Vapour pressure can be defined as pressure formed by the vapor of the liquid (or solid) over the surface of the liquid. This pressure is formed in a thermodynamic equilibrium state in a closed container at a certain temperature. Liquid’s evaporation rate is identified by the equilibrium vapor pressure. Vapour pressure increases with the temperature. The boiling point of the liquid is the point when the pressure exerted by surrounding equals to the pressure exerted by vapor. In this chapter, we will study more about the Vapour Pressure on liquids and its characteristics.

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## What is Vapour Pressure?

You have read about pressure. It is the force that a mass exerts on another. But, do you know vapour also exerts pressure, even if you can’t see it? You have of course felt it! Haven’t you?

We form a liquid Solution by dissolving a solid, liquid or gas in a specific liquid solvent. Vapour Pressure of liquid solutions is the amount of pressure that the vapours exert on the liquid solvent when they are in equilibrium and a certain temperature.

Vapour pressure changes with the temperature of the surroundings and the nature of the liquid. Let us now have a look a some of the important characteristics of vapour pressure on liquid solutions.

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### Characteristics of Vapour Pressure

- A pure liquid experiences a greater amount of vapour pressure as against a liquid’s solution.
- It is inversely proportional to the forces of attraction existing between the molecules of a liquid.
- It increases with a rise in the temperature. This is because the molecules gain kinetic energy and thus, vapourise briskly.

A liquid solution might have a volatile solute and solvent as well. In the majority of the cases, we can see that the solvent is volatile, while the solute is not. Based on this, we will study it in two instances: Liquid-Liquid Solutions and Solid-Liquid Solutions.

### Vapour Pressure of Liquid-Liquid Solutions

To evaluate this, we take two volatile liquid solutions. Let us name their liquid components as A and B. After placing the volatile liquid with their components in a closed vessel, we find that equilibrium establishes between the liquid phase and vapour phase.

Suppose P_{total} is the overall vapour pressure at an equilibrium state and let P_{A} and P_{B} be the partial vapour pressures of components A and B respectively. Adding further, the mole fraction of respective components is x_{A} and x_{B }respectively. To evaluate vapour pressure of volatile liquids, we use Raoult’s Law.

## Raoult’s Law

The law states that the partial pressure is directly proportional to the mole fraction of the solute component. So, according to Raoult’s Law, the partial pressure of A will be

P_{A} ∝ x_{A}

P_{A} = P_{A}^{0} x_{A}

where P_{A}^{0} is the vapour pressure of pure liquid component A. Similarly partial pressure of B will be

P_{B} ∝ x_{B}

P_{B} = P_{B}^{0 }x_{B}

where P_{B}^{0} is the vapor pressure of pure liquid component B. Now, we will apply Dalton’s law of partial pressures. This law tells us that the total pressure ( P_{total}) of the solution placed in a container is the sum of partial pressures of its respective components. That is

P_{total} = P_{A} + P_{B}

P_{total} = P_{A}^{0} x_{A} + P_{B}^{0 }x_{B}

Also since, x_{A} + x_{B} = 1 , we can write the relation as:

P_{total} = P_{A}^{0} + (P_{B}^{0} – P_{A}^{0}) x_{B}

### Vapour Pressure of Solutions of Solids in Liquids

Let us now look at the other type of solutions i.e. solids in liquid solution. Here we consider a solid as the solute, while the solvent is a liquid. In these cases, the solute is non-volatile in nature. The vapour pressure is less than the pure vapor pressure of the solution. Let us now see how we can find the overall vapour pressure of such a solution.

We consider a solution in which A is the solvent and B is the solute. Applying the Raoult’s Law, we know that partial vapour pressure of individual component (solute/solvent) is directly proportional to its mole fraction.

Now, if we add a non-volatile solute, it is imperative that vapour pressure comes only from the solvent part. This is because they are the only available component in the vapour phase. Hence, if P_{A} is the vapor pressure of the solvent, x_{A} is its mole-fraction and P_{A}^{0} is the vapor pressure of the pure solvent, then by Raoult’s law, the relation will come out to be:

P_{A} ∝ x_{A}

P_{A} = P_{A}^{0} x_{A}

When we plot a graph between mole fraction of solvent and vapour pressure, we find its nature to be linear.

## Solved Example For You

Question: Why does the Vapour Pressure decrease when we add a non-volatile solute in the solvent?

Answer: We know that evaporation is a surface phenomenon. The greater the surface area, the more is the evaporation. Hence, the vapour pressure is more. In a pure liquid, we can see that there is more surface area available for the molecules to get evaporated. Therefore, they have more vapour pressure.

On adding a non-volatile solute, we cause the solvent molecules to get less surface to escape. Hence, it decreases.