A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(i) What is the initial effect of the change on vapour pressure?
(ii) How do rates of evaporation and condensation change initially?
(iii) What happens when equilibrium is restored finally and what will be the final vapour pressure?

(i) Vapour pressure will decrease initially.

(ii) The rate of evaporation remains constant at a temperature in a closed vessel. However, the rate of condensation will be low initially due to the presence of lesser molecules per unit volume in the vapour phase and hence the number of collisions per unit time with the liquid surface decreases.

(iii) When equilibrium is restored finally, then the rate of evaporation becomes equal to the rate of condensation. The final vapour pressure will be the same as it was initial.

(i) The initial effect on the vapour pressure of increasing the volume of the container will be that the vapour pressure is lowered. This is because the same amount of vapours now occupy large space.

(ii) Due to a sudden increase in volume, the pressure inside the sealed container suddenly decreases. According to Le-Chatelier’s principle for the equilibrium,



A decrease in pressure will shift the equilibrium in the forward direction. Thus the rate of evaporation will increase initially. The rate of condensation decreases initially because vapour pressure per unit volume decreases. However, due to increase in the rate of evaporation, the amount of vapour begins to increase and so the rate of condensation also begins to increase.

(iii) When the equilibrium is again reached, the rate of evaporation becomes equal to the rate of condensation. The final vapour pressure will remain the same as vapour pressure before increasing the volume of the container. This is because the vapour pressure of a liquid does not depend on upon the amount of the liquid or the space above it but depends on only upon temperature.