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MHT CET 2024 (11 May, Shift 2) · Molecular Speeds & Maxwell Distribution

Molecular Speeds & Maxwell Distribution question from MHT CET 2024 (11 May, Shift 2)

Q28MHT CET 2024 · 11 MAY · SHIFT 2Medium

Molecular Speeds & Maxwell Distribution

' $N$ ' molecules of gas $A$ , each having mass ' $m$ ' and ' 2 N ' molecules of gas B, each of mass ' 2 m ' are contained in the same vessel which is at constant temperature ' T '. The mean square velocity of $B$ is $V^{2}$ and mean square of x -component of A is $\omega^{2}$ . The value of $\frac{\omega^{2}}{V^{2}}$ is

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Answer: (b)

For B (mass $2m$ ): $V^2 = \overline{v^2} = \frac{3kT}{2m}$
For A (mass $m$ ): $\overline{v^2} = \frac{3kT}{m}$ , so $\omega^2 = \overline{v_x^2} = \frac{1}{3}\cdot\frac{3kT}{m} = \frac{kT}{m}$
$\frac{\omega^2}{V^2} = \frac{kT/m}{3kT/2m} = \frac{2}{3}$
So the answer is (b) $2 : 3$ .

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