The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are $3 \times 10^{-7} \mathrm{~m}$ and $600 \mathrm{~m} / \mathrm{s}$, respectively. Find the frequency of its collisions.

(C) $2 \times 10^{9} / \mathrm{s}$

(A) $2 \times 10^{10} / \mathrm{s}$

(B) $9 \times 10^{5} / \mathrm{s}$

(D) $5 \times 10^{8} / \mathrm{s}$

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Heat & Thermodynamics

The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are

3 \times 1 0^{- 7} m

and

600 m / s

, respectively. Find the frequency of its collisions.

(A) $2 \times 1 0^{10} / s$
(B) $9 \times 1 0^{5} / s$
(C) $2 \times 1 0^{9} / s$ ✓ Correct
(D) $5 \times 1 0^{8} / s$

Answer: (C)

Solution

The collision frequency is the number of collisions a molecule makes each second, found by dividing the average speed by the distance covered between successive collisions, which is the mean free path.

ν = \frac{v _{a v g}}{λ} = \frac{600}{3 \times 1 0 ^{- 7}} = 2 \times 1 0^{9} s^{- 1} .

The frequency of collisions is

2 \times 1 0^{9}

per second.

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Single correct option

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