Ideal Vapour Pressure of a Liquid Blend

Introduction

Two miscible volatile liquids both contribute molecules to the vapour above their blend. For an ideal liquid solution, each component's equilibrium vapour partial pressure equals its liquid mole fraction times its pure-liquid vapour pressure at the same temperature. Adding those partial pressures gives the total vapour pressure. The method requires liquid mole fractions, not mass fractions, and both pure-component values at one common temperature.

Core explanation

For liquid components A and B at temperature T, Raoult's law gives P A = x A,liquid P A and P B = x B,liquid P B , where P A and P B are vapour pressures of the pure liquids at that T. If the liquid contains only A and B, x A + x B = 1, and P total = P A + P B = x AP A + (1 − x A)P B . The vapour pressure has units inherited from the supplied pure-component values. This is an equilibrium relation for an ideal solution; it is not a universal description of all liquid pairs.

As a hypothetical numerical example at one fixed temperature, let P A = 100 kPa and P B = 40.0 kPa. If liquid mole fractions are x A = 0.300 and x B = 0.700, then P A = 30.0 kPa and P B = 28.0 kPa. Total equilibrium vapour pressure is 58.0 kPa. Component A has a smaller liquid mole fraction but a slightly larger vapour partial pressure because its pure vapour pressure is much higher. Using 0.300 × P total instead of 0.300 × P A would incorrectly assume vapour and liquid compositions are identical.

When P A > P B , increasing x A makes P total rise linearly in the ideal model, from P B at pure B to P A at pure A. The formula can be written P total = P B + x A(P A − P B ). The pure endpoints are useful checks: x A = 0 gives P B , and x A = 1 gives P A . For values between, ideal P total lies between those endpoints.

Real mixtures may show positive or negative deviations from Raoult's law because unlike-molecule interactions differ from like-molecule interactions. A sufficiently strong deviation can produce non-linear vapour-pressure curves. If a problem says “ideal solution,” use the linear model; if it provides activity coefficients or measured partial pressures, follow that information instead. The simple Raoult expression also needs attention when a component is effectively nonvolatile, reacts, or changes liquid composition as vapour is removed.

These pressures are equilibrium values above a liquid at the stated temperature. They do not equal the ambient pressure unless the vapour is the only gas present and the conditions make that equality appropriate. Air above a beaker adds its own pressure to a pressure gauge reading, while the liquid components still have equilibrium partial pressures in a suitable idealised open-system picture.

Step-by-step reasoning

1. Confirm that the blend is treated as ideal and both components are volatile. 2. Convert composition data to liquid mole fractions that sum to one. 3. Read P A and P B at the same stated temperature and put them in matching pressure units. 4. Multiply each pure pressure by its own liquid fraction, then sum the two partial pressures. 5. Check that the ideal total lies between the pure-component vapour pressures.

Visual explanation

Draw a liquid bar split into 30% A and 70% B by mole. Above it, draw two upward arrows: A labelled 0.30 × 100 = 30 kPa and B labelled 0.70 × 40 = 28 kPa. A brace around the arrows gives 58 kPa total. The arrows represent vapour contributions, not separate compartments above the liquid.

Real-world analogy

Imagine two teams contributing to a fundraiser. Each team's contribution equals its fraction of volunteers multiplied by its own average fundraising rate. A smaller team with a higher rate may contribute more than a larger team. Liquid fraction is like team share, while pure-component vapour pressure reflects each component's tendency to escape.

Real-world example

In designing a simple distillation exercise, a chemist estimates the vapour pressure above a binary liquid blend at a chosen temperature. The more volatile component contributes disproportionately to the vapour. This first estimate helps predict which component may be enriched in the vapour, though real-mixture deviations and temperature changes can matter.

Why?

Why multiply by liquid mole fraction? In the ideal-solution model, a component's escaping tendency is reduced in proportion to its share of molecules at the liquid interface relative to its pure liquid. The proportionality recovers the pure vapour pressure when that component's liquid fraction is one.

Common misconception

“Total vapour pressure is the arithmetic average of the pure pressures.” That occurs only for equal liquid mole fractions in the ideal binary case. In general, it is a mole-fraction-weighted sum. Another error is to use vapour mole fractions as Raoult-law inputs; the x values here belong to the liquid.

Worked example

At a stated temperature, ideal liquid A/B has P A = 80.0 kPa and P B = 20.0 kPa. A sample contains 0.750 mol A and 0.250 mol B in the liquid, so x A = 0.750 and x B = 0.250. Partial vapour pressures are P A = 0.750(80.0) = 60.0 kPa and P B = 0.250(20.0) = 5.00 kPa. Total vapour pressure is 65.0 kPa. It lies between the pure values 20.0 and 80.0 kPa and is closer to A's because A is more abundant in liquid. The calculation assumes equilibrium at that temperature and ideal liquid behaviour.

Quick check

1. If x A = 0.40, P A = 50 kPa and P B = 30 kPa in an ideal binary liquid, find P total. Answer: x B = 0.60; P total = 0.40(50) + 0.60(30) = 38 kPa.

Exam focus

Label x as liquid mole fraction and P as pure-liquid vapour pressure at the same T. Calculate each partial pressure before adding. Use endpoint checks and recognise that measured nonideal behaviour needs more than Raoult's simple law.

Advanced insight

For an ideal binary blend, the total-pressure line has slope P A − P B when plotted against x A. That slope is positive if A is more volatile. Vapour composition is found from y A = P A/P total, which generally differs from x A and is developed in the next calculation.

Summary

For an ideal binary liquid, each volatile component obeys Pᵢ = xᵢ,liquidPᵢ and total vapour pressure is the sum of those partial pressures. Use common-temperature pure values and liquid mole fractions, then check the ideal total against the pure endpoints.

Practice questions

1. In an ideal binary liquid with x A = 0.25, P A = 100 kPa and P B = 40 kPa, find P A. Answer: P A = 0.25(100) = 25 kPa. 2. For the same mixture, find P B and P total. Answer: x B = 0.75, so P B = 0.75(40) = 30 kPa and P total = 55 kPa. 3. If P A exceeds P B , what happens to ideal P total as x A increases? Answer: It increases linearly because P total = P B + x A(P A − P B ) has a positive slope.