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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 x 10' mm and 6.51 xx 10^7mm, respectively, calculate the composition of these gases in water. |
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Answer» <P> Solution : Total pressure of air in equilibrium with water = 10 atmAir contains 20% oxygen and 79% NITROGEN by volume. According to Raoult.s LAW, PARTIAL pressure of oxygen `(P_(O_2)) = 20/100 xx 10 atm = 2 atm = 2 xx 760 mm = 1520mm` Partial pressure of nitrogen `(P_(N_2)) =79/100 xx 10 atm = 7.9 atm = 7.9 xx 760 mm= 6004mm` `K_H(O_2)= 3.30 xx 10^7 mm , K_H (N_2) = 6.51xx 10^7 mm` Applying Henry.s law and substituting the values, we have `P_(O_2) = K_H xx x_(O_2) " or " x_(O_2) = (P_(O_2))/(K_H) = (1520mm)/(3.30 xx 10^7) = 4.61 xx 10^(-5)` `P_(N_2) = K_H xx x_(N_2) " or " x_(N_2) = (P_(N_2))/(K_H) = (6004 mm)/(6.51 xx 10^7) = 9.22 xx 10^(-5) ` |
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