1. Chapter 15
Thermodynamics

2. AP Learning Objectives
Kinetic theory and thermodynamics
 Ideal gases
 Students should know how to apply the ideal gas law and
thermodynamic principles, so they can:
 Relate the pressure and volume of a gas during an
isothermal expansion or compression.
 Relate the pressure and temperature of a gas during
constant-volume heating or cooling, or the volume and
temperature during constant-pressure heating or
cooling.
 Calculate the work performed on or by a gas during an
expansion or compression at constant pressure.
 Understand the process of adiabatic expansion or
compression of a gas.
 Identify or sketch on a PV diagram the curves that
represent each of the above processes.

3. AP Learning Objectives
Laws of Thermodynamics
 Students should know how to apply the first law of thermodynamics,
so they can:
 Relate the heat absorbed by a gas, the work performed by the
gas, and the internal energy change of the gas for any of the
processes above.
 Relate the work performed by a gas in a cyclic process to the area
enclosed by a curve on a PV diagram.
 Students should understand the second law of thermodynamics, the
concept of entropy, and heat engines and the Carnot cycle, so they
can:
 Determine whether entropy will increase, decrease, or remain the
same during a particular situation.
 Compute the maximum possible efficiency of a heat engine
operating between two given temperatures.
 Compute the actual efficiency of a heat engine.
 Relate the heats exchanged at each thermal reservoir in a Carnot
cycle to the temperatures of the reservoirs.

4. Table of Contents
1. Thermodynamic Systems & Surroundings
2. The Zeroth Law of Thermodynamics
3. The First Law of Thermodynamics
4. Thermal Processes
5. Thermal Processes Using an Ideal Gas
6. Specific Heat Capacities
7. The Second Law of Thermodynamics
8. Heat Engines
9. Carnot’s Principle and the Carnot Engine
10. Refrigerators, Air Conditioners, and Heat Pumps
11. Entropy
12. The Third Law of Thermodynamics

5. Chapter 15
Thermodynamics
Section 1:
Thermodynamic Systems & Surroundings

6. Thermodynamic Systems and Their
Surroundings
 Thermodynamics is the branch of physics
that is built upon the fundamental laws that
heat and work obey.
 The collection of objects on which attention is
being focused is called the system, while
everything else in the environment is called the
surroundings.
 Walls that permit heat flow are called
diathermal walls, while walls that do not
permit heat flow are called adiabatic walls.
 To understand thermodynamics, it is necessary
to describe the state of a system.

7. Chapter 15
Thermodynamics
Section 2:
The Zeroth Law of Thermodynamics

8. Zeroth Law of Thermodynamics
 The forgotten Law of Science
 Two systems are said to be in
thermal equilibrium if there is
no heat flow between them
when they are brought into
contact.
 Temperature is the indicator of
thermal equilibrium in the
sense that there is no net flow
of heat between two systems
in thermal contact that have
the same temperature.

9. Two systems individually in thermal equilibrium
with a third system are in thermal equilibrium
with each other.
The Zeroth Law of Thermodynamics

10. 15.2.1. Which one of the following situations is described by the zeroth law of
thermodynamics?
a) An air conditioner transfers heat from the inside of a house to the outside of the
house.
b) A monatomic gas is held within a container that has a moveable piston. The gas
absorbs heat from the surroundings and expands at constant pressure and
temperature.
c) A container with adiabatic walls holds boiling water. A thermometer is calibrated by
inserting it into the boiling water and allowing it to reach thermal equilibrium with
the water.
d) A pot contains oil at 175 °C. When frozen sliced potatoes are dropped into the oil,
heat is transferred from the oil to the potatoes.
e) A physicist removes energy from a system in her laboratory until it reaches a
temperature of 3 × 10−10
K, a temperature very close to (but still greater than)
absolute zero.

