 So the Clausius-Clapeyron equation tells us the details of the pressure-temperature behavior, how the pressure changes with temperature along these coexistence lines that involve the gas phase on a pressure-temperature phase diagram. So, for example, sticking with our example of water, we know water boils at 100 degrees C in one atmosphere, so that's this point on the phase coexistence line, the liquid-gas coexistence line, that allows us to answer questions about how much the boiling point would change as we change the pressure or, equivalently, how much the pressure will change if we modify the temperature. So, to see how that works, let's work an example. Let's suppose, again, sticking with water, we're talking about boiling of water, vaporization of water. I'll tell you that the heat of vaporization of water near its normal boiling point is about 41 kilojoules per mole, cost me 41 kilojoules of enthalpy to vaporize one mole of water at the normal boiling point. And we know, normally, when the pressure is one atmosphere, give a couple of sig figs here, so when the pressure is 1.00 atmospheres, the boiling point is 373 Kelvin. Let's suppose, we go to somewhere where the pressure is lower, like the top of a mountain. The mountain, tall mountain closest to me is Mount Mitchell. The elevation at the top of Mount Mitchell is a little bit above 6,600 feet, or in metric units, that's a little over, what is that? I think that's 2,040 meters, roughly. So, let's say I'm going to camp near the top of Mount Mitchell, where the pressure, atmospheric pressure is reduced because I'm at a little bit of altitude, atmospheric pressure is reduced down to only 0.78 atmospheres. So, I'm giving you that number, but we could, in fact, have calculated that number itself. Remember, from the Boltzmann distribution, the population of molecules in the air at some altitude h is related to the Boltzmann factor e to the minus energy, Mgh divided by kT. But I've done that work for you. I've told you that the barometric pressure at the top of Mount Mitchell is 0.78 atmospheres, and I would like to know at what temperature water boils. I'm going to boil myself some macaroni and cheese at the top of Mount Mitchell. What temperature will that water boil under that reduced pressure? In other words, at this somewhat reduced pressure, the boiling point is going to be lower as well. How much will that boiling point be lower? The Clausius-Clapeyron equation tells us how to calculate that. So, I'm looking for what's called these T1 and P1. I'm looking for T2. So, if I rearrange this equation to solve for T2, I'll have to do that in a couple of steps. I'll say that quantity in parentheses 1 over T2 minus 1 over T1 is equal to minus r over the heat of vaporization multiplied by log of P2 over P1. I can then add 1 over T1 to both sides. So, over here, I'll just write 1 over T1. On this side, I'll just add 1 over T1 and subtract those two. And then the number I'm looking for T2 is the reciprocal of this. So, I have to write the reciprocal of everything on the right side, r over heat of vaporization with a negative sign, log P2 over P1 plus 1 over T1. I'll take in the reciprocal, so 1 over that quantity in parentheses. So, we can plug numbers in doing that somewhat carefully to make sure we treat the units correctly. The gas constant, I'll use that in units of joules per mole Kelvin. I'll divide that by heat of vaporization, which I've given you as 41 kilojoules per mole. To cancel the joules here, I'll write it now as 41,000 joules per mole. That looks good. Joules per mole is going to cancel. I multiply that by the natural log of P2 over P1, the lower pressure divided by the higher pressure. In this case, so 0.78 atmospheres divided by the original pressure of one atmosphere. Add that to 1 over this temperature, 373, and take 1 over all of that. So, units look good. Joules cancel. 1 over moles cancel. These atmospheres cancel. I've got a 1 over Kelvin adding to a 1 over Kelvin, and when I take the reciprocal, that's going to end up in units of Kelvin, which is good. And when I do the arithmetic, that works out to be 366 Kelvin. So, what does that mean? 366 Kelvin is 7 degrees cooler than the initial temperature of 373 Kelvin, so the boiling point has reduced a bit. It's been reduced from 100 degrees Celsius to 93 degrees Celsius when I boil water at 0.78 atmospheres rather than 1 atmosphere. So, we are able to calculate how much the boiling point has been reduced when I'm in an environment with lower atmospheric pressure. We can do that, of course, not only up here near the boiling point. We can do it at colder temperatures as well. We don't typically call it boiling when the water changes from the liquid phase to the gaseous phase at a colder temperature. We normally just call that evaporation. If I put a glass of water out in the room, much of the water will evaporate over