Chapter 20– Second Law of Thermodynamics Hints-4/26/11
1. The 2nd Law of Thermodynamics can be stated in many different ways. The following statements are all
versions of the 2nd Law. Show how these are indeed equivalent statements. That is, if one of these is
violated, the others would be violated as well. They have to be either all true or all false!
a) Heat tends to “flow”* from hot to cold and not the reverse.
b) It is impossible to build an engine that is 100% efficient.
c) It is impossible to build a “perfect” refrigerator with an infinite coefficient of performance K.
The above were discussed in class. It’s pretty obvious that (c), a “perfect refrigerator” that makes heat flow
from cold to hot without effort, directly violates (a). If a “perfect engine” (b) could exist it would be used to
run a real refrigerator and turn it into a “perfect refrigerator”, so logically, such an engine is also
impossible.
d) The state of disorder (entropy) of a closed system always tends to increase. (You will have to wait till the
end of the chapter to answer this one.)
*We use the phrase “heat flows” here for historical reasons going back to a time when heat was considered
a kind of fluid. In reality heat is not a substance so it cannot “flow”, but is rather a process that transfers
energy between objects in thermal contact. We will continue to say “heat flows” because it is traditional
and is well understood that the phrase is not meant to be literally true.
Engines, refrigerators, and the Carnot cycle:
2. A device called a thermoelectric converter was demonstrated in class. It has two metal legs and a
propeller and it uses a series of semiconductor cells to convert thermal energy to electrical energy. The
device will not work unless one leg is at a higher temperature than the other.
a) Why doesn’t this work if the legs are at the same temperature even if that temperature is very high?
a) All engines require a hot heat source and a cold heat source to work; the absolute temperatures are not
relevant, it is the temperature difference that counts.
b) In what sense does this intriguing experiment demonstrate the second law of thermodynamics?
b) Heat engines require a flow of energy to work; heat flows naturally from hot to cold temperatures making
it possible to intercept this flow and make a heat engine.
3. Suppose you try to cool the kitchen of your house by leaving the refrigerator door open. What happens?
Why? Would the result be the same if you left open a picnic cooler full of ice? Explain the reason for any
differences. The room will get warmer because the refrigerator expels more heat into the outside than is
takes out of the inside of the refrigerator. Opening a picnic cooler full of ice is quite different, since there is
no motor doing work to make the ice melt. In this situation the melting ice would be removing heat from the
surrounding and really cooling the room.
4. Review the definition of efficiency for an engine and coefficient of performance for a refrigerator and a
heat pump.
e =W/Qhot =1-(Qc/Qh);
Krf = Qcold /W= Qc/(Qh - Qc);
Khp= Qhot /W= Qh/(Qh - Qc)
a) What are the maximum and minimum values of these quantities for each device?
a) emax=1 and Kmax=∞ (both of which are not possible). The minimum values would be zero.
b) Compare the e and K of a reversible engine-refrigerator (one that works equally well as one or the other
device between the same hot and cold reservoirs), and show they are related by K = (1-e)/e.
b) A reversible engine-refrigerator would have the same W, Qhot and Qcold requirements so you can
substitute out these values combining the equationseK=(Qc/Qh)=(1-e)…
c) A reversible engine takes in 300 J from a hot source and does 100 J of work. Determine the efficiency of
this engine and the coefficient of performance of its refrigerator version.
c) e=1/3; K=2
5. Carnot’s theory shows that in a Carnot reversible engine-refrigerator Qh/Qc= Th/Tc. Derive the efficiency
(e) formula and the coefficient of performance (K) formula for a Carnot engine and refrigerator. For
maximum engine efficiency do you want the temperatures to be as far apart as possible or as close as
possible? Answer the same question for the K of a Carnot refrigerator.
Doing the algebra gives: eCarnot=1-(Tc/Th). For refrigerator KCarnot=[(Th/Tc)-1] -1, for heat pump KCarnot=[1(Tc/Th)] -1. It should be clear that e increases as the temperatures get farther apart, while K increases as the
temperatures get closer together. If you think about this a bit you can agree that these make sense….