11. Chapter 15
Thermodynamics
Section 3:
The First Law of Thermodynamics

12.  Suppose that a system gains heat Q and that is the only effect
occurring.
 Consistent with the law of conservation of energy, the internal
energy of the system changes:
 Heat is positive when the system gains heat and negative
when the system loses heat.
QUUU if =−=∆
Effect of Heat on a System

13.  If a system does work W on its surroundings and there is no
heat flow, conservation of energy indicates that the internal
energy of the system will decrease:
 Work is positive when it is done on the system and negative
when it is done by the system.
WUUU if =−=∆
Effect of Work on a System

14. The First Law of Thermodynamics
 The internal energy of a system changes due to heat and work:
 Heat is positive when the system gains heat and negative when
the system loses heat.
 Work is positive when it is done on the system and negative
when it is done by the system.
if UUU −=∆
WQU +=∆

15. Example 1 Positive and Negative Work
In part a of figure, the system gains 1500J of heat
and 2200J of work is done by the system on its
surroundings.
In part b, the system also gains 1500J of heat, but
2200J of work is done on the system.
In each case, determine the change in internal energy
of the system.
(a)
(b)
WQU +=∆
WQU +=∆
( ) ( )J2200J1500 −++=∆U J700−=
( ) ( )J2200J1500 +++=∆U J3700+=

16. Example 2 An Ideal Gas
The temperature of three moles of a monatomic ideal gas is reduced
from 540K to 350K as 5500J of heat flows into the gas.
Find (a) the change in internal energy and (b) the work done by the
gas.
TnRU ∆=∆ 2
3
WQUUU if +=−=∆
if nRTnRTU 2
3
2
3
−=∆
QUW −∆=
(a)
(b)
( ) ( )( )( )K540K350KmolJ31.8mol0.32
3
−⋅=∆U J7100−=
J5500J7100 −−= J12600−=

17. 15.3.1. An insulated container is filled with a mixture of water and
ice at zero °C. An electric heating element inside the container
is used to add 1680 J of heat to the system while a paddle does
450 J of work by stirring. What is the increase in the internal
energy of the ice-water system?
a) 450 J
b) 1230 J
c) 1680 J
d) 2130 J
e) zero J

18. 15.3.2. The internal energy of a system increases during some time interval.
Which one of the following statements concerning this situation must be
true?
a) The increase in internal energy indicates that work was done on the system.
b) The increase in internal energy indicates that heat was added to the system.
c) The increase in internal energy indicates that work was done by the system.
d) The increase in internal energy indicates that heat was removed from the
system.
e) The information given is insufficient to indicate the reason for the increase.

19. 15.3.3. A gas is enclosed in a cylinder by a piston. The volume of the gas
is then reduced to one half its original value by applying a force to the
piston. Which one of the following statements concerning the internal
energy of the gas is true?
a) The internal energy of the gas will decrease.
b) The internal energy of the gas will increase.
c) The internal energy of the gas will neither increase nor decrease.
d) The internal energy of the gas will equal the work done in moving the
piston.
e) The internal energy of the gas may increase, decrease, or remain the
same depending on the amount of heat that is gained or lost by the
gas.

20. 15.3.4. Which one of the following statements is not consistent with the
first law of thermodynamics?
a) The internal energy of a finite system must be finite.
b) An engine may be constructed such that the work done by the machine
exceeds the energy input to the engine.
c) An isolated system that is thermally insulated cannot do work on its
surroundings nor can work be done on the system.
d) The internal energy of a system decreases when it does work on its
surroundings and there is no flow of heat.
e) An engine may be constructed that gains energy while heat is
transferred to it and work is done on it.

21. Chapter 15
Thermodynamics
Section 4:
Thermal Processes

22. A quasi-static process is one that occurs slowly enough that a uniform
temperature and pressure exist throughout all regions of the system at all
times.
isobaric: constant pressure
isochoric: constant volume
isothermal: constant temperature
adiabatic: no transfer of heat
Thermal Processes

23. An isobaric process is one that occurs at
constant pressure.
FsW −=
Isobaric process: ( )if VVPVPW −−=∆−=
Isobaric Process
( )AsP−= VP∆−=
Sign Convention of AP is opposite that of our book!

