JEE Main 2018 — Thermodynamics (C) Question with Solution
JEE Main 2018 (15 Apr Shift 1 Online)
Question
An ideal gas undergoes a cyclic process as shown in Figure.
$\begin{aligned} &\Delta \mathrm{U}_{\mathrm{BC}}=-5 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{q}_{\mathrm{AB}}=2 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ &\mathrm{~W}_{\mathrm{AB}}=-5 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{~W}_{\mathrm{CA}}=3 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$ Heat absorbed by the system during process CA is:
$\begin{aligned} &\Delta \mathrm{U}_{\mathrm{BC}}=-5 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{q}_{\mathrm{AB}}=2 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ &\mathrm{~W}_{\mathrm{AB}}=-5 \mathrm{~kJ} \mathrm{~mol}^{-1}, \mathrm{~W}_{\mathrm{CA}}=3 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$ Heat absorbed by the system during process CA is:
Choose an option
Show full solutionCorrect option: B
Correct answer
B
Step-by-step explanation
For cyclic process,
$\begin{aligned}
&\Delta \mathrm{U}_{\mathrm{AB}}+\Delta \mathrm{U}_{\mathrm{BC}}+\Delta \mathrm{U}_{\mathrm{CA}}=0 \\
&\Delta \mathrm{U}_{\mathrm{CA}}=-\Delta \mathrm{U}_{\mathrm{AB}}-\Delta \mathrm{U}_{\mathrm{BC}} \\
&\Delta \mathrm{U}_{\mathrm{CA}}=-(-3)-(-5)=8 \mathrm{~kJ} / \mathrm{mol} \\
&\Delta \mathrm{U}_{\mathrm{CA}}=\mathrm{q}_{\mathrm{CA}}+\mathrm{W}_{\mathrm{CA}} \\
&8=\mathrm{q}_{\mathrm{CA}}+3 \\
&\mathrm{q}_{\mathrm{CA}}=+5 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}$
Heat absorbed has positive sign.
Practice this on the real CBT interface
Solve this JEE Main question (and the rest of the Thermodynamics (C) chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.
Solve interactively →About this question
This is a previous-year question from JEE Main 2018, 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.