Vapour Composition from Partial Pressures

Introduction

The liquid and vapour above a mixture usually have different compositions. The component with a higher pure-liquid vapour pressure tends to be enriched in the vapour. First use liquid mole fractions to calculate component partial pressures; then divide each partial pressure by their sum to obtain the equilibrium vapour composition. Keeping x for liquid and y for vapour makes the two stages clear.

Core explanation

For an ideal binary liquid at one temperature, Raoult's law gives P A = x A P A and P B = x B P B . The vapour is treated as an ideal gas mixture, so Dalton's law gives y A = P A/P total and y B = P B/P total, with P total = P A + P B. Combining them yields y A = x A P A /[x A P A + (1 − x A)P B ]. The liquid fractions x A and x B sum to one, and the vapour fractions y A and y B also sum to one, but x A need not equal y A.

As a hypothetical example at a fixed temperature, let P A = 100 kPa and P B = 40.0 kPa. Liquid x A = 0.300 gives x B = 0.700, P A = 30.0 kPa and P B = 28.0 kPa. Total vapour pressure is 58.0 kPa. Therefore y A = 30.0/58.0 = 0.517 and y B = 28.0/58.0 = 0.483. Component A is only 30.0% of liquid moles but 51.7% of vapour moles. Its greater pure-component vapour pressure produces this enrichment.

For an ideal binary mixture with P A > P B , one can show y A > x A for any interior composition 0 < x A < 1. Let α = P A /P B > 1. Then y A = αx A/[1 + (α − 1)x A]. Since the denominator is smaller than α for x A < 1, y A/x A > 1. At pure A or pure B endpoints, liquid and vapour fractions coincide because only one component remains. The equation is a model result, not a guarantee for strongly nonideal mixtures.

Distillation uses repeated liquid–vapour equilibrations to exploit composition differences. A single equilibrium step does not usually yield pure A: in the example the vapour still contains 48.3% B. As vapour is withdrawn, the remaining liquid composition shifts, so later vapour portions need a new x and perhaps a new temperature. Nonideal liquids can form azeotropes, at which liquid and vapour compositions may coincide at a particular composition despite both components being present. The simple ideal model cannot represent that behaviour.

The pure-component pressures must refer to the same temperature as the liquid equilibrium. The total pressure above the liquid is a sum of component partial pressures, but an inert gas above it adds a separate contribution to an overall gauge reading. For the A/B condensable vapour composition , divide A and B partial pressures by P A + P B, not by a total that includes inert carrier gas, unless the problem explicitly asks for fractions in the entire gas mixture.

Step-by-step reasoning

1. Determine liquid mole fractions x A and x B and pure vapour pressures P A , P B at one temperature. 2. Calculate P A = x AP A and P B = x BP B . 3. Sum P A + P B to get the total A/B vapour pressure. 4. Divide each component pressure by that sum to get y A and y B. 5. Verify y A + y B = 1 and compare y of the more volatile component with its liquid x.

Visual explanation

Draw two stacked bars. The liquid bar has 30% A and 70% B. The vapour bar has 51.7% A and 48.3% B. Between them put the partial-pressure calculation 30 kPa A plus 28 kPa B. The bar change shows that evaporation selects A disproportionately without eliminating B.

Real-world analogy

Suppose a crowd has two groups that enter a room at different rates. The group with a faster entry rate may make up a larger fraction inside the room than in the crowd outside. The liquid fraction is the starting crowd share, while pure vapour pressure reflects an escape tendency; partial pressures determine the vapour-room share.

Real-world example

An introductory distillation calculation predicts the composition of vapour just above an ideal binary solvent. The vapour is richer in the more volatile solvent, so condensation of that vapour gives a different mixture from the original liquid. Repeating the process can improve separation, but composition changes at every step.

Why?

Why divide by total vapour pressure rather than pure A's pressure? A vapour mole fraction is the share of all vapour molecules attributed to A. Dalton's law makes that share P A/(P A + P B). Pure A's pressure is an input to the liquid equilibrium relation, not the denominator of vapour composition.

Common misconception

“The vapour has the same mole fractions as the liquid.” This would hold only in special cases, such as equal pure vapour pressures in the ideal model or a pure-component endpoint. In general, use Raoult's law and Dalton's law in sequence.

Worked example

An ideal liquid at a stated temperature has x A = 0.400 and x B = 0.600. The pure-liquid vapour pressures are P A = 80.0 kPa and P B = 20.0 kPa. Thus P A = 0.400(80.0) = 32.0 kPa and P B = 0.600(20.0) = 12.0 kPa. Total is 44.0 kPa. Vapour fractions are y A = 32.0/44.0 = 0.727 and y B = 12.0/44.0 = 0.273. A is enriched from 40.0% of liquid moles to 72.7% of vapour moles, but the vapour is still a mixture.

Quick check

1. Partial pressures of A and B are 15 kPa and 5 kPa. What fraction of their ideal vapour mixture is A? Answer: y A = 15/(15 + 5) = 0.75, or 75% of the A/B vapour moles.

Exam focus

Use x for liquid and y for vapour; show both sets of fractions. Apply Raoult's law to find partial pressures, then Dalton's law to find gas composition. Clarify whether the requested vapour fraction includes an inert carrier gas.

Advanced insight

Relative volatility α = P A /P B is constant with composition at fixed temperature only in the ideal binary model. The relation y A = αx A/[1 + (α − 1)x A] makes enrichment quantitative: when α approaches one, y A approaches x A and a simple equilibrium separation becomes weak.

Summary

Liquid fractions produce partial pressures through Raoult's law; partial pressures produce vapour fractions through Dalton's law. The more volatile component is enriched in the vapour of an ideal binary mixture, but one equilibrium step generally leaves both components present.

Practice questions

1. An ideal liquid has x A = 0.25, P A = 100 kPa and P B = 40 kPa. What are P A and P B? Answer: P A = 25 kPa and P B = 0.75(40) = 30 kPa. 2. What is y A for that mixture? Answer: Total A/B pressure is 55 kPa, so y A = 25/55 ≈ 0.455. 3. Does y A exceed x A in that example, and why? Answer: Yes, 0.455 > 0.25 because A's pure vapour pressure is higher than B's at that temperature.