24. ( )if VVPVPW −−=∆−=
Isobaric Process
Work by the system is the area under a PV graph.
Work is path dependent

25. Example 3 Isobaric Expansion of Water
One gram of water is placed in the cylinder and
the pressure is maintained at 2.0x105
Pa. The
temperature of the water is raised by 31o
C. The
water is in the liquid phase and expands by the
small amount of 1.0x10-8
m3
.
Find the work done and the change in internal
energy.
VPW ∆−=
WQU +=∆
TmcQ ∆=
J0020.0−=( )( )385
m100.1Pa100.2 −
××−=W
J130=J0020.0J130 −=
( ) ( )[ ]( )
C31CkgJ4186kg0010.0 ⋅= J130=

26. isochoric: constant volume
Q=
0=
Isochoric Process
VPW ∆−=
WQU +=∆

27. Example 4 Work and the Area Under a
Pressure-Volume Graph
Determine the work for the process in
which the pressure, volume, and
temperature of a gas are changed along the
straight line in the figure.
Since the volume increases, the work
is positive.
( )( )345
m100.1Pa100.29.8 −
××−=W
Estimate that there are 8.9 colored
squares in the drawing.
J180−=W

28. 15.4.1. Consider the pressure-versus-volume plot shown. There are
eight points labeled and the choices below indicate possible multi-
step processes. In which one of the processes does the work done
have the largest value?
a) G−H−B−D
b) G−F−B−D
c) H−A−B−D
d) E−D−F−H
e) C−B−F−G

29. 15.4.2. Consider the pressure-versus-volume plot shown. There are eight
points labeled and the choices below indicate possible multi-step
processes. If the initial state of the system is at A and the final state
is at E, which of the following paths between these two states results
in the largest increase in internal energy of the system?
a) A−H−D−E
b) A−B−F−E
c) A−G−E
d) A−C−E
e) All paths between A and E are equivalent for internal energy.

30. 15.4.3. An isobaric process is represented by which one of the following
graphs?
a) A
b) B
c) C
d) D
e) E

31. 15.4.4. An insulated container with rigid walls has two compartments
within. One compartment contains n moles of an ideal gas and the
other compartment has been evacuated. A valve connecting the two
chambers is opened at time t = 0 s. Which one of the following
statements concerning this situation is true?
a) There is no change in the internal energy of the gas.
b) There is no change in the pressure of the gas.
c) The temperature of the gas decreases with time.
d) Work is done by the gas as it fills the previously evacuated
compartment.
e) The gas will remain in the first compartment unless heat is added to the
system.

32. 15.4.5. In which of the following cases is a system undergoing an isobaric
process?
a) The system is placed within a thermal bath held at constant temperature.
b) The system is an ideal gas enclosed in a container with a piston that may
move up or down. A heavy object is placed on top of the piston.
c) The system is an ideal gas enclosed in a container that is in contact with
an object that is continually kept warmer or cooler than the gas within the
system.
d) The system is an ideal gas enclosed in a container has a constant volume.
e) The system is an ideal gas enclosed in a container that is connected to a
source of the gas from which gas may be added or removed to maintain a
constant pressure.

33. 15.4.6. In which of the following cases is a system undergoing an
adiabatic process?
a) The system is placed within a thermal bath held at constant
temperature.
b) The system is an ideal gas enclosed in a container with a piston that
may move up or down. A heavy object is placed on top of the piston.
c) The system is an ideal gas enclosed in a container that is in contact
with an object that is continually kept warmer or cooler than the gas
within the system.
d) The system is an ideal gas enclosed in a container has a constant
volume.
e) The system volume is changed rapidly.

34. Chapter 15
Thermodynamics
Section 5:
Thermal Processes Using an Ideal Gas

35. Isothermal
expansion or
compression of
an ideal gas








=
f
i
V
V
nRTW ln
Isothermal Expansion or Compression of an Ideal Gas
Calculus Alert!
∫−=
f
i
V
V
PdVW
∫−=
f
i
V
V
dV
V
nRT
W
∫−=
f
i
V
V V
dV
nRTW
if
x
x
xx
x
dxf
i
lnln −=∫ 





=
i
f
V
V
ln
Work is the area under
a PV graph  integral

36. Example 5 Isothermal Expansion of an Ideal Gas
Two moles of the monatomic gas argon expand isothermally at 298K
from and initial volume of 0.025m3
to a final volume of 0.050m3
. Assuming
that argon is an ideal gas, find (a) the work done by the gas, (b) the
change in internal energy of the gas, and (c) the heat supplied to the
gas.
(a)








=
f
i
V
V
nRTW ln
if nRTnRTU 2
3
2
3
−=∆(b)
WQU +=∆(c)
J3400−=
( ) ( )( )( ) 





