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Question

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

  1. -5J
  2. -10J
  3. -15J
  4. -20J


Solution

The correct option is A -5J
For cyclic process. Total work done
=WAB+WBC+WCA
ΔVWAB=pΔV=10(21)=10J
and ΔWAB=0(as V=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

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