Raoult's Law and Vapour Pressure: Quick Revision

A brief note on Raoult's Law and Vapour Pressure. Read it before a test. Try each question at the end before you open its answer.

Vapour pressure

• The vapour pressure of a liquid is the pressure of its vapour when the liquid and the vapour are in equilibrium.
• It depends on the liquid and the temperature. It does not depend on the amount of liquid or the size of the space.
• A liquid boils when its vapour pressure equals the pressure outside.

Two volatile liquids

• Raoult's law: $p_A = x_A\,p_A^\circ$ and $p_B = x_B\,p_B^\circ$.
• Total pressure: $P = x_A p_A^\circ + x_B p_B^\circ = p_B^\circ + (p_A^\circ - p_B^\circ)\,x_A$. It is a straight line in $x_A$.
• The vapour: $y_A = \dfrac{p_A}{P}$. It is richer in the more volatile liquid.
• Mole fractions use moles. Divide each mass by its molar mass first.

A non-volatile solute

• Vapour pressure: $p = x_{solvent}\,p^\circ$.
• Relative lowering: $\dfrac{p^\circ - p}{p^\circ} = x_{solute}$.
• Molar mass: $\dfrac{p^\circ - p}{p} = \dfrac{n_B}{n_A}$, then $M_B = \dfrac{\text{mass}}{n_B}$.
• The lowering depends on the number of solute particles, not on their kind.

Ideal and non-ideal solutions

• Ideal: obeys Raoult's law everywhere. $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$. Examples: hexane and heptane, benzene and toluene.
• Positive deviation: pressure above the Raoult line, weaker A–B pull, $\Delta_{mix}H > 0$, $\Delta_{mix}V > 0$. Example: ethanol and acetone.
• Negative deviation: pressure below the line, stronger A–B pull, $\Delta_{mix}H < 0$, $\Delta_{mix}V < 0$. Example: acetone and chloroform.

Azeotropes

• An azeotrope boils at one temperature, and its vapour has the same composition as its liquid. Distillation cannot separate it.
• Positive deviation: a minimum-boiling azeotrope, which comes off at the top. Example: ethanol and water.
• Negative deviation: a maximum-boiling azeotrope, which stays at the bottom. Example: nitric acid and water.
• An azeotrope is a mixture. Its composition changes with pressure.

A worked example

Liquids A and B form an ideal solution with $p_A^\circ = 90\ \text{kPa}$ and $p_B^\circ = 30\ \text{kPa}$. Find the total pressure and the vapour composition for $x_A = 0.30$.

1. Partial pressures. $p_A = 0.30 \times 90\ \text{kPa} = 27\ \text{kPa}$. $p_B = 0.70 \times 30\ \text{kPa} = 21\ \text{kPa}$.
2. Total. $P = 27\ \text{kPa} + 21\ \text{kPa} = 48\ \text{kPa}$.
3. The vapour. $y_A = \dfrac{27}{48} = 0.5625$, so the vapour is richer in A than the liquid.

Practice with new numbers

A program to try

The program finds the pressure above an ideal mixture and the pressure above a solution of a non-volatile solute.

Mistakes to avoid

• Adding the pure pressures. Use the partial pressures.
• Masses in a mole fraction. Divide by the molar mass first.
• The solute's mole fraction in $p = x\,p^\circ$. Use the solvent's.
• Mixing up the signs. Positive deviation: pressure above the line, $\Delta_{mix}H > 0$, minimum-boiling azeotrope.
• Calling an azeotrope a compound.

Check yourself

1. In an ideal solution, what is the partial pressure of A when $p_A^\circ = 50\ \text{kPa}$ and $x_A = 0.40$?

2. A solution shows a pressure above the Raoult line. Is the A–B attraction weaker or stronger than the others?

3. What does the vapour above an azeotrope look like compared with the liquid?

4. A non-volatile solute has $x_B = 0.10$. By what fraction does the vapour pressure fall?