Question

QUESTION 2.39

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 pressure of 10 atm. At 298 K if the Henry's law constants for oxygen and nitrogen at 298 K are $3.30\times {10}^{7}mm$ and $6.51\times {10}^{7}mm$ respectively, calculate the composition of these gases in water.

Answer 1

Step I : Calculation of partial pressure of oxygen and nitrogen.

Partial pressure of $O_2\left(p_{O_2}\right)=\left(10atm\right)\times \frac{20}{100}=2atm=2\times 760mm$

Partial pressure of $O_2\left(p_{N_2}\right)=\left(10atm\right)\times \frac{79}{100}=79atm=79\times 760mm$

Step II: Composition of $O_2$ and $N_2$ dissolved in water.

The amount of gases dissolved in water is calculated in terms of their mole fractions.

$x_{O_2}=\frac{p_{O_2}}{K_H}$

$=\frac{(2\times 760mm)}{(3.30\times {10}^{7}mm)}$

$=4.6\times {10}^{-5}$

$x_{N_2}=\frac{P_{N_2}}{K_H}=\frac{(7.9\times 760mm)}{(6.51\times {10}^{7}mm)}$

$=9.22\times {10}^{-5}$