ΔUAB=qAB+WAB=2+(−5)=−3 kJ/mol ΔUBC=−5 kJ/mol For cyclic process, ΔU=0 $\begin{aligned}
&\Delta \mathrm{U}{\mathrm{AB}}+\Delta \mathrm{U}{\mathrm{BC}}+\Delta \mathrm{U}{\mathrm{CA}}=0 \
&\Delta \mathrm{U}{\mathrm{CA}}=-\Delta \mathrm{U}{\mathrm{AB}}-\Delta \mathrm{U}{\mathrm{BC}} \
&\Delta \mathrm{U}{\mathrm{CA}}=-(-3)-(-5)=8 \mathrm{~kJ} / \mathrm{mol} \
&\Delta \mathrm{U}{\mathrm{CA}}=\mathrm{q}{\mathrm{CA}}+\mathrm{W}{\mathrm{CA}} \
&8=\mathrm{q}{\mathrm{CA}}+3 \
&\mathrm{q}{\mathrm{CA}}=+5 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}$ Heat absorbed has positive sign.
JEE Main 2018 — Chemistry Physical Chemistry
2018mcqmedium
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:
Official previous-year question
Held on 15 Apr 2018 · Verified 6 Jul 2026.
Options
- A
−5 kJ mol−1
- B
+5 kJ mol−1
- C
18 kJ mol−1
- D
−18 kJ mol−1
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