6. In light of the above question discuss the following:
a) A steam-driven turbine is one major component of an electric power plant. Why is it advantageous to
have the temperature of the steam as high as possible?
a) You want to maximize the temperature difference to improve e…
b) A heat pump is to be installed in a region where the average outdoor temperature in the winter months is
- 20°C. In view of this, why would it be advisable to place the outdoor compressor unit deep in the
ground? Why are heat pumps not commonly used for heating in cold climates?
b) Underground temperatures vary less than above-ground temperatures, so in winter, underground
temperatures would be closer to inside room temperatures (in summer it’s the reverse). Having the hot
and cold sources closer in temperature would improve K…
c) Why is your automobile likely to burn more gas in winter than in summer?
c) One would think the opposite would be true since in the winter there would be a larger difference
between the working temperatures of the car engine, but here we need to consider other factors. Colder
temperatures make lubricants thicker and so there are larger heat loses due to friction between the
moving parts…This illustrates that besides temperature differences the reversibility of an engine is an
important factor in determining efficiency.
Thot
7. “Heat flow diagrams” are useful in mastering the concepts of engines and the 2nd law.
e=1
a) Use a heat flow diagram to show that if you could build a perfect engine (e=1),
K<∞
then you could build a perfect refrigerator (K=∞), and vice-versa. This was done
Tcold
in class. See if you can reconstruct the logical argument to go with the diagram.
b) Use a heat flow diagram to show that a reversible engine-refrigerator is the most
Thot
efficient device that can be built between two given temperatures.
ec
This was done in class. The logical argument is that a reversible engine connected
Kc
to its refrigerator version would be able to return all the heat it expelled to the cold
reservoir back to the hot reservoir and make up for the original heat supplied by
Tcold
Engine-refrigerator
the hot reservoir to run the engine. The net Qhot=0 and the net Qcold=0, so the heat
reversible pair
flow would have been effectively stopped, which is the best one can do in battling the
natural flow of heat.
Thot
Now consider an engine that is more efficient than a reversible engine. This engine
“Better” ec
would be able to do the work needed to run the refrigerator above with less heat
than
K
input from the hot reservoir. The net effect would be some heat flowing from cold to reversible
engine
Tcold
hot in violation of the 2nd law of thermodynamics. Therefore a reversible engine is
the most efficient engine allowed by nature.
c) For a numerical example consider the reversible engine-refrigerator described in 4c (e=1/3; K=2).
Assume that it is possible to make a more efficient engine that would produce 100 J of work by taking in
only 200 J of heat using the same temperature sources. What would be the result if you used the “superengine” to run the original reversible refrigerator (K=2)? How does that square with the 2nd Law?
The result would be a net flow of heat of 100 J from cold to hot, an direct violation of the 2nd Law.
8. Imagine a special air filter placed in a window of a house. Assume this filter is made such that only air
molecules moving faster than a certain speed can exit the house, and only air molecules moving slower than
that speed can enter the house from outside. Explain why such an air filter would cool the house, and why
the second law of thermodynamics makes building such a filter an impossible task.
Such a filter would cool the house because slower molecules would lower the average kinetic energy of the
air and therefore lower temperature. Such a filter couldn’t be built because it would violate the 2nd law
reversing the natural flow of heat without external work.
9. Real heat engines, like the gasoline engine in a car, always have some friction between their moving
parts, although lubricants keep the friction to a minimum, would a heat engine with completely frictionless
parts be 100% efficient? Why or why not? Does the answer depend on whether or not the engine runs on the
Carnot cycle? Again, why or why not? No, even a frictionless engine could not violate the 2nd law because
friction has nothing to do with the natural flow of heat. A Carnot engine is not 100% efficient; it’s simply
the most efficient engine for a particular temperature difference. Real engines have efficiencies lower than
the Carnot engine because of non-conservative force (such as friction) and other irreversible features of the
engine’s cycle.
10. An engine absorbs 1600 J from a hot reservoir and expels 1000 J to a cold reservoir in each cycle. a)
What is the efficiency of the engine? e=0.375
b) How much work is done in each cycle? W=600 J
c) What is the average power output of the engine if each cycle lasts for 0.30 s? P= 2 kwatts
d) What fraction of the heat absorbed is expelled to the cold reservoir? |Qc/Qh|=0.625
e) If this were a Carnot reversible engine, what would be the extreme temperatures ratio? Would you expect
the actual extreme temperatures ratio (Tc/Th) of this engine to be larger or smaller? If we assume this engine
is as efficient as a Carnot engine then |Qc/Qh|=(Tc/Th)Carnot=0.625. Since the efficiency of a real engine is
always less than the Carnot efficiency, |Qc/Qh|>(Tc/Th)real, and the actual (Tc/Th) ratio in the real engine
would be smaller (this means that ∆T is larger in the real engine).
