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Chapter 11 Fluids
Chapter 11
FLUIDS
PREVIEW
A fluid is any substance that flows, typically a liquid or a gas. Hydrostatics is the study of
fluids at rest, such as the pressure of a fluid at a particular depth, or the buoyant force
acting on an object in a fluid. Archimedes principle states that the buoyant force acting on
an object in a fluid is equal to the weight of the fluid displaced by the object
Hydrodynamics is the study of fluids in motion. As a fluid flows through a pipe, the flow
rate through the cross section is the same at any point in the pipe. Bernoulli’s equation
relates static pressure of a fluid to its dynamic (moving) pressure.
The content contained in sections 1 – 4, 6 – 10, and 12 of chapter 11 of the textbook is
included on the AP Physics B exam.
QUICK REFERENCE
Important Terms
absolute pressure
the total static pressure at a certain depth in a fluid, including the pressure at the
surface of the fluid
Archimedes principle
the buoyant force acting on an object in a fluid is equal to the weight of the fluid
displaced by the object
Bernoulli’s principle
the sum of the pressures exerted by a fluid in a closed system is constant
density
the ratio of the mass to the volume of a substance
flow rate continuity
the volume or mass entering any point must also exit that point
fluid
any substance that flows, typically a liquid or a gas
gauge pressure
the difference between the static pressure at a certain depth in a fluid and the
pressure at the surface of the fluid
hydrodynamics
the study of fluids in motion
hydrostatics
the study of fluids at rest
136
Chapter 11 Fluids
ideal fluid
a noncompressible, nonviscous fluid which exhibits steady flow, that is, the
velocity of the fluid particles is constant
liquid
substance which has a fixed volume, but retains the shape of its container
pressure
force per unit area
the SI unit for pressure equal to one newton of force per square meter of area
Equations and Symbols
where
F
A
m

V
Pdepth  P0  gh
P
P = pressure
F = force perpendicular to a surface
A = area
ρ = density
m = mass
V = volume
FB = buoyant force
W = weight
g = acceleration due to gravity
v = speed or velocity
y = height above some reference level
FB  W fluid  gV fluid
1 A1v1   2 A2 v 2 (mass flow rate)
A1v1  A2 v 2 (volume flow rate)
P1 
1
1
v1 2  gy1  P2  v 2 2  gy 2
2
2
Ten Homework Problems
Chapter 11 Problems 21, 36, 39, 50, 52, 59, 61, 65, 77, 87
DISCUSSION OF SELECTED SECTIONS
11.1 - 11.2 Mass Density, and Pressure
The mass density  of a substance is the mass of the substance divided by the volume it
occupies:
m

V
A fluid is any substance that flows and conforms to the boundaries of its container. A
fluid could be a gas or a liquid; however on the AP Physics B exam fluids are typically
liquids which are constant in density. An ideal fluid is assumed
137
Chapter 11 Fluids



to be incompressible (so that its density does not change),
to flow at a steady rate,
to be non-viscous (no friction between the fluid and the container through which
it is flowing), and
flows irrotationally (no swirls or eddies).

Any fluid can exert a force perpendicular to its surface on the walls of its container. The
force is described in terms of the pressure it exerts, or force per unit area:
p
F
A
11.3 Pressure and Depth in a Static Fluid
The SI unit for pressure is the Newton per meter squared, or the Pascal. Sometimes
pressure is measured in atmospheres (atm). One atmosphere is the pressure exerted on us
every day by the earth’s atmosphere. The relationship between one atmosphere and
Pascals is
1 atm = 1.013 x 105 Pa
This is approximately equal to 15 lbs/in2. In mechanics, it is often convenient to speak in
terms of mass and force, whereas in fluids we often speak of density and pressure.
A static (non-moving) fluid produces a pressure within itself due to its own weight. This
pressure increases with depth below the surface of the fluid. Consider the containers of
water with the surface exposed to the earth’s atmosphere.
p1
p1
p1
h
p2
h
p2
h
p2
The pressure p1 on the surface of the water is 1 atm, or 1.013 x 105 Pa. If we go down to a
depth h below the surface, the pressure becomes greater by the product of the density of
the water , the acceleration due to gravity g, and the depth h. Thus the pressure p2 at this
depth is
p2  p1  gh
In this case, p2 is called the absolute pressure. The difference in pressure between the
surface and the depth h is
p2  p1  gh
138
Chapter 11 Fluids
This difference in pressure is called the gauge pressure. Note that the pressure at any
depth does not depend of the shape of the container, only the pressure at some reference
level (like the surface) and the vertical distance below that level.
11.6 Archimedes Principle
Archimedes principle allows us to calculate the buoyant force acting on an object in a
fluid. The buoyant force is the upward force exerted by the fluid on the object in the
fluid, and is equal to the weight of the fluid which is displaced by the object. For
example, if a floating object displaces one liter of water, the buoyant force acting on the
object is equal to the weight of one liter of water, which is about 10 N.
The buoyant force acting on an object in a fluid can be found by the equation
Fbuoyant   gV fluid displaced
where  is the density of the fluid, g is the acceleration due to gravity, and V is the
volume of the displaced fluid. If the buoyant force acting on an object in a fluid is equal
to the weight of the object, the object will float.
Example 1
3m
A large container of water (ρ = 1000 kg/m3 ) contains a thin, light plate at a depth of 3 m
below the surface of the water. Neglect the mass and volume of the thin plate. The plate
can be elevated by a jack without disturbing the water in the container.
(a) What is the gauge pressure at the depth of the plate?
(b) What is the absolute pressure at the depth of the plate?
139
Chapter 11 Fluids
A solid aluminum cylinder (ρ = 2700 kg/m3 )
of radius 0.25 m and height 1 m is lowered by
a cable in the water until half the cylinder is
beneath the surface of the water where it
remains at rest.
(c) What is the tension in the cable?
3m
(d) The cylinder is then lowered onto the light plate, and the cable is removed. Find the
force exerted by the plate on the cylinder if the jack lifts the plate upward at
i. a constant speed of 2 m/s
ii. an acceleration of 1 m/s2.
3m
Solution
(a) p gauge  gh  1000 kg / m3 10 m / s 2 3 m  3x10 4 Pa






(b) p 2  p1  gh  1.013x10 5 Pa  1000 kg / m 3 10 m / s 2 3 m  1.313x10 5 Pa
(c) The tension in the cable is equal to the weight of the cylinder minus the buoyant force
acting on the cylinder.
FT  m Al g  gVdisplaced water
FT   AlV Al g  gVdisplaced water
The volume of the aluminum is
2
VAl  r 2 h   0.25 m 1m  0.20 m3
The volume of the displaced water is half of the volume of the aluminum, or 0.10 m3 .
Substituting the known values into the equation for the tension, we get
FT  4400 N
(d) i. For the jack to lift the aluminum cylinder it must apply a force equal to the apparent
weight of the cylinder.
F  mg  FBuoyant   AlV Al g  gVdisplaced water  gV  Al   water 