⋅= 3
3
m050.0
m250.0
lnK298KmolJ31.8mol0.2W J3400−=
0=
WQ =

37. ( )if TTnRW −= 2
3
Adiabatic Expansion/Compression of a
Monatomic Ideal Gas
Adiabatic: no heat transfer
WQU +=∆
TnRU ∆=∆
2
3
γγ
ooii VPVP =
If you want to see the proof, go here

38. 15.5.1. A cylinder with a moveable piston contains an ideal gas. The gas
is subsequently compressed adiabatically. Which of the following
choices correctly identifies the signs of (1) the heat exchanged with
the environment, (2) the work done, and (3) the change in the
internal energy?
a) (1) is zero, (2) is negative, and (3) is negative
b) (1) is negative, (2) is positive, and (3) is negative
c) (1) is zero, (2) is negative, and (3) is positive
d) (1) is zero, (2) is positive, and (3) is positive
e) (1) is positive, (2) is negative, and (3) is zero

39. 15.5.2. Two moles of an ideal gas have an initial Kelvin temperature T0
and absolute pressure P0. The gas undergoes a reversible
isothermal compression from an initial volume V0 to a final volume
0.5 V0. How much heat is exchanged with the environment,
specifying whether it is absorbed or released?
a) Heat is released to the environment and its value is Q = 0.5P0V0.
b) Heat is absorbed from the environment; and its value is
Q = 0.5P0V0.
c) No heat is exchanged with the environment.
d) Heat is released to the environment; and its value is Q = P0V0 ln 2.
e) Heat is absorbed from the environment; and its value is
Q = P0V0 ln 2.

40. 15.5.3. Consider the pressure-volume graph shown for an ideal gas that may
be taken along one of two paths from state A to state B. Path “1” is
directly from A to B via a constant volume path. Path “2” follows the
path A−C−B. How does the amount of work done along each path
compare?
a) W1 = W2; and the value is not equal
to zero
b) W1 = W2 = 0
c) W1 > W2
d) W1 < W2
e) It is not possible to compare the work done along each path without
knowing the values of the temperature, pressure, and volume for each
state.

41. 15.5.4. Consider the following pressure-volume graphs. Which of these
graphs represents the behavior of a gas undergoing free expansion?
a) A
b) B
c) C
d) D
e) None of the
graphs represent a
gas undergoing free
expansion.

42. 15.5.5. A gas is enclosed in a cylinder by a piston. The volume of the gas
is then reduced to one half its original value by applying a force to the
piston. Which one of the following statements concerning the internal
energy of the gas is true?
a) The internal energy of the gas will decrease.
b) The internal energy of the gas will increase.
c) The internal energy of the gas will neither increase nor decrease.
d) The internal energy of the gas will equal the work done in moving the
piston.
e) The internal energy of the gas may increase, decrease, or remain the
same depending on the amount of heat that is gained or lost by the gas.

43. Chapter 15
Thermodynamics
Section 6:
Specific Heat Capacities

44. To relate heat and temperature change in solids and liquids, we
used:
TmcQ ∆=
specific heat
capacity
The amount of a gas is conveniently expressed in moles, so we write the
following analogous expression:
TnCQ ∆=
molar specific
heat capacity
Specific Heat Capacities

45. For gases it is necessary to distinguish between the molar specific heat
capacities which apply to the conditions of constant pressure and constant
volume:
PV CC ,
WUQ +∆=pressureconstant
constant pressure
for a monatomic
ideal gas RCP 2
5
=
Specific Heat Capacities – Constant P
( ) ( )ifif TTnRTTnR −+−= 2
3 TnR∆= 2
5
first law of
thermodynamics
TnRU ∆=∆ 2
3
VPW ∆=
Tn
Q
CP
∆
=
Tn
TnR
∆
∆
= 2
5

46. constant volume
for a monatomic
ideal gas
RCV 2
3
=
monatomic
ideal gas
3
5
2
3
2
5
===
R
R
C
C
V
P
γ
any ideal gas RCC VP =−
Specific Heat Capacities – Constant V
WUQ +∆=pressureconstant ( ) 02
3
+−= if TTnR TnR∆= 2
3
first law of
thermodynamics
TnRU ∆=∆ 2
3
γ is the ratio of the
molar specific heat
capacities at constant
pressure and volume

