 Hi, well, I'm Stephen Nesheba, and I'm here to talk to you a little bit about the calculation of phase boundaries. Phase boundaries are these lines that I've drawn on this phase diagram, and the idea here is that if we could come up with an expression for the slope of any one of these phase boundaries, like say from here to here, that slope, then I could use calculus to integrate it to reproduce the rest of that curve and same thing here and here. So what we're interested here then, therefore, is that slope, which would be the rise dp divided by the run dt. Okay, that's what we want to relate to each other. Now, we do have an equation that has dp and dt in it, of course, it's the Gibbs differential equation of state, and it looks like this. Now, I've labeled two phases i and j, i could be a liquid and j could be a vapor, for example, and here's the idea. In going from here to here, since the two phases were in equilibrium here and they were in equilibrium here, that means however much they give energy of the liquid change, that's exactly the same amount that the Gibbs energy of the vapor change because they're still in equilibrium. So that means that is equal to that, so we can take this entire equation and subtract it from the first one. So on the left-hand side, of course, we get zero. I've written delta v, which is, you know, vi minus vj. That's the volume change upon going from, in this case, from liquid to gas or it could have been any of these other phase transitions. And same thing here, I have the entropy of the liquid minus the entropy of the gas, and so that would be that delta s is the delta s of the phase transition. So now what we can do is we can rearrange this. I just isolated dp on the left-hand side and put the dt term on the right, change in entropy divided by change in volume, and what we use, what we know about the entropy, namely the change in entropy at a given temperature is equal to q rev over t, and since we're also at a constant pressure for these transitions, then we would say that that's equal to delta h over t. So the equation of this to this term here is called the Clapeyron equation. Now, and that tells us the slope, obviously, that at any point in the phase diagram along the phase boundary. In order to reproduce an algebraic expression for the entire curve, we will need to integrate, and before we integrate, we have to decide on how all these terms depend on the pressure and temperature. So one approximation that we're going to make is that the enthalpy, delta h, doesn't change very much the enthalpy of the transition. You know, we know that that's about 44 kilojoules per mole for water at room temperature and 41 at boiling, so that's maybe not such a bad approximation. On the other hand, the change in volume might be substantial, especially if we're dealing with the gas. So let's just take on that one first, and I imagine I'm going from liquid to gas, so we're talking about that part of the curve. I've got the enthalpy of vaporization, which, you know, we could say, well, maybe that's about 44 kilojoules per mole. That's its room temperature value. And now, what are we going to do about the change in volume and going from a liquid to a gas? Well, you know, we know gases have about a thousand times the volume of their condensed phases at ambient temperature and pressure. So I say that that's, that minus that is about the same as the volume of the gas. Then when you use the ideal gas law, and so we say that the volume of the gas is equal to RT over P. And so that term comes down here. You notice they have a pressure term here. We can do a separation of variables then, bring the pressure over this side. Okay, and we can integrate both sides separately. If you do that, the result of it is all called the Clausius-Clapeyron equation. What would we do if we were dealing with this sort of transition from solid to gas? Well, it's almost identically, almost the same. There's now a different enthalpy of vaporization. For water, that's about 50 kilojoules per mole. And the change in the volume actually turns out to be about the same, because we still have the volume of the gas, volume of ices, much, much smaller than the volume of the same number of moles of water. Same thing, end up with the Clausius-Clapeyron equation again. But of course, the enthalpy term that appears in there would be the enthalpy of sublimation. Okay, and then finally, what are we going to do for this transition right here from solid to liquid? Well, one can look up the enthalpy of fusion. Turns out that's about six kilojoules per mole for water. And now the delta B term that appears downstairs, that's the volume of the liquid divided by minus the volume of the solid. Now, one of those is super much, much bigger than the others, but we can't do the trick that we just used. But you can look up tabulated values of the volume. Now, this would be the volume of one mole, assuming that we had the delta H per mole. If you can't find that, then what you can often do is find the density of a substance, and then the density in kilograms per cubic meter would be the SI unit. The molar volume is just the molar mass divided by the density. So often you can look up these values and then you can calculate the delta B. Then what we say is that the delta B is about a constant somewhere along there. It doesn't change very much, and therefore a lot of that just comes straight out of the integral. And the result of that, the algebraic equation that results from that is called the Thomson equation, and that would apply here. So we have got Thomson equation here, Clausius-Clapeyron here, and Clausius-Clapeyron there. Okay.