An ideal gas undergoes the cyclic process abc | vedclass

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(A) For a cyclic process, the change in internal energy $\Delta U$ over the complete cycle is zero.According to the first law of thermodynamics, $\Del

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$30$

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(A) For a cyclic process, the change in internal energy $\Delta U$ over the complete cycle is zero.According to the first law of thermodynamics, $\Delta Q = \Delta U + W$.For the complete cycle $abca$, $\Delta U_{net} = 0$, so $\Delta Q_{net} = W_{net}$.The net heat exchanged is $\Delta Q_{net} = \Delta Q_{ab} + \Delta Q_{bc} + \Delta Q_{ca}$.Given:$\Delta Q_{ab} = -50 \,J$ (heat rejected)$\Delta Q_{bc} = 0 \,J$ (adiabatic process)$\Delta Q_{ca} = 80 \,J$ (heat absorbed)Thus, $\Delta Q_{net} = -50 + 0 + 80 = 30 \,J$.Since $\Delta Q_{net} = W_{net}$, the net work done by the gas in the cycle is $30 \,J$.The area enclosed by the $P-V$ diagram represents the net work done in the cycle.Therefore, the area of the closed curve $abca$ is $30 \,J$.

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