The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 Kif the Henry’s law constants for oxygen and nitrogen are 3.30 × 107 mm and 6.51 × 107 mm respectively, calculate the composition of these gases in water.

The air is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 Kif the Henry’s law constants for oxygen and nitrogen are 3.30 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.

Solution

Given that

K_H for O₂  =3.30 × 10⁷ mm Hg

K_H for N₂  =6.51 × 10⁷ mm Hg

Percentage of oxygen (O₂) = 20 %

Percentage of nitrogen (N₂)= 79%

Total pressure = 10 atm,

1atm = 760 mm Hg so  we get

Total pressure = 10 × 760 = 7600 mm Hg

The partial pressure of oxygen

= 1520 mm Hg

The partial pressure of nitrogen

= 6004 mmHg

Use the formula of Henry law(K_H is given in problem )

P_A = K_H  ×  X_A

Molar fraction of oxygen  X_O2 = P_O2 / K_H

X_O2 = 1520 /(3.30 × 10⁷)  = 4.61 × 10 ^{– 5}  

Similarly for nitrogen

X_N2 = 6004 /(6.51 × 10⁷)  = 9.22 × 10 ^{– 5}  

Answer

Mole fractions of oxygen = 4.61 ×10⁻⁵

Mole fraction of nitrogen= 9.22 ×10⁻⁵