16 mL of He gas effuses through a pin hole in 4 sec from a container having `P_(He)` equal to 1 atm. If same container is filled with `CH_(4)` having pressure 2 atm, how much volume (in mL) of `CH_(4)` will be leaked through same pin hole in 2 sec?

Explanation

To solve the problem, we will use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. We will also consider the relationship between the rates of effusion of the two gases involved (He and CH4) and the pressures they are under.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Volume of He gas effused, \( V_{He} = 16 \, \text{mL} \)
- Time for He effusion, \( t_{He} = 4 \, \text{s} \)
- Pressure of He, \( P_{He} = 1 \, \text{atm} \)
- Pressure of CH4, \( P_{CH4} = 2 \, \text{atm} \)
- Molar mass of He, \( M_{He} = 4 \, \text{g/mol} \)
- Molar mass of CH4, \( M_{CH4} = 16 \, \text{g/mol} \)

2. **Calculate the Rate of Effusion for He:**
\[ \text{Rate}_{He} = \frac{V_{He}}{t_{He}} = \frac{16 \, \text{mL}}{4 \, \text{s}} = 4 \, \text{mL/s} \]

3. **Apply Graham's Law of Effusion:**
According to Graham's law:
\[ \frac{\text{Rate}_{He}}{\text{Rate}_{CH4}} = \frac{P_{He}}{P_{CH4}} \cdot \sqrt{\frac{M_{CH4}}{M_{He}}} \]

4. **Substitute the Values:**
\[ \frac{4 \, \text{mL/s}}{\text{Rate}_{CH4}} = \frac{1 \, \text{atm}}{2 \, \text{atm}} \cdot \sqrt{\frac{16 \, \text{g/mol}}{4 \, \text{g/mol}}} \]
\[ \frac{4 \, \text{mL/s}}{\text{Rate}_{CH4}} = \frac{1}{2} \cdot \sqrt{4} = \frac{1}{2} \cdot 2 = 1 \]

5. **Calculate Rate of Effusion for CH4:**
\[ \text{Rate}_{CH4} = 4 \, \text{mL/s} \]

6. **Calculate Volume of CH4 that will Effuse in 2 seconds:**
\[ V_{CH4} = \text{Rate}_{CH4} \cdot t_{CH4} = 4 \, \text{mL/s} \cdot 2 \, \text{s} = 8 \, \text{mL} \]

### Final Answer:
The volume of CH4 that will be leaked through the same pinhole in 2 seconds is **8 mL**.

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