11. A refrigerator has a coefficient of performance equal to 3 and consumes 400 watts of power when it’s
running. For each second that it runs, find
a) the work done by the motor. W=P∆t=400 J
b) the thermal energy removed from the inside of the refrigerator, and the thermal energy expelled out into
the room. Qc=KW=1200 J and Qh=1600 J
c) if it were a Carnot refrigerator, and the outside temperature was a typical 25 ºC, what would be the
temperature of the inside of the refrigerator. Is this a realistic temperature? If this were a Carnot refrigerator
then (Qc/Qh)=(Tc/Th)12/16=[Tc /(25+273)]  Tc =224 K=-50ºC. This is very unrealistic since typical
temperatures inside a refrigerator are more like 10ºC, even freezer temperature are only about -5 ºC. This
shows how far from the Carnot ideal real refrigerators actually are.
12. A Carnot engine operates between two heat reservoirs at temperatures Th and Tc. An inventor proposes
to increase the efficiency by running one engine between Th and an intermediate temperature T and a second
engine between T and Tc . The second engine uses the heat expelled by the first engine as its own input
energy. Compute the efficiency of this composite system, and compare it to that of the original engine.
Using the definition of efficiency: e1=W1/Qh and e2=W2/Q’, where Q’ is the heat expelled by the first engine
and taken up by the second engine. By definition the combined efficiency e12= (W1+W2)/Qh. Since W is
equal to the difference between heat in and out of each engine, e=[(Qh -Q’)+(Q’- Qc)]/Qh=(Qh- Qc)/Qh
The combined “e” is the same as if a single engine were running between the extreme temperatures;
there’s no gain in efficiency from this arrangement.
13. At point A in a Carnot cycle, 2.30 moles of a monatomic gas have a pressure of 1400 kPa, a volume of
10.0 liters, and a temperature of 720 K. It expands isothermally to point B and then expands adiabatically to
point C, where its volume is 24.0 liters. An isothermal compression brings it to point D, where its new
volume is 15.0 liters. An adiabatic process returns the gas to point A.
a) Determine all the unknown pressure, volumes, and temperatures by filling in the following table:
a) You can determine the unknown values by setting up constant ratios using the isothermal condition
(PV=Constant) and the various adiabatic conditions (PV=C; TV=C; PT=C). The calculations can be
tedious but are straightforward.
P
V
T
A
1400 kPa
10.0 liters
720 K
B
875 kPa
16.0 l
720 K
C
445 kPa
24.0 liters
550 K
A
D
711 kPa
15.0 liters
550 K
B
b) Draw a P vs. V graph.
D
c) Find the Q, W, and ∆U for each of the steps, AB, BC, CD, and DA.
C
c) Use U=3nRT/2 to determine internal energies and at all points, use the appropriate expressions to
determine the work done in each process, then find Q’s from the 1st law:
AB
BC
CD
DA
Q
+6.58 kJ
0
-5.03 kJ
0
W
14ln(16/10) kJ= +6.58 kJ
+5 kJ
-5.03 kJ
- 5 kJ
∆U
0
(15.7-20.7) kJ= -5 kJ
0
+5 kJ
d) Show that Wnet/Qin = 1 – (TCD/TAB), the Carnot efficiency.
d) Wnet/Qin =(6.58-5.03)/ 6.58 =1 – (550/720)=0.236
14. This is just a variation on the problem above. One mole of an ideal gas ( = 1.4) is carried through the
Carnot cycle similar to the one described above. At point A, the pressure is 25 atm and the temperature is
600 K. At point C, the pressure is 1 atm and the temperature is 400 K.
Similar problem to the one above. You must use the gas law as well as the isothermal and adiabatic
conditions to find the needed quantities.
a) Determine the pressures and volumes at points A, B, C, and D.
a) Many calculations later: VA=1.98 l; VB=11.9 l; VC=32.9 l; VD=5.44 l; PB=4.14 atm; PD=6.05 atm.
b) Calculate the net work done per cycle.
b) Wnet=nRTABln(VB/VA) + nRTCDln(VD/VC)=(8.31)[600 ln(11.9/1.98) – 400 ln(32.9/5.440]=2.96 kJ
c) Determine the efficiency of an engine operating in this cycle. e=0.33
15. An experimental power plant at the Natural Energy Laboratory of Hawaii generates electricity from the
temperature gradient of the ocean. The surface and deep-water temperatures are 27°C and 6°C, respectively.
a) What is the maximum theoretical efficiency of this power plant? eCarnot=0.07
b) If the power plant is to produce 210 kW of power, at what rate must heat be extracted from the warm
water? At what rate must heat be absorbed by the cold water? Assume the maximum theoretical
efficiency. dQh/dt = 210/0.07 = 3 MW; dQc/dt = 3-0.21=2.79 MW
c) The cold water that enters the plant leaves it at a temperature of 10 ºC. What must be the flow rate of
cold water through the system? Give your answer in kg/h and liters/h.
c) dQc/dt =2.79 MW = (dm/dt)c ∆Tdm/dt=2.79 x 106/4186(4)=0.167 kg/sec= 601 kg/hr601 l/hr.
d) What compensating factor makes this a worthwhile experiment despite the low efficiency?
d) Even though the efficiency is low the vast size of the oceans could make this practical. Also there are
no fuel requirements, pollution would be low, maybe designs are simpler…there are a lot of other things
to consider besides efficiency in engine design.
