# Why does gas solubility increase at higher pressure?

Mar 23, 2017

From Henry's law (at least, this version),

$s = {k}_{H} {P}_{g a s}$,

where:

• $s$ is the solubility in $\text{M}$.
• ${k}_{H}$ is the Henry's law constant in $\text{M/atm}$.
• ${P}_{g a s}$ is the vapor pressure of the gas above the solution.

With increased external pressure, the gas above the solution has an increased vapor pressure above the solution, since ${P}_{g a s} = {\chi}_{\left(g\right)} {P}_{e x t}$, where ${\chi}_{\left(g\right)}$ is the mole fraction of the gas above the solution and we've set the external pressure as the total pressure above the solution.

That mathematically tells you that the solubility increases in the solution for higher pressures.

That is, ${P}_{e x t} \uparrow \implies {P}_{g a s} \uparrow \implies s \uparrow$. The Henry's law constant is obviously going to stay the same, because it's constant at constant temperature.

A more conceptual way of explaining this is that higher pressures means greater force of collision between gas particles above the solution (pressure is force per unit area). Therefore, their average kinetic energy is greater, and their average speeds are greater.

As a result, they are more likely to knock some of the particles down into the solution, thereby dissolving them.