 So, we've seen now this equation that tells us the slope of one of these coexistence lines on a phase diagram. That equation is called the Clapeyron equation and we can determine that slope if we know the enthalpy and the change in volume and the temperature of a phase transition. So we'll see how that works with an example. We'll take a substance that we're fairly familiar with, water. So if I tell you that the heat of vaporization, the molar enthalpy of vaporization, or actually not vaporization, we'll do fusion. So the molar enthalpy of fusion, when I melt solid water to make it liquid, is 6 kJ per mole. I can also tell you the molar volume of the liquid, the molar volume of the solid. Probably the easiest way to think about those is we could look up pretty easily the density of the liquid. The density of liquid is pretty close to 1 gram per cubic centimeter. Look up the density of solid, the density of ice. And then convert those densities to molar volumes relatively easily. I'll just give you the molar volumes. One mole of water, 18 grams of water, takes up about 18 milliliters. So that molar volume is 18 and if I need a few more sig figs, that turns out to be 18.0124 cubic centimeters is the volume occupied by one mole of liquid water at its freezing point. The molar volume of ice, we know that ice is less dense than water. Ice cubes float in water, so the molar volume is going to be a little bit larger. If we look at what the number actually is, the density of ice at the freezing point is a little over 19 and a half cubic centimeters per mole. So that same mole of water expands and takes up a slightly larger volume when it freezes. That's enough information combined with the fact that we already know the freezing point for us to calculate the slope of that solid liquid coexistence line using the Clapperon equation. So if that's our question, what is the slope dp dt for the solid liquid coexistence line? Clapperon equation tells us how to calculate that. Enthalpy of fusion, 6.01 kilojoules per mole or 6,010 joules per mole. The temperature I need is the temperature of fusion, the temperature of the melting point. So that's 273.15 Kelvin, freezing point of water. And then we need the change in the molar volume for fusion. So remember the fusion process, when we say enthalpy of fusion, temperature of fusion, molar volume change upon fusion, the fusion process is melting, so solid becoming liquid. So it requires an input of 6 kilojoules per mole to convert lower energy, lower enthalpy solid into higher enthalpy liquid. So we have to make sure we're doing this in the right order. The change in volume upon fusion is the change when we go from solid to liquid. So the delta V is volume of the liquid, 18 cubic centimeters per mole, minus the molar volume of the solid, 19 and change, cubic centimeters per mole. And that would be that, except we need to be a little bit careful about units. Moles are fine, the moles in the denominator of the denominator, cancel moles in the denominator of the numerator. But we're left with joules and Kelvin and cubic centimeters, none of which seem like they'll cancel each other. We can, however, do some unit conversion. So let's multiply this whole thing by, a joule is not equivalent to a cubic centimeter, but a joule is equivalent to a liter, some number of liter atmospheres. Remember, energy can be expressed in units of joules or as a pressure volume product as liter atmospheres. So recalling the values of the gas constant, the gas constant is either 8.314 in units of joules per mole Kelvin, or .08206 in units of liter atmospheres per mole Kelvin. So that unit conversion is going to allow us to convert joules to liter atmospheres. And now at least we've got a volume here, and if I write down volume as liters in the denominator and cubic centimeters in the numerator, these liters will cancel, cubic centimeters will cancel, and I'm in good shape, and I need to know that there's a thousand cubic centimeters in one liter. So then by the time we're done those unit conversions, we've got an answer that's going to be in units of atmospheres in the numerator, and I've still got a Kelvin that hasn't canceled in the denominator. So that's what I expect. That's a ratio of a pressure to a temperature should come out in something like atmospheres per Kelvin. The units have worked out fine. If I do the arithmetic of calculating these numbers, what I'll find is that that works out to negative 133. Notice the negative sign came because I took a small volume of the liquid, 18, and I subtracted a larger volume for the solid, 19. So this quantity in parentheses is a negative number, which is where that negative sign came from. So what we've learned from the Clapperon equation is the slope of the fusion coexistence curve, the slope of the solid-liquid coexistence line on a pressure-temperature phase diagram, solid-liquid gas. The line we're talking about is the solid-liquid coexistence line. The slope of that line, I've drawn it leaning backwards because Clapperon told us that the slope of that line is negative 133 atmospheres per Kelvin. When I change the temperature by positive one Kelvin, the pressure drops by 133 atmospheres. So that's quite a steep curve. It's leaning backwards, but it's pointing upwards quite close to vertical. So that's one thing that this Clapperon equation tells us. We can calculate numerically what the slopes of these lines are. So near the normal freezing point of water, the slope of that line is this particular numerical value. The other thing it tells us that's interesting is in fact the sign of this equation. The one we've just calculated, the solid-liquid coexistence line, we now know slopes backwards when we're talking about water. On the phase diagram for water, that solid-liquid coexistence line has a negative slope. For other substances, it won't work that way. So for most other substances, it worked this way in particular, as I pointed out, because the volume of the liquid is a smaller number than the volume of the solid. So when the volume of the liquid is less than the volume of the solid like it is for water, or conversely, if the density of the liquid is greater than the density of the solid, then the curve will look this way. If those things are reversed as they are for most substances, if we have a substance where the molar volume of the liquid is larger than that of the solid, or the density of the liquid is less than that of the solid, that would be a substance whose ice cubes sink rather than float. So a substance whose denser in the solid phase than the liquid phase, so the solid form would sink in its own liquid. The phase diagram for that substance is still going to have a solid and a liquid and a gas. But that phase diagram is going to have a solid liquid coexistence curve, a melting line that leans forward. It's going to have a positive slope. Most likely it'll still be quite a steep curve, quite a steep slope, because the volumes won't be all that different from one another, but sometimes it'll slope negatively, sometimes it'll slope positively depending on the relative densities or the relative molar volumes of the solid and liquid phase. The last thing we can learn from this type of calculation, in addition to the slope of this curve, we can think about what that slope means physically. It's perhaps a little difficult to think about the slope in this direction. If I say, if I were to change the melting point of water by a Kelvin, that's going to change its vapor pressure, not its vapor pressure, sorry, its coexistence pressure by 133 atmospheres, that's kind of difficult to imagine. It's a little bit easier to understand what that means if I do it the other way around. If I ask, by exerting pressure on the substance, how much do I change its melting point? So that's just the reciprocal. I've turned this derivative upside down. So that's going to be 1 over negative 133 atmospheres per Kelvin. And that's going to work out to still negative, but now a fairly small number instead of a large number. And that one's easier to think about physically. So the change in the temperature of fusion, the change in the melting point is negative, let's call that negative a little less than a hundredth of a Kelvin per atmosphere. If I change the pressure by one atmosphere, I'm going to change the melting point by negative 0.00754 Kelvin. So instead of melting ice at ambient pressure, one atmosphere, if I were to melt that same ice by exerting two atmospheres of pressure on it or ten atmospheres of pressure on it, if I increase the pressure by a few atmospheres, I'm going to decrease, because of the negative sign, I'm going to decrease the melting point by a few hundredths of a Kelvin. So that's instead of melting at one atmosphere and the normal melting point, if I increase the pressure to a larger pressure, the melting point will have shifted slightly negative. It'll be a few hundredths of a degree less than 273 Kelvin. So we can decrease the melting point of ice by exerting pressure on it. If it weren't water, but some other substance, I could increase the melting point of that substance by exerting higher pressures on it. So those give you some examples of how to use the Clapeyron equation as well as the significance of the results we achieve by using it.