Physics · Kinetic Theory of Gases

JEE Main 2024 — 9 April, Shift 2 — Question 32

The temperature of a gas is $-78^{\circ} \mathrm{C}$ and the average translational kinetic energy of its molecules is K . The temperature at which the average translational kinetic energy of the molecules of the same gas becomes 2 K is :

Option A:
$-39^{\circ} \mathrm{C}$
Option B:
$117^{\circ} \mathrm{C}$
Correct
Option C:
$127^{\circ} \mathrm{C}$
Option D:
$-78^{\circ} \mathrm{C}$

Answer: B

Step-by-step solution

$K . E=\frac{\mathrm{nf}_{1} \mathrm{RT}}{2}$

$\mathrm{T}_{\mathrm{i}}=-78^{\circ} \mathrm{C} \rightarrow 273+\left[-78^{\circ} \mathrm{C}\right]=195 \mathrm{~K}$

K.E $\alpha$ T

To double the K.E energy temp also become double

$\mathrm{T}_{\mathrm{f}}=390 \mathrm{~K}$

$\mathrm{T}_{\mathrm{f}}=117^{\circ} \mathrm{C}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Kinetic Theory of Gases
Topic: Equipartition Law of Energy and Degrees of Freedom