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Physics · Thermodynamics

JEE Main 2026 — 22 January, Morning Shift — Question 39

The volume of an ideal gas increases 8 times and temperature becomes (1/4)th (1 / 4)^{\text {th }} of initial temperature during a reversible change. If there is no exchange of heat in this process ( ΔQ=0\Delta \mathrm{Q}=0 ) then identify the gas from the following options (Assuming the gases given in the options are ideal gases) :

  1. Option A:

    CO2\mathrm{CO}_{2}

  2. Option B:

    O2\mathrm{O}_{2}

  3. Option C:

    NH3\mathrm{NH}_{3}

  4. Option D:

    He

    Correct

Answer: D

Step-by-step solution

PVγ= \mathrm{PV}^{\gamma}= constant TVγ−1=\mathrm{TV}^{\gamma-1}= constant TVγ−1=(T4)(8 V)(γ−1)\mathrm{TV}^{\gamma-1}=\left(\frac{\mathrm{T}}{4}\right)(8 \mathrm{~V})^{(\gamma-1)} 4=8(γ−1)4=8^{(\gamma-1)} 22=23γ−32^{2}=2^{3 \gamma-3} 2=3(γ−1)2=3(\gamma-1) γ=53\gamma=\frac{5}{3} Gas is a monoatomic gas Answer is He .

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
22 January, Morning Shift
Subject
Physics
Chapter
Thermodynamics
Topic
Efficiency of Processes and Miscellaneous Problems

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