 Okay, so now we're in a position to combine a couple of things we have pointed out and obtained something really useful from them, which is, first of all, we've seen that we can calculate the chemical potential in a solution, a mixture of two different substances in the liquid phase. Relative to that in the liquid, if we know the pressure above the solution and the vapor pressure of that component, the only problem is we don't know much about how to calculate that partial pressure above the solution, and we'd like to be able to. The other thing we've reminded ourselves of recently is that for an ideal solution, one where the energy doesn't change when I mix two substances together, the entropy of mixing can be calculated from the mole fractions of the components. So just to make sure we have a decent picture in our head of what these are talking about, an ideal solution would be one for which, and we've thought about this as a lattice model, but we've got molecules of one type mixed in with molecules of a different type, and when I change the configuration of that system, the energy doesn't change. Mixing, this process where I'm talking about the entropy of mixing, if I have a beaker of A and I have another beaker of the different component B and I mix those together to make a larger system with both A and B in it. So B molecules and A molecules both in this liquid. When I combine those two solutions together, they go from being pure liquids to a mixture of two different liquids, and the entropy of mixing is the entropy change when I go from those reactants, the two pure liquids to the product, the impure solution. So we know enough now about ideal solutions to calculate not just the entropy of mixing, but also the Gibbs free energy of mixing. For example, we know for a pure solution, I'm sorry, for an ideal solution, because it's ideal, the enthalpy of mixing must be zero. The energy doesn't change when I mix the solutions together, but the free energy, the Gibbs energy, when I mix two solutions together, must be delta H minus T delta S. So since I know the enthalpy of mixing and the entropy of mixing, delta H is zero. If I subtract that from that T times the entropy of mixing, so this negative sign and this negative sign cancel each other, and I've got kT, which I'll choose to write as RT times this term in parentheses, the sum of the mole fractions times log of mole fractions. That is the Gibbs free energy of mixing for two pure components mixed together with some mole fractions. So, so far so good. We can actually, that is the molar Gibbs free energy of mixing I should have at the beginning specified that this was the molar entropy of mixing, and I know that because there's no n's anywhere, it would be n times k if I were talking about the extensive entropy of mixing. So, if I do want to write this as an extensive entropy of mixing without that bar on top, I need to multiply by the total number of moles in the system. So, since these mole fractions, actually those are not the ones I'm interested in, since that mole fraction looks like moles of B over total moles. Likewise, this mole fraction looks like moles of A over total moles. If I multiply on the left and on the right of this expression by total number of moles, that's just going to get rid of the denominator in these mole fractions. So, the extensive Gibbs free energy of mixing is going to be RT times not mole fraction, but just the numerator from that mole fraction, just n sub A and the XB is going to turn into just n sub B times natural log of XB. All right, so far so good. That is a thermodynamic description of what the Gibbs free energy will change will be when I mix together two different liquids to make an ideal solution. That delta G of mixing I can think about in a slightly different way. Delta G of mixing is the free energy of the product, the free energy of the ideal solution minus the free energy of reactants. So, minus free energy of the pure liquid A and pure liquid B in the liquid phase. Those are absolute free energies. What I can use to write those is to remind ourselves where this comes from. We have seen previously that when we derive the Gibbs-Duhem expression that the Gibbs free energy can be written as the sum of the moles times chemical potentials. So, in some cases that's going to be quite easy. In the solution, in the two component solution, that's going to be equal to moles of A times chemical potential of A in the solution phase plus moles of B times chemical potential of B in the solution phase. What I've written here, that's the Gibbs free energy in the solution. This two-component solution that I've written with this intermediate result we got from the Gibbs-Duhem derivation. I have to subtract from that Gibbs free energy in the liquid phase, the pure liquid A phase. So, that's just moles of A times chemical potential of A in the liquid. And I also need to subtract Gibbs free energy of B, which is moles of B times chemical potential of B in the liquid phase. So, I have moles of A showing up a couple different places, moles of B showing up in a couple different places. So, that I can rewrite as moles of A times chemical potential of solution minus chemical potential in the liquid. And then the same thing for component B, moles of A, chemical potential of A of B in solution minus chemical potential of B in the liquid. But that difference, chemical potential in the solution minus the chemical potential in the liquid, how big is the chemical potential in the solution relative to that of the liquid? That's exactly what this expression tells us. We know how to calculate the chemical potential in the solution relative to that in the liquid. The difference is just this term RT log pressure over vapor pressure. So, I'll rewrite this again. I've got N sub A. This difference in brackets is equal to RT natural log of pressure of A relative to its vapor pressure in this solution. The second term looks like the same thing with B's, moles of B, RT natural log partial pressure of B relative to its vapor pressure, partial pressure above that particular ideal solution. So, now we have a completely different expression for the free energy of mixing. I've derived this result and this result for the free energy of mixing, they must be equal to the same thing. One of them looks like N RT log pressure over pressure. The other one looks like N RT log mole fraction. One term for component A, a second term for component B. So, what that tells us, if I combine these two expressions, is that this ratio, pressure of A over vapor pressure of A, has to be equal to that expression, just mole fraction. Or, if I rearrange that a little bit, the partial pressure of A above this solution, the thing that we don't know, the reason we can't use this expression usefully is that we don't know what the partial pressure of A is above a solution. Rearranging this expression tells me what that is. Partial pressure of A above that solution is equal to the mole fraction, multiplied by the vapor pressure. So, that's the version worth putting in a box because we will use that over and over. If I have a solution, an ideal solution, let me draw a bigger picture of this so we can understand what's going on. If I have my ideal solution with some B molecules and some A molecules mixed in some mole fraction, so I'm only drawing the ones at the surface, but there's molecules throughout this solution. I've got a mixture of A molecules and B molecules. Above the pure substance, there's some vapor pressure. Liquid in equilibrium with its vapor, when the liquid is pure, we call that pressure the vapor pressure. There's a vapor pressure of B, there's a vapor pressure of A. What this expression tells us is the partial pressure of A above this particular solution in equilibrium with the solution is not the pure substance vapor pressure, it's mole fraction times vapor pressure. If, in my case, only three out of seven of the molecules at the surface of the solution are A molecules, I guess the pink ones I've drawn here are B molecules, but if some fraction of the molecules at the surface are A molecules, that limits the fraction of them that can escape from the liquid and enter the vapor phase. The fraction of them entering and becoming vapor in this solution is reduced to a fraction representing the number of them that appear at the surface. This relatively simple expression tells us if I have a 50-50 mixture of A and B, then the pressure of A above the solution is 50% of the vapor pressure. If it's only 10% A and 90% B, then the pressure above the solution is only 10% of the vapor pressure. So this linear relationship that's a nice, easy mathematical relationship turns out to have plenty of important consequences that we can calculate about the properties of these solutions. And keep in mind that this particular result that we've obtained called Raoult's Law, we've derived that specifically for the case of an ideal solution, only when the entropy of mixing is equal to this result did the results follow in the way that they did. So for an ideal solution, equivalently a solution that obeys Raoult's Law, we can predict the partial pressure above a solution if we know just what the mole fraction is and what the vapor pressure of the pure substances.