An ideal gas is taken through the cycle A→B→C→A, as shown in the figure. If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process C→A is
A
−5J
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B
−10J
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C
−15J
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D
−20J
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Solution
The correct option is A−5J ΔWAB=PΔV=(10)(2−1)=10J ΔWBC=0 (as V=constant)
From first law of thermodynamics ΔQ=ΔW+ΔU ΔU=0 (process ABCA is cyclic) ∴ΔQ=ΔWAB+ΔWBC+ΔWCA ΔWCA=ΔQ−ΔWAB−ΔWBC=(5−10−0) J =−5J