AB→ isobaric
BC→ isochoric
CA→ not defined
For cyclic process, ΔU=0
ΔUAB=q+W
=2−5=−3
ΔUABC=ΔUAB+ΔUBC
=−3−5=−8KJ
ΔUCA=+8
=Q+W
8=Q+3
Q=+5kJ
Therefore, the heat absorbed by the system during process CA is +5KJmol−1.
JEE Main 2018 — Chemistry Physical Chemistry
An ideal gas undergoes a cyclic process as shown in the figure

ΔUBC=−5KJmol−1,qAB=2KJmol−1
WAB=−5KJmol−1,WCA=3KJmol−1
Heat absorbed by the system during process CAis
Held on 15 Apr 2018 · Verified 6 Jul 2026.
−5KJmol−1
+5KJmol−1
−18KJmol−1
18KJmol−1
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