Question
At a particular temperature and at a pressure of 1 atm, the mass and volume of O3 that can be dissolved in 1 L of H2O is ‘m’ gram and ‘V’ ml, respectively. Assuming that the solution obeys Henry’s law, the mass and volume of O3 dissolved in 2 L of water at the same temperature and at 5 atm will be
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Solution
$p = K_{H}.X \approx K_{H}.\frac{n_{\text{gas}}}{n_{\text{liq}}} \Rightarrow m_{\text{gas}} \propto P.V_{\text{liq}}$
$\therefore\frac{m_{2}}{m_{1}} = \frac{P_{2}.V_{2}}{P_{1}.V_{1}} \Rightarrow \frac{m_{2}}{m_{1}} = \frac{5 \times 2}{1 \times 1} \Rightarrow m_{2} = 10\text{ m}$
Now, $P = K_{H}.\frac{n_{\text{gas}}}{n_{\text{liq}}} = \frac{K_{H}}{n_{\text{liq}}}.\frac{PV}{RT}$
∴ Volume of gas dissolved, $V = \frac{RT}{K_{H}}.n_{\text{liq}}$
(Volume of gas dissolved is independent of pressure of gas)
$\therefore\frac{V_{2}}{V_{1}} = \therefore\frac{V_{2}.\text{liquid}}{V_{1}.\text{liquid}} \Rightarrow \frac{V_{2}}{V\text{ ml}} = \frac{2}{1} \Rightarrow V_{2} = 2V\text{ ml}$
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