The cycle shown in the figure represents an engine (the engine consists of one mole of gas in a cylinder with a piston). $A$ to $B$ is isochoric,$B$ to $C$ is isothermal,$C$ to $D$ is isochoric,and $D$ to $A$ is isothermal. Also,$V_C = V_D = 2V_A = 2V_B$.
$(a)$ In which part of the cycle is heat supplied to the engine from the outside?
$(b)$ In which part of the cycle can the engine give energy to its surroundings?
$(c)$ How much work is done by the engine during one cycle? Give your answer in terms of $P_A, P_B$ and $V_A$.
$(d)$ What is the efficiency of the engine? (For the gas,$\gamma = 5/3$,and for one mole,$C_V = 3/2 R$)

(N/A) Heat is supplied during the processes where the internal energy and work done increase. In the isothermal expansion $B \to C$,heat is absorbed $(Q_{BC} = nRT_B \ln(V_C/V_B) > 0)$. In the isochoric heating $A \to B$,heat is absorbed $(Q_{AB} = nC_V(T_B - T_A) > 0)$. Thus,heat is supplied in $A \to B$ and $B \to C$.
$(b)$ The engine gives energy to the surroundings during the cooling processes: isochoric cooling $C \to D$ and isothermal compression $D \to A$.
$(c)$ Work done $W = W_{AB} + W_{BC} + W_{CD} + W_{DA}$. Since $AB$ and $CD$ are isochoric,$W_{AB} = W_{CD} = 0$. $W_{BC} = nRT_B \ln(V_C/V_B) = P_B V_B \ln(2)$. $W_{DA} = nRT_A \ln(V_A/V_D) = P_A V_A \ln(1/2) = -P_A V_A \ln(2)$. Total work $W = (P_B - P_A) V_A \ln(2)$.
$(d)$ Efficiency $\eta = W / Q_{in}$. $Q_{in} = Q_{AB} + Q_{BC} = C_V(T_B - T_A) + P_B V_B \ln(2) = \frac{3}{2}(P_B - P_A)V_A + P_B V_A \ln(2)$. $\eta = \frac{(P_B - P_A) V_A \ln(2)}{\frac{3}{2}(P_B - P_A)V_A + P_B V_A \ln(2)} = \frac{(P_B - P_A) \ln(2)}{\frac{3}{2}(P_B - P_A) + P_B \ln(2)}$.