One mole of a monatomic ideal gas executes a rectangular cycle on a P–V diagram: from a(P0,V0) isochorically to b(2P0,V0), then isobarically to…

Temperatures in units of $T_0=P_0V_0/R$ are $a:1,\ b:2,\ c:4,\ d:2$. Net work is the rectangle area $(2P_0-P_0)(2V_0-V_0)=P_0V_0$: (A) true. Heat absorbed: $a\to b$ is $C_V\Delta T=\tfrac32R(T_0)=1.5P_0V_0$ and $b\to c$ is $C_P\Delta T=\tfrac52R(2T_0)=5P_0V_0$, so $Q_{in}=6.5P_0V_0$ and $\eta=1/6.5=2/13\approx15.4\%$: (B) true. In $d\to a$ the gas is compressed isobarically from $2T_0$ to $T_0$, rejecting $\tfrac52R\,T_0=2.5P_0V_0$: (C) true. The extreme temperatures are $T_0$ and $4T_0$, so $\eta_{Carnot}=0.75$; half of it is $0.375$, which exceeds $0.154$: (D) false. As a cross-check, the heat taken in is $6.5P_0V_0$ and the heat rejected is $3P_0V_0+2.5P_0V_0=5.5P_0V_0$; their difference is the net work $P_0V_0$, in agreement with the area of the loop on the diagram.