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For N_{2} : Initially, P_{1} = 2 atm w_{1} = 14 kg (diffusion time 1 hr) After diffusion, P_{2} = \frac{1}{2} atm w_{2} = ? Since at constant V and temperature, for a gas P α w Therefore, \frac{P_{1}}{P_{2}} = \frac{w_{1}}{w_{2}} or, \frac{2}{\frac{1}{2}} = \frac{14}{w_{2}} or, w_{2} = \frac{14}{4} kg N_{2} Therefore, weight of N_{2} diffused in 1 hr = 14 - \frac{14}{4} = \frac{21}{2} kg Similarly for H_{2} Initially P_{1} = 2 atm w_{1} = 1 kg (diffusion time t hr) After diffusion, P_{2} = \frac{1}{2} atm w_{2} = ? Again, \frac{P_{1}}{P_{2}} = \frac{w_{1}}{w_{2}} or, \frac{2}{\frac{1}{2}} = \frac{1}{w_{2}} or, w_{2} = \frac{1}{4} kg H_{2} Therefore, weight of H_{2} diffused = 1 - \frac{1}{4} = \frac{3}{4} For diffusion of N_{2} and H_{2} t_{H_{2}} = 1 hr = 60 \min \frac{w_{H_{2}}}{t_{H_{2}}} \times \frac{t_{N_{2}}}{w_{N_{2}}} = \sqrt{\frac{M_{H_{2}}}{M_{N_{2}}}} \frac{(\frac{3}{4})}{(\frac{21}{2})} \times \frac{60}{t} = \sqrt{\frac{2}{28}} Therefore, t = 16 minute

here,t be the required time.

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