Q

# Solve it, - Heat and Thermodynamics - BITSAT

The mean free path of nitrogen molecules at 27°C is 3 × $10^{-7}$ m/s. if the average speed of nitrogen molecules at the same temperature is 600 m/s then the collision frequency will be

• Option 1)

$10^{9}/ sec$

• Option 2)

$1.5 *10^{9}/sec$

• Option 3)

$2 *10^{9}/sec$

• Option 4)

$3 *10^{9}/sec$

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As we learnt in

Mean free path -

The average distance travelled  by a molecule between two successive collision.

-

Mean free path =$3\times 10^{-7}m$

Speed = $600\ m/s$

Collision Frequency =$\frac{1}{T}=\frac{1}{(L/V)}=\frac{V}{L}=\frac{600 }{3 \times 10^{-7}}=2 \times 10^9 /sec$

Option 1)

$10^{9}/ sec$

Incorrect

Option 2)

$1.5 *10^{9}/sec$

Incorrect

Option 3)

$2 *10^{9}/sec$

Correct

Option 4)

$3 *10^{9}/sec$

Incorrect

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