There is an endless number of variations of the “cycle problem”. Here are a few:
16. One mole of an ideal diatomic gas is taken through the cycle shown.
Assume that the processes are reversible. Calculate
a) the net work done by the gas. Wnet= WAB +WBC=8.13-4.04=4.09 kJ
b) the thermal energy (heat) added to the gas.
b) Qin=QCA + QAB =8.13 + 10.1=18.2 kJ
c) the thermal energy (heat) expelled by the gas. Qout=QBC =-14.1 kJ
d) the efficiency of the cycle. e=W/Qin=4.09/18.2 =0.22
P (atm)
5
A
C
1
B
10
50
V(liters)
e) Why is this cycle not as efficient as the Carnot ideal? The Carnot efficiency results from extracting the
heat at the highest temperature possible and expelling the heat at the lowest temperature possible (not just
because it’s reversible). In this cycle heat is added from C to A at less than the highest temperature and heat
is expelled from B to C at more that the lowest temp…If all the heat were extracted at TAB (608 K) and all
the heat expelled at TC (122 K), then the efficiency would be ec=1- (1/5)=0.8, but that would require a
Carnot cycle.
17. A gas is taken through the cyclic process ABCA shown.
a) If Q is negative for the process BC and ∆U is negative for the process CA,
determine the signs of Q, W, and ∆U associated with each process.
a) Positive: QAB ,WAB , and ∆UAB ; all others are negative, WBC is zero.
b) Find the net heat transferred to the system during one complete cycle.
b) Qnet = Wnet=Area within cycle= 18 kJ
c) If the cycle is reversed, that is, the process follows the path ACBA,
what is the net heat transferred per cycle? -18 kJ
P (kPa)
B
6
2
A
C
6
15
m3
18. One mole of a monatomic ideal gas is taken through the
Q2
reversible cycle shown. At point A the pressure,
3Po
B
C
volume, and temperature are Po, Vo, and To, respectively.
In terms of R and To, find First determine TB=3To , TC=6To , TD=2To.
2Po
Q1
Q3
a) the total heat entering the system per cycle,
Po
a) Q1+ Q2=3R(2To)/2 +5R(3To)/2 =(21/2)RTo
A
Q4
D
b) the total heat leaving the system per cycle,
b) Q3 +Q4=3R(-4To)/2 +5R(-To)/2 =(-17/2)RTo
Vo
2Vo
c) the efficiency of an engine operating in this reversible cycle, e=1-(17/21)=0.19
d) Consider this same cycle in reverse so it represents a refrigerator. What is its coefficient of performance?
Is this cycle reversible? Hint: Review problem 4b. Here K=4.25. In problem 4b you were asked to prove
that K=(1-e)/e for a reversible engine. This cycle satisfies this criterion so it is reversible, even though it’s
not a Carnot cycle.
Note: This problem shows that any cycle is reversible if it’s well-defined and one knows the state of the gas
at every instant of the cycle. In fact, that is always the assumption when we draw a solid line in a PV graph,
and all the formulas we have derived for Q, W, and ∆U take reversibility for granted. What is different
about the Carnot cycle is that there are only two extremes temperatures involved and the Qh is extracted at
the highest temp and all the Qc expelled at the lowest temp.
You should realize that the efficiencies derived by analyzing a cycle are always ideal. The efficiency or real
engines are determined experimentally, by actually testing the engine, not by analyzing a cycle.
19. Your textbook has a nice illustration of a basic refrigerator. The mechanism consists of two sets of coils
(a condenser outside the refrigerator and an evaporator inside) containing a refrigerant fluid that is partially
liquefied and vaporized as it flows through the coils. A compressor compresses the fluid adiabatically when
it is mostly gaseous, raising its temperature and pressure and delivers it to the condenser at a temperature
higher than the surrounding room temperature. As the fluid passes through the condenser it gives up heat to
the outside and the fluid liquefies. Then an expansion valve allows the fluid to
P
expand adiabatically, lowering its pressure and cooling it below the temperature
C
B
inside the refrigerator. As the cooled fluid flows through the evaporator it
absorbs heat from the inside of the refrigerator and vaporizes just before it
enters the compressor where the cycle starts all over again.
