Question

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.

Answer 1

Percentage of oxygen (O₂) in air = 20 %

Percentage of nitrogen (N₂) in air = 79%

Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is, (10 × 760) mm Hg = 7600 mm Hg

Therefore,

Partial pressure of oxygen,

= 1520 mm Hg

Partial pressure of nitrogen,

= 6004 mmHg

Now, according to Henry’s law:

p = K_H.x

For oxygen:

For nitrogen:

Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10⁻⁵and 9.22 × 10⁻⁵ respectively.