A certain ideal gas has a temperature 300 K and a pressure `5.0 xx 10^4 ` Pa. The molecules have a mean free path of `4.0 xx 10^(-7)`m. If the temperature is raised to 350 K and the pressure is reduced to `1.0 xx 10^4` Pa the mean free path is then

Options

A. `6.9 xx 10^(-8) m`
B. `9.3 xx 10^(-8) m`
C. `3.4 xx 10^(-7) m`
D. `2.3 xx 10^(-6) m`

Answer

To find the new mean free path (λ₂) of the gas when the temperature is raised to 350 K and the pressure is reduced to \(1.0 \times 10^4\) Pa, we can use the relationship between mean free path, temperature, and pressure.

### Step-by-Step Solution:

1. **Understand the Relationship**: The mean free path (λ) of a gas is directly proportional to the temperature (T) and inversely proportional to the pressure (P). This can be expressed as:
\[ \lambda \propto \frac{T}{P} \]
Therefore, we can write the relationship for two different states as:
\[ \frac{\lambda_2}{\lambda_1} = \frac{T_2}{T_1} \cdot \frac{P_1}{P_2} \]

2. **Identify Given Values**:
- Initial temperature (T₁) = 300 K
- Final temperature (T₂) = 350 K
- Initial pressure (P₁) = \(5.0 \times 10^4\) Pa
- Final pressure (P₂) = \(1.0 \times 10^4\) Pa
- Initial mean free path (λ₁) = \(4.0 \times 10^{-7}\) m

3. **Substitute the Values into the Equation**:
\[ \lambda_2 = \lambda_1 \cdot \frac{T_2}{T_1} \cdot \frac{P_1}{P_2} \]
Plugging in the values:
\[ \lambda_2 = (4.0 \times 10^{-7} \, \text{m}) \cdot \frac{350 \, \text{K}}{300 \, \text{K}} \cdot \frac{5.0 \times 10^4 \, \text{Pa}}{1.0 \times 10^4 \, \text{Pa}} \]

4. **Calculate the Ratios**:
- Calculate \(\frac{T_2}{T_1}\):
\[ \frac{350}{300} = \frac{7}{6} \approx 1.1667 \]
- Calculate \(\frac{P_1}{P_2}\):
\[ \frac{5.0 \times 10^4}{1.0 \times 10^4} = 5 \]

5. **Combine the Results**:
\[ \lambda_2 = (4.0 \times 10^{-7} \, \text{m}) \cdot 1.1667 \cdot 5 \]
\[ \lambda_2 = (4.0 \times 10^{-7} \, \text{m}) \cdot 5.8335 \approx 2.3334 \times 10^{-6} \, \text{m} \]

6. **Final Calculation**:
\[ \lambda_2 \approx 2.33 \times 10^{-6} \, \text{m} \]

### Conclusion:
The new mean free path (λ₂) when the temperature is raised to 350 K and the pressure is reduced to \(1.0 \times 10^4\) Pa is approximately \(2.33 \times 10^{-6}\) m.