Total Vapour Pressure of an Ideal Binary Solution

Introduction

For a binary mixture of volatile liquids A and B, the pressure above the liquid includes contributions from both vapours. If the liquid mixture is ideal, Raoult's law gives each contribution and Dalton's law adds them. A plot of total pressure against liquid composition is then a straight line connecting the pure-component vapour pressures at a fixed temperature. The line's slope reveals which liquid is more volatile.

Core explanation

Write pA = xA pA° and pB = xB pB°. The liquid is binary, so xB = 1 − xA. Adding partial pressures gives ptotal = pA + pB = xA pA° + (1 − xA)pB°. Rearranging, ptotal = pB° + xA(pA° − pB°). At xA = 0, the liquid is pure B and total pressure equals pB°. At xA = 1, it is pure A and total pressure equals pA°. The straight line between these endpoints follows algebraically from ideal behaviour, not from an arbitrary sketching convention.

If pA° > pB° at the same temperature, A is the more volatile component in this comparison. As xA increases, the total ideal pressure rises; its slope pA° − pB° is positive. If the pure pressures are equal, the ideal total pressure is constant across composition even though each partial pressure changes. A pressure increase with xA does not mean chemical reaction is creating gas; it means the liquid composition favours a component with a higher equilibrium vapour pressure.

For example, let pA° = 60 kPa, pB° = 20 kPa and xA = 0.25. Then pA = 15 kPa and xB = 0.75 gives pB = 15 kPa. Total pressure is 30 kPa. Although A occupies only one quarter of the liquid mole amount, it contributes half the vapour pressure because its pure vapour pressure is three times B's. The gas-phase mole fraction yA = pA/ptotal = 0.50 under ideal gas behaviour. Liquid composition and vapour composition differ.

The total pressure expression needs both components to be volatile and the vapour to be at equilibrium with the liquid. If B is nonvolatile, pB is negligible and the formula reduces to ptotal ≈ xA pA° for a nearly pure A vapour. If a third volatile component exists, a two-component denominator is incomplete; total pressure is the sum over all components. If air or another inert gas is present, the measured absolute headspace pressure also contains its partial pressure, so subtract or account for it before comparing with the liquid vapour prediction.

Nonideal liquid interactions can bend the pressure–composition curve. A curve above the ideal line is a positive deviation; one below is a negative deviation under comparable conditions. A pressure measurement alone at one composition may show a deviation if reliable pure-component data are known, but a full curve gives stronger evidence of how the deviation varies. The ideal line is a reference model, not a claim that all real solvent blends behave linearly.

As vapour forms, the remaining liquid composition can change because the vapour is enriched in the more volatile component. Thus a calculation made at initial xA applies to that stated liquid composition; an open or continuously distilled mixture may not retain it. For a closed vessel with enough liquid, phase equilibrium has a definite pair of compositions at fixed conditions, and a mass balance may be needed if the vapour amount is substantial.

Step-by-step reasoning

1. Read pure vapour pressures for A and B at the same temperature. 2. Read or calculate liquid xA and use xB = 1 − xA. 3. Calculate pA and pB separately with Raoult's law. 4. Add them for the ideal total vapour pressure; include other gases only if the question asks total headspace pressure. 5. Check that an ideal ptotal lies between pA° and pB° for a genuine binary composition.

Visual explanation

Plot pressure vertically and liquid xA from zero to one horizontally. Draw pA rising from zero to pA°, pB falling from pB° to zero, and their sum as a straight line from pB° to pA°. At xA = 0.25 mark the two partial pressures as stacked contributions to the total. This shows that one line alone is not the whole vapour pressure.

Real-world analogy

Two taps fill a tank at rates that depend on how far each is opened. The combined flow is the sum of the two contributions, even if one tap is opened less but has greater capacity. Partial pressures likewise add, while a more volatile component can contribute strongly at a smaller liquid fraction. The taps are only an arithmetic analogy; vapour pressure is equilibrium, not a liquid flow rate.

Real-world example

In distillation planning, pure-component vapour pressures provide a first ideal estimate of vapour pressure across blend compositions. A mixture richer in the higher-pressure component tends to have higher total pressure at the same temperature and a vapour enriched in that component. Actual design needs measured vapour–liquid data when the blend is nonideal.

Why?

Why is the ideal total-pressure plot a straight line? Substituting xB = 1 − xA into ptotal = xA pA° + xB pB° leaves a constant intercept plus xA times a constant difference of pure pressures. That is the equation of a line at fixed temperature.

Common misconception

“The vapour pressure of a blend equals the average of pure pressures.” The arithmetic average applies only at xA = xB = 0.50 in the ideal binary model. Other liquid compositions require mole-fraction weighting, and nonideal mixtures may differ even at 0.50.

Worked example

At one temperature, pA° = 80 kPa and pB° = 40 kPa. A binary ideal liquid has xA = 0.30, so xB = 0.70. Compute pA = 0.30 × 80 = 24 kPa and pB = 0.70 × 40 = 28 kPa. Therefore ptotal = 52 kPa, between 40 and 80 kPa as expected. Direct substitution into pB° + xA(pA° − pB°) gives 40 + 0.30(40) = 52 kPa, an independent arithmetic check.

Quick check

1. For an ideal binary volatile liquid, what is total vapour pressure if xA = 0? Answer: It equals pure B's vapour pressure pB°, because the liquid contains only B.

Exam focus

Use two separate Raoult-law partial pressures, then add. Keep the pure pressures at one temperature and check the result lies between them for an ideal binary liquid. Do not include inert-gas pressure unless asked for total headspace pressure.

Advanced insight

The pressure line alone does not give the vapour composition until each partial pressure is known. From Dalton's relation yA = pA/ptotal, a vapour curve can be calculated. The vapour and liquid curves bound a two-phase region in pressure–composition diagrams and help explain separation by repeated vaporisation and condensation.

Summary

An ideal binary volatile mixture has ptotal = xA pA° + xB pB°. Because xB = 1 − xA, total pressure varies linearly between the two pure-component pressures at fixed temperature. Partial pressures and vapour composition must still be kept distinct from liquid mole fractions.

Practice questions

1. Pure A and B have pressures 70 and 30 kPa; xA = 0.50. Find ideal total. Answer: pA = 35 kPa, pB = 15 kPa, so ptotal = 50 kPa. 2. If pA° is greater than pB°, what happens to ideal ptotal as xA increases? Answer: It rises linearly at fixed temperature because the higher-pressure component has a larger liquid fraction. 3. Why can a measured sealed-vessel pressure exceed the sum of predicted vapour partial pressures? Answer: An inert gas such as air may also contribute a partial pressure to the headspace.