Vapour Pressure of a Pure Liquid

Introduction

Place a pure liquid in a closed container with empty space above it. Some liquid molecules escape into the vapour, and some vapour molecules return. Eventually the opposing rates can match, producing a stable vapour pressure while molecules continue to move. This equilibrium vapour pressure depends mainly on temperature and the substance, not simply on how much empty space was left, provided liquid remains and equilibrium can be reached.

Core explanation

Evaporation occurs when surface molecules have enough energy to overcome attractions and leave the liquid. In a closed container, the number of vapour molecules initially rises. As vapour builds, more molecules strike and re-enter the liquid. At equilibrium, evaporation and condensation rates are equal on average. The number of vapour molecules can then remain roughly steady despite continuous exchange; equilibrium is dynamic, not motionless.

The pressure contributed by that vapour at equilibrium is the liquid's vapour pressure at the given temperature. In a simple pure-substance two-phase system, it is determined by temperature as long as enough liquid remains to coexist with the vapour. If the headspace volume increases, some additional liquid evaporates to restore the equilibrium pressure. If the liquid runs out, the two-phase condition fails and pressure can depend on the remaining gas amount and volume instead.

Temperature strongly affects vapour pressure. At higher temperature, a greater fraction of liquid molecules can escape and equilibrium is reached with more vapour, giving higher pressure. The exact relation is not usually linear over a broad range. A graph of vapour pressure against temperature curves upward for many liquids; extrapolating from only two points with a straight line can be unreliable.

Intermolecular attractions influence the pressure at a common temperature. Stronger attractions generally make escape more difficult and lower vapour pressure, if other molecular factors are comparable. Water and a more volatile liquid can have very different vapour pressures at room temperature. This qualitative comparison does not mean the liquid with lower vapour pressure must be denser or more viscous; those properties depend on additional factors.

Vapour pressure is a partial pressure when other gases are present. In a closed container containing air plus liquid water, the total pressure can be approximated as dry-air partial pressure plus water-vapour partial pressure under an ideal-mixture model. The water component can approach its equilibrium vapour pressure if liquid water remains and enough time passes. This is why gas collected over water needs a vapour-pressure correction.

In an open container, escaping vapour can be carried away by air, so the space above the liquid may not reach saturation. The liquid can keep evaporating. Vapour pressure remains a useful equilibrium property of the pure liquid at that temperature, but the actual water-vapour partial pressure in moving air may be lower. Confusing ambient humidity with equilibrium vapour pressure leads to incorrect interpretations.

Boiling begins when equilibrium vapour pressure equals the pressure exerted on the liquid by its surroundings. Evaporation can occur below the boiling point because surface molecules can escape even when bulk vapour bubbles cannot persist. The next lesson uses this distinction to explain why a liquid can slowly disappear without ever boiling.

The vapour-pressure concept assumes a pure liquid and a well-defined temperature. Solutions can have different vapour pressures, and changing composition during evaporation can change the value. A student should not take a table for pure water and apply it unchanged to every salt solution or solvent mixture.

Step-by-step reasoning

1. Identify the pure liquid, temperature and whether a closed container allows equilibrium. 2. Check that liquid remains in coexistence with its vapour. 3. Describe evaporation and condensation continuing at equal average rates. 4. Use a stated vapour-pressure value for that temperature or compare trends qualitatively. 5. Separate vapour partial pressure from total pressure when other gases are present.

Visual explanation

Draw liquid at the bottom of a sealed box and vapour dots above. Use upward arrows for molecules evaporating and downward arrows for molecules condensing, with equal arrow counts at equilibrium. Draw a pressure gauge connected to the headspace and label its vapour component. A second box with larger headspace but remaining liquid shows the same equilibrium vapour pressure at the same T after adjustment.

Real-world analogy

A room with people entering and leaving at equal average rates can have a steady occupancy while movement continues. A liquid-vapour equilibrium is similarly dynamic. The analogy does not specify molecular energy or why temperature sets the equilibrium pressure.

Real-world example

A partially filled sealed bottle of a volatile liquid develops vapour in its headspace. At fixed temperature and while liquid remains, the vapour can reach a characteristic equilibrium pressure. Opening the bottle lets vapour escape and can restart net evaporation, especially if air circulation carries it away.

Why?

Why does enlarging a closed headspace not necessarily permanently lower vapour pressure? With liquid still present, additional molecules evaporate until the vapour returns to its equilibrium pressure at the same temperature. If no liquid remains, this restoring mechanism stops.

Common misconception

“At equilibrium, evaporation stops.” Evaporation and condensation continue at equal average rates. The macroscopic amounts remain steady because the two ongoing flows balance.

Worked example

At a specified temperature, pure water's equilibrium vapour pressure is supplied as 3.2 kPa. A sealed vessel contains liquid water and a headspace large enough for equilibrium. Increasing headspace volume while holding temperature fixed initially lowers water-vapour partial pressure, but more water evaporates until it returns near 3.2 kPa, provided liquid remains. If dry air contributes 95.0 kPa at the final state, total ideal-mixture pressure is about 95.0 + 3.2 = 98.2 kPa. The dry-air contribution may itself change during expansion, so 95.0 kPa must refer to the final state for that sum.

Quick check

1. At liquid-vapour equilibrium, are individual molecules still crossing the phase boundary? Answer: Yes. Evaporation and condensation continue at equal average rates.

Exam focus

Define vapour pressure for a pure liquid with coexisting vapour at a stated temperature. Mention liquid remaining and dynamic equilibrium. Distinguish vapour partial pressure from total pressure and evaporation from boiling.

Advanced insight

The Clausius–Clapeyron relation gives an approximate connection between the slope of ln vapour pressure versus inverse absolute temperature and enthalpy of vaporisation over a limited range. It explains why vapour pressure rises rapidly with temperature and why a simple linear P-versus-T extrapolation is often poor.

Summary

In a closed pure-liquid system with liquid remaining, evaporation and condensation can reach dynamic equilibrium. The resulting vapour pressure is chiefly a function of substance and temperature. It rises with temperature and contributes a partial pressure when other gases are present.

Practice questions

1. Why can a closed bottle have steady vapour pressure while molecules keep moving? Answer: Evaporation and condensation occur at equal average rates, maintaining a steady vapour population. 2. If a pure liquid's temperature rises, what general change occurs in its equilibrium vapour pressure? Answer: It generally rises because more molecules can escape into the vapour. 3. When can headspace volume affect pressure despite a stated equilibrium vapour pressure? Answer: If all liquid evaporates, the two-phase condition ends and vapour pressure is then governed by gas amount, volume and temperature.