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(D) For a thermodynamic cycle,the total change in internal energy is zero: $\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 0$.Using the first law of

Vedclass · Accepted Answer

$\Delta U_{CA} = - 160 \, J$

Vedclass · Answer

(D) For a thermodynamic cycle,the total change in internal energy is zero: $\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 0$.Using the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,we can find the missing values:$1$. For process $AB$: $\Delta U_{AB} = \Delta Q_{AB} - \Delta W_{AB} = 150 - 40 = 110 \, J$.$2$. For process $BC$: $\Delta W_{BC} = \Delta Q_{BC} - \Delta U_{BC} = 10 - 50 = - 40 \, J$.$3$. Since $\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 0$,we have $110 + 50 + \Delta U_{CA} = 0$,which gives $\Delta U_{CA} = - 160 \, J$.$4$. For process $CA$: $\Delta Q_{CA} = \Delta U_{CA} + \Delta W_{CA} = - 160 + 30 = - 130 \, J$.Comparing these results with the given options,option $D$ is correct.

	$AB$	$BC$	$CA$
$\Delta W$	$40 \, J$		$30 \, J$
$\Delta U$		$50 \, J$
$\Delta Q$	$150 \, J$	$10 \, J$

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