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

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$P_{0} V_{0}$

$(\frac{13}{2}) p_{0} V_{0}$

$(\frac{11}{2}) p_{0} V_{0}$

$4 P_{0} V_{0}$

answer is B.

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Detailed Solution

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$\begin{array}{l} = {nC}_{V} (2 T_{0} - T_{0}) + {nC}_{p} (2 T_{0} - T_{0}) \\ = \frac{n 3 {RT}_{0}}{2} + \frac{n 5 R}{2} 2 T_{0} \\ = \frac{13}{2} {nRT}_{0} \\ = \frac{13}{2} p_{0} V_{0} \end{array}$

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