Physics · Kinetic Theory of Gases

JEE Main 2025 — 28 January, Evening Shift — Question 51

The kinetic energy of translation of the molecules in 50 g of $\mathrm{CO}_{2}$ gas at $17^{\circ} \mathrm{C}$ is :

Option A:
3986.3 J
Option B:
4102.8 J
Correct
Option C:
4205.5 J
Option D:
3582.7 J

Answer: B

Step-by-step solution

(\text{KE})_{\text{Translational}}=\left(\frac{3}{2}kT\right)\times (\text{number \;of\; molecules})

\text{Number \;of\; molecules} =\frac{50}{44}\times 6.023\times 10^{23}

(\text{KE})_{\text{Translational}}=4102.844\ \text{J}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Kinetic Theory of Gases
Topic: Equipartition Law of Energy and Degrees of Freedom

The ratio of vapour densities of two gases at the same temperature is 4/25, then the ratio of r.m.s. velocities will be :

The kinetic energy of translation of the molecules in 50 g of CO 2… | JEE Main 2025 PYQ with Solution · DhiX AI