F  10 m / s 2 0.20 m 3 2700 kg / m 3  1000 kg / m 3
F  3400 N
140

Chapter 11 Fluids
ii. Drawing the free-body diagram for the cylinder:
FB
FP
FP  FB  mg  ma
FP   FB  mg  ma
a
where the mass of the aluminum cylinder is ρAl VAl = 540 kg.
Then
FP   gVdisplaced water  mg  ma
mg
Substituting, we get
FP  3940 N
11.8 The Equation of Continuity
Consider a fluid flowing through a tapered pipe:
A1
A2
v1
v2
The area of the pipe on the left side is A1, and the speed of the fluid passing through A1 is
v1. As the pipe tapers to a smaller area A2, the speed changes to v2. Since mass must be
conserved, the mass of the fluid passing through A1 must be the same as the mass of the
fluid passing through A2. If the density of the fluid is 1, and the density of the fluid at A2
is 2, the mass flow rate through A1 is 1A1v1, and the mass flow rate through A2 is 2A2v2.
Thus, by conservation of mass,
1 A1 v1 = 2 A2 v2
This relationship is called the equation of continuity. If the density of the fluid is the same
at all points in the pipe, the equation becomes
A1 v1 = A2 v2
The product of area and the velocity of the fluid through the area is called the volume
flow rate.
141
Chapter 11 Fluids
11.9 and 11.10 Bernoulli’s Equation and Applications of Bernoulli’s
Equation
Recall that in the absence of friction or other nonconservative forces, the total mechanical
energy of a system remains constant, that is,
U1 + K1 = U2 + K2
mgy1 + ½ mv12 = mgy2 + ½ mv22
Bernoulli’s principle states that the total pressure of a fluid along any tube of flow
remains constant. Consider a tube in which one end is at a height y1 and the other end is
at a height y2:
v2
v1
y1
y2
Let the pressure at y1 be p1 and the speed of the fluid be v1. Similarly, let the pressure at
y2 be p2 and the speed of the fluid be v2. If the density of the fluid is , Bernoulli’s
equation is
p1 
1
1
v1 2  gy1  p 2  v 2 2  gy 2
2
2
This equation states that the sum of the pressure at the surface of the tube, the dynamic
pressure caused by the flow of the fluid, and the static pressure of the fluid due to its
height above a reference level remains constant. Note that if we multiply Bernoulli’s
equation by volume, it becomes a statement of conservation of energy.
If a fluid moves through a horizontal pipe (y1 = y2), the equation becomes
1
1
p1  v1 2  p 2  v 2 2
2
2
This equation implies that the higher the pressure at a point in a fluid, the slower the
speed, and vice-versa. The equation of continuity and Bernoulli’s principle are often used
together to solve for the pressure and speed of a fluid, as the following review questions
illustrate.
142
Chapter 11 Fluids
CHAPTER 11 REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
Unless otherwise noted, use g = 10 m/s2.
1. Gauge pressure at a certain depth
below the surface of a fluid is equal to
(A) the pressure at the surface of the
fluid
(B) the difference between the absolute
pressure and the pressure at the
surface of the fluid
(C) the sum of the absolute pressure and
the pressure at the surface of the fluid
(D) the absolute pressure
(E) the density of the fluid
4. If the beaker, water, and the ball in the
water are placed on a Newton scale, the
scale will read
(A) 16 N
(B) 15 N
(C) 11 N
(D) 10 N
(E) 6 N
A1
2. The pressure at the surface of the
ocean is 1 atm (1 x 105 Pa). At what
approximate depth in the ocean water
(ρ = 1025 kg/m3) would the absolute
pressure be 2 atm?
(A) 1 m
(B) 5 m
(C) 10 m
(D) 100 m
(E) 1000 m
v1
A2
A3
v3
v2
Questions 5-6: The three sections of the
pipe shown above have areas A1, A2, and
A3. The speeds of the fluid passing
through each section of the pipe are v1,
v2, and v3, respectively. The areas are
related by A2 = 4A1 = 8A3. Assume the
fluid flows horizontally.
Questions 3 – 4: A ball weighing 6 N in
air and having a volume of 5 x 10-4 m3 is
fully immersed in a beaker of water and
rests on the bottom. The combined
weight of the beaker and water without
the ball is 10 N.
5. Which of the following is true of the
speeds of the fluid in each section in the
pipe?
(A) v3 = 2v1
(B) v3 = 8v2
(C) v2 = ½ v1
(D) v2 = 16v1
(E) v3 = 64v2
3. The buoyant force acting on the ball is
most nearly
(A) 1 N
(B) 2 N
(C) 3 N
(D) 4 N
(E) 5 N
6. Which of the following is true of the
pressures in each section of the pipe?
(A) p1 > p2 > p3
(B) p2 > p1 > p3
(C) p3 > p2 > p1
(D) p2 > p3 > p1
(E) p1 > p3 > p2
143
Chapter 11 Fluids
Questions 8-9:
v1=2m/s
7. The large container above is filled
with water. Three small spouts near the
bottom of the container are of equal size
and are initially corked. If the corks are
removed from the spouts, which of the
following best represents the path of the
water stream from each spout?
(A)
v2=6m/s
A glass pipe containing two vertical
tubes of equal size is filled with water so
that the level of the water is the same in
the two pipes. Air (ρ = 1.3 kg/m3) is
blown across the end of the left tube
with a speed of 2 m/s and air is blown
across the right tube with a speed of 6
m/s.
8. Which of the following statements is
true of the water in the pipe as the air is
blown across the vertical tubes?
(A) The water level in each pipe does
not change.
(B) The water level on the left rises and
the water level on the right is lowered.
(C) The water level on the left is lowered
and the water level on the right rises.
(D) The water level on both sides rises.
(E) The water level on both sides is
lowered.
(B)
(C)
9. The magnitude of the difference in
pressure between the two ends of the
pipe is most nearly
(A) 40 Pa
(B) 32 Pa
(C) 24 Pa
(D) 21 Pa
(E) 16 Pa
(D)
(E)
144
Chapter 11 Fluids
Free Response Question
Directions: Show all work in working the following question. The question is worth 15
points, and the suggested time for answering the question is about 15 minutes. The parts
within a question may not have equal weight.
1. (15 points)
y1
b
c
a
y2
h
d
Note: Figure not drawn to scale.
A cylindrical-shaped pipe can carry water from a very large elevated container on the left
to a lower container on the right. The area of the wider portion of the pipe containing the
point b has a cross-sectional area Ab = 7.80 x 10-3 m2 , and the narrower section of the
pipe containing both points c and d has a cross-sectional area of Ac = 3.14 x 10-4 m2.
Point C is at a height of y2 = 2 m above point d. A water valve closes the elevated
container at point a, and thus there is initially only water in the upper container, and none
in the pipe. The rectangular block in the lower container above has dimensions 10 cm x 3
cm x 3 cm and mass 0.075 kg, and it rests on the bottom of the lower container before
any water enters the lower container.
(a) If the pressure at the surface of the water is 1 atm, what is the absolute pressure at
point a which is at a depth of y1 = 2 meters below the surface of the water in the
tank?
145
Chapter 14 The Ideal Gas Law and Kinetic Theory
The valve at point a is opened to create an opening equal to the area of the pipe
containing the point b so that water flows from the elevated container through the pipe,
and into the lower container.
(b) Consider the pressure at points band c. At which of these points is the pressure the
least? Justify your answer.
(c) If the speed of the water at point b is vb = 6 m/s, what is the speed of the water at
point c?
(d) Determine vd, the speed at which the water initially enters the lower container.
(e) As the water level rises in the lower container, the block eventually begins to
float. What is the height h of the water level at the instant the block is lifted off
the bottom of the container, that is, the block just begins to float?
y1
b
c
a
y2
h
d
158
Chapter 14 The Ideal Gas Law and Kinetic Theory
ANSWERS AND EXPLANATIONS TO CHAPTER 11 REVIEW QUESTIONS
Multiple Choice
1. B
p gauge  p absolute  p surface  gh
2. C
The gauge pressure is the difference between the absolute pressure and the pressure at the
surface of the water:
p gauge  pabsolute  psurface  2atm  1atm  1atm  1 x105 Pa
p gauge  gh
1 x 10 5 Pa
h

