 So there's one more feature of these pressure-temperature phase diagrams that's worth talking about. And that arises from the fact that as we've seen recently, if I were to tell you the enthalpy of vaporization, let's say for water, if I tell you the enthalpy of vaporization at 373 Kelvin, that value is, with one more sig fig than I used the last time I told it to you, that's 40.7 kilojoules per mole. At 298 Kelvin, that value is different. So at room temperature, the enthalpy of vaporization for water is a little higher, 44.0 kilojoules per mole. So the enthalpy of vaporization depends on the temperature is what I'm emphasizing here. And that's not really surprising when we think about it. Vaporization is the process of turning liquid into gas. So enthalpy of vaporization is the enthalpy for that process, enthalpy change for that process. And we know that, well first of all, enthalpy of vaporization, that's the enthalpy of the products minus the enthalpy of the reactants. So enthalpy of gaseous water minus the enthalpy of liquid water. And it's not a surprise to us that that difference will change with temperature, because of course the enthalpy of gas changes with the temperature, the enthalpy of the liquid changes with temperature. That's what the heat capacity tells us about, is how quickly those enthalpy just change. And of course the enthalpy of the gas, I'm sorry, the heat capacity of the gas and the heat capacity of the liquid are different numbers. We've talked about that in some detail when we talked about the equipartition theorem. In fact, Cp for the liquid is bigger than Cp for the gas. That's one of the important things we learned by considering the equipartition theorem. So it's not surprising that the enthalpy of vaporization is different at different temperatures. We've even got some equations. If we wanted to calculate specifically how the enthalpy of vaporization is changing as we change the temperature, we won't use this equation, but I'll remind you that we know the change in the enthalpy with temperature would just be the sum of stoichiometric coefficients times heat capacities for the species involved. So in our case, that would just be heat capacity of the gas minus heat capacity of the liquid. That's what's over here on the right-hand side. That's telling us how quickly the enthalpy of vaporization changes as we increase the temperature. So if we wanted to, we could numerically, quantitatively predict what the enthalpy of vaporization is going to be at different temperatures. We won't do that. We'll settle for just trying to understand it qualitatively, trying to understand why it is that the enthalpy of vaporization drops as I increase the temperature. When I increase from 298 to 373, the enthalpy of vaporization decreases from 44 to about 41. So one way of thinking about why that happens is related to the fact that the heat capacity of the liquid is larger than the heat capacity of the gas. So this reaction at room temperature cost me 44 kilojoules per mole. If I heat both reactants and products up to 373, it costs me more enthalpy to heat up the liquid than it does to heat up the vapor. So when I heat the reaction up to 373 Kelvin, I've had to put more enthalpy into the reactants, the liquid, to get them to 373 Kelvin. So it doesn't cost me as much enthalpy to turn them into the gas, which has increased its enthalpy by less. So this statement does help us explain why the heat of vaporization drops as the temperature increases. In fact, if we try to understand on a graph how that's happening, let's say I plot the heat of vaporization for water as a function of temperature. Just to put some numbers on this graph, we've got room temperature a little above the triple point. We've got a boiling point and then the curve continues in that direction. If I use the same axis down here and the numbers I've given you, 44 is the enthalpy of vaporization at room temperature, a little bit lower at 41 is the enthalpy of vaporization at the boiling point. So there's some curve connecting these points. We understand why the enthalpy of vaporization is dropping. If we continue to increase the temperature above the normal boiling point, the enthalpy of vaporization will continue to drop. One important observation is if it continues to drop, and certainly if it continues to drop at the rate that it's dropping so far, it's eventually going to cross this axis. In fact, that does happen. The enthalpy of vaporization does continue to drop. It actually begins to do that a little bit faster as time goes on. If I try to draw that somewhat more to scale, it becomes very steep. At some particular temperature that we'll talk about in just a second, the enthalpy of vaporization will drop to zero. That's an interesting phenomenon. That means the enthalpy change when I convert liquid to gas is zero. What that means on the phase diagram, we can think about as well. So we've gone from room temperature to the boiling point of water. If we continue to move upward along this phase coexistence curve, two things are happening. We've got liquid in equilibrium with gas, boiling at 373 or boiling at higher temperatures or even higher temperatures. But at any one of these points on the phase coexistence curve, liquid is in coexistence with gas. It's in equilibrium with the gas. As we raise the temperature, making the system hotter, the liquid increases in temperature, becomes more energetic, so the liquid's becoming a very hot liquid. On the gas side, as we increase the temperature, the pressure is increased. If I have a fixed volume of gas, also we can see along this phase coexistence curve, as I increase the temperature, the pressure increases. So the pressure of the gas in equilibrium with the liquid is continuing to increase from 24 toward a 1 atmosphere to 2, 5, 10 atmospheres. The pressure is climbing on the gas side. So if you picture what that gas looks like as I increase its temperature and pressure, the gas is becoming more and more dense. A gas at 10 atmospheres occupies less volume than a gas at 1 atmosphere. In other words, this hot random disordered gas phase is becoming more like the hot random disordered liquid phase. The only difference between the liquid and the gas is the liquid is more dense and the molecules are closer to one another. As I climb this coexistence curve, the gas molecules are becoming closer to one another. At some point, the hot disordered gas will become so much like the hot disordered liquid that there's no difference between the two. And in fact, that happens at the same temperature position. So at the point where the enthalpy of vaporization becomes zero, the reason that happens, the reason there's no difference in enthalpy between the gas and the liquid is because a dense hot gas is no different than a dense hot liquid. We've compressed the gas to the point where it's liquid like and there's no difference between the two. So at this point that we call the critical point is the point at which there's no point distinguishing between liquid and gas anymore. They're both just very hot, very dense disordered substances. So above the critical point, we don't distinguish between liquid and gas. We say that the substance up here is called a supercritical fluid, SCF. So supercritical just means it's above that critical point. Fluid is the word we use when we intentionally are not distinguishing between a liquid and a gas. Both liquids and gases are fluids. We can have fluids at cold temperatures, which could be either liquid or gas. We can have fluids at high temperatures and high pressures, which we call supercritical fluids, because they're above that critical point. So this T sub C is the temperature above which the enthalpy is no different for the liquid and the gas. In fact, above the at or above the critical temperature, not only is the molar enthalpy of the liquid equal to the gas, but in fact the entropy of the liquid is equal to the gas. The free energy of the liquid is equal to the free energy of the gas and so on. All the properties of the liquid and gas become the same as you approach that critical point and above that critical point, we don't distinguish between the two anymore. We just call them supercritical fluid. So when we draw a phase diagrams separating the solid liquid gas on this pressure temperature curve, the solid liquid line continues essentially indefinitely. It may run into another phase, a different solid phase up above, but there's no reason this solid liquid phase coexistence line needs to stop it. We need to stop drawing it because there is a fundamental difference between an ordered solid and a disordered liquid. But because the only difference between a liquid and a gas is their density, then this curve ends at some point. It ends at the critical point and we in fact don't continue drawing the curve up beyond the critical point. So understanding critical points is important to being able to understand these phase diagrams. It's also worth pointing out that the phase diagrams we've talked about so far are pressure temperature phase diagrams, distinguishing between the solid liquid and gas phases on the pressure and temperature axes. But these are not the only variables that we can use to describe the thermodynamic state of the system. So the next thing we'll do is talk about phase diagrams if we choose a different set of variables for the phase diagram.