Physics · Kinetic Theory of Gases

JEE Main 2024 — 29 January, Shift 2 — Question 47

$N$ moles of a polyatomic gas $(f=6)$ must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of N is :

Option A:
6
Option B:
3
Option C:
4
Correct
Option D:
2

Answer: C

Step-by-step solution

$\mathrm{f}_{\mathrm{eq}}=\frac{\mathrm{n}_{1} \mathrm{f}_{1}+\mathrm{n}_{2} \mathrm{f}_{2}}{\mathrm{n}_{1}+\mathrm{n}_{2}}$

For diatomic gas $\mathrm{f}_{\text {eq }}=5$

$5=\frac{(\mathrm{N})(6)+(2)(3)}{\mathrm{N}+2}$

$5 \mathrm{~N}+10=6 \mathrm{~N}+6$

$\mathrm{N}=4$

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Kinetic Theory of Gases
Topic: Equipartition Law of Energy and Degrees of Freedom

The temperature of a gas having 2.0 × 10^25 molecules per cubic meter at 1.38 atm (Given, k= .1.38 × 10^-23 JK^-1 ) is :

Answer: C

Step-by-step solution

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