For a gas dissolved in a liquid, Henry's law can be written as $P = K_H x$, where $x$ is the mole fraction of the gas in solution. If the pressure of the gas above the liquid is increased at constant temperature, what happens to the solubility of the gas?
AIt decreases because $x$ is inversely proportional to $P$
BIt increases because $x$ is directly proportional to $P$
CIt remains unchanged because $K_H$ changes to keep $x$ constant
DIt first increases and then decreases
Answer & Solution
Correct answer: B. It increases because $x$ is directly proportional to $P$
From Henry's law, $P = K_H x$, so at constant temperature for a given gas-liquid pair, $K_H$ is constant and $x = P/K_H$. Therefore, as pressure increases, the mole fraction solubility of the gas increases.
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