Air is a mixture gases. It contains 20% by volume of O₂ gas and 79% by volume of N₂ gas at 298K. Water is in quilibrium with air at a pressure of 10 atmospheres. KH for O₂ and N₂ at 298 K are 3.30x10⁷mm and 6.51x 10⁷ mm respectively. Determine the composition of these gases in water.

We are given that the total pressure of air above water = P = 10 atm.
partial pressure of O₂ in air above water = P1
partial pressure of N₂ above water = P2

Volume of air above water = V
Volume of O₂ = V1 = 0.20 V
Volume of N₂ = V2 = 0.79 V
let n₁ moles of Oxygen and n₂ moles of N₂ be present in air in volume V.
Let n = total number of moles of gases in air

From Daltons' law of partial pressures and the law of ideal gas mixture:

V1/V = P1/P = n1/n
V2/V = P2/P = n2/n
P1=(V1/V)*P=0.20*10=2.0 atm
P2=P*(V2/V)=0.79*10=7.9 atm


Henry's law for solubility : p=k_H/ c
where p = partial presssure of the gas above the liquid in equilibrium with solution
[k_H] Henry's constant
c = molar concentration of the gas in moles/Litre in the solution
On applying th above mentioned formula, one can easily find the concentration of oxygen and nitrogen in water.