JEE Main 2025 — Thermodynamics Question with Solution
JEE Main 2025 (3 Apr Shift 1)
Question
A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of and temperature . The change in temperature when the gas is adiabatically compressed to is :
(Take is the ratio of specific heats at constant pressure and at constant volume)
(Take is the ratio of specific heats at constant pressure and at constant volume)
Choose an option
Show full solutionCorrect option: D
Correct answer
D300 K
Step-by-step explanation
$\begin{aligned}
& \mathrm{V}_1=800 \mathrm{~cm}^3 \quad \mathrm{~V}_2=200 \mathrm{~cm}^3 \\ & \mathrm{~T}_1=300 \mathrm{~K}
\end{aligned}\begin{aligned}
& \mathrm{TV}^{\gamma-1}=\text { const. } \\ & (300)(800)^{1.5-1}=\mathrm{T}_2(200)^{1.5-1} \\ & \mathrm{~T}_2=300\left[\frac{800}{200}\right]^{0.5}=300 \times\left(2^2\right)^{1 / 2} \\ & \mathrm{~T}_2=600 \mathrm{~K} \\ & \Delta \mathrm{~T}=600-300=300 \mathrm{~K}
\end{aligned}$
& \mathrm{V}_1=800 \mathrm{~cm}^3 \quad \mathrm{~V}_2=200 \mathrm{~cm}^3 \\ & \mathrm{~T}_1=300 \mathrm{~K}
\end{aligned}\begin{aligned}
& \mathrm{TV}^{\gamma-1}=\text { const. } \\ & (300)(800)^{1.5-1}=\mathrm{T}_2(200)^{1.5-1} \\ & \mathrm{~T}_2=300\left[\frac{800}{200}\right]^{0.5}=300 \times\left(2^2\right)^{1 / 2} \\ & \mathrm{~T}_2=600 \mathrm{~K} \\ & \Delta \mathrm{~T}=600-300=300 \mathrm{~K}
\end{aligned}$
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This is a previous-year question from JEE Main 2025, covering the Thermodynamics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.