 OK, so we understand now that the slope of the various coexistence lines on pressure-temperature phase diagram is given by the Clapperon equation that tells how the pressure changes with respect to temperature along any one of these phase coexistence lines. So the equation right here tells how that works. We've also mentioned the fact that gas volumes, molar volumes for gas tend to be much larger than the molar volumes for a liquid or for a solid. Just to give you a sense of what I mean and why there's a double-equal sign there for a gas that we can have some everyday experience with, we talk about water at its boiling point. So liquid-vapor coexistence line somewhere around 373 Kelvin and one atmosphere is the normal boiling point of water. So liquid and gas are in coexistence at that temperature, the gas for any, if water behaves like an ideal gas, you might be able to tell me what the molar volume is at standard temperature and pressure. At 273 Kelvin, it's about 22.4 liters for every mole. Room temperature goes up to about 25 liters per mole. At 373 Kelvin, the molar volume is increased further to 30 point, I believe it's 30.6 liters per mole. The liquid on the other hand, again at room temperature you can tell me how much volume a mole of liquid water takes up. A mole of water weighs 18 grams, so if it's density were 18 grams per milliliter, it would take up 18 milliliters. At 373 Kelvin, it's expanded a little bit. That high temperature of the liquid on this side of the coexistence curve takes up about 19 milliliters, so I'll write that in liters as 0.019 liters per mole rather than 19 milliliters per mole. So that's what I mean when I say the volume of the gas is much larger than the volume of the same number of moles of a liquid. 30 liters is taken up by a mole of gaseous water, 0.019 liters is taken up by a mole of liquid water. So this volume of the liquid doesn't even approach the size of the volume of the gas to as many sig figs as I've given you the volume of the gas here. So essentially what that means is when we write the change in volume of a phase change, if we have a phase change like liquid turning into water, so this phase change right here, vaporization. So that phase change, the difference, so the final state gas minus the initial state of the liquid, volume of the gas minus volume of the liquid, 30.6 minus this number that's only out in the hundredths of a liter per mole, that's going to be about the same as the volume of the gas. Essentially the volume of the liquid is negligible compared to the volume of the gas. So what that means is we can rewrite the Clapeyron equation instead of writing the change in molar volume for the phase change, just write the molar volume of the gas, reflecting the fact that the liquid is negligible. So what that then allows us to do if we assume that the gas is at least somewhat like an ideal gas, that molar volume is RT over P. So we can rewrite the Clapeyron equation now as enthalpy of vaporization divided by the temperature at which the vaporization takes place, and now I'll write the molar volume of the gas as RT, that's the temperature of the phase change divided by pressure. So that quotient is in the denominator. Let's go ahead and rewrite that as, notice now there's two temperatures in the denominator, so I've got an R and I've got a T squared, and that's one over P in the denominator is like a P in the numerator. So we've got four of the specific case where either a liquid or a solid is turning into gas, whether it's a liquid vaporizing, boiling, or a solid subliming, the volume of the solid and liquid are going to be negligible compared to the gas, and the Clapeyron equation simplifies down to this form, but now it's in a form where we can actually rearrange it a little bit and write, if I bring all the things related to pressure over to the left hand side, so I'll bring this P over to the left hand side and write Dp with a P in the denominator, and I'll bring the DT over to the right hand side, so that should be a molar enthalpy of vaporization. My equation now looks like this, and I've been a little bit sloppy with my subscripts, but these pressures and temperatures that I'm talking about, these are the pressures and temperatures of the phase change, pressure of the phase change, temperature of the phase change, so this P, Dp, as well as the pressure or the pressure at which the vaporization takes place, this T of the phase change, temperature of vaporization, and DT are both for temperature of vaporization, so those are all the same variables. So I've got all the T's on one side, all the P's on the other. Now, what that lets me do is do an integral, so I can integrate both of those expressions, and notice that the way I've written that, I've pulled the delta H outside of the integral, so I've made an assumption at that step that the enthalpy of vaporization is constant, that it's not going to depend on temperature, so I'm only making a small change in this direction or this direction that doesn't change the enthalpy of vaporization too much. So we know how to do both of those integrals, integral of 1 over P is natural log of P, integral of 1 over T squared, I'll go ahead and write the delta H of vaporization over R, integral of 1 over T squared is 1 over T, but with a negative sign, and I have to evaluate that integral between some initial state if I start here and move to here, I've started at T1 and P1 and I've moved to some T2 and P2, so I'm integrating from P1 to P2 and T1 to T2. So now if I insert these limits into those expressions, I'll get on the left side natural log of P2 minus natural log of P1, which I'll write as ln of P2 over P1, and on the right side I've got minus enthalpy of vaporization 1 over T2 minus 1 over T1. So that equation is actually in a form that's perfectly valid that we can use many times, I'll rewrite it in a couple of different forms that are also useful, but that equation, if I put that in a box, that is an equation that we call the Clausius-Clapeyron equation. Let me just remind you what that is, what we've done, we started with the Clapeyron equation, the equation that tells us the slope of these various lines, coexistence curves on a pressure-temperature phase diagram. We said if we assume the volume of the liquid and solid is negligible compared to the gas, and if we assume the gas is behaving like an ideal gas, and if we assume the enthalpy of vaporization is relatively constant, then we arrive at this expression. So under certain special conditions the Clapeyron equation takes this form, which we call the Clausius-Clapeyron equation. If we rearrange this equation to get rid of the natural log, taking E to the both sides, this equation ends up looking like E to the minus enthalpy of vaporization over R 1 over T2 minus 1 over T1. There's a couple other ways we can write that expression as well, but this is a convenient one when we want to solve for pressure at different temperatures. So for example, if we know the normal boiling point of water, 373 Kelvin in one atmosphere, we can calculate the boiling point at some different temperature. We can calculate the, if we have a different pressure, we can calculate a T2, or if we have a different T2, we can calculate a P2. So that's just a different form of the Clausius-Clapeyron equation. So as we said, what that allows us to do is calculate specific points along these phase coexistence curves, as long as one of the lines, one of the phases involved is the gas. So liquid to gas or solid to gas. And so we'll work some examples using the Clausius-Clapeyron equation coming up next.