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, respectively. The water is in equilibirum with air at a pressure of 10 atm. At 298 K, if the Henry.s law constant for oxygen and nitrogen are `3.30 xx 10^(7)mm` and `6.51 xx 10^(7) mm` respectively, calculate the composition of these gases in water.

Explanation

Given that, total pressure of air in equilibirum with water = 10 atm
As air contains 20% oxygen and 79% nitrogen by volume.
`therefore` Partial pressure of oxygen
`(p_(O_(2))) = (20)/(100) xx 10` arm = 2 atm `= 2 xx 760 mm`
`= 1520 mm`
Partial pressure of nitrogen,
`(p_(N_(2))) = (79)/(100) xx 10` atm = 7.9 atm `= 7.9 xx 760 mm`
= 6004 mm
Given that, `K_(H)(O_(2)) = 3.30 xx 10^(7)mm`,
`K_(H) (N_(2)) = 6.51 xx 10^(7) mm`
According to Henry.s law,
`p_(O_(2)) = K_(H) xx chi_(O_(2))`
or `chi_(O_(2)) = (p_(O_(2)))/(K_(H)) = (1520 mm)/(3.30 xx 10^(7)mm) = 4.61 xx 10^(-5)`
`p_(N_(2)) = K_(H) xx chi_(N_(2))`
or `chi_(N_(2)) = (p_(N_(2)))/(K_(H)) = (6004 mm)/(6.51 xx 10^(7)mm)`
`= 9.22 xx 10^(-5)`