Raoult's Law for Two Volatile Liquids

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Take a liquid mixture of benzene and toluene. The liquid is forty percent benzene, by moles. So how much benzene is in the vapour just above it? Forty percent, the same? Surprise: it is almost seventy percent! How can the vapour be so much richer than the liquid? Raoult's law answers exactly this, and it comes up often in JEE, so let's begin!

0:29Raoult's law: each liquid's share of the pressure

So, take two volatile liquids, A and B: both of them evaporate. Mix them in a closed flask, and say they form an ideal solution, as on screen. Both kinds of molecules escape into the vapour, okay? But think. At the surface, only some of the molecules are now A molecules. So will A still push as hard as it did when pure? No, less! Its pressure is cut down by its share of the moles. That is Raoult's law: p A equals x A times p A nought. p with the subscript A is the partial vapour pressure of A; x with the subscript A is its mole fraction in the liquid. The small circle on top, read as nought, means the pure liquid. The same for B: p B equals x B times p B nought. Right? So, p A nought and p B nought are the vapour pressures of the pure liquids, taken at the same temperature as the solution, the capital T on screen. With me? Now the total pressure. By Dalton's law of partial pressures, the two simply add: capital P equals p A plus p B, nothing more. Clear? Since x B is one minus x A, substitute: capital P equals p B nought, plus, p A nought minus p B nought, times x A, okay? Look at the graph: the rising thin line is p A, the falling one is p B, and their sum, the thick line on top, runs straight from p B nought, for pure B, up to p A nought, for pure A, see?

2:03What is in the vapour?

Now, the vapour. One more symbol: y with the subscript A, the mole fraction of A in the vapour. Small x for the liquid, small y for the vapour, okay? Now tell me: is y A the same as x A? Is the vapour the same mix as the liquid? By Dalton's law, a gas's mole fraction is its share of the total pressure. So y A equals p A over capital P, that is, x A times p A nought, over capital P, right? That is not x A in general. So no, the vapour is a different mix! Mixing up x and y is the most common trap here. Don't fall for it! One more relation, handy when the vapour is given. One over capital P equals y A over p A nought, plus y B over p B nought. It comes from the last line, using x A plus x B equals one. Got it? Now, compared with the liquid, which of the two liquids will the vapour be richer in? Yes, the more volatile one! The liquid with the higher pure vapour pressure escapes more easily, so the vapour has a bigger share of it than the liquid has. We will prove it in a moment. Clear? And here is why it matters. Condense that vapour: the new liquid is richer in the volatile one. Evaporate it, condense again, and repeat. This is fractional distillation: the way we separate two such liquids. Okay?

3:31Topper's check: P lies between the two

Now, a topper's trick. Write capital P as x A times p A nought, plus x B times p B nought. Since x A and x B add up to one, capital P is a weighted average of the two pure vapour pressures. With me? So, if p A nought is greater than p B nought, capital P always lies between them: greater than p B nought, less than p A nought, as the less-than signs on screen show. On the graph, the thick line stays in that band. So check every answer: a total pressure outside this range is a mistake. Okay? And here is the proof I promised. From the vapour formula, y A over x A equals p A nought over capital P, which is more than one, because capital P is less than p A nought. The double arrow means which gives: y A is greater than x A, so the vapour is richer in the more volatile liquid, for every mixture. Right?

4:25Example 1: benzene-toluene vapour pressure and composition

Let's try one. At twenty degrees Celsius, the little circle before C, benzene and toluene have vapour pressures of seventy five and twenty two torr, a unit of pressure. Find the total vapour pressure, and the vapour composition, above a solution where x benzene, the mole fraction of benzene, is zero point four. Pause the video, and try it first. Done? Okay, let's check. On screen, the subscript B means benzene, and T means toluene, okay? Benzene's partial pressure is zero point four times seventy five: thirty torr. Right? Now toluene. What is its mole fraction in the liquid? One minus zero point four: zero point six, yes! So p T is zero point six times twenty two: thirteen point two torr. Clear? Add them: capital P is forty three point two torr. Topper's check: it lies between twenty two and seventy five. Perfect! Now the vapour. The liquid is forty percent benzene, so the vapour is too, right? No, no, careful! That is the x and y trap. y B is benzene's share of the pressure: thirty over forty three point two, that is zero point six nine four. And there is our puzzle, solved! Forty percent benzene in the liquid, but about sixty nine percent in the vapour. Got it?

5:48Example 2: liquid composition from equal-mole vapour

Now a harder one. p A nought is one hundred and p B nought is fifty, in m m H g, millimetres of mercury. The vapour contains equal moles of A and B, okay? Find the total pressure and the liquid composition. Pause the video, and try it. Done? Alright. Equal moles, so x A is zero point five, and the pressure is just the average, seventy five, right? No, no, wait! Equal moles in the vapour, not in the liquid. So y A equals y B equals zero point five. Now, which of our formulas uses y directly? Yes, exactly, the one over capital P formula! Zero point five over one hundred, plus zero point five over fifty: zero point zero zero five plus zero point zero one, that is zero point zero one five. With me? So capital P is one over zero point zero one five: about sixty six point seven millimetres of mercury; the wavy equals sign means approximately. Between fifty and one hundred, and not seventy five. See? Now the liquid. Turn the vapour formula around: x A equals y A times capital P, over p A nought. Zero point five times sixty six point seven, over one hundred: about zero point three three three. Okay? That is one third! So the liquid is only one third A, but the vapour is half A, just as our proof said: richer in A, the more volatile liquid. That's it!

7:17Quick revision: p = x p0, y = p/P and the 1/P formula

So, quick revision. Raoult's law: p A equals x A times p A nought, and capital P, p A plus p B, is a straight line in x A, okay? For the vapour, y A equals p A over capital P; when y is given, use the one over capital P formula. And never mix up x and y: the vapour is richer in the more volatile liquid. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!