A 4: 1 molar mixture of He and CH4 is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially?

Explanation

To solve the problem of determining the composition of the gas mixture that is effusing out initially, we will follow these steps:

### Step 1: Understand the Molar Ratio
We are given a 4:1 molar mixture of helium (He) and methane (CH₄). This means that for every 4 moles of He, there is 1 mole of CH₄.

### Step 2: Determine the Total Moles
Let’s denote the moles of He as 4x and the moles of CH₄ as x. Therefore, the total moles of the gas mixture is:
\[ \text{Total moles} = 4x + x = 5x \]

### Step 3: Calculate the Partial Pressures
The total pressure of the gas mixture is given as 20 bar. The partial pressures can be calculated using Dalton's Law of Partial Pressures:
- The partial pressure of He (P₁) is:
\[ P_{He} = \frac{\text{moles of He}}{\text{total moles}} \times \text{total pressure} = \frac{4x}{5x} \times 20 \text{ bar} = \frac{4}{5} \times 20 = 16 \text{ bar} \]

- The partial pressure of CH₄ (P₂) is:
\[ P_{CH₄} = \frac{\text{moles of CH₄}}{\text{total moles}} \times \text{total pressure} = \frac{x}{5x} \times 20 \text{ bar} = \frac{1}{5} \times 20 = 4 \text{ bar} \]

### Step 4: Apply Graham's Law of Effusion
According to Graham's Law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. We denote the rates of effusion of He and CH₄ as R₁ and R₂ respectively:
\[ \frac{R_{He}}{R_{CH₄}} = \frac{P_{He}}{P_{CH₄}} \times \sqrt{\frac{M_{CH₄}}{M_{He}}} \]

Where:
- Molar mass of He (M₁) = 4 g/mol
- Molar mass of CH₄ (M₂) = 16 g/mol

### Step 5: Substitute the Values
Now substituting the values we calculated:
\[ \frac{R_{He}}{R_{CH₄}} = \frac{16 \text{ bar}}{4 \text{ bar}} \times \sqrt{\frac{16}{4}} \]
\[ = 4 \times \sqrt{4} = 4 \times 2 = 8 \]

### Step 6: Determine the Composition of the Effusing Mixture
The ratio of the composition of the mixture effusing out initially is therefore:
\[ \text{Ratio of He to CH₄} = 8:1 \]

### Final Answer
Thus, the composition of the mixture effusing out initially is **8:1** (He:CH₄).

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