 Alright, so here's a summary of what Raoult's law can tell us so far about binary solutions, ideal solutions. First we just have a sketch of our binary solution, A and B in the liquid phase in equilibrium with their vapors. Raoult's law, of course, tells us that the partial pressure of any one of those components above the solution is its mole fraction times the total vapor pressure, summarized in this somewhat complicated looking diagram that has a very simple interpretation, which is that as the mole fraction changes from 0%A up to 100%A from 0 up to 1, we change from a pure solution with pressure equal to the vapor pressure of B up to a pure solution with a pressure equal to the vapor pressure of A and we just move linearly from one to the other. So that has all been in terms of the mole fraction in the liquid phase if instead I want to think about the mole fraction in the vapor phase. So I'm going to use, instead of x, which is the mole fraction, I'll use x for the mole fraction of the liquid phase and I'll use y for the mole fraction of the vapor phase. That's a fairly standard convention so we can keep track of which mole fraction we're talking about. The mole fraction in the vapor phase is going to be, we can think of it as moles of A over total number of moles in the vapor phase, but since it's in the gas phase of the vapor phase that we're talking about, I can think of N as PV over RT. So it's PV over RT for A, so in other words partial pressure of A times V over RT divided by total number of moles and that would be total pressure. In this case, if we think back to the case let's say where we have a 50-50 mixture that had 100 tor of B and 200 tor of A just as an example, if we have 200 tor of A and 100 tor worth of B, then after some cancellation here the V, all these systems are in the same volume, they're at the same temperature, R is a constant, so the V and the R and the T cancel and we just have PA over P total. In this case, PA of 200 over total pressure of 300 for that 50-50 mixture of a solution where A is more volatile has a higher vapor pressure than B, it turns out that the mole fraction in the vapor phase is two thirds, it's bigger than one half. The important observation here is the vapor has been enriched in the more volatile component. The more volatile component with the higher vapor pressure is A, the mole fraction in the vapor phase is greater than the mole fraction in the liquid phase and that's going to turn out to be a general trend. If we want to draw a more general conclusion about not just for specific cases with specific numbers but if I want an expression for how to compute the mole fraction in the vapor phase in the general case, let's go back to this expression. So before I plug actual numbers in, the mole fraction in the vapor phase is pressure in the vapor phase of that component divided by the total pressure. So partial pressure of A in the vapor phase, Raoult's law tells us how to calculate that. That's mole fraction of A times vapor pressure of A and I want to divide that by total pressure. But we know something else about the mole fraction of A and or the total pressure. This expression tells us what the total pressure is as a function of the mole fraction of A. So in fact probably the easiest way to think about this is to take this expression and rewrite it to solve for mole fraction of A. So if I write mole fraction of A, so I'm going to insert that in right here. So if I'll solve this on the fly, I'll write mole fraction of A is equal to total pressure minus Pb star, moving this one to the other side and then I'll divide that by this quantity in parentheses, Pa star minus Pb star. So all I've done here is I've solved for mole fraction of A in terms of the total pressure, inserted it here for the mole fraction of A. I still have to multiply by these terms Pa star and P total. That, if I know these numbers, if I know enough to plug into the right side I can compute now the mole fraction of A. What I'm actually more interested in is drawing a graph like this one pressure as a function of mole fraction. So I'd really rather rearrange and solve this equation solving for P total which appears here twice as a function of mole fraction in the vapor phase. So let's work on that. To get rid of the denominator, I'll just take these two terms and multiply over to the left. So I've got P total times vapor pressure of A minus vapor pressure of B all multiplying mole fraction. That's on the left side. That's all equal to what's left on the right side is this numerator, P total times Pa star minus Pa star times Pb star. So that's equally ugly but it's no longer involves fractions. If I want to isolate the two terms involving P total, let's put the Pa star Pb star over on the left side and bring this term with the P total over to the right side so that everything on the right will involve P total. And I've got P total times Pa star. And once I bring this over, I've got P total times a negative y sub A Pa star minus Pb star. And if my goal is to rearrange this and solve for P total, I'll do that over here. I can take the term in brackets, move it over underneath the Pa star Pb star on the left. And I've got Pa star Pb star in the numerator. And then in the denominator, Pa star minus mole fraction in the vapor phase. And again, this difference of the mole, the vapor pressures, Pa star minus Pb star. All right. That's what I claim to be after. That's the total pressure of the solution as a function of the mole fraction if I also happen to know these vapor pressures. So I can draw a graph much like this one. If I draw a graph of total pressure as a function of mole fraction, but now mole fraction in the vapor phase, I'd have to graph this function. We can find the two limiting cases relatively easily. When the mole fraction is zero, if I put a zero in here, so the whole second half of this denominator disappears, the Pa star is cancel, then I'm going to have just as before. And unsurprisingly, when I have mole fraction zero, if the vapor above the solution is purely B, then the solution must also be pure B and the partial pressure is equal to the vapor pressure. On the other hand, if all of the vapor is A, if 100% of the vapor above that solution is A, because the mole fraction is 100%, if y sub A is equal to one, then let's see this negative one times Pa star is going to cancel this Pa star, negative one times negative Pb star, that Pb star cancels this Pb star, and all I'm left with is a vapor pressure of A Pa star. So far, this is looking exactly like it did in the case we considered before, but we can look at this equation and know that it's not going to be a straight line connecting these two. This is no longer the equation of a straight line. It doesn't look like this is a constant plus a slope times the variable we're interested in. It's a nonlinear function, and if we stared at it a little bit harder, we could convince ourselves that it's actually going to not only not be linear, it's going to dip below linear, so I'll draw this graph as a sagging line of sorts. It still is equal to vapor pressure of B on this side, vapor pressure of A on this side, but it's not a straight line. In fact, if I combine these two graphs onto one, if I plot total pressure as a function, so this upper curve is total pressure as a function of x of A, the lower curve is total pressure as a function of y sub A, the two curves are different. This expression is different than this expression. So I have two different equations, two different curves for how the total pressure depends on mole fraction, because mole fraction in the liquid and mole fraction in the vapor are not equal to each other. That itself has some interesting consequences, and if we think carefully about how to interpret this diagram, we'll learn a lot about the equilibrium between liquids and vapors above a solution. So that's what we'll do next.