Question

At 20°C, two balloons of equal volumes and porosity are filled to a pressure of 2 atm, one with 14 kg N₂ and other with 1 kg of H₂. The N₂ balloon leaks to a pressure of half atm in 1 hr. How long will it take for H₂ balloon to reach a pressure of half atm?

• A. 1 hour
• B. 30 minutes
• C. 16 minutes
• D. 3.74 hours

Answer 1

Answer: (C)

16 minutes

Answer 2

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Rate of pressure drop $\propto \frac{1}{\sqrt{M}}$

Molar mass of $N_2$ ($M_{N_2}$) = 28 g/mol. Molar mass of $H_2$ ($M_{H_2}$) = 2 g/mol.

The ratio of their rates of pressure drop is: $\frac{r_{H_2}}{r_{N_2}} = \sqrt{\frac{M_{N_2}}{M_{H_2}}} = \sqrt{\frac{28}{2}} = \sqrt{14}$ So, $H_2$ leaks $\sqrt{14} \approx 3.74$ times faster than $N_2$.

For $N_2$, the pressure drops from 2 atm to 0.5 atm (a drop of 1.5 atm) in 1 hour. Rate of pressure drop for $N_2$ ($r_{N_2}$) = $\frac{1.5 \text{ atm}}{1 \text{ hr}}$.

The $H_2$ balloon also needs to drop by 1.5 atm. Rate of pressure drop for $H_2$ ($r_{H_2}$) = $\sqrt{14} \times r_{N_2} = \sqrt{14} \times \frac{1.5 \text{ atm}}{1 \text{ hr}}$.

Time taken for $H_2$ to drop by 1.5 atm: $t_{H_2} = \frac{\text{Pressure Drop}}{r_{H_2}} = \frac{1.5 \text{ atm}}{\sqrt{14} \times \frac{1.5 \text{ atm}}{1 \text{ hr}}} = \frac{1}{\sqrt{14}} \text{ hours}$ Converting to minutes: $t_{H_2} = \frac{1}{\sqrt{14}} \text{ hours} \times 60 \frac{\text{minutes}}{\text{hour}} \approx 16.0356 \text{ minutes}$ This is approximately 16 minutes.