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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.
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.x
For oxygen:
`"p"_("O"_2) = "K"_"H" xx "x"_("O"_2)`
`=> "x"_("O"_2) = "p"_("O"_(2))/"K"_"H"`
= `(1520 "mm Hg")/(3.30xx10^(7) "mm Hg")` ...(Given KH = 3.30 × 107 mm Hg)
= 4.61 × 10−5
For nitrogen:
`"p"_("N"_2) = "K"_"H" xx "x"_("N"_(2))`
`=>"x"_("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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