Vapour Pressure Formula: Definition, Raoult’s law, Solved Examples

Vapour Pressure Formula: Vapour pressure is the pressure exerted by the gaseous molecules formed when a liquid undergoes evaporation. When the liquid is contained within a sealed enclosure, these gaseous molecules accumulate above the liquid’s surface, creating a pressure known as vapour pressure.

What is Vapour Pressure

Vapour pressure represents a substance’s inclination to transition into a gaseous form. It’s typically the pressure exerted by the vapor of a liquid in thermodynamic equilibrium with its condensed phases within a sealed environment. As temperature rises, so does the vapour pressure of a liquid. This pressure can be quantified through various methods, with one straightforward approach being the use of a manometer within a sealed container or flask.

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Vapour Pressure Formula

When a solid dissolves in a liquid, a solution is formed, and the vapour pressure of this solution is influenced by the presence of the solute. Raoult’s law provides a formula to calculate the vapour pressure of the solution:

P_solution = (X_solvent)(P°_solvent)

Where:

X_solvent = the mole fraction of the solvent in the solution

P_solution = the vapour pressure of the solution

P°_solvent = the vapour pressure of the pure solvent

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Factors Affecting Vapour Pressure

Vapour pressure is contingent on several factors:

Nature of Liquid: It’s determined by the intermolecular forces within the liquid. Stronger forces lead to lower vapour pressure.

Effect of Temperature: Higher temperatures increase kinetic energy, boosting the tendency of molecules to escape, thereby increasing vapour pressure.

Concentration of Solute: The presence of solute reduces vapour pressure, with the extent of reduction depending on solute concentration.

Independence from Humidity: Vapour pressure is solely influenced by temperature, not humidity, as long as other variables remain constant.

Volume: Vapour pressure remains unaffected by container volume due to equilibrium between liquid and vapour phases.

Surface Area: Generally, vapour pressure is not influenced by surface area.

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Vapour Pressure Formula Solved Examples 

Example 1. At 25°C, the vapour pressure of an aqueous solution is measured at 23.80 mmHg. Given that the vapour pressure of pure water at the same temperature is 25.756 mm Hg, calculate the mole fraction of the solute in this solution.

Solution:

Using Raoult’s Law:

23.80 = (X)(25.756)

Solving for X:

X = 0.92405 (the mole fraction of the solvent)

To find the mole fraction of the solute:

X_solute = 1 – 0.92405

X_solute = 0.07595

Example 2. In a laboratory experiment, a mixture of ethanol (C2H5OH) and water (H2O) is prepared. The vapor pressure of this mixture at 30°C is found to be 45 mmHg. Knowing that the vapor pressure of pure ethanol at this temperature is 43 mmHg and the vapor pressure of pure water is 31 mmHg, calculate the mole fraction of ethanol and water in the mixture.

Solution:

Using Raoult’s Law for ethanol:

45 = (X_ethanol)(43)

Solving for X_ethanol:

X_ethanol = 45 / 43 ≈ 1.0465

To find the mole fraction of water:

X_water = 1 – X_ethanol

X_water = 1 – 1.0465 ≈ -0.0465

Since mole fractions cannot be negative, it appears there might be an error in the experiment or calculation.

Example 3. A chemist is investigating a binary mixture of two volatile liquids, A and B. At a given temperature, the vapor pressure of this mixture is measured to be 85 mmHg. When pure liquid A is subjected to the same conditions, its vapor pressure is 60 mmHg, and pure liquid B has a vapor pressure of 40 mmHg under the same conditions. Determine the mole fraction of each component in the mixture.

Solution:

Using Raoult’s Law for liquid A:

85 = (X_A)(60)

Solving for X_A:

X_A = 85 / 60 ≈ 1.4167

To find the mole fraction of liquid B:

X_B = 1 – X_A

X_B = 1 – 1.4167 ≈ -0.4167

Again, mole fractions cannot be negative, indicating a potential error in the experiment or calculation.