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Thursday, 2 May 2013

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.


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
Given that
KH for O2  =3.30 × 107 mm Hg
KH for N2  =6.51 × 107 mm Hg

Percentage of oxygen (O2) = 20 %
Percentage of nitrogen (N2)= 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(KH is given in problem )
PA = KH  ×  XA
Molar fraction of oxygen  XO2 = PO2 / KH
XO2 = 1520 /(3.30 × 107)  = 4.61 × 10 – 5  
Similarly for nitrogen
XN2 = 6004 /(6.51 × 107)  = 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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