Question

At ${20}^{\circ }C$, two balloons of equal volume and porosity are filled to a pressure of 2 atm. One with 14 kg $N_2$ and other with 1 kg of $H_2$. The $N_2$ balloon leaks to a pressure of 0.5 atm in 1 hour. How long will it take for $H_2$ balloon to reach a pressure of 0.5 atm?

• A. 15 minutes
• B. 20 minutes
• C. 19 minutes
• D. 26 minutes

Answer 1

Answer: (A)

15 minutes

For gases at constant V and T

$P\alpha w$

For $N_2$:

$\frac{P_1}{P_2}=\frac{W_1}{W_2}$

$\frac{2}{0.5}=\frac{14}{W_2}$

$W_2=\frac{14}{4}=3.5kg\left(N_2\right)$

Weight of $N_2$ diffused $=(14-3.5)=10.5Kg$

For $H_2$

$\frac{2}{0.5}=\frac{1}{W_2}$

$W_2=0.25kg\left(H_2\right)$

Weight of $H_2$ diffused $1-0.25=0.75kg$

$\frac{r_{N_2}}{r_{H_2}}=\sqrt{\frac{M_{H_2}}{M_{N_2}}}$

$\frac{W_{H_2}{t}_{N_2}}{W_{N_2}t{H}_2}=\sqrt{\frac{M_{H_2}}{M_{N_2}}}$

$\frac{0.75}{10.5}\times \frac{60}{t_1}=\sqrt{\frac{2}{28}}$

$t_1=15.87min$