 as the argument of the log. What about the right-hand side with the integral of a constant times 1 over T squared Dt? Just turns into 1 over T, which this is again evaluated at the limits from T3 up to some arbitrary temperature. So that's what that leads to, and when we can equate those, of course, according to the Clapeyron equation, but it's more convenient often to just solve for the pressure itself instead of the log of the pressure, so what I've done is taken the exponent of both of those sides. So when I take the exponent of the log, I just get the argument there, and on the right-hand side I get e to all of this, and just for economy of notation, we're just going to call that the symbol chi, so e to the minus chi. So this is the, what's called the Clausius-Clapeyron equation, which again, describes a phase boundary between liquid and gas, and it's all rooted on, in reference to the triple point in this case. What often happens is that one might want to multiply both sides of this by p3, so that you just have the phase boundary expression in terms of p3 and t3, and so that's pretty much that derivation. One more note here is that if we wanted to get that part of the curve, which is the solid vapor phase boundary, you'll notice that all these considerations work pretty much the same. The only, the only real difference is that, and for talking about that part of the curve, we would be talking about the enthalpy of sublimation, and we're talking about delta v would be the volume of the gas minus the volume of the solid, and which in any case, that second, the volume of the solid, is also small compared to the volume of the gas. So, so those approximations are as valid as what we just derived. Okay.