The air is a mixture of a number of gases. The major components are oxygen and nitrogen with the 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 constant for oxygen and nitrogen at $298 K$ are $3.30 \times 10^{7} m m$ and $6.51 \times 10^{7} m m$ respectively. Calculate the composition of these gases in water.

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Solution

Given:

$K_{H} f o r O_{2} = 3.30 \times 10^{7} m m H g K_{H} f o r N_{2} = 6.51 \times 10^{7} m m H g$

$p e r c e n t a g e o f N_{2} = 79$

$p e r c e n t a g e o f O_{2} = 20$

$t o t a l p r e s s u r e = 10 a t m$

$1 a t m = 760 m m H g$

$t h e r e f o r e t o t a l p r e s s u r e 10 \times 760 = 7600 m m H g$

$T h e p a r t i a l p r e s s u r e o f o x y g e n$

$P_{O_{2}} = \frac{20}{100} \times 7600 m m H g = 1520 m m H g$

The partial presure of Nitrogen,

$P_{N_{2}} = \frac{79}{100} \times 7600 m m H g = 6400 m m H g$

By using the Henry law,

$P_{A} = K_{H} \times X_{A}$

Molar fraction of oxygen,

$X_{O_{2}} = \frac{P_{O_{2}}}{K_{H}}$

$X_{O_{2}} = \frac{1520}{3.30 \times 10^{7}} = 4.61 \times 10^{- 5}$

$X_{N_{2}} = \frac{6004}{6.5 \times 10^{7}} = 9.22 \times 10^{- 5}$ .

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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 $3.30 \times 10^{7} m m$ and $6.51 \times 10^{7} m m$ respectively, calculate the composition of these gases in water.