In the previous lecture, we developed the relationship between the entropy of the gas and the pressure of the gas. We discovered that if the pressure of the gas goes down, the entropy of the gas goes up because the volume of the gas is going up and that means there is more room for the gas molecules to move around. We also introduce the notion of the free energy as a means of determining where the location of equilibrium will be. Let's review those results very quickly. The entropy of the universe does not increase when we are at equilibrium, and we rewrote that in terms of the Gibbs free energy changes here at the bottom. To give us a criterion that says at equilibrium, delta H minus T delta S is 0, and correspondingly, delta G is equal to 0. In the specific case of the vaporization of water, we were able to use our expression for the entropy as a function of the pressure to calculate that the entropy of liquid water when we are at 23.8 tor is 217 joules per mole kelvin. The entropy of the liquid is 69.9. The enthalpy change for the vaporization is 44.0 kilojoules. If we take all of those numbers and plug them into the expression at the bottom here we actually found that we were at zero telling us that the system is at equilibrium at 23.8 tor, consistent with what we've observed experimentally. Our goal, in this particular lecture, is to generalize these results. So remember this is the expression that we derived in the previous lecture for the entropy as a function of the pressure of the gas. Remember what this says is that when P2 is smaller than P1, then the right side of the equation is positive, so the left side of the equation is positive. So the entropy is greater when the pressure is smaller. In addition, now, we can write, actually I wanted to write over here first. That at equilibrium we know that delta G is equal to 0, which is equal to delta H minus T delta S. I can rearrange this expression by simply moving the T delta S to the other side of the equation and dividing by T. I get that, at equilibrium delta S is equal to delta H divided by the temperature. Remember all these means is that the entropy change in the system is exactly offset by the entropy change in the surroundings resulting from the energy flow either into or out of the surroundings, dependent upon whether it's an endothermic or exothermic process. So here we're balancing the enthalpy change in this system with the enthalpy change in the universe. And that's all the expression says. But now we can actually calculate this delta s for the process of liquid going to gas. Because delta S is simply the entropy of the gas at whatever pressure that gas is at, minus the entropy of the liquid. And of course, from the previous expression, we can easily use the previous expression for the entropy at P2 and plug it in for the entropy of the gas at pressure P and that gives us the result. That delta s is equal to S at pressure P minus the entropy of the liquid, but that's s for the gas at pressure P1, minus this R logarithm P over P1 term. And now what we want to do is actually focus in on the possibility that P1 is equal to 1 atmosphere. So that we just have standard conditions for P1, if that's the case, then clearly this P1 here is just a 1, and this becomes as S0. So we wind up with the expression then that delta S becomes equal to delta S0, that is under standard conditions, minus RT logarithm P, subject only to the assumption that we have taken one of the pressures, the standard pressure could be 1 atmosphere. Let's now use that in our previous expression back here, if we, oops. If we take this delta S and plug it back into our expression for equilibrium which is over here on the pad, then we wind up with the very straightforward result that given the delta S, is as we derived on the previous slide, and is also equal to this change here. And we know that delta S is delta H over T. We can now write that delta H0 over T is delta S0 minus R logarithm of P. Or, just putting terms on the same side of the equation, We wind up with a fairly general result here, which relates the pressure at equilibrium and the temperature at equilibrium to the change in enthalpy for the process, and the change of entropy in the process, for the specific case that we are talking about, the vaporization of a liquid. This equation also would result for the sublimation of a solid as well. What we can do then, is relate for any given temperature, we can predict what the vapor pressure would be. That actually means that we can use this equation and graph it. I could, for example, plot the logarithm of the pressure versus 1 over the temperature. And what I would see on this graph is that I have a slope which is since delta H is greater than zero for any vaporization process, we get a straight line graph whose slope, is equal to minus delta H0 over R. And the alternative, we could actually plot a graph of the temperature versus the pressure and I've actually done that just by simply plugging in the values of delta H0 for the vaporization of water in Delta S0 for vaporization of water, we get this graph. That graph hopefully looks very familiar. That's in fact the standard shape of the vapor pressure as a function of temperature curve back when we were doing phase diagrams relating liquids and gasses. This is the boundary between the liquid and the gas. It's the vapor pressure and also the boiling point curve. So we have now, on the basis of the second law of Thermodynamics and a few simply assumptions, derived the relationship between vapor pressure and temperature. We can catch that in terms of Gibbs Free Energy relatively straight forwardly. We have our relationship here that delta S0 minus delta H0 over T is equal to the R logarithm of P, if all we do is multiply this expression by minus T, then minus T delta S0 plus delta H0 is equal to minus RT logarithm the P. And of course now I notice on the left side of the equation, that's actually equal to delta G0. That gives rise to this beautiful expression relating the changes in the thermodynamic properties to the equilibrium pressure and the equilibrium temperature. There's one other way we could actually rewrite this equation to to give us another beautiful result that is somewhat general. Let's take our expression written here as delta S0 minus delta H0 over T, equals R logarithm of P. And let's take that expression for two different pressures correspond to two different temperatures. For example, at temperature T2, we have equilibrium pressure, P2. At temperature T1, we have equilibrium pressure, P1. We can now actually take the difference between these two equations. Let's subtract the second equation from the first one. The delta S0's go away. We have minus delta H0. We'll push this up over here. Divided by T2 minus negative delta H0, over T1, is equal to R logarithm P2, minus R logarithm of P1. The left side of the equation can be simplified, by moving the minus delta H out in front and we'll go ahead and divide by R, while we're at it. We have minus delta H0 over R multiplied by 1 over T2 minus 1 over T1. The right side of the equation can be simplified. Remember I've already divided by the R, the logarithm of P2 minus the logarithm of P1 is logarithm of P2 divided by P1. This expression is actually an extremely important expression referred to as the Clausius-Clapeyron equation, that actually allows us to relate the vapor pressure as a function of the temperature. And it is, in fact, a graph of, or if we graph this equation, we get either this graph I've shown here, Or if I plot the pressure versus the temperature, it is the graph of this particular curve here. Let's do a quick double check here, let's take the circumstance that, for example, T2 is greater than T1. So we're increasing the temperature. In this expression, if I in fact increase T2 over T1, then this term here is a negative number, because 1 over T2 is going to be smaller than 1 over T1. But delta H vaporization is a positive number And then we have this minus sign here. So the right side of the equation in that circumstance is greater than zero, so the left side of the equation is greater than zero. And if a logarithm is greater than zero, then the argument is greater than one. That means that this stuff in parenthesis is a number greater than one. That means that the vapor pressure at temperature T2 is greater than the vapor pressure at T1. So P2 is greater than P1 exactly as we expected it to be on the basis of either of the graphs that we've shown here. This is an extremely powerful equation. It means that if I know the vapor pressure at one temperature I can find the vapor pressure at any temperature within a reasonable range. And I can actually make predictions from that for example of where the boiling point might be. If I know the vapor pressure at one temperature, I can find the boiling point at other pressure that I am interested in. This is an extremely powerful result resulting basically from the observation of the Second Law of Thermodynamics, and the condition of equilibrium that delta H minus T delta S has to be equal to 0 at equilibrium. Or alternatively the delta G has to be equal to 0 at equilibrium. Given the power of this for predicting equilibrium for, vapor pressures. It'll be also interesting to find out how this can work to predict equilibrium for chemical reactions. And we'll pick that up in the next lecture.