Vapour Pressure of Liquid-Liquid Solutions

When you mix two liquids that can both evaporate, something interesting happens to the vapour sitting above the mixture. Neither liquid evaporates as freely as it would on its own, and the total vapour pressure of the mixture falls somewhere between the individual vapour pressures of the two pure liquids. How exactly does the mixture’s composition control what happens in the vapour? That is the question Raoult’s law answers.

Setting the Scene: Two Volatile Liquids in a Closed Container

Think of a binary solution (a solution with exactly two components) made from two volatile liquids. Call them component 1 and component 2. Place this mixture inside a sealed container and wait.

Both liquids begin to evaporate. Molecules of component 1 escape into the space above the liquid, and so do molecules of component 2. Over time, the rate at which molecules leave the liquid matches the rate at which they return, and the system settles into equilibrium (a balanced state where the amount of each substance in the liquid and vapour no longer changes).

At equilibrium, the space above the liquid holds a mixture of vapours. Each component contributes its own partial vapour pressure (the portion of the total pressure that comes from that one component alone). Let $p_1$ and $p_2$ denote these partial pressures, and let $p_{\text{total}}$ be the total vapour pressure above the solution.

In the liquid itself, the composition is described by mole fractions (the fraction of the total number of moles that belongs to a given component). Denote these as $x_1$ for component 1 and $x_2$ for component 2. Since there are only two components:

$x_1 + x_2 = 1$

Raoult’s Law: How Composition Controls Partial Pressure

The French chemist Francois Marte Raoult worked out the precise connection between composition and vapour pressure in 1886. The relationship he discovered, known as Raoult’s law, states:

For a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution.

Written mathematically for component 1:

$p_1 \propto x_1$

The proportionality constant turns out to be $p_1^0$, the vapour pressure of pure component 1 (the pressure its vapour would exert if the liquid were 100% component 1, at the same temperature). So:

$p_1 = p_1^0 \, x_1 \qquad \text{(1.12)}$

The logic is straightforward: in a pure liquid, every molecule at the surface is of the same kind and evaporates at a rate that produces $p_1^0$. In a mixture, only a fraction $x_1$ of the surface molecules belong to component 1, so the evaporation rate, and therefore the partial pressure, drops in proportion.

For component 2, the identical reasoning gives:

$p_2 = p_2^0 \, x_2 \qquad \text{(1.13)}$

where $p_2^0$ is the vapour pressure of pure component 2 at the same temperature.

Finding the Total Vapour Pressure

With the partial pressures known from Raoult’s law, the next step is to find the total vapour pressure above the solution. This is where Dalton’s law of partial pressures comes in. It says that the total pressure of a gas mixture is simply the sum of the partial pressures of each gas present:

$p_{\text{total}} = p_1 + p_2 \qquad \text{(1.14)}$

Now substitute the Raoult’s law expressions for $p_1$ and $p_2$:

$p_{\text{total}} = x_1 \, p_1^0 + x_2 \, p_2^0$

This equation is correct, but it contains two variables ($x_1$ and $x_2$). Since the two mole fractions add up to 1, we can replace $x_1$ with $(1 - x_2)$:

$p_{\text{total}} = (1 - x_2) \, p_1^0 + x_2 \, p_2^0 \qquad \text{(1.15)}$

Expand the bracket:

$p_{\text{total}} = p_1^0 - x_2 \, p_1^0 + x_2 \, p_2^0$

Factor out $x_2$ from the last two terms:

$p_{\text{total}} = p_1^0 + (p_2^0 - p_1^0) \, x_2 \qquad \text{(1.16)}$

This is the key result. It is a straight-line equation in $x_2$, with intercept $p_1^0$ and slope $(p_2^0 - p_1^0)$.

What Equation (1.16) Tells Us

Three important conclusions follow directly from the total vapour pressure equation:

1. One variable is enough. Because $x_1 + x_2 = 1$, the total vapour pressure can be written entirely in terms of the mole fraction of just one component. Knowing $x_2$ (or $x_1$) is sufficient.

2. The relationship is linear. If you plot $p_{\text{total}}$ on the y-axis against $x_2$ on the x-axis, you get a straight line. The intercept (at $x_2 = 0$, meaning pure component 1) is $p_1^0$. The other end (at $x_2 = 1$, meaning pure component 2) is $p_2^0$.

3. The direction of change depends on which component is more volatile. If $p_2^0 > p_1^0$ (component 2 evaporates more easily), the slope $(p_2^0 - p_1^0)$ is positive, and total vapour pressure rises as $x_2$ increases. If $p_2^0 < p_1^0$, the slope is negative, and total vapour pressure falls as $x_2$ increases.

Reading the Vapour Pressure Graph

Fig 1.3: Vapour pressure versus mole fraction for an ideal binary solution at constant temperature. Line I shows $p_1$ decreasing linearly from $p_1^0$ to zero, line II shows $p_2$ increasing from zero to $p_2^0$, and line III shows $p_{\text{total}}$ as a straight line connecting $p_1^0$ to $p_2^0$.

