K_{H} values for H_2, CO_2, CH_4 and O_2 at 2 | vedclass

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(A) According to Henry's Law,the solubility of a gas in a liquid is inversely proportional to its Henry's Law constant $(K_{H})$.Mathematically,$S \pr

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$H_2 < CH_4 < O_2 < CO_2$

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(A) According to Henry's Law,the solubility of a gas in a liquid is inversely proportional to its Henry's Law constant $(K_{H})$.Mathematically,$S \propto \frac{1}{K_{H}}$.Given $K_{H}$ values are:$H_2 = 71.18 \ kbar$$CH_4 = 41.85 \ kbar$$O_2 = 34.86 \ kbar$$CO_2 = 1.67 \ kbar$Since the solubility is inversely proportional to $K_{H}$,the gas with the highest $K_{H}$ value will have the lowest solubility.The order of $K_{H}$ values is: $CO_2 Therefore,the increasing order of solubility is: $H_2 < CH_4 < O_2 < CO_2$.

$K_{H}$ values for $H_2, CO_2, CH_4$ and $O_2$ at $298 \ K$ are $71.18, 1.67, 41.85$ and $34.86 \ kbar$ respectively. Arrange these gases in the increasing order of their solubility.

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