Thermodynamics

JEE Main level, single correct, medium

32 of 190

Question 32 of 190

A thermodynamic system is taken through cyclic process. The total work done in the process is :

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Answer: (B) 300J300J

The work done is given by the area of the PVPV graph of the loop.
W=12(VE−VC)(PD−PC)=2×3002JW=\dfrac{1}{2}\left({V}_{E}-{V}_{C}\right)({P}_{D}-{P}_{C})=\dfrac{2\times 300}{2}J
=300J=300J

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