ChemicalThermodynamics / VLEmediumFE ExamCourseworkPE / Advanced

Vapour Composition over an Ideal Binary Liquid

An ideal binary liquid mixture has $x_A = 0.40$ (mole fraction of the more volatile component A).

At the system temperature, the pure-component vapour pressures are $P_A^{sat} = 120\ \text{kPa}$ and $P_B^{sat} = 60\ \text{kPa}$ .

Find the mole fraction of A in the vapour, $y_A$ .

Given

xA	0.4 -	Liquid mole fraction of A
PsatA	120 kPa	Vapour pressure of pure A
PsatB	60 kPa	Vapour pressure of pure B

Hint 1

Raoult's law gives each component's partial pressure from its liquid mole fraction.

Hint 2

Remember x_B = 1 − x_A. Total pressure is the sum of the partials.

Hint 3

y_A = p_A/P, and it should come out greater than x_A.

Worked solution — try the problem first

Raoult's law gives each partial pressure:

$p_A = x_A P_A^{sat} = 0.40(120) = 48\ \text{kPa}$ $p_B = x_B P_B^{sat} = 0.60(60) = 36\ \text{kPa}$

Note $x_B = 1 - x_A = 0.60$ .

Total pressure (Dalton):

$P = 48 + 36 = 84\ \text{kPa}$

Vapour composition:

$y_A = \frac{p_A}{P} = \frac{48}{84} = 0.571$

$\boxed{y_A = 0.571}$

The vapour is enriched in A, as it must be: $y_A = 0.571 > x_A = 0.40$ because A is the more volatile component. That enrichment is precisely what distillation exploits, and it is your check on the answer — if $y_A$ had come out below $x_A$ , something is wrong.

Relative volatility here is $\alpha = 120/60 = 2.0$ , a comfortable separation.

Concepts:Raoult's lawVapour-liquid equilibriumDalton's lawRelative volatility

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