Vapour pressure, Antoine equation and Raoult's law

Why it matters

Vapour pressure decides whether a liquid boils, how much solvent is lost from a storage tank, how a condenser must be cooled and how easily two liquids can be separated by distillation. The Antoine equation gives vapour pressure from temperature, and Raoult's law turns pure-component vapour pressures into the bubble and dew points of mixtures – the core of every flash and distillation calculation.

Key ideas

Vapour pressure P_sat. The pressure of a pure vapour in equilibrium with its liquid at a given temperature. It depends only on temperature (not on the amount of liquid or the vessel size) and rises steeply, roughly exponentially, with T. A liquid boils when its vapour pressure equals the surrounding pressure; the normal boiling point is the temperature at which P_sat = 1 atm.

Clausius–Clapeyron. Assuming ideal vapour, negligible liquid volume and constant latent heat, ln P_sat is linear in 1/T, with slope −ΔH_vap / R. This is useful to estimate ΔH_vap from two data points or extrapolate over a short range.

Antoine equation. An empirical three-constant fit, log₁₀ P_sat = A − B / (T + C), more accurate than Clausius–Clapeyron over its fitted range. The constants are tied to specific units (commonly mmHg with °C, or kPa with °C, or bar with K) and to a temperature range – always use them with the units of the source table and do not extrapolate far outside the range.

Raoult's law (ideal solutions). For an ideal liquid solution in equilibrium with an ideal-gas vapour, the partial pressure of each component is p_i = x_i·P_i_sat, so y_i·P = x_i·P_i_sat. It works well for chemically similar species (benzene–toluene, hexane–heptane). Strongly non-ideal mixtures (ethanol–water) need an activity coefficient γ_i: y_i·P = γ_i·x_i·P_i_sat.

Bubble and dew points.

• Bubble point: the liquid composition x is known; the first bubble of vapour forms when Σ x_i·P_i_sat = P. Its composition is y_i = x_i·P_i_sat / P.
• Dew point: the vapour composition y is known; the first drop of liquid forms when Σ y_i·P / P_i_sat = 1.
• At fixed P, finding a bubble or dew temperature needs trial and error because P_sat(T) is non-linear.

Relative volatility. α = (y_A / x_A) / (y_B / x_B), which for Raoult's law equals P_A_sat / P_B_sat. The more volatile component is enriched in the vapour; α close to 1 means a difficult separation.

Formulas

• log₁₀ P_sat = A − B / (T + C) – Antoine; P_sat and T in the units of the constants' source (e.g. mmHg and °C).
• ln(P₂ / P₁) = −(ΔH_vap / R)·(1/T₂ − 1/T₁) – Clausius–Clapeyron; T in K, ΔH_vap in J/mol, R = 8.314 J/(mol·K).
• y_i·P = x_i·P_i_sat – Raoult's law; x, y liquid and vapour mole fractions, P total pressure.
• P = Σ x_i·P_i_sat – bubble-point pressure (ideal solution).
• 1 / P = Σ y_i / P_i_sat – dew-point pressure (ideal solution).
• α_AB = P_A_sat / P_B_sat – relative volatility for an ideal system.
• y_i·P = γ_i·x_i·P_i_sat – modified Raoult's law for non-ideal liquids.

Worked examples

Example 1 (standard). Antoine constants for benzene (P in mmHg, T in °C) are A = 6.90565, B = 1211.033, C = 220.79. Find its vapour pressure at 80 °C in kPa and its normal boiling point.

1. log₁₀ P_sat = 6.90565 − 1211.033 / (80 + 220.79) = 6.90565 − 4.02617 = 2.87948.
2. P_sat = 10^2.87948 = 757.7 mmHg.
3. In kPa: 757.7 × 0.133322 = 101.0 kPa.
4. Normal boiling point: set P_sat = 760 mmHg, log₁₀ 760 = 2.88081, so T = B / (A − log₁₀ P) − C = 1211.033 / 4.02484 − 220.79 = 80.1 °C.

Answer: P_sat = 757.7 mmHg ≈ 101.0 kPa; T_b = 80.1 °C

Example 2 (GATE level). A liquid of 50 mol % benzene and 50 mol % toluene is heated at 760 mmHg. Assuming Raoult's law, find the bubble-point temperature and the composition of the first vapour. Toluene Antoine constants: A = 6.95464, B = 1344.8, C = 219.48 (mmHg, °C); benzene as in Example 1.

1. Condition: 0.5·P_B_sat(T) + 0.5·P_T_sat(T) = 760 mmHg. The answer lies between the pure boiling points (80.1 °C and 110.6 °C).
2. Trial T = 92.0 °C: P_B_sat = 1081.3 mmHg, P_T_sat = 433.7 mmHg; sum = 0.5 × 1081.3 + 0.5 × 433.7 = 757.5 mmHg (slightly low).
3. Trial T = 92.2 °C: P_B_sat = 1087.5, P_T_sat = 436.5; sum = 762.0 mmHg (slightly high).
4. Interpolate: T = 92.0 + 0.2 × (760 − 757.5) / (762.0 − 757.5) = 92.1 °C, where P_B_sat = 1084.7 mmHg and P_T_sat = 435.3 mmHg.
5. First vapour: y_B = x_B·P_B_sat / P = 0.5 × 1084.7 / 760 = 0.714; y_T = 0.286.
6. Relative volatility: α = 1084.7 / 435.3 = 2.49.

Answer: T_bubble ≈ 92.1 °C; y_benzene ≈ 0.714

Common mistakes

• Using Antoine constants with the wrong units (mmHg constants with kPa, or K instead of °C) – errors are huge, not small.
• Using log₁₀ constants with the natural log, or vice versa.
• Applying Raoult's law to strongly non-ideal mixtures such as ethanol–water.
• Computing a bubble point but reporting the vapour composition equal to the liquid composition.
• Using °C in Clausius–Clapeyron, which needs absolute temperature.
• Thinking vapour pressure depends on the amount of liquid or on the total pressure.

For GATE CH

Expect bubble/dew-point pressure and temperature numericals with Antoine constants supplied, vapour composition from Raoult's law, relative volatility, and Clausius–Clapeyron estimates of latent heat from two vapour-pressure points. Practise the trial-and-error bubble-point search and always read the units attached to the Antoine constants first.

Quick check

1. At its normal boiling point, what is a liquid's vapour pressure?
2. For an ideal binary with x_A = 0.4, P_A_sat = 20 kPa and P_B_sat = 10 kPa, what is the bubble-point pressure?
3. In Example 2, is the first vapour richer or poorer in benzene than the liquid?
4. What is the relative volatility of the binary in question 2?
5. What extra factor does the modified Raoult's law include?

Answers: 1. 101.325 kPa (1 atm); 2. 14 kPa; 3. richer (0.714 vs 0.5); 4. 2.0; 5. the activity coefficient γ.