Vapour Composition over a Solution

Introduction

The vapour above a liquid mixture need not have the same composition as the liquid. If two components are volatile and one has a higher pure-component vapour pressure, it often appears in a larger fraction of the vapour than in the liquid under ideal conditions. This difference makes distillation possible. The calculation requires two links: liquid composition to partial pressures, then partial pressures to vapour composition.

Core explanation

Use xA and xB for liquid mole fractions, yA and yB for vapour mole fractions. For an ideal binary liquid, Raoult's law gives pA = xA pA° and pB = xB pB°. Dalton's law for an approximately ideal gas mixture gives yA = pA/(pA + pB) and yB = pB/(pA + pB). The two y values sum to one because the denominator is total vapour pressure. Using xA directly as yA skips the essential volatility comparison.

Suppose pA° = 60 kPa and pB° = 20 kPa at one temperature. A is more volatile because its pure vapour pressure is higher. For a liquid with xA = 0.25 and xB = 0.75, pA = 15 kPa and pB = 15 kPa. The vapour has yA = 15/(15 + 15) = 0.50. A is 25% of liquid molecules by amount but 50% of the ideal vapour molecules. The higher volatility exactly offsets its lower liquid fraction in this chosen example.

The enrichment can be shown algebraically. For xA between zero and one and pA° > pB°, the ratio of vapour contributions is pA/pB = (xA/xB)(pA°/pB°). Since the pure-pressure ratio exceeds one, A's ratio to B is greater in vapour than in liquid. It follows that yA > xA for an ideal mixture away from pure endpoints. If pure pressures are equal, yA = xA; there is no volatility-based enrichment in this ideal model.

When some vapour leaves, the remaining liquid can become poorer in the more volatile A. The x used in the calculation then changes over time, so repeating the same original y calculation indefinitely would be wrong. Simple fractional distillation repeatedly forms vapour enriched in the more volatile component and condenses it. Real separation depends on heat transfer, column design and vapour–liquid equilibrium; a single ideal equilibrium step usually does not produce pure A.

For nonideal liquids, measured vapour composition may differ from the ideal calculation. An azeotrope can have the same composition in coexisting liquid and vapour at a particular condition despite different pure-component volatilities. That does not contradict the ideal algebra; it means the mixture's interactions cause deviation from the assumptions. Also, comparing only colour, smell or boiling point of a collected fraction is not a rigorous measurement of y without appropriate analysis.

State whether the problem gives a liquid or vapour composition. If yA is given and xA requested, one must solve the equilibrium relations rather than equating them. If an inert gas is present, use partial pressures of A and B and clarify whether yA is a fraction of just solvent vapours or of every headspace gas. In typical classroom vapour–liquid questions, y is defined for the two condensable components alone.

Step-by-step reasoning

1. Label x values as liquid fractions and y values as vapour fractions. 2. Use pure-component pressures at the same temperature to calculate pA and pB. 3. Add the two partial pressures for the total condensable vapour pressure. 4. Divide one partial pressure by that total to obtain its y value. 5. Compare yA with xA and explain the direction using pure-component volatility.

Visual explanation

Draw a vessel with a lower liquid layer containing one A dot for every three B dots. Above it, draw equal numbers of A and B dots for the numerical example. Label liquid xA = 0.25 and vapour yA = 0.50. Beside it write pA° = 60, pB° = 20 kPa, and show the two 15 kPa partial pressures that make the dot counts plausible.

Real-world analogy

Two runners start from a team in unequal numbers, but the faster runner type reaches a finish line more often. The finish-line fraction differs from the starting-team fraction because arrival tendencies differ. Vapour enrichment similarly combines liquid abundance with volatility, although molecules are in equilibrium rather than racing along a track.

Real-world example

A solvent recovery system begins with a liquid containing a smaller mole fraction of a more volatile solvent A. A first equilibrium vapour sample can already have a larger A fraction, which is why collecting and condensing vapour can enrich A. As separation proceeds, the liquid changes composition, so operation must use updated equilibrium data rather than a fixed initial calculation.

Why?

Why can xA = 0.25 still produce yA = 0.50? A's pure vapour pressure in the example is three times B's. Its smaller liquid fraction multiplied by its larger pure pressure gives the same partial pressure as B's larger fraction multiplied by B's smaller pure pressure.

Common misconception

“The more volatile component must be the majority of the vapour.” It is enriched relative to its liquid fraction under the ideal conditions, but it need not exceed 50%. If xA is extremely small, yA can remain below 0.50 despite enrichment.

Worked example

At a fixed temperature pA° = 90 kPa and pB° = 30 kPa. An ideal liquid has xA = 0.10, xB = 0.90. Then pA = 9 kPa and pB = 27 kPa, so total condensable vapour pressure is 36 kPa. Vapour yA = 9/36 = 0.25 and yB = 0.75. A is enriched from 10% in liquid to 25% in vapour but is not the vapour majority. The arithmetic illustrates why enrichment and dominance are distinct claims.

Quick check

1. If pA = 12 kPa and pB = 28 kPa, what is vapour yA? Answer: yA = 12/(12 + 28) = 0.30 under the two-component ideal-gas calculation.

Exam focus

Use x for liquid and y for vapour throughout. Calculate both Raoult-law partial pressures before finding y. Compare y and x quantitatively; do not infer that an enriched component necessarily dominates the vapour.

Advanced insight

In continuous distillation, the composition of both phases varies along a column, and local equilibrium is only an approximation to an operating tray or packing segment. The ideal x–y relationship provides a first limit on achievable separation. Deviations and azeotropes can fundamentally change the predicted behaviour, requiring measured data.

Summary

Vapour composition is obtained from partial pressures: yA = pA/(pA + pB). Raoult's law first links each partial pressure to the liquid x value. A more volatile component is enriched in the ideal vapour relative to its liquid fraction, while ongoing evaporation changes the remaining liquid composition.

Practice questions

1. Liquid xA = 0.40, pA° = 50 kPa and pB° = 25 kPa. Find vapour yA. Answer: pA = 20 kPa, pB = 0.60 × 25 = 15 kPa, so yA = 20/35 ≈ 0.571. 2. Does yA always equal xA in an ideal volatile binary mixture? Answer: No. They generally differ when pure-component vapour pressures differ. 3. Why must the starting x be updated during substantial distillation? Answer: Vapour preferentially removes the more volatile component, changing the composition of the liquid left behind.