JEE Main 2024 — Thermodynamics (C) Question with Solution
JEE Main 2024 (06 Apr Shift 1)
Question
An ideal gas, , is expanded adiabatically against a constant pressure of 1 atm untill it doubles in volume. If the initial temperature and pressure is and , respectively then the final temperature is _______ (nearest integer).
[ is the molar heat capacity at constant volume]
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Show full solutionCorrect answer: 274
Correct answer
274
Step-by-step explanation
$\begin{aligned}
& \Delta \mathrm{U}=\mathrm{q}+\mathrm{w}(\mathrm{q}=0) \\
& \mathrm{nC}_{\mathrm{V}} \Delta \mathrm{T}=-\mathrm{P}_{\mathrm{ext}}\left(\mathrm{V}_2-\mathrm{V}_1\right) \\
& \mathrm{V}_2=2 \mathrm{~V}_1 \\
& \frac{\mathrm{nRT}_2}{\mathrm{P}_2}=\frac{2 \mathrm{nRT}_1}{\mathrm{P}_1} \\
& \mathrm{P}_1=5, \mathrm{~T}_1=298 \\
& \mathrm{P}_2=\frac{5 \mathrm{~T}_2}{2 \times 298} \\
& \mathrm{n} \frac{5}{2} \mathrm{R}\left(\mathrm{T}_2-\mathrm{T}_1\right)=-1\left(\frac{\mathrm{nRT}_2}{\mathrm{P}_1}-\frac{\mathrm{nRT}_1}{\mathrm{P}_1}\right)
\end{aligned}\mathrm{T}_1=298$
and
Solve and we get
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This is a previous-year question from JEE Main 2024, covering the Thermodynamics (C) chapter of Chemistry. 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.