D
A
a) In the PV diagram shown, what sections correspond to each of the
processes described above? AB compression; BC condensation; CD expansion;
V
DA vaporization.
b) Explain why the flat sections in the graph are isothermal as well as isobaric. Why doesn’t this violate the
gas law PV=nRT? The fluid is undergoing a phase change during those processes and is not behaving like
an ideal gas…
c) How would you determine the K of this refrigerator? You would have to determine Qhot and Qcold. Since
the heat is transferred during the phase changes you would have to know the latent heat of vaporization of
this fluid “L” and the mass of the fluid undergoing the phase changes, then Q=mL.
d) What would the area under the graph mean in this diagram? Same as any P-V graph, the work done by
the fluid.
e) Is the 1st Law (Q-W=∆U) valid in this problem? Yes, the 1st Law is a version of the work-energy principle
that is a fundamental principle in all phenomena. However ∆U would not equal nC v∆T because the fluid is
not an ideal gas, and Q would depend on the latent heat.
The following problems feature some well-known engines:
20. Review the gasoline engine’s idealized Otto cycle discussed in the text and in class. Draw a graph of the
cycle and make sure you understand the derivation of the efficiency formula and the meaning of the
“compression ratio”. Here’s a typical gasoline engine problem.
a) A gasoline engine has a compression ratio of 6 and uses a gas for which  = 1.4. What is the
efficiency of the engine? The compression ratio is the larger initial volume V1 divided by the
smaller, compressed volume V2.
a) eOtto=1- (V1/ V2)1-= 1- (6)-0.4=0.51
b) If the actual efficiency is 15%, what fraction of the fuel is wasted as a result of friction and
unavoidable heat losses? (Assume complete combustion of the air-fuel mixture.) 36%
c) Each cycle the engine takes in 1.61 X 104 J of heat obtained by burning gasoline with a heat of
combustion of 4.60 X 104 J/g. How much heat is discarded in each cycle?
c) At 15 % efficiency, W=0.15(16.1 kJ)=2.42 kJ and Qc= Qh -W=16.1-2.42=13.7 kJ
d) What mass of fuel is burned in each cycle? 1.61/4.60=0.35 g
e) If the engine goes through 60.0 cycles per second, what is its power output in kilowatts and in
horsepower? P=W/∆t=2.42 x60=145 kW=194 hp.
21. An idealized Diesel engine operates in a cycle known as the
air-standard Diesel cycle. Fuel is sprayed into the cylinder at the
point of maximum compression, B. Combustion occurs during the
expansion B C, which is approximated as an isobaric process.
The rest of the cycle is the same as in the gasoline engine. Show
that the efficiency of an engine operating in this idealized
1 T  T 
Diesel cycle is: e  1  D A 
 TC  TB 
P
B
C
Adiabatic
D
Adiabatic
A
V
V
Here Qh =QBC =nCP(TC –TB) and Qc =QDA =nCV(TA –TD). Just plug into e =1- |Qc /Qh | and =CP/CV…
 to derive an expression for the efficiency of the Diesel engine in terms of the
Note: It is possible
compression ratios (VA/VB) and (VA/VC) but the proof is algebraically messy. It involves eliminating the
temperatures from the formula by relating them to the volumes using adiabatic and isobaric relationships. It
is not something I would ask you to do on a test, but if you’re interested check out Challenge Problem 32.
22. The graph illustrates the Stirling engine cycle that operates between two isotherms T1 and T2 (T2 >T1).
Assuming that the operating gas is an ideal monatomic gas, calculate the efficiency of this engine whose
constant-volume processes occur at the volumes Vo and Vo/r, where “r” is the compression ratio.
a) How does this cycle differ from the Otto cycle?
a) The adiabatics are replaced with isothermal processes.
b) Assuming an ideal monatomic gas, determine Q, W, and ∆U
for each of the processes: a b, b c, c d, and d a.
