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JEE Main 2007 (AIEEE) · P-V Diagrams & Cyclic Processes

P-V Diagrams & Cyclic Processes question from JEE Main 2007 (AIEEE)

Q88JEE Main 2007 · AIEEEMedium

P-V Diagrams & Cyclic Processes

When a system is taken from state ii to state ff along the path iaf, it is found that Q=50Q = 50 cal and W=20 calW = 20\text{ cal}. Along the path ibf Q=36 cal. Wibf\ Q = 36\text{ cal}.\ W along the path ibfibf is

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Answer: (b)

  1. Path iaf: ΔU=Q−W=50−20=30 cal\Delta U = Q - W = 50 - 20 = 30\ \text{cal}.
  2. Path ibf: W=Q−ΔU=36−30=6 calW = Q - \Delta U = 36 - 30 = 6\ \text{cal}
  3. So the answer is (b) 6 cal6\text{ cal}.
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