47. Chapter 15
Thermodynamics
Section 7:
The Second Law of Thermodynamics

48. THE SECOND LAW OF THERMODYNAMICS: THE HEAT FLOW STATEMENT
Heat flows spontaneously from a substance at a higher temperature to a substance
at a lower temperature and does not flow spontaneously in the reverse direction.
The second law is a statement about the natural tendency of heat to
flow from hot to cold, whereas the first law deals with energy conservation
and focuses on both heat and work.
Second Law of Thermodynamics

49. Chapter 15
Thermodynamics
Section 8:
Heat Engines

50. heatinputofmagnitude=HQ
heatrejectedofmagnitude=CQ
doneworktheofmagnitude=W
Heat Engine
 A heat engine is any device that uses heat to
perform work. It has three essential features.
 Heat is supplied to the engine at a relatively
high temperature from a place called the hot
reservoir.
 Part of the input heat is used to perform work
by the working substance of the engine.
 The remainder of the input heat is rejected to a
place called the cold reservoir

51. The efficiency of a heat engine is defined as
the ratio of the work done to the input heat:
HQ
W
e =
If there are no other losses, then
CH QWQ +=
H
CH
Q
QQ
e
−
=
Efficiency
CH QQW −=
H
C
Q
Q
−=1

52. Example 6 An Automobile Engine
An automobile engine has an efficiency of 22.0% and produces
2510 J of work. How much heat is rejected by the engine?
HQ
W
e =
CH QWQ +=
e
W
QH =
J8900=
WQQ HC −=
W
e
W
QC −=
( ) 





−= 1
220.0
1
J2510CQ






−= 1
1
e
W

53. 15.8.1. An automobile engine that burns gasoline has been
engineered to have a relatively high efficiency of 22 %. While a
car is being driven along a road on a long trip, 14 gallons of
gasoline are consumed by the engine. Of the 14 gallons, how
much gasoline was used in doing the work of propelling the car?
a) 14 gallons
b) about 11 gallons
c) about 8 gallons
d) about 3 gallons
e) about 1 gallon

54. Chapter 15
Thermodynamics
Section 9:
Carnot’s Principle & the Carnot Engine

55. A reversible process is one in which both the system and the
environment can be returned to exactly the states they were in
before the process occurred.
CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE SECOND
LAW OF THERMODYNAMICS
No irreversible engine operating between two reservoirs at constant temperatures
can have a greater efficiency than a reversible engine operating between the same
temperatures. Furthermore, all reversible engines operating between the same
temperatures have the same efficiency.
Carnot Principle

56. H
C
Q
Q
e −=1
H
C
H
C
T
T
Q
Q
=
Carnot Engine
 The Carnot engine is useful as an
idealized model.
 All of the heat input originates from a
single temperature, and all the
rejected heat goes into a cold
reservoir at a single temperature.
 Since the efficiency can only depend
on the reservoir temperatures, the
ratio of heats can only depend on
those temperatures
H
C
T
T
−=1
H
C
c
T
T
e −=1

57. Example 7 A Tropical Ocean as a Heat Engine
Water near the surface of a tropical ocean has a temperature of 298.2 K, whereas
the water 700 meters beneath the surface has a temperature of 280.2 K. It has
been proposed that the warm water be used as the hot reservoir and the cool water
as the cold reservoir of a heat engine. Find the maximum possible efficiency for
such and engine.
H
C
T
T
e −=1c
K298.2
K2.280
1−= 060.0=

58. Conceptual Example 8 Natural Limits on the Efficiency of a Heat Engine
Consider a hypothetical engine that receives 1000 J of heat as input from a
hot reservoir and delivers 1000J of work, rejecting no heat to a cold reservoir
whose temperature is above 0 K. Decide whether this engine violates the first
or second law of thermodynamics.

