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Question 39

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 x 107 mm and 6.51 x 107mm respectively, calculate the composition of these gases in water.

Answer

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, po2 = 20/100 *7600

=1520 mm Hg

Partial pressure of nitrogen,pN2 = 79/100 *7600

= 6004 mmHg

Now, according to Henry's law:

p = KH.x

For oxygen:

po2  = KH. xO2

xO2 po2 / KH

= 1520 / 3.30 X 107

= 4.61 10-5

For nitrogen:

pN2 = KH.xN2

xN2 pN2 / KH

= 6004 / 6.51 x 107

= 9.22 x 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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