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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 at 298 K are $3.30 \times 10^7$ mm and $6.51 \times 10^7$ mm respectively, calculate the composition of these gases in water.

$\begin{array}{1 1}4.61\times 10^{-5},9.22\times 10^{-5}\\3.61\times 10^{-5},8.22\times 10^{-5}\\5.61\times 10^{-5},7.22\times 10^{-5}\\2.61\times 10^{-5},6.22\times 10^{-5}\end{array} $

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Total pressure of air in equilibrium with water =10 atm
As air contains 20% oxygen and 79% nitrogen by volume
Partial pressure of oxygen $(PO_2)=\large\frac{20}{100}$$\times 10$ atm
$\Rightarrow 2$ atm =$2\times 760$ mm
$\Rightarrow 1520$ mm
Partial pressure of nitrogen $(P_{N_2})=\large\frac{79}{100}$$\times 10$ atm
$\Rightarrow 7.9\times 760$ mm =6004 mm
$K_H(O_2)=3.30\times 10^7$ mm
$K_H(N_2)=6.51\times 10^7$ mm
Applying Henry's law
$PO_2=K_H \times x_{O_2}$
$x_{O_2}=\large\frac{PO_2}{K_H}=\frac{520mm}{3.30\times 10^7mm}$
$\Rightarrow 4.61\times 10^{-5}$
$PN_2=K_H \times x_{N_2}$
$x_{N_2}=\large\frac{P_{N_2}}{K_H}$
$x_{N_2}=\large\frac{6004}{6.51\times 10^7}$
$\Rightarrow 9.22\times 10^{-5}$
answered Aug 6, 2014 by sreemathi.v
 
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