In maxwell's speed distribution curve, for `N_(2 )` gas, the average of |relative velocity| between two molecules at 300 k will be : -

Options

A. 300 m/sec
B. 610 m/sec
C. 920 m/sec
D. zero

Explanation

To find the average relative velocity between two nitrogen (N₂) gas molecules at 300 K using Maxwell's speed distribution, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Relative Velocity**:
The relative velocity \( |v_{rel}| \) between two molecules can be expressed as:
\[ |v_{rel}| = \sqrt{v_1^2 + v_2^2 - 2v_1v_2 \cos \theta} \]
where \( v_1 \) and \( v_2 \) are the speeds of the two molecules and \( \theta \) is the angle between their velocity vectors.

2. **Assuming Equal Speeds**:
For simplicity, we assume both molecules have the same speed \( v \). Thus, the expression simplifies to:
\[ |v_{rel}| = \sqrt{2v^2(1 - \cos \theta)} = v \sqrt{2(1 - \cos \theta)} \]

3. **Using Trigonometric Identity**:
We can use the identity \( 1 - \cos \theta = 2 \sin^2(\theta/2) \):
\[ |v_{rel}| = v \sqrt{2 \cdot 2 \sin^2(\theta/2)} = 2v |\sin(\theta/2)| \]

4. **Finding the Average Relative Velocity**:
To find the average of the relative velocity, we integrate over the angle \( \theta \) from 0 to \( \pi \):
\[ \langle |v_{rel}| \rangle = \frac{1}{\int_0^{\pi} d\theta} \int_0^{\pi} 2v |\sin(\theta/2)| d\theta \]

5. **Calculating the Integral**:
The integral can be computed as follows:
\[ \int_0^{\pi} |\sin(\theta/2)| d\theta = 2 \int_0^{\pi/2} \sin(\theta/2) d\theta = 2 \cdot 2 = 4 \]
Thus, the average relative velocity becomes:
\[ \langle |v_{rel}| \rangle = \frac{4v}{\pi} \]

6. **Relating Speed to Temperature**:
The average speed \( v \) of a gas molecule can be expressed in terms of the temperature \( T \) and the molar mass \( M \):
\[ v = \sqrt{\frac{8RT}{\pi M}} \]
where \( R \) is the universal gas constant.

7. **Substituting Values**:
For nitrogen (N₂), the molar mass \( M \) is 28 g/mol or \( 28 \times 10^{-3} \) kg/mol. At \( T = 300 \) K:
\[ v = \sqrt{\frac{8 \times 8.314 \times 300}{\pi \times 0.028}} \approx 478.6 \text{ m/s} \]

8. **Final Calculation**:
Now substituting back into the average relative velocity equation:
\[ \langle |v_{rel}| \rangle = \frac{4 \cdot 478.6}{\pi} \approx 606 \text{ m/s} \]

### Conclusion:
The average relative velocity between two nitrogen molecules at 300 K is approximately \( 606 \) m/s, which is closest to the option \( 610 \) m/s.