A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.

(a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume. 

(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially. 

(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.  

On increasing volume of the container, the vapour pressure will initially decrease because the same amount of vapours are now distributed over a larger space.

On increasing the volume of the container the rate of evaporation will increase initially because now more space is available. Since the amount of the vapour per unit volume decreases on increasing the volume, therefore, the rate of condensation will decrease initially.

Finally, equilibrium will be restored when the rates of the forward and backward processes become equal. However, the vapour pressure will remain unchanged because it depends upon the temperature and not upon the volume of the container.

(a) Vapour pressure decreases due to increase in volume. 

(b) Rate of evaporation remains same and rate of condensation decreases. 

(c) Finally the same vapour pressure is restored and the rate of evaporation becomes equal to the rate.