Which of the following gases will have the lowest solubility in water at pressure p and temperature 293 K?
(Given: Henry’s constant (kbar) at 293 K for He, H2, N2 and O2 are 144.97, 69.16, 76.48 and 34.86, respectively)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Helium
Henry’s law states that the partial pressure of a gas above a solution is proportional to the mole fraction of the gas dissolved in it:
\( \displaystyle p=K_H\,x \quad\Rightarrow\quad x=\frac{p}{K_H} \)
The mole fraction x is a measure of solubility. At the same pressure and temperature, the gas with the larger KH has the smaller mole fraction in solution, i.e. it is less soluble.
Arranging the given KH values (kbar, 293 K):
\( \displaystyle \ce{O2}\ (34.86) < \ce{H2}\ (69.16) < \ce{N2}\ (76.48) < \ce{He}\ (144.97) \)
So the order of solubility is the reverse: \( \ce{O2} \) > \( \ce{H2} \) > \( \ce{N2} \) > He. Helium, with the highest KH, has the lowest solubility.
NCERT (2026-27 edition):
Solubility of a Gas in a Liquid – …
