At $20^{0} C$ , two balloons of equal volume and porosity are filled to a pressure of 2 atm, one with $14 k g N_{2}$ and the other with $1 k g$ of $H_{2} .$ The $N_{2}$ balloon leaks to a pressure of $\frac{1}{2} a t m$ in 1 hour. How long (min) will it take for the $H_{2}$ balloon to reach a pressure of $\frac{1}{2} a t m$ ?

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answer is 16.

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Detailed Solution

We know that
rate of diffusion $\propto \sqrt{\frac{1}{D e n s i t y}} .......... (i)$
Rate of diffusion of $N_{2} (r_{N_{2}}) = \frac{\Pr e s s u r e l e a k e d}{T i m e} = \frac{1 / 2 a t m}{10 h}$
Rate of diffusion of $H_{2} (r_{H_{2}}) = \frac{\Pr e s s u r e l e a k e d}{T i m e} = \frac{1 / 2 a t m}{t}$
Using equation (i), we get
$\begin{array}{l} \frac{r_{N_{2}}}{r_{H_{2}}} = \sqrt{\frac{d_{H_{2}}}{d_{N_{2}}}} o r \frac{(1 / 2) / 60}{(1 / 2) / t} + \sqrt{\frac{1}{14}} \\ t = \sqrt{\frac{1}{14}} = 0.267 h o u r = 0.267 \times 60 = 16.0 \min \end{array}$