Question

Following data has been given for $C{O}_2$ for the concentration in $H_{2}O.$
$\begin{array}{ccc}\text{Temperature}& \text{Henry's constant}& \text{Pressure}\\ 273K& 728atm& 0.10atm\\ 333K& 3410atm& p_2\end{array}$
If solution of $C{O}_2$ in $H_{2}O$ is heated from $273K$ to $333K$, pressure $\left(p_2\right)$ needed to keep $C{O}_2$ in the solution is :

• A. $0.108atm$
• B. $0.805atm$
• C. $0.212atm$
• D. $0.468atm$

Answer 1

The correct option is D $0.468atm$

Henry Law : The solubility of a gas in a liquid at a given temperature is directly proportional to its partial pressure of the gas present above the solution.

Given,
$\text{At}273K$,
$K_H=728atm$
Partial pressure of gas, $p_1=0.10atm$
By Henry's law,
$p=K_H\times \chi$
Thus, ​
${\chi }_{C{O}_2}\left(\text{mole fraction of}C{O}_2\right)=\frac{p}{K_H}=\frac{0.10}{728}$
$\text{At}333K$
By Henry's law:

${\chi }_{C{O}_2}=\frac{p}{K_H}=\frac{p_2}{3410}$

$\therefore$ $\frac{p_2}{3410}=\frac{0.10}{728}$

$p_2=\frac{0.10\times 3410}{728}=0.468atm$
Thus, the partial pressure required to keep $C{O}_2$ in the solution at $333K$ is $0.468atm$