In a cyclic process, ΔU=0
From 1st law of thermodynamics
Q=ΔU+W
Q=W
Qab+Qbc+Qca=Wabc+Wca
Wabc=Qab+Qbc+Qca−Wca
Wabc=Qab+Qbc+Uca
Wabc=250+60−180=130J
A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path…
JEE Main 2019 — Physics Thermodynamics
2019mcqmedium
A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is −180J. The gas absorbs 250J of heat along the path ab and 60J along the path bc . The work done by the gas along the path abc is:
Official previous-year question
Held on 12 Apr 2019 · Verified 6 Jul 2026.
Options
- A
130J
- B
100J
- C
120J
- D
140J
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