59.  Reasoning The first law of thermodynamics is an expression of
energy conservation. The second law states that no irreversible
engine operating between two reservoirs at constant temperatures
can have a greater efficiency than a reversible engine operating
between the same temperatures. The efficiency of such a reversible
engine is eC the efficiency of a Carnot engine.
 From the point of view of energy conservation, nothing is wrong with
an engine that converts 1000 J of heat into 1000 J of work. Energy
has been neither created nor destroyed; it has only been transformed
from one form (heat) into another form (work). Therefore, this engine
does not violate the first law of thermodynamics.
 Since all of the input heat is converted into work, the efficiency of the
engine is 1, or 100%. But Equation 15.15, which is based on the
second law of thermodynamics, indicates that the maximum possible
efficiency is eC = 1 - TC/TH where TC and TH are the temperatures of the
cold and hot reservoirs, respectively. Since we are told that TC is
above 0 K, it is clear that the ratio TC/TH is greater than zero, so the
maximum possible efficiency is less than 1 (or less than 100%). The
engine, therefore, violates the second law of thermodynamics, which
limits the efficiencies of heat engines to values less than 100%

60. 15.9.1. Which one of the following statements is not consistent with the first
law of thermodynamics?
a) The internal energy of a finite system must be finite.
b) An engine may be constructed such that the work done by the machine
exceeds the energy input to the engine.
c) An isolated system that is thermally insulated cannot do work on its
surroundings nor can work be done on the system.
d) The internal energy of a system decreases when it does work on its
surroundings and there is no flow of heat.
e) An engine may be constructed that gains energy while heat is transferred
to it and work is done on it.

61. Chapter 15
Thermodynamics
Section 10:
Refrigerators, Air Conditioners, and Heat Pumps

62. Refrigerators, air conditioners, and heat pumps are devices that make
heat flow from cold to hot. This is called the refrigeration process.
Refrigeration

63. Refrigeration

64. Conceptual Example 9 You Can’t Beat the Second Law of
Thermodynamics
Is it possible to cool your kitchen by leaving the refrigerator door open or
to cool your room by putting a window air conditioner on the floor by the
bed?
Neither, the work done by the motor
would be added to the hot
reservoir, the room, and make it
hotter.

65. W
QC
=eperformancoftCoefficienRefrigerator or
air conditioner
“Efficiency”

66. The heat pump uses
work to make heat from
the wintry outdoors flow
into the house or to hot
Summer weather
outside.
Heat Pumps
heat
pump
W
QH
=Coefficient of performance

67. Example 10 A Heat Pump
An ideal, or Carnot, heat pump is used to heat a house at 294 K. How much
work must the pump do to deliver 3350 J of heat into the house on a day when
the outdoor temperature is 273 K?
H
C
H
C
T
T
Q
Q
=
CH QQW −=
H
C
HC
T
T
QQ =
H
C
HH
T
T
QQW −=






−=
H
C
H
T
T
QW 1 ( ) 





−=
K294
K273
1J3350 J240=

68. 15.10.1. A house that is heated using a heat pump with an ideal coefficient of
performance loses heat to its surroundings at a rate of Z1(Thouse − Tsurr.), where
Z1 is a constant, Thouse is the temperature inside the house; and Tsurr. is the
temperature of its surroundings. In this process, heat is taken from the
surroundings and heats the house at a rate of Z2(Tout − Thouse) where Tout is the
temperature of the air output from the heat pump, which has a constant value.
Which one of the following expressions is equal to the efficiency of the heat
pump?
a) d)
b) e)
c)
surr.out
out
TT
T
−
outsurr.
surr.
TT
T
−
surr.out
house
TT
T
−
housesurr.
house
TT
T
−
housesurr.
out
TT
T
−

69. 15.10.2. An air conditioner pumps heat from a cold room to the hot outdoors in a three step
cyclic process:
(1) Room temperature, low pressure refrigerant gas passes through a compressor and comes
out with increased temperature and increased pressure. The hot gas passes through
piping on the outside, where heat is rejected to the surroundings.
(2) The gas then passes through a narrower pipe before entering a compressor. Work is
done by the compressor to increase the pressure enough for the gas to turn into a liquid.
(3) The liquid then undergoes free expansion into a gas and cools. The cool gas passes
through pipes that are inside the house. The inside air is cooled by coming into contact
with these pipes. The refrigerant gas exits these pipes as a room temperature, low
pressure gas. The cycle is then repeated.
Why doesn’t this system violate the second law of thermodynamics?
a) The internal energy of the gas is constant.
b) Heat is normally taken from a warm place and transported to a warmer place.
c) The system involves a closed cycle.
d) Work is continually done on the system.
e) Since the compressor adds entropy, the total entropy increases.

