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

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$- P \frac{ V }{2}$

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(A) For a cyclic process,the change in internal energy is $\Delta U = 0$.According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,which implies $\Delta Q = \Delta W$.The work done $\Delta W$ in a cyclic process is equal to the area enclosed by the $P-V$ curve.The area of the triangle formed by the process $1 ightarrow 2 ightarrow 3 ightarrow 1$ is given by:Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$Base $= 2V - V = V$Height $= 2P - P = P$Area $= \frac{1}{2} 	imes V 	imes P = \frac{PV}{2}$.Since the cycle is in the clockwise direction,the work done by the system is positive,meaning heat is absorbed. Therefore,the amount of heat released by the system is $-\frac{PV}{2}$.

$A$ gas system is taken through the thermodynamic cyclic process $1 \rightarrow 2 \rightarrow 3 \rightarrow 1$ as shown in the figure. The amount of heat released by the system is

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