 10 m
g
1025 kg / m 3 10 m / s 2
p gauge



3. E
FB  gVdisp fluid  1000 kg / m3 10 m / s 2 5 x10 4 m3  5 N




4. A
The scale will read the actual weight of the beaker, the water, and the ball, since the
buoyant force is an internal force as far as the scale is concerned.
Weight on scale = 10 N + 6 N = 16 N
5. A
According to the equation of continuity, the speed of a fluid through a pipe is inversely
proportional to the area of the pipe. Since 4A1 = 8A3, 8v1 = 4v3, or v3 = 2v1.
6. B
According to Bernoulli’s principle, the higher the speed in a pipe, the lower the pressure
of the fluid. Since v3 > v1 > v2, then p2 > p1 > p3.
7. D
The lowest spout has the highest pressure since it is at the greatest depth. Thus, the lowest
spout will project the water the farthest.
8. C
The higher the speed of the air across the opening of a vertical pipe, the lower the
pressure in the pipe. Thus, the water in the pipe on the right will rise to fill the space and
the water in the pipe on the left will be lowered.
159
Chapter 14 The Ideal Gas Law and Kinetic Theory
9. D
If we neglect the small difference water level between the pipes, the Bernoulli equation
1
1
becomes p1  v1 2  p 2  v 2 2 . Solving for the pressure difference, we get
2
2
p1  p 2 
1
1
v2 2  v1 2  21 Pa
2
2
Free Response Question Solution
(a) 3 points
pa  psurface  gh  1.013x105 Pa  1000 kg / m3 10 m / s 2 2 m  1.213x105 Pa



(b) 3 points
The equation of continuity states that the speed in a pipe is inversely proportional to the
area of the pipe:
Ab vb = Ac vc
Since the area at b is greater than the area at c, the speed at c is greater than the speed at
b. According to the Bernoulli equation, a higher speed at a point indicates a lower
pressure at that point. Thus, the pressure at point c is a lower than at point b.
(c) 3 points
Ab vb  Ac vc
7.80 x10
3



m 2 6 m / s   3.14 x 10  4 m 2 vc
vc  149 m / s
(d) 2 points
As the water enters the lower container at point d it must have the same speed as the
water at point c. The water does not separate and is not compressed as it flows through
the pipe from point c to point d, and thus keeps a constant speed between the two points.
(e) 4 points
As the lower container fills with water, there is a height h at which the water will cause
the rectangular block to float. When the water reaches this height, the buoyant force
acting on the block is just equal to the weight of the block:
FB  mg
gVdisp water  mg
g lwh   mg
h
0.075 kg
m

 0.083 m  8.3 cm
lw 1000 kg / m 3 0.03 m 0.03 m 


Chapter 14
160
Chapter 14 The Ideal Gas Law and Kinetic Theory
THE IDEAL GAS LAW AND KINETIC THEORY
PREVIEW
Kinetic molecular theory involves the study of matter, particularly gases, as very small
particles in constant motion. Because of the motion of the particles, an ideal gas has
internal energy that can be transferred. We study gases by relating their pressure, volume,
number of moles, and temperature in the ideal gas law.
The content contained in sections 1, 2, 3, and 5 of chapter 14 of the textbook is included
on the AP Physics B exam.
QUICK REFERENCE
Important Terms
atomic mass unit
one-twelfth the mass of a carbon-12 atom
ideal gas law
the law which relates the pressure, volume, number of moles, and temperature of
an ideal gas
internal energy
the sum of the potential and kinetic energy of the molecules of a substance
kinetic theory of gases
the description of matter as being made up of extremely small particles which are
in constant motion
mole
one mole of a substance contains Avogadro’s number (6.02 x 1023) of molecules
or atoms
161
Chapter 14 The Ideal Gas Law and Kinetic Theory
Equations and Symbols
PV  nRT
PiVi  Pf V f (constant temperatur e)
where
P = pressure
V = volume
n = number of moles
R = universal gas constant
= 8.31 J / (mol K)
KEavg = average kinetic energy of
molecules
vrms = root-mean-square speed
kB = Boltzmann constant
= 1.38 x 10-23 J/K
T = Kelvin temperature
M = molecular mass
μ = mass of molecule
Vi Vf
(constant pressure)

