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 Kif 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.
Solution
Percentage of oxygen (O2) in air = 20 %
Percentage of nitrogen (N2) 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)) = 20/100xx7600` mm Hg
= 1520 mm Hg
Partial pressure of oxygen, `p_(N_(2)) = 79/100xx7600` mm Hg
= 6004 mmHg
Now, according to Henry’s law:
p = KH.x
For oxygen:
`p_(o_(2)) =K_H.x_(o_2)`
`=> x_(o_2) = p_(o_(2))/K_H`
`= (1520 "mm Hg")/3.30xx10^(7) mm Hg` (Given `K_H = 3.30 xx 10^(7)` mm Hg)
= 4.61 x 10-5
For nitrogen:
`p_(N_(2)) = K_H.x_(N_(2))`
`=>x_(N_(2)) = p_(N_(2))/K_H`
` = (6004 "mm Hg")/(6.51xx10^(7)"mm Hg")`
= 9.22 x 10-5
Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10−5and 9.22 × 10−5 respectively.
