Raoult’s law: the partial vapour pressure of each volatile component of a solution equals its pure vapour pressure times its mole fraction. A dissolved non-volatile solute therefore lowers the vapour pressure of the solvent.
Place a liquid in a closed container. Molecules at the surface with enough energy escape into the space above: the liquid evaporates. Some vapour molecules strike the surface and return to the liquid: they condense.
As vapour builds up, condensation becomes faster, until it equals the rate of evaporation. The amount of vapour then stays constant. The pressure of the vapour at this equilibrium is the vapour pressure of the liquid at that temperature. A liquid with weak attractions between its molecules evaporates easily, is called volatile, and has a high vapour pressure. At 25°C, water has a vapour pressure of about 24 mm Hg and acetone about 230 mm Hg.
In a solution of two volatile liquids, 1 and 2, both evaporate, and the vapour above is a mixture. Raoult’s law states that the partial vapour pressure of each component equals the vapour pressure of the pure component multiplied by its mole fraction in the solution:
p₁ = p₁° x₁ p₂ = p₂° x₂
p₁° and p₂° are the vapour pressures of the pure liquids at the same temperature. If component 1 makes up 60% of the molecules (x₁ = 0.6), its partial pressure is 0.6 times its pure vapour pressure.
By Dalton’s law of partial pressures, the total vapour pressure is the sum:
ptotal = p₁° x₁ + p₂° x₂
Raoult’s law holds exactly only for ideal solutions, in which the attractions between unlike molecules equal those between like molecules. The next session covers solutions that deviate from it.
Raoult’s law for an ideal solution. Each partial pressure is proportional to the component’s mole fraction, so each is a straight line. Their sum, the total pressure, is a straight line from p₂° to p₁°.
Two liquids have p₁° = 200 mm Hg and p₂° = 400 mm Hg. They form an ideal solution with x₁ = x₂ = 0.5. What is the total vapour pressure?
p₁ = 200 × 0.5 = 100 mm Hg.
p₂ = 400 × 0.5 = 200 mm Hg.
ptotal = 100 + 200 = 300 mm Hg.
With equal mole fractions, the total is the average of the two pure vapour pressures, because the total is a straight-line function of x₁.
A non-volatile solute, such as sugar or salt, has a negligible vapour pressure. In its solution, only the solvent contributes to the vapour. Raoult’s law becomes:
p₁ = p₁° x₁
With solute present, the solvent’s mole fraction x₁ is less than 1. So the vapour pressure of the solution is lower than that of the pure solvent. The fractional lowering is:
(p₁° − p₁) / p₁° = x₂
It equals the mole fraction of the solute. It depends on the number of solute particles, not on what they are. This is the basis of the colligative properties, two sessions from now.
Salt does not evaporate. Why does dissolving it lower the vapour pressure of water?
A simple model: evaporation happens at the surface. In pure water, every surface position is occupied by a water molecule that could escape. In salt water, some surface positions are occupied by ions, which do not escape. So fewer water molecules escape per second, and equilibrium is reached at a lower vapour pressure.
The model has a limit. It explains the direction of the effect, and why the effect depends on the number of solute particles. A more complete explanation uses thermodynamics: mixing increases the disorder of the liquid, which makes the solvent less likely to leave it. Both lead to the same result, p₁ = p₁°x₁, for an ideal solution.
1The vapour pressure of a liquid is…
2Raoult’s law states that the partial pressure of a volatile component equals…
3For p₁° = 200 and p₂° = 400 mm Hg, in an ideal solution with x₁ = x₂ = 0.5, the total vapour pressure is…
4For an ideal solution of two liquids, a graph of total vapour pressure against mole fraction is…
5Dissolving a non-volatile solute makes the vapour pressure of the solution…
6The relative lowering of vapour pressure, (p₁° − p₁)/p₁°, equals…
Next: ideal and non-ideal solutions, and deviations from Raoult’s law.
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