A cyclic process ABCA is shown in the T-V dia | vedclass

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(D) For a cyclic process $ABCA$, the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics, $\Delta Q = \Delta W$.Give

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$-6200$

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(D) For a cyclic process $ABCA$, the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics, $\Delta Q = \Delta W$.Given that heat is rejected, $\Delta Q = -1200 	ext{ J}$.Thus, $\Delta W_{AB} + \Delta W_{BC} + \Delta W_{CA} = -1200 	ext{ J}$.Process $AB$ is a straight line passing through the origin, so $T \propto V$, which means it is an isobaric process. However, work done in a $T-V$ diagram for a straight line through the origin is $W = nR\Delta T$.For process $AB$: $W_{AB} = nR(T_B - T_A) = 2 	imes 8.314 	imes (600 - 300) = 2 	imes 8.314 	imes 300 = 4988.4 	ext{ J}$.Process $CA$ is a vertical line, so it is an isochoric process, meaning $\Delta W_{CA} = 0$.Substituting these into the equation: $4988.4 + \Delta W_{BC} + 0 = -1200$.$\Delta W_{BC} = -1200 - 4988.4 = -6188.4 	ext{ J}$.Rounding to the nearest option, we get $-6200 	ext{ J}$.

$A$ cyclic process $ABCA$ is shown in the $T-V$ diagram. An ideal gas of $2 \text{ mole}$ undergoes this process. $A$ total of $1200 \text{ J}$ of heat is rejected by the gas in the complete cycle. Find the work done by the gas during the process $BC$. (in $\text{ J}$)

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