In the given elliptical P-V diagram,

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the work done is positive

the change in internal energy is non-zero

the work done =−(π/4)P2−P1V2−V1

the work done =πV1−V12−πP1−P12

answer is C.

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Detailed Solution

The work done = area of P-V diagrama=V2−V12, b=P2−P12W=πV2−V12P2−P12But the cyclic process is anticlockwise. Hence, the work done is negative.

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