The graph contains three straight lines:

• Line I ($p_1$ vs $x_1$): Starts at $p_1^0$ on the left (where $x_1 = 1$, pure component 1) and drops to zero on the right (where $x_1 = 0$). This is Raoult’s law for component 1.

• Line II ($p_2$ vs $x_2$): Starts at zero on the left (where $x_2 = 0$) and rises to $p_2^0$ on the right (where $x_2 = 1$, pure component 2). This is Raoult’s law for component 2.

• Line III ($p_{\text{total}}$): The sum of lines I and II. It connects $p_1^0$ on the left end to $p_2^0$ on the right end. If component 1 is less volatile than component 2 ($p_1^0 < p_2^0$), this line rises from left to right. Its minimum value is $p_1^0$ and its maximum value is $p_2^0$.

All three plots are straight lines because Raoult’s law produces a direct (first-power) proportionality between pressure and mole fraction.

Composition of the Vapour Phase

Knowing the partial pressures also lets us figure out the makeup of the vapour sitting above the liquid. If $y_1$ and $y_2$ are the mole fractions of components 1 and 2 in the vapour phase, then Dalton’s law gives:

$p_1 = y_1 \, p_{\text{total}} \qquad \text{(1.17)}$

$p_2 = y_2 \, p_{\text{total}} \qquad \text{(1.18)}$

In general, for any component $i$:

$p_i = y_i \, p_{\text{total}} \qquad \text{(1.19)}$

Rearranging to isolate $y_i$:

$y_i = \frac{p_i}{p_{\text{total}}}$

This formula is powerful: once you know each partial pressure (from Raoult’s law) and the total pressure (from Dalton’s law), you can calculate exactly how much of each component is present in the vapour.

An important general result emerges from this: the vapour phase is always richer in the more volatile component (the one with the higher $p^0$). The more volatile liquid contributes a larger share of the total vapour pressure, which translates into a larger mole fraction in the vapour compared to the liquid.

Solved Example 1.5: Chloroform and Dichloromethane Mixture

Problem: The vapour pressure of chloroform ($CHCl_3$) at 298 K is 200 mm Hg and that of dichloromethane ($CH_2Cl_2$) at 298 K is 415 mm Hg. A solution is prepared by mixing 25.5 g of $CHCl_3$ and 40 g of $CH_2Cl_2$. Calculate (i) the vapour pressure of the solution and (ii) the mole fractions of each component in the vapour phase.

Solution:

Part (i): Total vapour pressure of the solution

Step 1: Find the molar masses.

$M_{CH_2Cl_2} = 12 \times 1 + 1 \times 2 + 35.5 \times 2 = 85 \text{ g mol}^{-1}$

$M_{CHCl_3} = 12 \times 1 + 1 \times 1 + 35.5 \times 3 = 119.5 \text{ g mol}^{-1}$

Step 2: Convert masses to moles.

$n_{CH_2Cl_2} = \frac{40 \text{ g}}{85 \text{ g mol}^{-1}} = 0.47 \text{ mol}$

$n_{CHCl_3} = \frac{25.5 \text{ g}}{119.5 \text{ g mol}^{-1}} = 0.213 \text{ mol}$

Step 3: Calculate the total moles and mole fractions.

Total moles $= 0.47 + 0.213 = 0.683$ mol

$x_{CH_2Cl_2} = \frac{0.47}{0.683} = 0.688$

$x_{CHCl_3} = 1 - 0.688 = 0.312$

Step 4: Apply Equation (1.16) to get the total vapour pressure.

Treat $CHCl_3$ as component 1 ($p_1^0 = 200$ mm Hg) and $CH_2Cl_2$ as component 2 ($p_2^0 = 415$ mm Hg):

$p_{\text{total}} = p_1^0 + (p_2^0 - p_1^0) \, x_2$

$= 200 + (415 - 200) \times 0.688$

$= 200 + 215 \times 0.688$

$= 200 + 147.9 = 347.9 \text{ mm Hg}$

Part (ii): Mole fractions in the vapour phase

Step 5: Find each partial pressure using Raoult’s law.

$p_{CH_2Cl_2} = x_{CH_2Cl_2} \times p^0_{CH_2Cl_2} = 0.688 \times 415 = 285.5 \text{ mm Hg}$

$p_{CHCl_3} = x_{CHCl_3} \times p^0_{CHCl_3} = 0.312 \times 200 = 62.4 \text{ mm Hg}$

Step 6: Use Equation (1.19) to get vapour-phase mole fractions.

$y_{CH_2Cl_2} = \frac{p_{CH_2Cl_2}}{p_{\text{total}}} = \frac{285.5}{347.9} = 0.82$

$y_{CHCl_3} = \frac{p_{CHCl_3}}{p_{\text{total}}} = \frac{62.4}{347.9} = 0.18$

Key observation: $CH_2Cl_2$ is the more volatile component ($p^0 = 415$ mm Hg vs 200 mm Hg for $CHCl_3$). Its mole fraction jumps from 0.688 in the liquid to 0.82 in the vapour. Meanwhile, $CHCl_3$ drops from 0.312 in the liquid to just 0.18 in the vapour. This confirms the general rule: the vapour phase is always enriched in the more volatile component.