Ti Tf
KE avg 
v rms 
3
1
2
mvrms  k B T
2
2
v avg 2

3k B T
3RT


M
Ten Homework Problems
Chapter 14 Problems 9, 10, 11, 14, 24, 26, 28, 29, 34, 52
DISCUSSION OF SELECTED SECTIONS
14.1 The Mole, Avogadro’s Number, and Molecular Mass
When we are dealing with small particles like atoms and molecules, it is convenient to
express their masses in atomic mass units (u) rather than kilograms. Atomic mass units
and kilograms are related by the conversion 1 u = 1.6605 x 10-27 kg, which is
approximately the size of a proton. When we buy 12 eggs we say we have a dozen eggs,
but if we have 6.022 x 1023 atoms, we say we have a mole of those atoms. In other words,
a mole of a substance is Avogadro’s number (6.022 x 1023) of atoms or molecules of that
substance.
14.2 The Ideal Gas Law
All gases display similar behavior. When examining the behavior of gases under varying
conditions of temperature and pressure, it is most convenient to treat them as ideal gases.
An ideal gas represents a hypothetical gas whose molecules have no intermolecular
forces, that is, they do not interact with each other, and occupy no volume. Although
gases in reality deviate from this idealized behavior, at relatively low pressures and high
temperatures many gases behave in nearly ideal fashion. Therefore, the assumptions used
for ideal gases can be applied to real gases with reasonable accuracy.
162
Chapter 14 The Ideal Gas Law and Kinetic Theory
The state of a gaseous sample is generally defined by four variables:
 pressure (p),
 volume (V),
 temperature (T), and
 number of moles (n),
though as we shall see, these are not all independent. The pressure of a gas is the force
per unit area that the atoms or molecules exert on the walls of the container through
collisions. The SI unit for pressure is the pascal (Pa), which is equal to one newton per
meter squared. Sometimes gas pressures are expressed in atmospheres (atm). One
atmosphere is equal to 105 Pa, and is approximately equal to the pressure the earth’s
atmosphere exerts on us each day. Volume can be expressed in liters (L) or cubic meters
(m3), and temperature is measured in Kelvins (K) for the purpose of the gas laws. Recall
that we can find the temperature in K by adding 273 to the temperature in Celsius. Gases
are often discussed in terms of standard temperature and pressure (STP), which refers to
the conditions of a temperature of 273 K (0C) and a pressure of 1 atm.
These four variables are related to each other in the ideal gas law:
PV  nRT
where R is a constant known as the universal gas constant = 8.31 J / (mol K). If the
number of moles of a gas does not change during a process, then n and R are constants,
and we can write the equation as the combined gas law:
P1V1 P2V2

T1
T2
where a subscript of 1 indicates the state of the gas before something is changed, and the
subscript 2 indicates the state of the gas after something is changed.
Example 1
A
F
Fig. I
Fig. II
Fig. III
A cylinder is closed at one end with a piston which can slide to change the closed volume
of the cylinder. When the piston is at the end of the cylinder, as in Figure III, the volume
of the cylinder is 1.0 liter. The area of the piston is 0.01 m2. The piston is positioned at
half the length of the cylinder in Fig. I, and the cylinder is filled with an ideal gas. The
force F necessary to hold the piston in this position is 10 N, and the temperature of the
gas is 50˚ C.
163
Chapter 14 The Ideal Gas Law and Kinetic Theory
(a) Determine the following for the gas in Fig. I:
i. the volume of the gas
ii. the pressure of the gas
A force is applied to the piston so that it is now positioned at one-third the length of the
cylinder from its closed end, but the temperature of the gas remains at 50˚ C.
(b) Determine the following for the gas in Fig. II:
i. the volume of the gas
ii. the pressure of the gas
(c) The temperature of the gas is raised to 80˚ C between Fig. II and Fig. III. Determine
the pressure of the gas in Fig. III.
Solution
(a) i. The volume of the gas is half of the full cylinder, or 0.5 liter.
10 N
F
 100 Pa
ii. PI  
A 0.1 m 2
(b) i. The volume of the gas is one-third of the full cylinder, or 0.33 liter.
ii. At a constant temperature, the pressure and volume of the gas are inversely
proportional according to Boyle’s law:
PIVI  PIIVII
100 Pa(0.5 l )  PII 0.33 l 
PII  151.5 Pa
(c) Converting Celsius degrees to Kelvins:
TII  50 C  273  323 K
TIII  80 C  273  353 K
PIIVII PIIIVIII

TII
TIII
151.5 Pa 0.33 l  
323 K
PIII 1.0 l 
353 K
PIII  54.7 Pa
14.3 Kinetic Theory of Gases
As indicated by the gas laws, all gases show similar physical characteristics and behavior.
A theoretical model to explain why gases behave the say they do was developed during
the second half of the 19th century. The combined efforts of Boltzmann, Maxwell, and
others led to the kinetic theory of gases, which gives us an understanding of gaseous
behavior on a microscopic, molecular level. Like the gas laws, this theory was developed
in reference to ideal gases, although it can be applied with reasonable accuracy to real
gases as well.
164
Chapter 14 The Ideal Gas Law and Kinetic Theory
The assumptions of the kinetic theory of gases are as follows:

Gases are made up of particles whose volumes are negligible compared to the
container volume.
 Gas atoms or molecules exhibit no intermolecular attractions or repulsions.
 Gas particles are in continuous, random motion, undergoing collisions with other
particles and the container walls.
 Collisions between any two gas particles are elastic, meaning that no energy is
dissipated and kinetic energy is conserved.
 The average kinetic energy of gas particles is proportional to the absolute (Kelvin)
temperature of the gas, and is the same for all gases at a given temperature. As
listed in the list of equations, the average kinetic energy of each molecule is
related to Kelvin temperature T by the equation
3
K avg  k BT , where kB is the Boltzmann constant, 1.38 x 10-23 J/K. The root-mean
2
3k B T
square speed of each molecule can be found by v rms 
, where  is the mass of

each molecule. This equation is very seldom used on the AP Physics B exam, and is
provided on the exam if needed.
Example 2
The temperature of an ideal gas is 60˚ C.
(a) Find the average kinetic energy of the molecules of the gas.
(b) On the axes below, sketch a graph of
i. average kinetic energy Kavg vs Kelvin temperature T
ii. root-mean-square speed vrms of each molecule in the gas vs. Kelvin temperature T.
Kavg
vrms
T
T
165
Chapter 14 The Ideal Gas Law and Kinetic Theory
Solution
(a)
K avg


3
3
k B T  1.38 x 10  23 J / K 60 C  273
2
2
 21
 6.89 x 10 J
K avg 
(b) The average kinetic energy of each molecule is directly proportional to the Kelvin
temperature of the gas, and vrms is proportional to the square root of the Kelvin
temperature of the gas:
Kavg
vrms
T
T
CHAPTER 14 REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
1. A mole is to Avogadro’s number as
(A) kinetic energy is to temperature
(B) atomic mass unit is to kg
(C) gas is to liquid
(D) pressure is to volume
(E) decade is to ten years
3. A sample of argon occupies 50 liters
at standard temperature. Assuming
constant pressure, what volume will
argon occupy if the temperature is
doubled?
(A) 25 liters
(B) 50 liters
(C) 100 liters
(D) 200 liters
(E) 2500 liters
2. Which of the following is NOT true of
an ideal gas?
(A) Gas molecules have no
intermolecular forces.
(B) Gas particles are in random motion.
(C) Gas particles have no volume.
(D) The collisions between any two gas
particles are elastic.
(E) The average kinetic energy of the
gas molecules is proportional to the
temperature in Celsius degrees.
4. What is the final pressure of a gas that
expands from 1 liter at 10C to 10 liters
at 100C if the original pressure was 3
atmospheres?
(A) 0.3 atm
(B) 0.4 atm
(C) 3 atm
(D) 4 atm
(E) 30 atm
166
Chapter 14 The Ideal Gas Law and Kinetic Theory
5. Which of the following pressure vs.
volume graphs best represents how
pressure and volume change when
temperature remains constant?
(A)
6. Which of the following volume vs.
temperature graphs best represents how
volume changes with Kelvin temperature
if the pressure remains constant?
P
(A)
V
V
T
(B)
P
(B)
V
V
T
(C)
P
(C)
V
V
(D)
T
P
(D)
V
V
T
(E)
P
(E)
V
V
T
167
Chapter 15 Thermodynamics
Free Response Question
Directions: Show all work in working the following question. The question is worth 10
points, and the suggested time for answering the question is about 10 minutes. The parts
within a question may not have equal weight.
1. (10 points)
A special balloon contains 3 moles of an ideal gas and has an initial pressure of 3 x 105
Pa and an initial volume of 0.015 m3.
(a) Determine the initial temperature of the gas.
(b) The pressure in the balloon is changed in such a way as to increase the volume of the
balloon to 0.045 m3 but the temperature is held constant. Determine the pressure of the
gas in the balloon at this new volume.
(c) If the balloon contracts to one-third of its initial volume, and the pressure is increased
to twice its initial value, describe the change in temperature that would have to take place
to achieve this result.
ANSWERS AND EXPLANATIONS TO CHAPTER 14 REVIEW QUESTIONS
Multiple Choice
1. E
A mole is defined as a certain number of things (6 x 1023), and a decade is a certain
number of years (10).
2. E
All of the statements are true, except for “Celsius” should be replaced with “Kelvin”.
3. C
With constant pressure, volume and temperature are proportional to each other, so twice
the temperature would result in twice the volume.
167
Chapter 15 Thermodynamics
4. B
First, we must convert Celsius to Kelvin:
T1 = 10˚ C + 273 = 283 K
T2 = 100˚ C + 273 = 373 K
P1 = 3 atm
V1 = 1 l
V2 = 10 l
P1V1 P2V2

