Vapour Pressure of Liquid Solutions

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🔵Vapour Pressure of Liquid Solutions

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🌡️ Vapour Pressure of Liquid Solutions

Introduction:
Liquid solutions are formed when the solvent is a liquid. The solute in such solutions can be a gas, a liquid, or a solid. In this section, we will discuss binary solutions, i.e., solutions containing two components. These include:

• (i) Solutions of liquids in liquids

• (ii) Solutions of solids in liquids

Such solutions may contain one or more volatile components, with the solvent generally being volatile.

🔹 Vapour Pressure of Liquid–Liquid Solutions

Let us consider a binary solution made of two volatile liquids, denoted by components 1 and 2.

When this mixture is placed in a closed container, both liquids evaporate. Eventually, an equilibrium is established between the vapour phase and the liquid phase.

Let:

• $p_1$ = Partial vapour pressure of component 1

• $p_2$ = Partial vapour pressure of component 2

• $p_{\text{total}}$ = Total vapour pressure of the solution

• $x_1$ = Mole fraction of component 1 in the liquid phase

• $x_2$ = Mole fraction of component 2 in the liquid phase

📘 Raoult’s Law:

Given by: François-Marie Raoult (1886)

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

So,

$$p_1 \propto x_1 \quad \Rightarrow \quad p_1 = p_1^0 \cdot x_1 \tag{1.12}$$
$$p_2 \propto x_2 \quad \Rightarrow \quad p_2 = p_2^0 \cdot x_2 \tag{1.13}$$

Where:

• $p_1^0$ and $p_2^0$ = Vapour pressures of pure components 1 and 2 respectively.

🧪 Total Vapour Pressure:

According to Dalton’s Law of Partial Pressures, the total pressure is the sum of partial vapour pressures:

$$p_{\text{total}} = p_1 + p_2 \tag{1.14}$$

Substituting from equations (1) and (2):

$$p_{\text{total}} = x_1 p_1^0 + x_2 p_2^0 \tag{1.15}$$

Since $x_1 + x_2 = 1$, we can write:

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

📊 Interpretation of Equation :

• (i) Total vapour pressure depends on mole fraction of any one component.

• (ii) The relation is linear with respect to mole fraction $x_{2.}$

• (iii) If $p_1^0 < p_2^0$, then total vapour pressure increases with increasing $x_2$.

🧩 Graphical Representation:

A plot of:

• $p_1$ vs. $x_1$

• $p_2$ vs. $x_2$

• $p_{\text{total}}$ vs. $x_2$

gives straight lines, intersecting at the points where mole fractions = 1 (i.e., pure components).

The line for $p_{\text{total}}$ lies between those of $p_1$ and $p_2$.

🌫️ Composition of Vapour Phase at Equilibrium:

Let:

• $y_1$ = Mole fraction of component 1 in vapour phase

• $y_2$ = Mole fraction of component 2 in vapour phase

Using Dalton’s law:

$$p_1 = y_1 \cdot p_{\text{total}} \tag{1.17}$$
$$p_2 = y_2 \cdot p_{\text{total}} \tag{1.18}$$

In general, for any component $i$:

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

This shows that mole fraction in vapour phase ( $y_i$ ) is not necessarily equal to mole fraction in liquid phase ( $x_i$ ).

🔍 Important Points to Remember:

• Raoult’s law applies to ideal solutions.

• For non-ideal solutions, deviations occur due to intermolecular interactions.

• Mole fractions in liquid and vapour phases differ unless the solution is ideal and forms an azeotrope.

• Vapour pressure is a colligative property, depending on the number of particles, not their nature.

📘 Application in Daily Life:

• Perfumes and essential oils use volatile liquid mixtures.

• Distillation techniques (like in petrochemicals) rely on vapour pressure principles.

• Meteorology and humidity studies use vapour pressure data of water.

✍️ Practice Questions:

1. State Raoult’s law and write its mathematical form.

2. Derive the expression for total vapour pressure in a binary liquid solution.

3. Explain why total vapour pressure changes with mole fraction.

4. A solution has $x_1 = 0.6$, $p_1^0 = 80 \, \text{kPa}$, $p_2^0 = 60 \, \text{kPa}$. Calculate $p_{\text{total}}$.

5. Define ideal and non-ideal solutions with examples.

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