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 heat exchanged by the
engine, with the surroundings for each section of the cycle. [CV = (3/2)R]

(a) AB : constant volume

(b) BC : constant pressure

(c) CD : adiabatic

(d) DA : constant pressure

Hint: Use the first law of thermodynamics.
(a) Step 1: Find the heat exchanged by the engine in the process AB.
For process AB,
Volume is constant, hence work done dW = 0
Now, by the first law of thermodynamics,
dQ=dU+dW=dU+0=dU
    =nCv dT=nCv(TB-TA)
    =32R(TB-TA)                                    (n=1)
   =32(RTB-RTA )=32(PBVB-PAVA)
Heat exchanged =32(PBVB-PAVA)
(b) Step 2: Find the heat exchanged by the engine in the process BC.
For process BC, P=constant
dQ=dU+dW=32R(TC-TB)+PB(VC-VB)
    =32(PCVC-PBVB)+PB(VC-VB)=52PB(VC-VB)
Heat exchanged =52PB(VC-VA)                                  (PB=PC and PB=VA)
(c) Step 3: Find the heat exchanged by the engine in the process CD.
Because CD is adiabatic, dQ = the heat exchanged = 0
(d) Step 4: Find the heat exchanged by the engine in process AB.
DA involves compression of gas from VD to VA at constant pressure PA.
 Heat exchanged can be calculated in a similar way as BC.
 Hence, dQ=52PA(VA-VD)