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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?

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    $\begingroup$ read Dr Pippiks answer $\endgroup$
    – jimchmst
    Commented Nov 20, 2023 at 22:09

2 Answers 2

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You state, it seems that, logically, "increasing pressure leads to an increase in dissolved gas." Your logic is not wrong... an increased mass of gas dissolves.

The confusion is because here, Henry's law is stated in terms of the volume of gas. Since increasing pressure makes the gas occupy less volume, it has higher density. As a hypothetical example:

  • 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.

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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 $$V=\frac{KRT}{n_{solvent}}$$

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  • $\begingroup$ Yes, from the OP, I believe that is understood. However, the question from that intuitive approach implies concern about the increased mass, rather than constant volume. $\endgroup$ Commented Nov 22, 2023 at 20:18

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