Question

1 mol He and 3 mol N\(_2\) exert a pressure of 16 atm. Due to a hole in the vessel in which mixture is placed, mixture leaks out. What is the composition of mixture effusing out initially?

• A. 0.22
• B. 0.44
• C. 0.66
• D. 0.88

Hint:

Rate of effusion \(\propto \frac{P}{\sqrt{M}}\). Lighter gases effuse faster.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Graham's law of effusion: \(\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\). Also, \(r \propto \frac{P}{\sqrt{M}}\).
Step 2: Detailed Explanation:
Partial pressures: \(P_{\text{He}} = \frac{1}{4} \times 16 = 4\ \text{atm}\), \(P_{\text{N}_2} = \frac{3}{4} \times 16 = 12\ \text{atm}\).
\(\frac{r_{\text{He}}}{r_{\text{N}_2}} = \frac{P_{\text{He}}}{P_{\text{N}_2}} \sqrt{\frac{M_{\text{N}_2}}{M_{\text{He}}}} = \frac{4}{12} \times \sqrt{\frac{28}{4}} = \frac{1}{3} \times \sqrt{7} = \frac{1}{3} \times 2.645 = 0.88\).
Mole ratio effusing out = \(0.88 : 1\). Composition of He = \(\frac{0.88}{1.88} \approx 0.468\) (not asked). The ratio itself is 0.88.
Step 3: Final Answer:
Thus, \(\frac{r_{\text{He}}}{r_{\text{N}_2}} = 0.88\).