Question In a container of capacity 1-Litre, air and some liquid water is present in equilibrium at total pressure of 200 mm of Hg. This container is connected to another 1-Litre evacuated container. Find total pressure inside the container when equilibrium is again stabilized. Aqueous Tension at this temp. is 96 mm of Hg. Solution( Given in the book ) Total Pressure = 200mm Pressure of Gas = 200 - Aq. Tension ==> P1 = 200-96 = 104mm Hg Now the volume gets doubled ( Temp and moles of the gas is conserved, Hence Boyle's law holds good over here ) P1V1 = P2V2 ==> P2 = 52mm Hg After equilibrium is established Ptotal = 52 + Aq.Tension = 148mm Hg Problem In the solution we separated the pressure of the gas and the water vapour, but WHY? Won't the vapour of water(Aq. Tension) also follow the ideal gas equation, so that we can put the total pressure of the container(200mm Hg) in the equation. Moreover, the above statement implies that the aqueous tension depends merely on temp. and not on the volume of the container. But being a gas (water vapour) its pressure should reduce upon increase in the volume. I have searched the internet to clear my concepts about aqueous tension, vapour pressure and ideal gas equation but to no profit. Please help me out. Thanks for your time.