JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Heat & Thermodynamics

The number of air molecules per

cm^{3}

increased from

3 \times 1 0^{19}

12 \times 1 0^{19}

. The ratio of collision frequency of air molecules before and after the increase in number respectively is

(A)1.25
(B)0.25✓ Correct
(C)0.75
(D)0.50

Answer: (B)

Solution

The collision frequency is given by

f = 2 π d^{2} v n_{v}

, where

n_{v}

is the number density of molecules. Since all other parameters remain constant, the collision frequency is directly proportional to the number density:

f \propto n_{v}

Therefore:

\frac{f _{1}}{f _{2}} = \frac{n _{v_{1}}}{n _{v_{2}}} = \frac{3 \times 1 0 ^{19}}{12 \times 1 0 ^{19}} = \frac{1}{4} = 0.25

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Single correct option

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