Practice question
Question
A gas dissolves in a solvent with a Henry’s law constant of 250 bar at 25°C. If the mole fraction of the gas doubles when the temperature changes, and the new partial pressure is 10 bar, what is the new Henry’s law constant?
Explanation
Initial: p = 250 · x . New: 10 = KH' · 2x . Assume initial p = 5 bar , then x = (5/250) = 0.02 . New x = 0.04 , 10 = KH' × 0.04 , KH' = 250 bar . Recalculate: KH' = (10/0.04) = 250 bar (consistent).
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