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 × 10

^{7}mm and 6.51 × 10^{7}mm respectively, calculate the composition of these gases in water.
Solution

Given
that

K

_{H}for O_{2}=3.30 × 10^{7}mm Hg
K

_{H}for N_{2}=6.51 × 10^{7}mm Hg
Percentage
of oxygen (O

_{2}) = 20 %
Percentage
of nitrogen (N

_{2})= 79%
Total
pressure = 10 atm,

1atm
= 760 mm Hg so we get

Total
pressure = 10 × 760 = 7600 mm Hg

The
partial pressure of oxygen

=
1520 mm Hg

The
partial pressure of nitrogen

=
6004 mmHg

Use
the formula of Henry law(K

_{H}is given in problem )
P

_{A}= K_{H}× X_{A}
Molar
fraction of oxygen X

_{O2}= P_{O2}/ K_{H}
X

_{O2 }= 1520 /(3.30 × 10^{7}) = 4.61 × 10^{– 5 }
Similarly
for nitrogen

X

_{N2 }= 6004 /(6.51 × 10^{7}) = 9.22 × 10^{– 5 }
Answer

Mole
fractions of oxygen = 4.61 ×10

^{−5 }
Mole
fraction of nitrogen= 9.22 ×10

^{−5}
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