A 0.1 mole sample of an ideal diatomic gas ( first compressed at a constant pressure of 1 atm
from 2.46 liters to 1 liter, it then undergoes a sudden pressure increase to 3 atm constant
volume it then undergoes an isothermal expansion followed by an adiabatic expansion back to
its original state.
We can make the following figure for the process given above.
(a) Find the pressure and volume at Point D where the change from isothermal to adiabatic
expansion occurs.
1-D is isothermal expansion. Hence, PV = constant for 1-D. If P and V are pressure and volume at
D, we have
PV = 3
---- (1)
1.4
1.4
D-2 is adiabatic process. Hence,
PV = 1*2.46
-----(2)
0.4
1.4
Or,
3*V = 2.46 =3.5262
Or,
V = (3.5262/3)2.5= 1.4978, say 1.5 lites
P = 3/1.5 = 2 atm
(b) Find the change in internal energy for each part of this process and the change in internal
energy for the entire cycle.
For constant pressure process, ΔU = nCvΔT
But
nRΔT=PΔV
We have 1atm = 101325.01N/m2 and 1 litre = 0.001m3.
Hence, ΔT = 101325.01*1.46*0.001/(0.1*8.3145 = 177.9120K
Hence, ΔU =0.1* 19.9*177.912 = 354.04J
If air is initially at 200C or 2930K, temperature at the end of constant pressure process
= 293+177.912 = 470.9120K
For constant volume process, PΔV = 0.
Hence, ΔU = nCvΔT
Temperature at 1 = (3/1)*470.912 = 1412.7360K
ΔU = nCvΔT = 0.1*19.9*(1412.736-470.912) = 1874.23J
For isothermal process, ΔU = 0
For over all cycle, ΔU = 0.
Hence, ΔU = -1874.23-354.04 = -2228.27J for adiabatic process.
(d) Find the work done for each part of this process and the total work done during the cycle.
Total work done in isobaric process = 101325.01*1.46*0.001 = 147.934J. This should be –ve as
work is done on the system.
Total work done in constant volume process = 0
Total work done in isothermal process =0.1*8.3145*1412.7360ln2 = 814.184J
For adiabatic process,

W = PV
1 1
V21  V11
(2.46  0.001)0.4  1.5  0.0010.4
 2 101325.01 (1.5  0.001)1.4
1 
0.4
= 136.775J
Total work done in the cycle = -147.934+814.184+136.755= 803.005J
(e) Find the change in entropy for each part of the process and the entire cycle.
For constant volume process, ΔS =Cv ln(T2/T1) = 19.9ln3 = 21.862
For constant pressure process, ΔS =1.4Cv ln(T2/T1) = 1.4*19.9ln(470.912/293) = 13.219
For isothermal process, ΔS =Rln(P2/P1) = 8.3145ln2 = 5.763
For adiabatic process, ΔS = 0
(f) Is this an engine or a refrigerator?
Engine.
(g) If this is an engine compute the efficiency. If it is a refrigerator compute the coefficient of
performance instead.
Efficiency = 814.184/(147.934+814.184+136.755) = 74%