The diameter of a gas molecule is `2.4 xx 10^(-10) m`. Calculate the mean free path at NTP. Given Boltzmann constant `k = 1.38 xx 10^(-23) J molecule^(-1) K^(-1)`.

Explanation

To calculate the mean free path of a gas molecule at normal temperature and pressure (NTP), we can use the formula:

\[ \lambda = \frac{k_B \cdot T}{\sqrt{2} \cdot \pi \cdot d^2 \cdot P} \]

Where:
- \( \lambda \) = mean free path
- \( k_B \) = Boltzmann constant = \( 1.38 \times 10^{-23} \, \text{J molecule}^{-1} \text{K}^{-1} \)
- \( T \) = temperature in Kelvin = \( 273 \, \text{K} \) (at NTP)
- \( d \) = diameter of the gas molecule = \( 2.4 \times 10^{-10} \, \text{m} \)
- \( P \) = pressure = \( 1.01 \times 10^5 \, \text{N/m}^2 \) (at NTP)

### Step-by-Step Solution:

1. **Identify the given values**:
- Diameter of the gas molecule, \( d = 2.4 \times 10^{-10} \, \text{m} \)
- Boltzmann constant, \( k_B = 1.38 \times 10^{-23} \, \text{J molecule}^{-1} \text{K}^{-1} \)
- Temperature, \( T = 273 \, \text{K} \)
- Pressure, \( P = 1.01 \times 10^5 \, \text{N/m}^2 \)

2. **Substitute the values into the mean free path formula**:
\[ \lambda = \frac{(1.38 \times 10^{-23}) \cdot (273)}{\sqrt{2} \cdot \pi \cdot (2.4 \times 10^{-10})^2 \cdot (1.01 \times 10^5)} \]

3. **Calculate the denominator**:
- First, calculate \( d^2 \):
\[ d^2 = (2.4 \times 10^{-10})^2 = 5.76 \times 10^{-20} \, \text{m}^2 \]
- Now calculate \( \sqrt{2} \cdot \pi \cdot d^2 \cdot P \):
\[ \sqrt{2} \approx 1.414 \]
\[ \pi \approx 3.14 \]
\[ \sqrt{2} \cdot \pi \cdot d^2 \cdot P = 1.414 \cdot 3.14 \cdot (5.76 \times 10^{-20}) \cdot (1.01 \times 10^5) \]
\[ = 1.414 \cdot 3.14 \cdot 5.76 \times 10^{-15} \approx 2.54 \times 10^{-14} \]

4. **Calculate the numerator**:
\[ k_B \cdot T = (1.38 \times 10^{-23}) \cdot (273) \approx 3.77 \times 10^{-21} \]

5. **Calculate the mean free path**:
\[ \lambda = \frac{3.77 \times 10^{-21}}{2.54 \times 10^{-14}} \approx 1.48 \times 10^{-7} \, \text{m} \]

### Final Answer:
The mean free path at NTP is approximately \( \lambda \approx 1.48 \times 10^{-7} \, \text{m} \).