Work done in cycle =area under closed curve
W=P0V0
Heat given =QAB+QBC
=nCVΔTAB+nCpΔTBC
=23(nRTB−nRTA)+25(nRTC−nRTB)
=23(2P0V0−P0V0)+25(4P0V0−2P0V)
=213P0V0
Thermal efficiency, η=QabsorbedW=132=0.15
JEE Main 2017 — Physics Thermodynamics
An engine operates by taking nmoles of an ideal gas through the cycle ABCDA shown in figure. The thermal efficiency of the engine is:
(Take Cv=1.5R, whereR is gas constant)

Held on 8 Apr 2017 · Verified 6 Jul 2026.
0.24
0.15
0.32
0.08
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