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 K if 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

Total pressure of air in equilibrium with water = 10 atmosphere

As air contains 20% oxygen and 79% nitrogen by volume,

∴ Partial pressure of oxygen, `(p_(O_2)) = 20/100 xx 10` atm

= 2 atm

= 2 × 760 mm Hg

= 1520 mm Hg

Partial pressure of nitrogen, `(p_(N_2)) = 79/100 xx 10` atm

= 7.9 atm

= 7.9 × 760 mm Hg

= 6004 mm Hg

Now, according to Henry’s law:

p = K_H.x

For oxygen:

`p_(O_2) = K_H xx chi_(O_2)`

`=> chi_(O_2) = p_(O_(2))/K_H`

= `(1520  "mm Hg")/(3.30 xx 10^(7)  "mm Hg")`    ...(Given K_H = 3.30 × 10⁷ mm Hg)

= 4.61 × 10⁻⁵

For nitrogen:

`p_(N_2) = K_H xx x_(N_(2))`

`=> chi_(N_(2)) = p_(N_(2))/K_H`

= `(6004  "mm Hg")/(6.51xx10^(7)  "mm Hg")` ...(Given K_H = 6.51 × 10⁷ mm Hg)

= 9.22 × 10⁻⁵

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