Vapour pressure of a liquid
In a closed container, the fastest molecules at the surface escape into the space above. Some come back. Soon escaping and returning happen at the same rate: an equilibrium. The pressure of the vapour at that point is the vapour pressure.
- It depends only on the liquid and the temperature (hotter → higher).
- A liquid with weak attractions between its molecules (like acetone) has a high vapour pressure and is called volatile.
Raoult's law for a solution of two volatile liquids
Raoult's law (own words): in a solution of volatile liquids, the partial vapour pressure of each component equals its mole fraction in the liquid times its vapour pressure when pure.
pA = xA · pA° and pB = xB · pB°
By Dalton's law the total pressure is p = pA + pB = pB° + (pA° − pB°)·xA. This is a straight line against xA, from pB° (at xA = 0) to pA° (at xA = 1).
Composition of the vapour
The mole fraction of A in the vapour is yA = pA ÷ p. The vapour is always richer in the more volatile liquid. This idea is used in fractional distillation.
Solution with a non-volatile solute
If the solute does not evaporate (sugar, salt), only the solvent makes vapour: p = x₁·p₁°. Since x₁ < 1, the vapour pressure of the solution is lower than that of the pure solvent. This leads to colligative properties (next lesson).
Raoult's law as a special case of Henry's law
Henry's law says p = KH·x for a dissolved gas. Raoult's law says p = p°·x. Both say pressure ∝ mole fraction. So Raoult's law is Henry's law with KH = p° of the pure liquid.
Ideal solutions
An ideal solution obeys Raoult's law at every composition. This happens when the A–B pull is about the same as the A–A and B–B pulls. Then:
- ΔmixH = 0 (no heat taken or given on mixing)
- ΔmixV = 0 (volumes simply add)
Examples (nearly ideal): n-hexane + n-heptane, benzene + toluene, bromoethane + chloroethane.
Non-ideal solutions: positive and negative deviation
A non-ideal solution does not obey Raoult's law over the whole range. Its vapour pressure curve bends above or below the straight line.
Positive deviation
A–B attractions are weaker than A–A and B–B. Molecules escape more easily, so the vapour pressure is higher than Raoult predicts. ΔmixH > 0 (heat is absorbed), ΔmixV > 0. Examples: ethanol + acetone (acetone breaks some hydrogen bonds of ethanol), carbon disulphide + acetone, ethanol + water.
Negative deviation
A–B attractions are stronger. Fewer molecules escape, so the vapour pressure is lower. ΔmixH < 0 (heat is given out), ΔmixV < 0. Examples: chloroform + acetone (new hydrogen bond between them), phenol + aniline, nitric acid + water.
Azeotropes
An azeotrope is a liquid mixture that boils at a fixed temperature with vapour of the same composition as the liquid, so it cannot be separated by fractional distillation.
- Minimum boiling azeotrope: from large positive deviation. Ethanol–water at about 95% ethanol by volume.
- Maximum boiling azeotrope: from large negative deviation. Nitric acid–water at about 68% HNO₃ by mass.
Try it
At home: put one spoon of water and one spoon of hand sanitiser on two plates in the same room. Predict which dries first, then watch. The sanitiser (alcohol) goes first because its vapour pressure is higher. In the 3D: work out p by hand for xA = 0.4 in ideal mode, then set the slider and check. Now switch to positive and negative deviation and see how far the purple bar moves from the black ideal marker.
Key formulas and definitions
- Raoult's law: pA = xA · pA°
- Total: p = pA + pB = pB° + (pA° − pB°) xA
- Vapour mole fraction: yA = pA ÷ p
- Non-volatile solute: p = x₁ · p₁°
- Ideal: ΔmixH = 0, ΔmixV = 0
- Positive deviation: ΔmixH > 0, ΔmixV > 0; Negative: ΔmixH < 0, ΔmixV < 0
Worked examples
1. Pure A has vapour pressure 80 kPa. In an ideal solution xA = 0.25. Find pA.
pA = xA · pA° = 0.25 × 80 = 20 kPa.
2. pA° = 80 kPa, pB° = 30 kPa, xA = 0.4 (ideal). Find the total vapour pressure.
pA = 0.4 × 80 = 32 kPa. xB = 0.6, pB = 0.6 × 30 = 18 kPa. p = 32 + 18 = 50 kPa.
3. For the mixture in example 2, find the mole fraction of A in the vapour.
yA = pA ÷ p = 32 ÷ 50 = 0.64. The vapour (0.64) is richer in A than the liquid (0.4), because A is more volatile.
4. Benzene (p° = 12.8 kPa) and toluene (p° = 3.85 kPa) form an ideal solution. 78 g benzene (M = 78) is mixed with 46 g toluene (M = 92). Find the total vapour pressure.
n(benzene) = 1 mol, n(toluene) = 0.5 mol. x(benzene) = 1 ÷ 1.5 = 0.667, x(toluene) = 0.333. p = 0.667 × 12.8 + 0.333 × 3.85 = 8.53 + 1.28 = 9.81 kPa.
5. Two liquids have p° = 60 kPa and 20 kPa. What mole fraction of the first gives a total of 40 kPa (ideal)?
p = pB° + (pA° − pB°)xA → 40 = 20 + 40·xA → xA = 0.5.
6. Water at 298 K has vapour pressure 3.17 kPa. 1 mol of a non-volatile solute is dissolved in 9 mol of water. Find the new vapour pressure.
x(water) = 9 ÷ 10 = 0.9. p = 0.9 × 3.17 = 2.85 kPa (lowered by 0.32 kPa).
7. A mixture of A and B has total vapour pressure 55 kPa at xA = 0.5. Pure values: pA° = 70 kPa, pB° = 30 kPa. Is it ideal? If not, which deviation?
Ideal value = 0.5 × 70 + 0.5 × 30 = 50 kPa. Measured 55 kPa is higher, so positive deviation (A–B pull weaker).
Common mistakes
- Using the vapour mole fraction (y) in Raoult's law. pA = xA · pA° uses the LIQUID mole fraction x.
- Mixing up the signs: positive deviation has ΔmixH > 0 and ΔmixV > 0; negative deviation has both < 0.
- Thinking an azeotrope can be separated by fractional distillation. Its vapour has the same composition as the liquid, so it cannot.
- Forgetting that a non-volatile solute adds no vapour: only the solvent term x₁·p₁° remains.