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Physics · Kinetic Theory of Gases

JEE Main 2024 — 31 January, Shift 2 — Question 36

A gas mixture consists of 8 moles of argon and 6 moles of oxygen at temperature T. Neglecting all vibrational modes, the total internal energy of the system is

  1. Option A:

    29 RT

  2. Option B:

    20 RT

  3. Option C:

    27 RT

    Correct
  4. Option D:

    21 RT

Answer: C

Step-by-step solution

U=nCVTU=nC_{V} T

⇒U=n1CV1 T+n2CV2 T\Rightarrow \mathrm{U}=\mathrm{n}_{1} \mathrm{C}_{\mathrm{V}_{1}} \mathrm{~T}+\mathrm{n}_{2}\mathrm{C}_{\mathrm{V}_{2}} \mathrm{~T}

⇒8×3R2×T+6×5R2×T=27RT\Rightarrow 8 \times \frac{3 \mathrm{R}}{2} \times \mathrm{T}+6 \times \frac{5 \mathrm{R}}{2} \times \mathrm{T}=27 \mathrm{RT}

Answer key and solution verified before publishing.

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

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