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 nbitrogen are 3.30xx10^(7) and 6.51xx10^(7) respectively, calculate the composition of these gases in water.

The air is a mixture of a number of gases. The major components are ox

1.	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 nbitrogen are 3.30xx10^(7) and 6.51xx10^(7) respectively, calculate the composition of these gases in water.
Answer» Solution :Percentage of oxygen `(O_(2))` in air = 20% Percentage of NITROGEN `(N_(2))` in air= 79% Also, it is GIVEN that water is in equilibrium with air at a total PRESSURE of 10 atm, that is `(10xx760)` mm Hg = 7600 mm Hg Therefore, Partial pressure of oxygen, `P_(O_(2))=(20)/(100)xx7600` mm Hg = 1520 mm Hg Partial pressure of nitrogen, `P_(N_(2))=(79)/(100)xx7600` mm Hg = 6004 mm Hg 79 `= 79xx760` mm Now, according to Henry.s law : `p=K_(H)x` for oxygen : `p_(O_(2))=KH.x_(O_(2))` `x_(O_(2))=(P_(O_(2)))/(K_(H))` `= ("1520 mm Hg")/(3.30xx10^(7)mm Hg)` (Given `K_(H)=330xx10^(7)` mm Hg) `= 461xx10^(-5)` For nitrogen : `P_(N_(2))=KH.x_(N_(2))` `x_(N_(2))=(P_(N_(2)))/(K_(H))` `= ("6004 mm Hg")/(6.51xx10^(7)mm Hg)` `= 9.22xx10^(-5)` Hence, the mole fractions of oxygen and nitrogen in water are `4.61xx10^(-5)` and `9.22xx10^(-5)`.