$A$ cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in the figure. Find the heat exchanged by the engine with the surroundings for each section of the cycle. Given ${C_v} = \frac{3}{2}R$.
$(a)$ $A$ to $B$: constant volume
$(b)$ $B$ to $C$: constant pressure
$(c)$ $C$ to $D$: adiabatic
$(d)$ $D$ to $A$: constant pressure

(N/A) For process $A$ to $B$,the volume is constant,hence the work done $dW = 0$. From the first law of thermodynamics,$dQ = dU + dW = dU + 0 = dU = nC_V dT = nC_V(T_B - T_A) = (1)(\frac{3}{2}R)(T_B - T_A) = \frac{3}{2}(RT_B - RT_A)$. Using the ideal gas equation $PV = RT$,we get $dQ = \frac{3}{2}(P_B V_B - P_A V_A)$.
$(b)$ For process $B$ to $C$,the pressure is constant. The work done is $dW = P_B(V_C - V_B)$. From the first law of thermodynamics,$dQ = dU + dW = nC_V(T_C - T_B) + P_B(V_C - V_B) = \frac{3}{2}(P_C V_C - P_B V_B) + P_B(V_C - V_B)$. Since $P_B = P_C$,this simplifies to $dQ = \frac{3}{2}P_B(V_C - V_B) + P_B(V_C - V_B) = \frac{5}{2}P_B(V_C - V_B)$.
$(c)$ For process $C$ to $D$,it is an adiabatic process,so the heat exchanged $dQ = 0$.
$(d)$ For process $D$ to $A$,the pressure is constant at $P_A$. The gas is compressed from volume $V_D$ to $V_A$. Similar to process $(b)$,the heat exchanged is $dQ = \frac{5}{2}P_A(V_A - V_D)$.