# Is gas solubility independent of pressure?

My physical chemistry textbook poses this question:

Prove the statement that an alternative way to express Henry’s law of gas solubility is to say that the volume of gas that dissolves in a fixed volume of solution is independent of pressure at a given temperature.

My proof is thus:

$$P_2 = Kx_2$$, where $$P_2$$ is the partial pressure of a solute, $$K$$ is the Henry's law constant, and $$x_2$$ is the mole fraction of solute dissolved in solution. By the ideal gas law,

$$\frac{n_2RT}{V} = Kx_2 \rightarrow V = \frac{n_2RT}{Kx_2}$$.

Therefore, V depends on temperature, not pressure. Regardless of if my proof's correct, I'm missing part of the picture here. Intuitively, I want believe increasing pressure leads to an increase in dissolved gas. If you have a beaker of water and some gas in a metal box, and then you decrease the volume of that container, like a piston, at constant temperature, wouldn't more gas dissolve?

• At STP, i.e., 1 atm, Gas X has a density $${1 g/l}$$. It dissolves in water at the ratio $${1 l_{Gas X}/l_{H2O}}$$, so 1 gram of Gas X is absorbed in a liter of water.
• At 10 atm, Gas X has a density $${10 g/l}$$. Still, just one one liter dissolves in a liter of water. But that one liter is 10 grams of Gas X at that higher pressure, so ten times the mass of Gas X dissolves in a liter of water.
Of course, this assumes a perfect gas, and that there is no chemical reaction with the liquid. For example, $$\ce{CO2}$$ in water may form clathrates, throwing off a linear relation. The reference cited provides corrections to Henry's law to account for empirical results.
Let n be the number of moles of solute dissolved in $$n_{solvent}$$ moles of solvent (in dilute solution). Then, $$x=\frac{n}{n_{solvent}}$$So, $$P=\frac{n}{n_{solvent}}K$$or$$K=n_{solvent}\frac{P}{n}=n_{solvent}\frac{V}{RT}$$So, the volume of ideal gas at temperature T that dissolves is $$V=\frac{KRT}{n_{solvent}}$$