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(B) In a cyclic process,the change in internal energy $\Delta U$ is zero because the internal energy is a state function.According to the first law of

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$(p_1-p_2)(V_2-V_1)$

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(B) In a cyclic process,the change in internal energy $\Delta U$ is zero because the internal energy is a state function.According to the first law of thermodynamics,$\Delta Q = \Delta U + W$.Since $\Delta U = 0$,the heat absorbed by the gas in one cycle is equal to the work done by the gas,$\Delta Q = W$.The work done $W$ in a cyclic process is equal to the area enclosed by the cycle in the $p-V$ diagram.For a rectangular cycle with pressure limits $p_1$ and $p_2$ and volume limits $V_1$ and $V_2$,the area is given by the product of the change in pressure and the change in volume.$W = (p_1 - p_2) 	imes (V_2 - V_1)$.Therefore,the heat absorbed is $\Delta Q = (p_1 - p_2)(V_2 - V_1)$.

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