time. We call that evaporation. The terminology we there use there is in the boiling point. If I do, if I put a glass of water out at room temperature, let's say 25 Celsius, we talk about the vapor pressure of water. So, at room temperature, the vapor pressure of water is about 24 Torr. So, that's roughly a 30th of an atmosphere or so, 3% of an atmosphere in units of atmospheres, but the in equilibrium, liquid and gas are in equilibrium at room temperature. The gas will have a pressure of 24 Torr or 0.03 atmospheres or so. So, we can ask a similar question not about how the temperature changes, but what if I have a glass of water in equilibrium with 24 Torr of water vapor at room temperature, and then let's say I'm doing this outdoors, and overnight the temperature drops to something fairly chilly, 10 degrees Celsius, about 50 degrees Fahrenheit or so. So, if the temperature drops from 25 Celsius to 10 Celsius, what's going to happen to the vapor pressure, it's also going to drop, and we can ask how much the pressure would drop. So, we'll work another example just to illustrate that we can solve for pressures instead of solving for temperatures, pointing out also that the terminology differs when we're talking about vapor pressures, rather than boiling points. So, the question I've set up is if I've got temperature 1 is equal to 25 Celsius or 298 Kelvin, the vapor pressure is 24 Torr at a lower temperature of 10 Celsius, 283 Kelvin, that's not T1, that's T2. What would be the vapor pressure under those conditions? And we also need still to know the enthalpy of vaporization for water. That's actually not the same as it is over here. The enthalpy of vaporization, it cost me 41 kilojoules per mole to boil water up near its boiling point. It cost me a little bit more. It cost me about 44 kilojoules per mole to evaporate water down near room temperature. So, we'll talk in the next video actually about why that is, why the enthalpy of vaporization is different at different temperatures. But for now, we'll just take that as a number that we've looked up or been given 44 kilojoules per mole to evaporate water at 298 Kelvin in this vicinity. So, we have everything we need now to solve for P2. If I rearrange this equation, the form I'd rather use it in is the form I get by exponentiating both sides of this equation. P2 is equal to P1 times the exponential of minus enthalpy of vaporization over R 1 over T2 minus 1 over T1. So, now everything I know is over here on the right side. If we plug numbers in to see how the units work out, P1, we were given it in units of tour. I've got an exponential up in the exponent of that exponential. I've got an enthalpy of vaporization, 44 kilojoules per mole or 44,000 joules per mole divided by the gas constant, 8.314 joules per mole Kelvin. Multiply that by 1 over T2. So, 1 over 283 minus 1 over T1. And again, just to double check the units, those are going to work the same as before. Joules per mole, cancel. This 1 over Kelvin outside the parentheses, sorry, 1 over 1 over Kelvin outside the parentheses will perfectly cancel the 1 over Kelvin inside the parentheses. This is the 1 over Kelvin in the numerator. This is the 1 over Kelvin in the denominator. So, the units in the exponent all go away, which is good. That's what they should do. And now this arithmetic works out to be 24 tour times e to the sum negative number. So, it's going to reduce 24 tour down to with the number of sig figs we deserve, 9 tour. So, physically speaking, what that tells us about the real world is if water is in equilibrium with its vapor at 24 tour, at room temperature, when the temperature drops down to 10 degrees Celsius, the vapor pressure of water drops as well to only 10 tour. So, that's why, for example, dew forms on the grass overnight when it gets cold is if there was 24 tour of water vapor in the air during the day. When the temperature was in the 70s Fahrenheit, when the temperature drops to 50 degrees overnight, only 10 tours worth of that water vapor can stay in the air. The rest of it has to condense down into the liquid phase. So, the Clausius-Clapeyron equation can allow us to do calculations. Both of these are the same thing. They're just telling us how P1 and T1 change to P2 and T2 as we slide along this coexistence curve. We can phrase that in terms of how the boiling point changes if I drop the pressure or equivalently if I were to increase the pressure or we could phrase it in terms of how the vapor pressure changes as I increase or decrease the temperature. So, we can do that for boiling points, vapor pressures, we can do it for increases or decreases, we could do it for substances other than water as long as we know the heats of vaporization. So, the Clausius-Clapeyron equation is quite useful for this type of calculation, but remember we can only use it for the phase coexistence lines that involve the gas phase.