Explain liquid-vapour equilibrium.

(N/A) Process: We consider the example of a transparent box carrying a $U$-tube with mercury (manometer). $A$ drying agent like anhydrous calcium chloride $(CaCl_2)$ or phosphorus penta-oxide $(P_4O_{10})$ is placed for a few hours in the box. By doing this,the air in the box becomes free from vapour (moisture).
Now,after removing the drying agent by tilting the box on one side,a watch glass (or petri dish) containing water is quickly placed inside the box.
Observation: It will be observed that the mercury level in the right limb of the manometer slowly increases and finally attains a constant value.
Conclusion: The observation shows that:
$(i)$ In the beginning,the pressure inside the box increases due to the evaporation of water.
$(ii)$ After some time,the pressure inside the box becomes constant,indicating that the rate of evaporation equals the rate of condensation.
$(iii)$ The volume of water in the watch glass decreases initially and then becomes constant,representing the liquid-vapour equilibrium state: $H_2O(l) \rightleftharpoons H_2O(vap)$.