T1
T2
Solving for P2 and substituting, we get P2 = 0.4 atm.
5. E
Pressure and volume are inversely proportional to each other at constant temperature.
6. A
Volume and temperature are proportional to each other at constant temperature.
Free Response Question Solution
(a) 4 points
PV  nRT



3x10 5 Pa 0.015 m 3
PV

180.5 K
nR 3 moles 8.31 J / mol K 
(b) 3 points
For a constant temperature,
P1V1  P2V2
T
3x10
5


Pa (0.015 m 3 )  P2 0.045 m 3

P2  1x10 5 Pa
(c) 3 points
P1V1 P2V2

T1
T2
In order to keep the ratios constant on both sides,
1 1
  P  V
P1V1  3  1  2  1

T1
1
 T1
6
Thus, the temperature would have to decrease to 1/6 its initial value for these changes in
the pressure and volume to take place.
168
Chapter 15 Thermodynamics
Chapter 15
THERMODYNAMICS
PREVIEW
Thermodynamics is the study of heat transfer. Two of the laws that govern the flow of
heat in or out of a system are called the first and second laws of thermodynamics. These
laws relate to conservation of energy, the direction of heat flow from one system to
another, and the amount of entropy (disorder) in a system. Often we analyze the energy
transfer of a system using a pressure-volume (PV) diagram.
The content contained in sections 1 – 5, 7 – 13 of chapter 15 of the textbook is included
on the AP Physics B exam.
QUICK REFERENCE
Important Terms
adiabatic
the expansion or compression of a gas without a gain or loss of heat.
Carnot principle
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.
entropy
the measure of the amount of disorder in a system
first law of thermodynamics
the heat lost by a system is equal to the heat gained by the
system minus any work done by the system; conservation of energy
heat engine
device which changes internal energy into mechanical work
isobaric
any process in which the pressure of a gas remains constant
isochoric (or isovolumetric)
any process in which the volume of a gas remains constant
isothermal
any process in which the temperature of a gas remains constant
169
Chapter 15 Thermodynamics
pressure-volume (PV) diagram
a graph of pressure vs. volume which gives an indication of the work done by or
on a system, and the energy transferred during a process
reversible process
a process in which both the system and its environment can be returned to exactly
the states they were in before the process occurred
second law of thermodynamics
heat flows naturally from a region of higher temperature to a lower
temperature;all natural systems tend toward a state of higher disorder
thermodynamics
the study of heat transfer
Equations and Symbols
U  U f  U i  Q  W
where
W  PV  P(V f  Vi )
ΔU = change in internal energy
Q = heat
W = work
P = pressure
V = volume
T = Kelvin temperature
R = universal gas constant
= 8.31 J / (mol K)
e = efficiency
QH = input heat
TH = temperature of the hot reservoir
TC = temperature of the cold reservoir
W
e
QH
e
TH  TC
T
 1 C
TH
TH
Ten Homework Problems
Chapter 15 Problems 3, 5, 7, 8, 9, 12, 24, 27, 43, 58
DISCUSSION OF SELECTED SECTIONS
15.1 – 15.2 Thermodynamics
Thermodynamics is the study of heat flow and the work done on or by a system. There
are four laws of thermodynamics, of which two appear on the AP Physics B exam by
name (1st and 2nd). However, the concepts involved in the other two (0th and 3rd) are
important to the understanding of thermodynamics as well. The zeroth law of
thermodynamics states that if two systems are in equilibrium, that is, they have the same
temperature, there is no net heat flow between them. Further, if systems A and B are each
in thermal equilibrium with system C, they must be in equilibrium with each other.
170
Chapter 15 Thermodynamics
The third law of thermodynamics states that it is impossible to reach a temperature of
absolute zero.
15.3 The First Law of Thermodynamics
As we’ve discussed in previous chapters, energy can be transformed in many forms, but
is conserved, that is, the total amount of energy must remain constant. This is true of a
system only if it is isolated. Since energy can neither go in nor go out, it has to be
conserved. A system can exchange energy with its surroundings in two general ways: as
heat or as work. The first law of thermodynamics states that the change in the internal
energy ΔU of a system is equal to the heat Q added to the system plus the work W
done ON the system:
U = Q + W
On the AP Physics B exam, if work is done ON a system, the system gains energy
and W is positive.
If work is done BY the system, the system loses energy and W is negative.
This convention is consistent with the energy transfers you studied in earlier chapters. If
you do work on an object, you do positive work on that object. If the object does work on
something else, we say the object has done negative work.
Note that in your textbook, W is defined as the work done ON, rather than BY the system,
in which case the equation is written as U = Q – W, and the work done BY the system is
considered positive. Regardless of which convention is used, if work is done ON a
system, its energy would increase. If work is done BY the system, its energy would
decrease. Work is generally associated with movement against some force. For ideal gas
systems, for example, expansion against some external pressure means that work is done
BY the system, while compression implies work being done ON the system.
For example, if a system has 60 J of heat added to it, resulting in 20 J of work being done
BY the system, the change in internal energy of the system is U = Q – W = 60 J – 20 J =
40 J.
If 60 J of heat is added to a system AND work of 20 J is done ON the system, the internal
energy of the system would increase all the more:
Change in internal energy ΔU = Heat Q added + work W done ON the system
= 60 J + 20 J = 80 J
If heat is added to a system and no work is done, then the heat lost by one element in the
system is equal to the heat gained by another element. For example, if sample of metal is
heated and then dropped into a beaker containing water, then the first law implies
Q lost by the metal = Q gained by the water
171
Chapter 15 Thermodynamics
15.4 - 15.5 Thermal Processes, and Thermal Processes That Utilize an
Ideal Gas
We can study the changes in pressure, volume, and temperature of a gas by plotting a
graph of pressure vs. volume for a particular process. We call this graph a PV diagram.
For example, let’s say that a gas starts out at a pressure of 4 atm and a volume of 2 liters,
as shown by the point A in the PV diagram below:
P(atm)
A
B
4
2
C
V(liters)
2
4
If the pressure of the gas remains constant but the volume changes to 4 liters, then we
trace a line from point A to point B. Since the pressure remains constant from A to B, we
say that the process is isobaric. The if we decrease the pressure to 2 atm but keep the
volume constant, we trace a line from B to C. This constant-volume process is isochoric,
or isovolumetric. If we want the gas to return to its original state without changing
temperature, we must trace a curve from point C to A along an isotherm. Note that an
isotherm on a PV diagram is not a straight line. The work done during the process ABCA
is the area enclosed by the graph, since W = PV. In this case the work done on the
system is positive.
Any process which is done without the transfer of heat is called an adiabatic process.
Since there is no heat lost or gained in an adiabatic process, then the first law of
thermodynamics states that the change in internal energy of a system is simply equal to
the work done on or by the system, that is, U = W.
172
Chapter 15 Thermodynamics
The processes discussed above are summarized in the table below:
Process
Definition
PV diagram
P
isobaric
constant pressure
V
P
isochoric
constant volume
V
isothermal
P
constant temperature
V
adiabatic
Example 1
no heat added or taken
away (ΔU = W)
P(x105 Pa)
A
B
8
C
4
E
D
V(m3)
2
4
Four separate processes (AB, AC, AD, and AE) are represented on the PV diagram above
for an ideal gas.
(a) Determine the work done by the gas during process AB.
(b) Determine the work done on the gas during process AE.
173
Chapter 15 Thermodynamics
(c) Is work done on the gas during any other process on the diagram? If so, identify
which ones and explain how you know work is done.
(d) Identify the process which could be
i. isothermal ________
ii. adiabatic ________
(e) Calculate the heat lost during the process AE.
Solution
5
3
3
5
(a) W AB  PV   8 x10 Pa 4 m  2 m  1.6 x10 J
(b) There is no work done during the process AE since there is no change in volume.
(c) There is work done during processes AB, AC, and AD, since there is a change in
volume in each of these processes.
(d) i. Process AD is isothermal, since the pressure and change inversely proportionally,
and the temperature remains constant.
ii. Process AC could be adiabatic, since the pressure, volume, and temperature change,
indicating that heat may not added or removed.
(e) The heat lost during process AE is equal to the change in energy of the system during
the process.
Q  P V   4 x10 5 Pa  8x10 5 Pa 2 m 3   8x10 5 J






