An ideal gas is taken through the cycle A→B→C→A as shown in figure. If the heat supplied to the gas in the cycle is 5J. The work done by the gas in the process C→A
A
−25J
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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−25J For cyclic process ΔU=0 Q=W=WAB+WBC+WCA
WBC=0 as process is isochoric WAB=10(5−2)=30 i.e. Area of line AB with V axis ⇒5=30+0+WCA