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 5 J, 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 For cyclic process. Total work done =WAB+WBC+WCA ΔVWAB=pΔV=10(2−1)=10J and ΔWAB=0(asV=constant) For first law of thermodynamics, ΔQ=ΔU+ΔW ΔU = 0(Process ABCA is cyclic) ⇒ΔQ=ΔWAB+ΔWBC+ΔWCA ⇒5=10+0+ΔWCA ⇒ΔWCA=–5J