Given :
Solubility of O2
= S1 = 2.6 × 10-3 mol dm-3
Initial pressure of O2 = P1 = 2 atm
Final pressure of O2 = P2 = 8.4 atm
Solubility of O2 = S2 = ?
(i) By Henry’s law,
S1 = KH x P1
∴ Henry’s law constant KH is,
KH = \(\frac{S_1}{P_1}\) = \(\frac{2.6 \times 10^{-3}}{2}\)
= 1.3 × 10-3 mol dm-3 atm-1
(ii) Now,
S2 = KH x P2
= P2 = 1.3 x 10-3 x 8.4
= 10.92 × 10-3
= 1.092 × 10-2 mol dm-3
∴ Solubility of O2 = 1.092 × 10-2 mol dm-3
