Physics · Thermodynamics

JEE Main 2025 — 3 April, Morning Shift — Question 58

A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800 cm3800 \mathrm{~cm}^{3} and temperature 27∘C27^{\circ} \mathrm{C}. The change in temperature when the gas is adiabatically compressed to 200 cm3200 \mathrm{~cm}^{3} is : (Take γ=1.5\gamma=1.5; γ\gamma is the ratio of specific heats at constant pressure and at constant volume)

  1. Option A:

    300 K

    Correct
  2. Option B:

    522 K

  3. Option C:

    327 K

  4. Option D:

    600 K

Answer: A

Step-by-step solution

Being adiabatic process PV′=P V^{\prime}= constant,

where γ=\gamma= 1.5 and also TV−1=T V^{-1}= constant

300{800}0.5=T{200}0.5300\{800\}^{0.5}=T\{200\}^{0.5}

T=300×2=600 KT=300 \times 2=600 \mathrm{~K}

ΔT=600−300=300 K\Delta T=600-300=300 \mathrm{~K}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Thermodynamics
Topic
Different Thermodynamic Processes
A gas is kept in a container having walls which are thermally… | JEE Main 2025 PYQ with Solution · DhiX AI