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(A) In a cyclic process,the change in internal energy is zero,i.e.,$\Delta U = 0$.According to the first law of thermodynamics,$\Delta Q = \Delta U +

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$2PV$

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(A) In a cyclic process,the change in internal energy is zero,i.e.,$\Delta U = 0$.According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$. Since $\Delta U = 0$,we have $\Delta Q = \Delta W$.The work done in a cyclic process is equal to the area enclosed by the cycle on the $P-V$ diagram.The cycle $ABCD$ is traversed in an anticlockwise direction,so the work done by the gas is negative.Area of the rectangle $ABCD = (	ext{change in volume}) 	imes (	ext{change in pressure}) = (3V - V) 	imes (2P - P) = (2V) 	imes (P) = 2PV$.Since the cycle is anticlockwise,$\Delta W = -2PV$.Thus,$\Delta Q = -2PV$.The negative sign indicates that heat is rejected by the system.Therefore,the heat rejected by the gas is $2PV$.

$A$ thermodynamic system is taken through the cycle $ABCD$ as shown in the figure. Heat rejected by the gas during the cycle is:

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