N moles of a monatomic ideal gas, initially at pressure P0, volume V0 and temperature T0 (state A), execute the cycle A→B→C→A. A→B is an isothermal…

Use $P_0V_0=nRT_0$. In A$\to$B (isothermal, $\Delta U=0$) $Q_{AB}=W_{AB}=nRT_0\ln2=P_0V_0\ln 2$, so (A) is correct. At B the pressure is $P_0/2$ and $T_B=T_0$; the isobaric compression to $V_0$ cools the gas to $T_C=T_0/2$. Then $Q_{BC}=nC_p\Delta T=\tfrac52nR(-\tfrac{T_0}{2})=-\tfrac54P_0V_0$, i.e. heat $\tfrac54P_0V_0$ is released, so (B) is correct. In C$\to$A the gas is heated at constant volume: $Q_{CA}=\tfrac32nR\cdot\tfrac{T_0}{2}=\tfrac34P_0V_0$, not $\tfrac32P_0V_0$, so (C) is wrong. Net work $=(\ln2-\tfrac12)P_0V_0\approx0.193P_0V_0$ and heat absorbed is $Q_{AB}+Q_{CA}=(0.693+0.75)P_0V_0=1.443P_0V_0$, giving efficiency $\approx0.134$, i.e. about 13%, so (D) is correct. Answer: A, B and D. Because the net work is positive and the cycle runs A, B, C, A, it is a heat engine, consistent with a positive efficiency.