Q
W
∆U
a b
nRT1ln(1/r) nRT1ln(1/r) 0
b c
nCV(T2-T1) 0
nCV(T2-T1)
c d
nRT2ln(r)
nRT2ln(r)
0
d a
nCV(T1-T2) 0
nCV(T1-T2)
P
c
T2
d
b
T1
a
Vo/r
Vo
c) Compare the heat transfers during the constant volume processes to show that they cancel out and do not
require an external source of energy. (This is called regeneration.)
c) It’s obvious from the chart that the isochoric processes are equal in value and opposite in sign. This
means that these energies can be exchanged within the gas itself and require no external heat source or heat
sink. This allows us to leave them out of the efficiency calculation altogether.
d) Determine the efficiency of this engine and compare to a Carnot engine operating between the
same two extreme temperatures.
d) Leaving out the heat exchanges in the isochoric processes, e =1-[nRT1|ln(1/r)|/nRT2)|ln(r)|]=1-(T1/T2),
just like the Carnot cycle, but this assumes the isochoric energy exchanges are truly internal and do not
affect the outside environment.
e) If Sterling engines are so efficient, why are they not used more often, to power cars, for example?
Isotheral processes are necessarily slow and that makes the Sterling engine less powerful. We want cars to
go fast so we sacrifice efficiency to power. The same is true about the Carnot cycle which is why we don’t
see any real engines using it.
Note of interest: The Stirling cycle is used in external combustion engines where heat is supplied by burning
fuel outside the cylinder (instead of explosively inside the cylinder as in the Otto cycle). For this reason
Stirling-cycle engines are quieter than Otto-cycle engines, since there are no intake and exhaust valves (a
major source of engine noise). While small Stirling engines are used for a variety of purposes, Stirling
engines for automobiles have not been successful because they are larger, heavier, more expensive, and less
powerful than conventional automobile engines given their size.
Finally…Entropy!
23. A couple of conceptual questions about entropy.
a) If you run a movie film backwards, it is as if the direction of time were reversed. What scenes would
provide clues that the movie was running in reverse and what scenes would not? What law(s) are clearly
violated when the movie is run in reverse? Only the 2nd law!...
b) Some critics of biological evolution claim that it violates the 2nd Law of Thermodynamics, since
evolution involves simple life forms developing into more complex and more highly ordered organisms.
Explain why this is not a valid argument against evolution. The 2nd law applies to the entire Universe and
other “closed systems”. A single living organism or species is not a closed system; it is constantly
exchanging energy with its environment and the entropy decreases due to the evolution of complex life
forms is more than off-set by the increases in entropy of the rest of the environment.
24. In theory any process can be performed reversibly or very close to reversibly (not just isothermal and
adiabatic) if the pressure, volume, and temperature can be infinitesimally changed at every state of the
process.
a) Describe how one could conduct (approximate) reversible isochoric, isobaric, isothermal, and adiabatic
processes.
a) Reversible processes require complete control, so the state of the system (P,V, &T) is well-known at all
times and change only infinitesimally. A very important point is that in reversible processes heat is
transferred between objects at the same temperature (or varying in temperature by an infinitesimal
amount). In an engine the heat is exchanged between the gas going through the cycle and the external heat
sources, Thot and Tcold. In the reversible isothermal and adiabatic processes, heat flows in or out of the gas
as the pressure is changed infinitesimally by adding or subtracting weight from the piston. In the reversible
isochoric and isobaric processes the temperature of the gas and the heat sources must not vary more than
an infinitesimal amount even as the temperature changes; this can be done by having a fine thermostat
control that changes the temperature of the heat source slowly as the heat flows in or out of the container…
b) Determine expressions for calculating the entropy changes in each of those reversible processes and put
them in the chart below: Use the definition of ∆S=Q/T or ∫dQ/T…For example: ∆SV=∫nCV dT/T=
nCVln(T/To)…
Constant
Ratio
P vs. V
Graph
Isochoric
∆V=0
P/T= constant
Isobaric
∆P=0
V/T= constant
P
P
Isotherms
V
Isothermal
∆T=0
PV= constant
P
IsoV
therms
Adiabatic
Q=0

PV = constant
P
IsoV
therms
IsoV
therms
∆S
nCVln(T/To)
nCPln(T/To)
nRln(V/Vo)
0
and T/To=P/Po
and T/To=V/Vo
and V/Vo=Po/P
c) Show that for any process, the change entropy is due to either a volume change and/or a temperature
change.
c) As you can see from the expressions in the chart, the terms all have either a volume ratio or a temp. ratio.
The temperature and volume ratios can also be related to pressure ratios of course.
