The P-V diagram shows a cyclic process for a | vedclass

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(B) In a cyclic process, the net heat absorbed $(\Delta Q)$ is equal to the net work done $(\Delta W)$ because the change in internal energy $(\Delta

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$P_0V_0$

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(B) In a cyclic process, the net heat absorbed $(\Delta Q)$ is equal to the net work done $(\Delta W)$ because the change in internal energy $(\Delta U)$ over a complete cycle is zero $(\Delta Q = \Delta U + \Delta W$, where $\Delta U = 0)$.The net work done $\Delta W$ is equal to the area enclosed by the $P-V$ cycle $ABCD$.The cycle is a rectangle with sides of length $(2P_0 - P_0) = P_0$ and $(2V_0 - V_0) = V_0$.Therefore, the area $\Delta W = 	ext{length} 	imes 	ext{width} = P_0 	imes V_0 = P_0V_0$.Since the cycle is clockwise, the work done is positive, meaning heat is absorbed from the source.Thus, the energy absorbed is $\Delta Q = P_0V_0$.

The $P-V$ diagram shows a cyclic process for a thermodynamic engine containing a monoatomic ideal gas. The energy absorbed from the heat source during one complete cycle is .......

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