JEE Main 2026 — Jan 22, Shift 2 — Physics solution

JEE Main2026Jan 22, Shift 2Physics

Consider two boxes containing ideal gases A and B such that their temperatures, pressures and number densities are same. The molecular size of A is half of that of B and mass of molecule A is four times that of B. If the collision frequency in gas B is $32 \times 1 0^{18} / s$ then collision frequency in gas A is _____ /s.

A
$32 \times 1 0^{18}$
B
$4 \times 1 0^{18}$
C
$2 \times 1 0^{18}$
D
$8 \times 1 0^{18}$

Solution

Collision frequency $Z = 2 π d^{2} N \frac{8 R T}{π M}$ .

With same $T$ and $N$ : $Z \propto \frac{d ^{2}}{M}$ .

Given $d_{A} = \frac{d _{B}}{2}$ and $M_{A} = 4 M_{B}$ :

$\frac{Z _{A}}{Z _{B}} = (\frac{d _{A}}{d _{B}})^{2} \frac{M _{B}}{M _{A}} = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$ .

$Z_{B} = 32 \times 1 0^{18} / s$ (Given)

So, $Z_{A} = \frac{32 \times 1 0 ^{18}}{8}$
$Z_{A} = 4 \times 1 0^{18} / s$

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