It’s also interesting to note that, since CP= CV +R ∆SP=∆SV +∆ST.
d) Show that in an adiabatic process, the entropy change due to the volume change is equal and opposite to
the entropy change due to the temperature change (hence ∆S=0).
d) Since entropy is a state function, any process can be replaced with an equivalent series of processes. The
adiabatic process is equivalent to an isochoric process from To to T and an isothermal process form Vo to V,
so ∆Sadia=∆SV +∆ST= nCVln(T/To) + nRln(V/Vo) Since (T/To) =(Vo/V) -1 for adiabatic processes,
∆Sadi=nCVln(Vo/V) -1 +nRln(V/Vo)∆Sadi=n[CV(1-)+R]ln(Vo/V)=n[CV -CP +R]ln(Vo/V)=0
25. Consider 1 gram of super cold ice at -10 ºC in a insulated container at 1 atm. Heat from an external
source is added to the container and the ice goes through all the phase changes until you end up with 1 gram
of water vapor at 110ºC. In this problem I will use cal/K as entropy units for convenience.
a) Determine the entropy changes throughout the entire process. Recall: dQ = mcdT and L=Q/m.
∆Sice=mc ln(T/To)=(1)(0.5) ln(273/263)=0.0187 cal/K=0.078 J/K;
∆Smelt=mL/T=80/273=0.293 cal/K;
∆Swater=mc ln(T/To)=(1)(1) ln(373/273)=0.312 cal/K;
∆Svap=mL/T=540/373=1.45 cal/K;
∆Ssteam=mc ln(T/To)=(1)(0.5) ln(383/373)=0.0132 cal/K; or ∆Ssteam=nCP ln(T/To)=(1/18)(9R/2)
ln(383/373) =0.0550 J/K=0.0132 cal/K
b) If the heat is being transferred from a heat source that has a constant temperature of 110ºC, what is the
entropy change of the heat source? Does the combined entropy change satisfy the 2nd law of
thermodynamics?
∆Ssource= -Qtotal/T= -730/383 = -1.91 cal/K and ∆Sice/water/steam=+2.09 cal/K, so the overall entropy change is
+ which is consistent with the 2nd law…
26. One mole of H2 gas is contained in the left-hand side of the container
shown, which has equal volumes left and right. The right-hand side is evacuated.
H2
vacuum
When the valve is opened, the gas streams into the right side, without opposition.
valve
a) Does the temperature of the gas change? Explain.
a) The gas expands but it does NO work, so no heat is enters the system and its temperature doesn’t change.
It’s pressure does decease to ½ the original pressure.
b) This process is called an irreversible “free expansion”. Compare this process to reversible isothermal
expansion and the reversible adiabatic expansion for the same volume change. What are the similarities and
differences?
b) This process resembles the isothermal reversible process more than any other in that the final state of the
system has the same temperature and pressure. It has only a superficial resemblance to the reversible
adiabatic process in that no heat enters the system, but in the reversible adiabatic expansion the final
temperature and pressure are different.
c) How would you find the entropy change of this irreversible “free expansion”?
c) To find the entropy change it’s necessary to find a reversible process with the same initial and final
states. In this case, that’s an isothermal process. So, ∆Sfree expansion =∆ST = nRln(V/Vo)=Rln2=5.76 J/k
27. A 2.0-liter container has a center partition that divides it into two equal parts,
as shown. The left side contains H2 gas and the right side contains O2 gas. Both
0.5 mole 0.5 mole
gases are at room temperature and at atmospheric pressure. The partition is
H2
O2
removed and the gases are allowed to mix freely.
a) What is the entropy increase?
a) There is no temperature change but each gas now has a larger volume to move in. The entropy change
can be determined from the volume change of each gas, independent of the other gas. That is, ∆ST
=2nRln(V/Vo) =2(0.5)(8.31)ln2 =5.76 J/k
b) What is the entropy increase if the temperatures were different? Let’s say hydrogen had twice the
temperature of the oxygen.
b) The entropy change above would still apply, but in addition there would be entropy changes due to the
temperature changes of the gases. So we need the equilibrium temperature, which is 1.5Toxygen, you can
think of this as a two-step process, first let the gases come to thermal equilibrium, then remove the partition
and let them mix. The entropy changes due to the temperature changes will be: ∆SO2=nCV
ln(T/To)=(0.5)(5R/2)ln(1.5) =4.21 J/K, plus ∆SH2=nCV ln(T/To)=(0.5)(5R/2)ln(0.75) =-3.00 J/K The net
∆S=5.76 +4.21-3.00=6.97 J/K.
c) What is the entropy increase if the gases were the same but the temperatures of the samples were
different?
c) In this case the volume increase would not change the entropy so (a) would not apply. Only the
temperature changes would change the entropy net ∆S=4.21-3.00=1.21 J/K.
