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JEE Main 2019 (12 Apr, Shift 1) · P-V Diagrams & Cyclic Processes

P-V Diagrams & Cyclic Processes question from JEE Main 2019 (12 Apr, Shift 1)

Q63JEE Main 2019 · 12 APR · SHIFT 1Medium

P-V Diagrams & Cyclic Processes

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 –180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is:

Figure

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Answer: (b)

  1. ΔUca=−180\Delta U_{ca} = -180 J, so ΔUabc=ΔUac=+180\Delta U_{abc} = \Delta U_{ac} = +180 J.
  2. Qabc=250+60=310Q_{abc} = 250 + 60 = 310 J
  3. Wabc=Qabc−ΔUabc=310−180=130W_{abc} = Q_{abc} - \Delta U_{abc} = 310 - 180 = 130 J
  4. So the answer is (b) 130 J.
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