15.7 The Second Law of Thermodynamics
Entropy S is a measure of the disorder, or randomness, of a system. The greater the
disorder of a system, the greater the entropy. If a system is highly ordered, like the
particles in a solid, we say that the entropy is low. At any given temperature, a solid will
have a lower entropy than a gas, because individual molecules in the gaseous state are
moving randomly, while individual molecules in a solid are constrained in place.
Entropy is important because it determines whether a process will occur spontaneously.
The second law of thermodynamics states that all spontaneous processes proceeding in
an isolated system lead to an increase in entropy.
In other words, an isolated system will naturally pursue a state of higher disorder. If you
watch a magician throw a deck of cards into the air, you would expect the cards to fall to
the floor around him in a very disorderly manner, since the system of cards would
naturally tend toward a state of higher disorder. If you watched a film of a magician, and
his randomly placed cards jumped off the floor and landed neatly stacked in his hand, you
would believe the film is running backward, since cards do not seek this state of order by
themselves. Thus, the second law of thermodynamics gives us a direction for the passage
of time.
174
Chapter 15 Thermodynamics
15.8 Heat Engines
A heat engine is any device that uses heat to perform work. There are three essential
features of a heat engine:



Heat is supplied to the engine at a high temperature from a hot reservoir.
Part of the input heat is used to perform work.
The remainder of the input heat which did not do work is exhausted into a cold
reservoir, which is at a lower temperature than the hot reservoir.
Work
Input Heat
(High Temp
Reservoir)
Exhausted Heat
(Low Temp Reservoir)
In the diagram above, heat is used to do work in lifting the block which is sitting on the
piston in the gas-filled cylinder. Any heat left over after work is done is exhausted into
the low temperature reservoir. This diagram is used in Example 2 below.
The percent efficiency % e of the heat engine is equal to the ratio of the work done to the
amount of input heat:
%e 
Work
x 100
QHot
Example 2
Fig. I
Fig. II
Fig III
In the figures above, heat is added to the cylinder in Fig. I where the gas occupies half the
total volume of the cylinder, raising it to the top of the cylinder in Fig. II. Heat is then
removed and the block is lowered so that the gas occupies ¼ of the total volume of the
cylinder. The total volume of the cylinder is 2 x 10-3 m3, and the area of the piston is
0.05 m2. The mass of the block is 2.0 kg.
(a) If the block is at rest in Fig. I, determine the pressure of the gas in the cylinder.
175
Chapter 15 Thermodynamics
(b) It is determined that the efficiency of this heat engine between Fig. I and Fig. II is
60%. How much heat was added to the cylinder in Fig. I to cause piston to rise to the
level in Fig. II?
(c) Is the process between Fig. I and Fig. II isothermal, isobaric, or isochoric? Explain.
(d) If the temperature of the gas in Fig. I is 40º C, what is the temperature of the gas in
Fig. II?
(e) i. Between Figs. II and III,is the system acting as a heat engine or a refrigerator?
Explain.
ii. Determine the temperature of the gas in Fig. III
Solution
(a) The pressure the gas applies to the piston, block, and atmosphere is equal and
opposite to the pressure the piston, block, and atmosphere apply to the gas. Neglecting
the mass of the piston, we can write
2 kg 10 m / s 2  1x10 5 Pa  1.004 x10 5 Pa
F
mg
P   1 atm 
1 atm 
A
A
0.05 m 2

 

(b) The work done in lifting the block is
1

W  PV  1.004 x 10 5 Pa  2 x10 3 m 3   1.004 x10 2 J
2

This work done represents 60% of the heat input to the gas. Thus,
W 1.004 x10 2 J
Qinput 

 167.3J
%e
0.60
and the heat exhausted is 167.3 J – 100.4 J = 66.9 J.
(c) The process between Figs. I and II is isobaric, since the pressure the block and
atmosphere apply to the piston and gas does not change during the process.
(d) For constant pressure, volume and Kelvin temperature are proportional by the
combined gas law.
TI = 40º C +273 = 313 K
VI VII

