The above P-V diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is

(\frac{13}{2}) P_{o} V_{0}

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(\frac{11}{2}) P_{o} V_{0}

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4 P_{o} V_{0}

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P_{o} V_{0}

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Solution

The correct option is D $(\frac{13}{2}) P_{o} V_{0}$
Heat is extracted from the source would be in path $D A$ and $A B$ .
Thus heat supplied= $n C_{v} (2 T_{o} - T_{o}) + n C_{p} (4 T_{o} - T_{o}) = \frac{n 3 R T_{o}}{2} + \frac{n 5 R (2 T_{o})}{2} = \frac{13 n R T_{o}}{2} = \frac{13}{2} P_{o} V_{o}$

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The above P-V diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat extracted from the source in a single cycle is

The correct option is D (132)PoV0Heat is extracted from the source would be in path DA and AB. Thus heat supplied= nCv(2To−To)+nCp(4To−To)=n3RTo2+n5R(2To)2=13nRTo2=132PoVo