$A$ cycle followed by an engine (made of one mole of perfect gas in a cylinder with a piston) is shown in the figure.
$A$ to $B$: volume constant
$B$ to $C$: adiabatic
$C$ to $D$: volume constant
$D$ to $A$: adiabatic
$V_C = V_D = 2V_A = 2V_B$
$(a)$ In which part of the cycle is heat supplied to the engine from outside?
$(b)$ In which part of the cycle is heat given to the surrounding by the engine?
$(c)$ What is the work done by the engine in one cycle? Write your answer in terms of $P_A, P_B, V_A$.
$(d)$ What is the efficiency of the engine?
$(\gamma = 5/3, C_v = 3/2 R$ for one mole of the gas$)$

(N/A) In process $AB$, volume is constant $(dV = 0)$, so work done $dW = 0$. From the first law of thermodynamics, $dQ = dU + dW = dU$. Since pressure increases at constant volume, temperature increases, so $dU > 0$. Thus, heat is supplied in process $AB$.
$(b)$ In process $CD$, volume is constant and pressure decreases, so temperature decreases. Thus, heat is released to the surroundings in process $CD$.
$(c)$ Work done $W = W_{AB} + W_{BC} + W_{CD} + W_{DA}$. Since $W_{AB} = 0$ and $W_{CD} = 0$, $W = W_{BC} + W_{DA}$.
For adiabatic processes, $W = \frac{P_i V_i - P_f V_f}{\gamma - 1}$.
$W_{BC} = \frac{P_B V_B - P_C V_C}{\gamma - 1}$ and $W_{DA} = \frac{P_D V_D - P_A V_A}{\gamma - 1}$.
Given $V_C = V_D = 2V_A = 2V_B$, and adiabatic relations $P_B V_B^\gamma = P_C V_C^\gamma$ and $P_A V_A^\gamma = P_D V_D^\gamma$:
$P_C = P_B(1/2)^{5/3}$ and $P_D = P_A(2)^{5/3}$.
$W = \frac{1}{\gamma - 1} [P_B V_B - P_B(1/2)^{5/3}(2V_B) + P_A(2)^{5/3}(2V_A) - P_A V_A]$
$W = \frac{V_A}{2/3} [P_B(1 - 2^{-2/3}) + P_A(2^{8/3} - 1)] = \frac{3V_A}{2} [P_B(1 - 2^{-2/3}) + P_A(2^{8/3} - 1)]$.
$(d)$ Efficiency $\eta = 1 - \frac{|Q_{out}|}{Q_{in}} = 1 - \frac{C_v(T_C - T_D)}{C_v(T_B - T_A)} = 1 - \frac{P_C V_C - P_D V_D}{P_B V_B - P_A V_A} = 1 - \frac{2(P_C - P_D)}{P_B - P_A} = 1 - \frac{2(P_B 2^{-5/3} - P_A 2^{5/3})}{P_B - P_A}$.