TI TII
1x10 3 m 3 2 x10 3 m 3

313 K
TII
TII  616 K
For constant pressure, if the volume of a gas doubles, the temperature also doubles.
176
Chapter 15 Thermodynamics
(e) i. The system is acting as a refrigerator, removing heat from the cylinder and reducing
the volume of the gas.
ii. Since the pressure remains constant between Figs. II and III,
VII VIII

TII TIII
2 x10 3 m 3 0.5 x10 3 m 3

616 K
TIII
TIII  154 K
15.9 Carnot’s Principle and the Carnot Engine
The French engineer Sadi Carnot suggested that a heat engine has maximum efficiency
when the processes within the engine are reversible, that is, both the system and its
environment can be returned to exactly the states they were in before the process
occurred. In other words, there can be no dissipative forces, like friction, involved in the
Carnot cycle of an engine for it to operate at maximum efficiency. All spontaneous
processes, such as heat flowing from a hot reservoir to a cold reservoir, are not reversible,
since work would have to be done to force the heat back to the hot reservoir from the cold
reservoir (a refrigerator), thus changing the environment by using some of its energy to
do work. A reversible engine is called a Carnot engine.
CHAPTER 15 REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
1. The first law of thermodynamics is a
form of
(A) the law of conservation of energy.
(B) the law of specific heat.
(C) the ideal gas law.
(D) the law of entropy.
(E) the law of conservation of
temperature.
2. A system has 60 J of heat added to it,
resulting in 15 J of work being done by
the system, and exhausting the
remaining 45 J of heat. What is the
efficiency of this process?
(A) 100%
(B) 60%
(C) 45%
(D) 25%
(E) 15%
177
Chapter 15 Thermodynamics
3. The law of entropy states that
(A) heat always flows spontaneously
from a colder body to a hotter one.
(B) every natural system will tend
toward lower entropy.
(C) heat lost by one object must be
gained by another.
(D) the specific heat of a substance
cannot exceed a certain value.
(E) every natural system will tend
toward disorder.
6. If 2 J of heat is added during process
AB, and the total amount of work done
in the cycle ABCA is 3 J, how much
heat is added or removed during process
BCA?
(A) 2 J of heat is added
(B) 2 J of heat is removed
(C) 1 J of heat is added
(D) 1 J of heat is removed
(E) 3 J of heat is added
7. Which of the following statements
about a Carnot engine is true?
(A) Any Carnot engine has an efficiency
of 100%.
(B) Irreversible engines have the greatest
possible efficiency.
(C) Heat can spontaneously flow from a
cold reservoir to a hoter reservoir.
(D) If a process is reversible, the
efficiency of an engine is maximum.
(E) All engines are reversible.
Questions 4, 5, 6: Gas in a chamber
passes through the cycle ABCA as
shown below.
P
B
3
2
1
C
A
1
2
3
4
8. Which of the following best illustrates
the second law of thermodynamics?
(A) a refrigerator cools warm food
(B) a piston in a cylinder is forced
upward by expanding gas in the
cylinder
(C) your bedroom gets cleaner as the
week progresses
(D) a tadpole grows into a frog
(E) cards thrown from the top of a
stairway land in a stack in numerical
order.
V
4. In which process is no work done on
or by the gas?
(A) AB
(B) AC
(C) BC
(D) CB
(E) CA
5. At which point is the temperature of
the gas the highest?
(A) A
(B) B
(C) C
(D) A and B
(E) the temperature is the same at points
A, B, and C.
178
Chapter 15 Thermodynamics
Free Response Question
Directions: Show all work in working the following question. The question is worth 15
points, and the suggested time for answering the question is about 15 minutes. The parts
within a question may not have equal weight.
1. (15 points)
A cylinder contains 3 moles of a monatomic gas that is initially at a state A with a
pressure of 8 x 105 Pa and a volume of 2 x 10 –3 m3. The gas is then brought isochorically
to state B, where the pressure is 2 x 105 Pa. The gas is then brought isobarically to state C
where its volume is 4 x 10 –3 m3 and its temperature is 300 K. The gas is then brought
back isothermally to state A.
(a) On the axes below, sketch a graph of the complete cycle, labeling points A, B, and C.
179
Chapter 12 Temperature and Heat
(b) Determine the work done by the gas during the process ABC.
(c) Determine the change in internal energy during the process ABC.
(d) Determine the temperature of the gas at state B.
(e) State whether this device is a heat engine or a refrigerator, and justify your answer.
ANSWERS AND EXPLANATIONS TO CHAPTER 15 REVIEW QUESTIONS
Multiple Choice
1. A
The 1st law simply states that the energy of a thermodynamic system is constant.
2. D
%e 
W
15 J

 25%
QH 60 J
3. E
The law of entropy states that any system will spontaneous ly go from a state of order to
disorder.
4. A
No work is done on or by the gas in process AB since there is no change in volume.
5. B
The temperature is highest at point B, since all of the energy gained is a result of heat
added to the gas without changing the volume.
6. C
2 J of heat is added in the process AB, and since 3J of work is done in the cycle, 1 J of
additional heat must have been added.
7. D
A reversible engine is always more efficient than an irreversible engine, since more
energy is lost in an irreversible engine.
8. B
In all the other choices, each system is going from a state of disorder to order.
150
Chapter 12 Temperature and Heat
Free Response Question Solution
(a) 4 points
A
B
C
(b) 3 points
For the processes ABC, work is only done in the process BC:



W  PV  2 x10 5 Pa 2 x10 3 m3  4 x10 2 J
Since work is being done BY the gas, the work is considered negative on the AP Physics
B exam, and thus we would write W = - 4 x 102 J.
(c) 3 points
Heat Q is removed in process AB and work is done BY the gas in process BC, both of
which reduces the internal energy of the gas:
U  Q  W  P V   P V   2 x105 Pa  8x105 Pa 2 x10 3 m3  400 J  1600 J

 


(d) 3 points
The combined gas law gives
PAV A PBVB

TA
TB
Since the temperature at A is the same as at C, TA = 300 K. From the graph,
8 x10 5 Pa 2 x10 3 m 3
2 x10 5 Pa 2 x10 3 m 3

300 K
TB


 