70. Chapter 15
Thermodynamics
Section 11:
Entropy

71. In general, irreversible processes cause us to lose some, but not necessarily
all, of the ability to do work. This partial loss can be expressed in terms of
a concept called entropy.
H
C
H
C
T
T
Q
Q
=
H
H
C
C
T
Q
T
Q
=
R






=∆
T
Q
Sentropy
change
reversible
Carnot
engine
Entropy

72. Entropy, like internal energy, is a function of the state of the system.
Consider the entropy change of a Carnot engine. The entropy of the
hot reservoir decreases and the entropy of the cold reservoir increases.
R






=∆
T
Q
S
0=−+=∆
H
H
C
C
T
Q
T
Q
S
Reversible processes do not alter the entropy of the universe.
Entropy

73. Entropy
 What happens to the entropy change of the universe in an
irreversible process is more complex.
 Since ΔS is independent of the path, it does not matter what
path is used.

74. Example 11 The Entropy of the Universe Increases
The figure shows 1200 J of heat spontaneously flowing through
a copper rod from a hot reservoir at 650 K to a cold
reservoir at 350 K. Determine the amount by which
this process changes the entropy of the
universe.
H
H
C
C
T
Q
T
Q
S −+=∆ universe
K650
J1200
K350
J1200
−+= KJ6.1+=

75. Any irreversible process increases
the entropy of the universe. 0universe >∆S
THE SECOND LAW OF THERMODYNAMICS STATED
IN TERMS OF ENTROPY
The total entropy of the universe does not change when a
reversible process occurs and increases when an irreversible
process occurs.
Restating the Second Law

76. Example 12 Energy Unavailable for Doing Work
Suppose that 1200 J of heat is used as input for an engine
under two different conditions (as shown on the right).
Determine the maximum amount of work that can be obtained
for each case.
HQ
W
e =
The maximum amount of work will be achieved when the
engine is a Carnot Engine, where
(a)
H
C
T
T
e −=1carnot
( ) HQeW carnot=
(b)
H
C
T
T
e −=1carnot
( ) HQeW carnot=
The irreversible process of heat through the copper
rod causes some energy to become unavailable.
77.0=
K650
K150
1−=
( )( )J120077.0= J920=
K350
K150
1−= 57.0=
( )( )J120057.0= J680=

77. universeeunavailabl STW o∆=
More on Entropy

78. 15.11.1. A box with five adiabatic sides contains an ideal gas with an initial
temperature T0. The sixth side is diathermal and is placed in contact with a
reservoir with a constant temperature T2 > T0. Assuming the specific heat
capacity of the system does not change with temperature, why must the entropy
change of the universe always be increasing as the box warms?
a) Entropy will always be increasing since the work done on the gas in the box is
negative.
b) Entropy will always be increasing since the temperature of the box is always
less than or equal to T2.
c) Entropy will always be increasing since this process is reversible.
d) Entropy will always be increasing since the temperature of the box is always
greater than absolute zero.
e) Entropy will always be increasing since in any process entropy increases.

79. 15.11.2. A leaf is growing on a tree. Does this growth process violate the
second law of thermodynamics when it is stated in terms of entropy?
a) Yes, but the law does not apply to living things. It only applies to
inanimate objects.
b) Yes, because this law is not applicable in situations involving radiant
energy from the Sun.
c) No, because the entropy of the Sun has decreased while the entropy of the
leaf increases as it grows.
d) No, because while the entropy of the leaf is decreasing as it grows, there
is a net increase in entropy because of the light emitted from the leaf.
e) No, because there is no net increase in the energy of the leaf.

80. 15.11.3. While watching a fantasy film, you observe a wizard wave his
arms and six potion vials that had fallen to the floor suddenly piece
themselves back together with the potions inside and rise up with a
table. In the end, the table is upright and the six vials with their potions
are sitting on the table as if nothing had happened. Which of the
following principles or laws of physics is disobeyed by this scene from
the movie?
a) conservation of energy
b) second law of thermodynamics
c) Newton’s laws of motion
d) time dilation
e) the work-energy theorem

81. Chapter 15
Thermodynamics
Section 12:
The Third Law of Thermodynamics

82. It is not possible to lower the temperature of any system to absolute
zero in a finite number of steps.
The 3rd
Law of Thermodynamics