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

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(D) For a cyclic process,the change in internal energy $\Delta U = 0$. According to the first law of thermodynamics,$Q = \Delta U + W$. Since $\Delta U = 0$,the total heat absorbed $Q = W$,where $W$ is the total work done in the cycle.The work done in an isothermal process is given by $W = nRT \ln(V_f / V_i)$.Process $ab$ is isothermal at $T_1 = 500 	ext{ K}$ with volume changing from $V_0$ to $2V_0$:$W_{ab} = nRT_1 \ln(2V_0 / V_0) = nRT_1 \ln 2$Process $bc$ is isochoric ($V$ is constant),so $W_{bc} = 0$.Process $cd$ is isothermal at $T_2 = 300 	ext{ K}$ with volume changing from $2V_0$ to $V_0$:$W_{cd} = nRT_2 \ln(V_0 / 2V_0) = nRT_2 \ln(1/2) = -nRT_2 \ln 2$Process $da$ is isochoric ($V$ is constant),so $W_{da} = 0$.Total work $W = W_{ab} + W_{bc} + W_{cd} + W_{da} = nR(T_1 - T_2) \ln 2$.Substituting the given values ($n = 2.0 	ext{ mol}$,$R = 8.3 	ext{ J/mol-K}$,$T_1 = 500 	ext{ K}$,$T_2 = 300 	ext{ K}$,$\ln 2 = 0.69$):$Q = 2.0 	imes 8.3 	imes (500 - 300) 	imes 0.69$$Q = 2.0 	imes 8.3 	imes 200 	imes 0.69$$Q = 3320 	imes 0.69 = 2290.8 	ext{ J} \approx 2291 	ext{ J}$.

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