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, p_{o2=20/100x7600mmHg}

 = 1520 mm Hg

 Partial pressure of nitrogen,

                                       p_{N2=79/100X7600mmHg}

 = 6004 mmHg

 Now, according to Henry’s law: 

p = KH.x

 For oxygen: 

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