Calculate the mean free path of nitogen at `27^(@)C` when pressure is 1.0 atm. Given, diameter of nitogen molecule = `1.5 Å, k = 1.38 xx 10^(-23) JK^(-1)`. If the average speed of nitrogen molecules is `675 ms^(-1)`, find the time taken by the molecule between two successive collsions and the frequency of collisions.

Options

A. 0.6 ns
B. 0.4 ns
C. 0.8 ns
D. 0.3 ns

Explanation

To solve the problem step by step, we will calculate the mean free path of nitrogen molecules, the time taken between two successive collisions, and the frequency of collisions.

### Given Data:
- Temperature (T) = 27°C = 300 K (273 + 27)
- Pressure (P) = 1 atm = 1.013 × 10^5 N/m²
- Diameter of nitrogen molecule (d) = 1.5 Å = 1.5 × 10^(-10) m
- Average speed of nitrogen molecules (v) = 675 m/s
- Boltzmann constant (k) = 1.38 × 10^(-23) J/K

### Step 1: Calculate Mean Free Path (λ)
The formula for mean free path (λ) is given by:

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

Substituting the values:

\[ \lambda = \frac{(1.38 \times 10^{-23} \, \text{J/K}) \cdot (300 \, \text{K})}{\sqrt{2} \cdot \pi \cdot (1.5 \times 10^{-10} \, \text{m})^2 \cdot (1.013 \times 10^5 \, \text{N/m}^2)} \]

Calculating the denominator:

- \( d^2 = (1.5 \times 10^{-10})^2 = 2.25 \times 10^{-20} \, \text{m}^2 \)
- \( \sqrt{2} \approx 1.414 \)
- \( \pi \approx 3.14 \)

Now calculating:

\[ \lambda = \frac{(1.38 \times 10^{-23}) \cdot (300)}{1.414 \cdot 3.14 \cdot (2.25 \times 10^{-20}) \cdot (1.013 \times 10^5)} \]

Calculating the numerator:

\[ 1.38 \times 10^{-23} \cdot 300 = 4.14 \times 10^{-21} \]

Calculating the denominator:

\[ 1.414 \cdot 3.14 \cdot 2.25 \times 10^{-20} \cdot 1.013 \times 10^5 \approx 1.01 \times 10^{-14} \]

Now substituting back to find λ:

\[ \lambda \approx \frac{4.14 \times 10^{-21}}{1.01 \times 10^{-14}} \approx 4.1 \times 10^{-7} \, \text{m} \]

### Step 2: Calculate Time Between Successive Collisions (t)
The time taken between two successive collisions can be calculated using the formula:

\[ t = \frac{\lambda}{v} \]

Substituting the values:

\[ t = \frac{4.1 \times 10^{-7} \, \text{m}}{675 \, \text{m/s}} \approx 6.07 \times 10^{-10} \, \text{s} \]

### Step 3: Calculate Frequency of Collisions (f)
The frequency of collisions can be calculated using the formula:

\[ f = \frac{1}{t} \]

Substituting the value of t:

\[ f = \frac{1}{6.07 \times 10^{-10}} \approx 1.65 \times 10^9 \, \text{collisions/s} \]

### Final Results:
- Mean Free Path (λ) ≈ \( 4.1 \times 10^{-7} \, \text{m} \)
- Time between collisions (t) ≈ \( 6.07 \times 10^{-10} \, \text{s} \)
- Frequency of collisions (f) ≈ \( 1.65 \times 10^9 \, \text{collisions/s} \)