Solutions • Topic 2 of 3

Vapour Pressure & Raoult's Law

Every liquid in a closed container establishes an equilibrium with its vapour; the pressure exerted by the vapour at this point is the vapour pressure. When a solute is added, the behaviour of the solution's vapour pressure is governed by Raoult's law.

For a solution of two volatile liquids A and B, Raoult's law states that the partial vapour pressure of each component equals its vapour pressure in the pure state multiplied by its mole fraction in solution:

$p_A=p_A^0\,x_A$ and $p_B=p_B^0\,x_B$,
so the total vapour pressure is $p_{total}=p_A^0 x_A+p_B^0 x_B$.

Because $x_B=1-x_A$, this can be written $p_{total}=p_B^0+(p_A^0-p_B^0)x_A$ — a straight line when plotted against $x_A$. The composition of the vapour is richer in the more volatile component than the liquid is.

For a non-volatile solute dissolved in a volatile solvent, only the solvent contributes to the vapour pressure: $p_{solution}=p_{solvent}^0\,x_{solvent}$. Since $x_{solvent}<1$, the vapour pressure is always lowered — this leads directly to the colligative properties.

An ideal solution obeys Raoult's law over the entire composition range. In an ideal solution the A–B interactions are essentially identical to the A–A and B–B interactions, so $\Delta_{mix}H=0$ and $\Delta_{mix}V=0$ (e.g. benzene + toluene, n-hexane + n-heptane).

Non-ideal solutions deviate from Raoult's law:

Positive deviation: A–B forces are weaker than A–A/B–B forces, so molecules escape more easily — the observed vapour pressure is higher than predicted. Here $\Delta_{mix}H>0$ and $\Delta_{mix}V>0$ (e.g. ethanol + water, acetone + carbon disulphide).
Negative deviation: A–B forces are stronger (often hydrogen bonding), molecules escape less easily — the vapour pressure is lower than predicted, with $\Delta_{mix}H<0$ and $\Delta_{mix}V<0$ (e.g. acetone + chloroform, $\text{HNO}_3$ + water).

Solutions showing large deviations form azeotropes — constant-boiling mixtures that distil unchanged in composition. Minimum-boiling azeotropes arise from large positive deviation (ethanol–water at 95%), and maximum-boiling azeotropes from large negative deviation ($\text{HNO}_3$–water). Azeotropes cannot be separated into pure components by simple fractional distillation.

Solved Examples

Worked Example

At 350 K, the vapour pressures of pure A and pure B are 0.80 bar and 0.60 bar. A solution has $x_A=0.40$. Assuming ideal behaviour, find the total vapour pressure.

Solution

$x_B=1-0.40=0.60$.
$p_A=p_A^0 x_A=0.80\times0.40=0.32$ bar.
$p_B=p_B^0 x_B=0.60\times0.60=0.36$ bar.
$p_{total}=0.32+0.36=0.68$ bar.

Answer: $p_{total}=0.68\ \text{bar}$.

Worked Example

For the solution in Example 1, find the mole fraction of A in the vapour phase.

Solution

Mole fraction in vapour $y_A=\frac{p_A}{p_{total}}$.
$y_A=\frac{0.32}{0.68}=0.47$.
The vapour ($y_A=0.47$) is richer in A than the liquid ($x_A=0.40$), since A is more volatile.

Answer: $y_A\approx0.47$.

Worked Example

Heptane $(p^0=105.2\ \text{kPa})$ and octane $(p^0=46.8\ \text{kPa})$ form an ideal solution. A mixture has 25 g heptane (M = 100) and 35 g octane (M = 114). Find the total vapour pressure.

Solution

$n_{hep}=25/100=0.25$; $n_{oct}=35/114=0.307$.
$x_{hep}=0.25/0.557=0.449$; $x_{oct}=0.551$.
$p=105.2\times0.449+46.8\times0.551$.
$p=47.2+25.8=73.0\ \text{kPa}$.

Answer: $p_{total}\approx73.0\ \text{kPa}$.

Worked Example

Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (M = 60) is dissolved in 850 g of water. Calculate the vapour pressure of the solution.

Solution

$n_{urea}=50/60=0.833$; $n_{water}=850/18=47.22$.
$x_{water}=47.22/(47.22+0.833)=0.9827$.
$p_{solution}=p^0 x_{water}=23.8\times0.9827$.
$p_{solution}=23.39\ \text{mm Hg}$.

Answer: $p_{solution}\approx23.4\ \text{mm Hg}$.

Worked Example

A mixture of acetone and chloroform shows a vapour pressure lower than predicted by Raoult's law. Identify the type of deviation and the sign of $\Delta_{mix}H$.

Solution

Acetone and chloroform form hydrogen bonds between unlike molecules, strengthening A–B interactions.
Stronger A–B forces lower the escaping tendency, so vapour pressure is below the Raoult prediction — a negative deviation.
Bond formation releases energy, so $\Delta_{mix}H<0$ (exothermic) and $\Delta_{mix}V<0$.

Answer: Negative deviation; $\Delta_{mix}H<0$.

Worked Example

A 95% ethanol–water mixture distils at a constant temperature without change in composition. What is this mixture called, and which deviation produces it?

Solution

A mixture that boils at constant temperature and distils unchanged is an azeotrope.
Ethanol–water shows large positive deviation, giving a vapour-pressure maximum and hence a boiling-point minimum.
Therefore it is a minimum-boiling azeotrope and cannot be purified beyond 95% by simple fractional distillation.

Answer: A minimum-boiling azeotrope (large positive deviation).

Key Points

Raoult's law for volatile components: $p_A=p_A^0 x_A$, so $p_{total}=p_A^0 x_A+p_B^0 x_B$ — a straight line versus composition.
For a non-volatile solute, $p_{solution}=p_{solvent}^0 x_{solvent}$, so vapour pressure is always lowered.
Ideal solutions obey Raoult's law throughout with $\Delta_{mix}H=0$ and $\Delta_{mix}V=0$ (e.g. benzene + toluene).
Positive deviation: weaker A–B forces, higher vapour pressure, $\Delta_{mix}H>0$ (ethanol + water); negative deviation: stronger A–B forces, $\Delta_{mix}H<0$ (acetone + chloroform).
Large positive deviation gives minimum-boiling azeotropes; large negative deviation gives maximum-boiling azeotropes — neither can be separated by fractional distillation.

Quick Check — 4 MCQs

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Q1.For a solution of two volatile liquids obeying Raoult's law, the plot of $p_{total}$ versus $x_A$ is:

Explanation: $p_{total}=p_B^0+(p_A^0-p_B^0)x_A$ is linear in $x_A$.

Q2.Which pair shows positive deviation from Raoult's law?

Explanation: Ethanol + water have weaker A–B interactions than the pure liquids, giving higher-than-ideal vapour pressure.

Q3.In a solution showing negative deviation, the sign of $\Delta_{mix}H$ is:

Explanation: Stronger A–B attractions release heat on mixing, so $\Delta_{mix}H<0$.

Q4.A maximum-boiling azeotrope is formed by mixtures showing:

Explanation: Large negative deviation gives a vapour-pressure minimum, hence a boiling-point maximum (e.g. nitric acid–water).

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