28. An object of mass m1, specific heat c1, and temperature T1 is placed in contact with a second object of
mass m2, specific heat c2, and temperature T2 > T1.
a) Determine an expression for the equilibrium temperature T.
a) An old calorimetry problem, T=(m1c1T1 + m2c2T2)/(m1c1+ m2c2).
b) Show that the entropy change ∆S of the system is: ∆S= m1c1 ln(T/T1) + m2c2 ln(T/T2).
b) This follows straight from the entropy expression of ∆S = ∫mcdT/T…
c) Can the expression above ever give a value less that 0? Explain.
c) This is an irreversible process so entropy must increase; that means the net entropy change described by
the expression in (b) must always give a + answer. This is hard to prove algebraically, but it’s easy to see in
numerical examples like the ones in the next problem.
29. The following are a couple of numerical versions of the problem above.
a) Calculate the increase in entropy of the Universe when you add 20 g of 5°C cream to 200 g of 60°C
coffee. The specific heat of cream and coffee is 4.2 J/g·Cº.
a) Te=55 ºC net ∆S= 20 ln[(55+273)/278] + 200 ln(328/333)=13.9-12.7=1.19 cal/K
b) A 1.0-kg iron horseshoe is taken from a furnace at 900°C and dropped into 4.0 kg of water at 10°C. If no
heat is lost to the surroundings, determine the total entropy change.
b) You need ciron=470 J/kg and cwater=4190 J/kg to be able to get Te=34 ºC net ∆S= 470 ln[(34+273)/(900
+273)] + 4(4190) ln(307/283)=-630+1363 =733 J/K
30. Another useful graph is a TS-diagram. This problem shows how a different
graph from a PV-graph can be useful…
a) Graph a Carnot cycle, plotting Kelvin temperature vertically and entropy
horizontally. This is called a temperature-entropy diagram, or TS-diagram.
b) Show that the area under any curve representing a reversible path in a
temperature-entropy diagram represents the heat transferred during the process.
b) The graph’s area =∫TdS =∫T(dQ/T) =∫dQ =Q
c) Derive from your diagram the expression for the thermal efficiency of
a Carnot cycle.
c) Qhot=TAB∆S and Qcold=-TCD∆Se=1-(TCD∆S /TAB∆S)=1-(TCD/TAB)=1-(Tc/Th)
T
A
B
C
D
S
d) Draw a temperature-entropy diagram for the Stirling cycle, described
in Problem 22. Use this diagram to relate the efficiency of the Carnot and
Stirling cycles. The two cycles have the same “e”…
T
A
C
Challenge problems:
B
D
S
31. The maximum power that can be extracted by a wind turbine from an air stream is approximately P=kd2
v3, where d is the blade diameter, v is the wind speed, and the constant k = 0.5 W•s3/m5.
a) Explain the dependence of P on d and on v by considering a cylinder of air that passes over the
turbine blades in time t. This cylinder has diameter d, length L = v t, and density .
b) The Mod-5B wind turbine at Kahaku on the Hawaiian island of Oahu has a blade diameter of 97 m
(slightly longer than a football field) and sits atop a 58-m tower. It can produce 3.2 MW of electric
power. Assuming 25% efficiency, what wind speed is required to produce this amount of power?
Give your answer in m/s and in km/h.
c) Commercial wind turbines are commonly located in or downwind of mountain passes. Why?
a) The kinetic energy/time of the volume of wind passing through the blades
d
is (∆K/∆t)= ½(∆m/∆t)v2= ½( πd2∆x/4∆t)v2= ½( π/4) d2v3.
v
6
2 3
2 3
b) 32x10 = 0.25Pmax= 0.25(k d v )= 0.25(0.5)( 97 )v  v=30 m/s=108 km/hr.
∆x
c) Strong winds develop through mountain passes…
L=vt
32. It is possible to derive an expression for efficiency for a Diesel cycle in terms of the compression ratio.
Consider that the cycle starts at point A (see Problem 21) with air at temperature TA.
a) If the temperature at point C is TC , derive an expression for the efficiency of the cycle in terms of the
compression ratios, r1 =(VA/VB) and r2 =(VA/VC). Assume an ideal gas. This is a messy and long algebraic
problem. You have to eliminate the temperatures in the expression derived in problem 21 using adiabatic
and isobaric relationships.
a) The answer I got is: e= 1- --1[(r2-- r1-)/(r2-- r1-)], where r1=VA/VB and r2=VA/VC, are the compression
and expansion ratios for the adiabatic portions of the Diesel cycle illustrated in problem 21. In addition the
compression ratios can be related as follows: r2-= r1(TA/TC)]. Going on the web you can find a detail
solution. It’s mostly messy algebra.
b) What is the efficiency if TA ~ 300 K, Tc = 950 K,  = 1.40, and r = 21? Formula plug…