Raoult's Law for a Volatile Component

Introduction

If a liquid mixture contains volatile components, each can contribute molecules to the vapour. Raoult's law gives a simple ideal-solution rule for one component at a time: its vapour partial pressure equals its mole fraction in the liquid multiplied by its pure-liquid vapour pressure at the same temperature. The rule is powerful precisely because its assumptions and phase labels are explicit.

Core explanation

For component A, pA = xA pA°. Here xA is A's mole fraction in the liquid phase, pA° is the equilibrium vapour pressure of pure liquid A at the same temperature, and pA is A's partial pressure above the mixture. If xA = 0.40 and pA° = 50 kPa, the ideal partial pressure is 20 kPa. The result is not necessarily the total pressure, because another volatile component B may supply its own pB.

For a binary liquid, xA + xB = 1. If both components obey Raoult's law, then pA = xA pA° and pB = xB pB°. This is an ideal-solution model over the specified composition range. Unlike attractions between A and B can cause observed pressure to lie above or below the ideal prediction. The law is also a limiting approximation for a solvent near the pure-solvent end in many solutions. A problem should say or imply ideal behaviour before a numerical ideal result is treated as exact.

The pure vapour pressure is a property at temperature, not an amount of liquid. Doubling the amount of pure liquid in a closed vessel does not double pA° if liquid and vapour still coexist at the same temperature. The liquid mole fraction does change when composition changes, and that is what scales the ideal component partial pressure. Before using a tabulated pA°, verify the table temperature matches the mixture temperature.

A nonvolatile solute is the special case in which psolute is negligible. Then solvent pressure psolvent = xsolvent psolvent° can also represent almost all the solution's vapour pressure. In a mixture of two volatile liquids, applying the solvent-only calculation to the total pressure is wrong because both partial pressures must be added. The term “volatile” means the component has an appreciable vapour contribution under the stated conditions, not that it must have a boiling point below room temperature.

Raoult's law connects composition to equilibrium, not to how quickly evaporation happens. If the liquid is stirred, the system may reach equilibrium faster but the ideal pressure relation at the same temperature and composition is unchanged. Open containers with moving air do not maintain the same simple closed vapour equilibrium. Do not infer a partial pressure from how rapidly a beaker dries on a windy bench.

Finally, pA does not equal the vapour mole fraction yA. In a gas mixture that behaves ideally, yA = pA/ptotal. If A is more volatile than B, yA may exceed xA. A calculation using xA for the liquid and yA for the vapour can therefore predict enrichment on evaporation or distillation. Keeping symbols distinct prevents one of the most common solution-chemistry errors.

Step-by-step reasoning

1. Identify liquid component A and read its liquid mole fraction xA. 2. Obtain pA° for pure A at the mixture temperature. 3. Confirm that an ideal or suitable limiting model is intended. 4. Multiply xA pA° to find partial pressure pA, preserving pressure units. 5. If total pressure or vapour composition is asked, include all volatile components separately.

Visual explanation

Draw a horizontal liquid-composition axis from xA = 0 to xA = 1 and a vertical pA axis. For ideal behaviour, the line runs from zero to pA°. Mark xA = 0.4 and pA = 0.4pA°. Beside it show two layers, liquid and vapour, with labels xA inside liquid and yA inside vapour; no equality sign joins them.

Real-world analogy

Imagine two independent performers contributing to total sound. If each performer contributes a fraction of their full-volume sound according to their share of stage time, hearing one performer is not the same as hearing the whole room. A component's Raoult-law partial pressure is likewise only its contribution to total vapour pressure. Actual molecules can deviate because interactions are not independent performers.

Real-world example

A technician blends two volatile solvents for a controlled evaporation process. Pure-liquid vapour pressures are known at the operating temperature. Liquid mole fractions supply ideal starting estimates for each component's vapour contribution. Measured pressure can then be compared with the ideal sum; a deviation flags that interactions in the blend matter.

Why?

Why must pA° be measured or specified at the same temperature as xA's mixture? Vapour pressure is strongly temperature-dependent. Multiplying a composition at one temperature by a pure-component pressure at another combines incompatible equilibrium states.

Common misconception

“If liquid xA = 0.40, vapour A is automatically 40%.” The vapour proportion depends on pA relative to every volatile component's pressure. Different pure vapour pressures can make yA quite different from liquid xA.

Worked example

An ideal binary liquid has xA = 0.30. Pure A has pA° = 80 kPa at the stated temperature. Then pA = 0.30 × 80 = 24 kPa. If pure B has pB° = 20 kPa, xB = 0.70 and pB = 14 kPa, giving total 38 kPa. A's vapour mole fraction would be yA = 24/38 ≈ 0.632 under an ideal gas approximation, much greater than its liquid fraction 0.30. The partial-pressure law was applied separately to each liquid component.

Quick check

1. In pA = xA pA°, does xA describe the liquid or the vapour? Answer: It describes the liquid mixture; vapour composition needs a separate calculation from partial pressures.

Exam focus

Label pA, pA° and xA with component, phase and temperature. Do not treat one component's partial pressure as total pressure when another liquid is volatile. State the ideal-solution assumption.

Advanced insight

Chemical potential provides the equilibrium basis for the law. In an ideal liquid mixture, μA = μA° + RT ln xA; equality of chemical potential across liquid and vapour phases yields pressure proportional to xA. In a nonideal solution, activity aA replaces xA. The elementary multiplication remains useful as a benchmark for measured deviations.

Summary

Raoult's law predicts pA = xA pA° for an ideal volatile component at a fixed temperature. It gives one partial pressure, not necessarily total pressure or vapour mole fraction. Apply it component by component and keep phase labels clear.

Practice questions

1. If xA = 0.25 and pA° = 64 kPa, what is ideal pA? Answer: 0.25 × 64 = 16 kPa. 2. Can pA° be taken from a table at another temperature without adjustment? Answer: No. The reference pure-liquid vapour pressure must correspond to the mixture temperature. 3. Why might measured pA differ from xA pA°? Answer: The mixture may be nonideal because interactions among unlike molecules differ from those in the pure liquids.