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Q. Relation between the volume of gas $(2)$ that dissolves in a fixed volume of solvent $(1)$ and the partial pressure of gas $(2)$ is $\left( n _{ t }=\right.$ total moles, $K _{1}$ and $K _{2}$ are Henry's constants)

Solutions

Solution:

By Henry's law, $p=\chi K _{ H }$

For solute, $p_{2}=\chi_{2} K _{2}$

$=\frac{n_{2}}{n_{1}+n_{2}} K _{2}=\frac{n_{2} K _{2}}{n_{1}}$

Also, $p_{2}=\frac{n_{2} R T}{V}$

$\therefore \frac{n_{2} R T}{V}=\frac{n_{2}}{n_{1}} K _{2}$

$\therefore \frac{R T}{V}=\frac{ K _{2}}{n_{t}}$

$\therefore V=\frac{n_{t} R T}{ K _{2}}$