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2.39 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 $298 \; K$ . if the Henry’s law constants for oxygen and nitrogen at $298 \; K$ are $3.30\times 107 \; mm$ and $6.51\times 107 \; mm$ respectively, calculate the composition of these gases in water

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We have been given that the water is in equilibrium with air at a pressure of 10 atm or 7600 mm of Hg.

So the partial pressure of oxygen :

$\frac{20}{100}\times7600 = 1520\ mm\ of\ Hg$

and partial pressure of nitrogen :

$\frac{79}{100}\times7600 = 6004\ mm\ of\ Hg$

Now, by Henry's Law :

$P = K_h.x$

For oxygen :

$x = \frac{1520}{3.30\times10^{7}} = 4.61\times 10^{-5}$

For nitrogen :

$x = \frac{6004}{6.51\times10^{7}} = 9.22\times 10^{-5}$

Hence the mole fraction of nitrogen and oxygen in water is $9.22\times 10^{-5}$ and $4.61\times 10^{-5}$ respectively.

Posted by

Devendra Khairwa

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