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 at 298 K are $3.30 × 107 mm$ and $6.51 × 107 mm$ respectively, calculate the composition of these gases in water

Answer 1

Percentage of oxygen $(O_2)$ in air = $20 \%$
Percentage of nitrogen $(N_2)$ 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, $p^o_2=\frac{20}{100} \times 7600mmHg$
= 1520 mm Hg
Partial pressure of nitrogen, $p_{N_2}=\frac{79}{100} \times 7600mmHg$
= 6004 mmHg
Now, according to Henry's law:
$p = K_H.x$
For oxygen:
$p^o_2=K_H.x_{O_2}$
$⇒x_{O_2}=\frac{p_{O_2}}{K_H}$
$=\frac{1520mmHg}{3.30 \times 10^7mmHg}$ (Given $K_H=3.30 \times 10^7mmHg$)
$=461 \times 10^{-5}$
For nitrogen:
$p_{N_2}=K_H.x_{N_2}$
$⇒x_{N_2}=\frac{p_{N_2}}{K_H}$
$=\frac{6004mm\,Hg}{6.51 \times 107mm\,Hg}$
$=9.22 \times 10^{-5}$
Hence, the mole fractions of oxygen and nitrogen in water are $4.61 ×10^{-5}$ and $9.22 × 10^{-5}$ respectively.