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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.
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Solution
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 = KH . χ
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 KH = 3.30 × 107 mm Hg)
= 4.61 × 10−5
For nitrogen:
`P_(N_2) = K_H xx chi_(N_(2))`
⇒ `chi_(N_(2)) = P_(N_(2))/K_H`
= `(6004 "mm Hg")/(6.51xx10^(7) "mm Hg")` ...(Given KH = 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.
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