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 step by step, 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.

### Step 1: Write down the given data.
- Volume of He gas effused (V_He) = 16 mL
- Time for He gas effusion (t_He) = 4 seconds
- Pressure of He (P_He) = 1 atm
- Pressure of CH₄ (P_CH₄) = 2 atm
- Molar mass of He (M_He) = 4 g/mol
- Molar mass of CH₄ (M_CH₄) = 16 g/mol

### Step 2: Calculate the rate of effusion of He.
The rate of effusion can be defined as the volume of gas effused per unit time. For He:
\[ \text{Rate of effusion of He} = \frac{V_{He}}{t_{He}} = \frac{16 \text{ mL}}{4 \text{ s}} = 4 \text{ mL/s} \]

### Step 3: Apply Graham's law of effusion.
According to Graham's law:
\[ \frac{\text{Rate of effusion of He}}{\text{Rate of effusion of CH₄}} = \frac{P_{He}}{P_{CH₄}} \cdot \sqrt{\frac{M_{CH₄}}{M_{He}}} \]

Substituting the known values:
\[ \frac{4 \text{ mL/s}}{\text{Rate of effusion of CH₄}} = \frac{1 \text{ atm}}{2 \text{ atm}} \cdot \sqrt{\frac{16 \text{ g/mol}}{4 \text{ g/mol}}} \]

### Step 4: Simplify the equation.
Calculating the right side:
\[ \frac{1}{2} \cdot \sqrt{4} = \frac{1}{2} \cdot 2 = 1 \]
Thus, we have:
\[ \frac{4 \text{ mL/s}}{\text{Rate of effusion of CH₄}} = 1 \]

### Step 5: Solve for the rate of effusion of CH₄.
From the equation:
\[ \text{Rate of effusion of CH₄} = 4 \text{ mL/s} \]

### Step 6: Calculate the volume of CH₄ that will effuse in 2 seconds.
Using the rate of effusion of CH₄:
\[ \text{Volume of CH₄ in 2 seconds} = \text{Rate of effusion of CH₄} \times t_{CH₄} = 4 \text{ mL/s} \times 2 \text{ s} = 8 \text{ mL} \]

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

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