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(B) For a cyclic process,the change in internal energy $\Delta U = 0$. Therefore,the total heat absorbed $\Delta Q$ equals the total work done $\Delta

Vedclass · Accepted Answer

$30$

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(B) For a cyclic process,the change in internal energy $\Delta U = 0$. Therefore,the total heat absorbed $\Delta Q$ equals the total work done $\Delta W$ by the gas.$\Delta Q_{	ext{cycle}} = \Delta Q_{AB} + \Delta Q_{BC} + \Delta Q_{CA} = 40 + 0 - 60 = -20 \ J$.Since $\Delta Q_{	ext{cycle}} = \Delta W_{	ext{cycle}}$,we have $\Delta W_{	ext{cycle}} = -20 \ J$.The total work done is $\Delta W_{	ext{cycle}} = W_{AB} + W_{BC} + W_{CA}$.From the figure,the process $AB$ is isochoric (vertical line),so $W_{AB} = 0$.Given $W_{BC} = -50 \ J$ (work done on the gas).Thus,$-20 = 0 + (-50) + W_{CA}$.$W_{CA} = -20 + 50 = 30 \ J$.

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