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

Answers (1)

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.

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