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Q.

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:

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a

- 5 J

b

-10 J

c

-15 J

d

-20 J

answer is A.

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Detailed Solution

Q=W+∆U⇒Wcycle=Q-∆U=5-0=5J But                     Tcycle=WAB+WBC+WCA So,                             5=10(2-1)+0+WCA ⇒                              WCA=-5 J
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