Rms speed of `O_(2)` molecule is `200m//s` at `T=300 K` and `P=3`atm. If diameter of molecule is `0.3nm` then collision frequency is :

Options

A. `2.9xx10^(7)S^(-1)`
B. `2.9xx10^(6)S^(-1)`
C. `2.9xx10^(8)`
D. `2.9xx10^(5)`

Answer

To find the collision frequency of an O₂ molecule, we can use the formula for collision frequency given by:

\[ f = \sqrt{\frac{8 k T}{\pi m}} \cdot \sqrt{2 \pi d^2 \rho} \]

Where:
- \( f \) = collision frequency
- \( k \) = Boltzmann constant = \( 1.38 \times 10^{-23} \, \text{J/K} \)
- \( T \) = temperature in Kelvin = \( 300 \, \text{K} \)
- \( m \) = mass of one O₂ molecule
- \( d \) = diameter of the molecule = \( 0.3 \, \text{nm} = 0.3 \times 10^{-9} \, \text{m} \)
- \( \rho \) = number density of the gas

### Step 1: Calculate the mass of one O₂ molecule
The molar mass of O₂ is approximately \( 32 \, \text{g/mol} \). To find the mass of one molecule, we convert this to kilograms and divide by Avogadro's number (\( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)):

\[ m = \frac{32 \times 10^{-3} \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 5.31 \times 10^{-26} \, \text{kg} \]

### Step 2: Calculate the number density \( \rho \)
Using the ideal gas law, we can find the number density:

\[ P = \rho k T \implies \rho = \frac{P}{k T} \]

Given \( P = 3 \, \text{atm} = 3 \times 101325 \, \text{Pa} \):

\[ \rho = \frac{3 \times 101325}{1.38 \times 10^{-23} \times 300} \approx 1.06 \times 10^{25} \, \text{m}^{-3} \]

### Step 3: Substitute values into the collision frequency formula
Now we can substitute all the values into the collision frequency formula:

\[ f = \sqrt{\frac{8 \times (1.38 \times 10^{-23}) \times 300}{\pi \times (5.31 \times 10^{-26})}} \cdot \sqrt{2 \pi (0.3 \times 10^{-9})^2 (1.06 \times 10^{25})} \]

### Step 4: Calculate the first part of the formula
Calculating the first part:

\[ \frac{8 \times (1.38 \times 10^{-23}) \times 300}{\pi \times (5.31 \times 10^{-26})} \approx 3.14 \times 10^{6} \]

Taking the square root gives:

\[ \sqrt{3.14 \times 10^{6}} \approx 1776.5 \]

### Step 5: Calculate the second part of the formula
Calculating the second part:

\[ 2 \pi (0.3 \times 10^{-9})^2 (1.06 \times 10^{25}) \approx 5.94 \times 10^{-17} \]

Taking the square root gives:

\[ \sqrt{5.94 \times 10^{-17}} \approx 7.71 \times 10^{-9} \]

### Step 6: Combine both parts to find \( f \)
Now, we multiply the two results together:

\[ f \approx 1776.5 \times 7.71 \times 10^{-9} \approx 2.77 \times 10^{7} \, \text{s}^{-1} \]

### Final Answer
The collision frequency \( f \) is approximately:

\[ f \approx 2.77 \times 10^{7} \, \text{s}^{-1} \text{ or } 28.77 \times 10^{6} \, \text{s}^{-1} \]