TB  75 K
(e) 2 points
This device is a refrigerator, since it removes heat from the system.
Chapter 12
TEMPERATURE AND HEAT
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Chapter 12 Temperature and Heat
PREVIEW
The total internal energy of the molecules of a substance is called thermal energy. The
temperature of a substance is a measure of the average kinetic energy of the molecules in
the substance, and gives an indication of how hot or cold the substance is relative to some
standard. The energy transferred between two substances because of a temperature
difference is called heat. Many substances expand when heated.
The content contained in sections 1, 2, 4, 6 – 8, and 11 (not including volume expansion)
of chapter 12 of the textbook is included on the AP Physics B exam.
QUICK REFERENCE
Important Terms
absolute zero
the lowest possible temperature, at which all molecular motion would cease and a
gas would have no volume.
calorie
the amount of heat required to raise the temperature of one gram of water by one
Celsius degree
Celsius (C)
temperature scale in which the freezing point of water is 0 and the boiling point of
water is 100 .
heat
the energy which is transferred from one body to another because of a
temperature difference
Kelvin (absolute) temperature scale
scale in which zero Kelvins is defined as absolute zero, the temperature at which
all molecular motion ceases
temperature
the property of a body which indicates how hot or cold a substance is with respect
to a standard
thermal energy
the sum of the internal potential and kinetic energy of the random motion of the
molecules making up an object
thermal equilibrium
state between two or more objects in which temperature doesn’t change
thermal expansion
increase in length or volume of a material due to an increase in temperature
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Chapter 12 Temperature and Heat
Equations and Symbols
TK  TC  273.15
where
L  Lo T
TK = Kelvin temperature
TC = Celsius temperature
ΔT = change in temperature
ΔL = change in length
Lo = initial length
α = coefficient of linear expansion
Ten Homework Problems
Chapter 12 Conceptual Questions 1, 3, 5, 11, 15, Problems 9, 14, 17, 19, 21
DISCUSSION OF SELECTED SECTIONS
12.1 and 12.2 Common Temperature Scales and The Kelvin
Temperature Scale
Temperature is the measure of how hot or cold a substance is relative to some standard. It
is the measure of the average kinetic energy of the molecules in a substance. The two
temperature scales used most widely in scientific applications is the Celsius scale and the
Kelvin scale. The only difference between them is where each starts. On the Celsius
scale, the freezing point of water is 0 C, and the boiling point of water (at standard
pressure) is 100 C. The Kelvin scale has temperature units which are equal in size to the
Celsius degrees, but the temperature of 0 Kelvin is absolute zero, defined as the
temperature at which all molecular motion in a substance ceases. Zero Kelvin is equal to
- 273.15 C, so we can convert between the Kelvin scale and the Celsius scale by the
equation
TK = TC + 273
Note that we have rounded 273.15 to 273. The boiling point of water in Kelvins would be
TK = 100 C + 273 = 373 K.
12.4 Linear Thermal Expansion
When a solid is heated, it typically expands. Different substances expand at different
rates, which is why you might heat the lid of a jar when the lid is too tight. The metal lid
will expand more than the glass jar when it is heated, making it easier to loosen. Solids
undergo two types of expansion when heated: linear thermal expansion, which is the
increase in any one dimension of the solid, and volume thermal expansion, which results
in an increase in the volume of the solid. Volume expansion is not typically covered on
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Chapter 12 Temperature and Heat
the AP Physics B exam. In the case of linear expansion, the change in length L is
proportional to the original length Lo and the change in temperature T of the solid:
L   Lo T
where  is the coefficient of linear expansion.
Example 1 The ends of a copper bar and a steel bar, each of length 0.20 m, are separated
by a gap of 0.50 mm, as shown. The other ends of the bars are attached to a rigid frame
which does not expand significantly when heated. If the two bars are heated from 0 C to
100 C, determine whether or not the bars will come into contact with each other.
Gap
Figure not drawn to scale
Solution
Since the frame is rigidly attached to the floor, we can assume the expansion of each bar
is toward the other bar. Finding the change in length of each bar:
Copper:
L   Lo T  17 x 10 6 C 1 0.20 m 100C  0C   3.4 x 10 4 m
Steel:
L   Lo T  12 x 10 6 C 1 0.20 m 100C  0C   2.4 x 10 4 m
Adding the two changes in length, we get 5.8 x 10-4 m = 0.58 mm. Thus the 0.50 mm gap
will close and the bars will come into contact with each other.
12.6 Heat and Internal Energy
In any state of matter, the molecules are moving and therefore have energy. They have
potential energy because of the bonds between them and kinetic energy because the
molecules have mass and speed. The sum of the potential and kinetic energies of the
molecules in a substance is called the internal energy of the substance. When a warmer
substance is brought in contact with a cooler substance, some of the kinetic energy of the
molecules in the warmer substance is transferred to the cooler substance. The energy
representing the kinetic energy of molecules that is transferred spontaneously from a
warmer substance to a cooler substance is called heat energy. Heat is generally given the
symbol Q, and, since it is a form of energy, is measured in Joules (J) or calories (cal).
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Chapter 12 Temperature and Heat
CHAPTER 12 REVIEW QUESTIONS
For each of the multiple choice questions below, choose the best answer.
1. The average kinetic energy of the
molecules in a substance is most closely
associated with
(A) heat
(B) temperature
(C) expansion
(D) absolute zero
(E) potential energy
5. A brass spring has a spring constant k.
When the spring is heated, the spring
constant will
(A) increase
(B) decrease
(C) remain the same
(D) increase, then decrease
(E) decrease then increase
2. The Celsius temperature at absolute
zero is equal to
(A) 0 C
(B) 100 C
(C) 273 C
(D) – 273 C
(E) – 100 C
6. Which of the following statement(s)
is/are true?
I. Every substance contains heat.
II. For heat to flow between two
substances, they must be at different
temperatures.
III. The internal energy of a substance is
equal to the kinetic energy of the
molecules in the substance.
3. Which of the following is true of the
Celsius and Kelvin temperature scales?
(A) Both the Celsius and Kelvin
temperature scales have negative
values.
(B) A Kelvin degree and a Celsius
degree are equivalent in size.
(C) A Kelvin degree is larger in size
than a Celsius degree.
(D) A Kelvin degree is smaller in size
than a Celsius degree.
(E) The Kelvin scale reaches much
higher temperatures than the Celsius
scale.
(A) I and II only
(B) II and III only
(C) II only
(D) III only
(E) I, II, and III
4. In general, when a solid is heated, it
(A) expands proportionally to the change
in temperature
(B) contracts proportionally to the
change in temperature
(C) expands inversely proportionally to
the change in temperature
(D) contracts inversely proportionally to
the change in temperature
(E) does not expand nor contract.
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Chapter 12 Temperature and Heat
ANSWERS AND EXPLANATIONS TO CHAPTER 12 REVIEW QUESTIONS
Multiple Choice
1. B
The temperature of a substance is proportional to the average kinetic energy of its
molecules.
2. D
TC = TK – 273 = 0 – 273 = – 273 C
3. B
The Kelvin and Celsius degrees are equivalent in size, they are simply offset by 273.
4. A
In the equation L   Lo T , the coefficient of linear expansion α and the initial length
of the metal Lo are both constants. Thus the change in length is proportional to the change
in temperature.
5. B
A heated spring will lengthen, causing the spring to be less stiff, and the spring constant
to decrease.
6. C
Heat can only be transferred between substances of different temperatures. It is not
proper to say that a substance contains heat, but heat is the energy transferred between
two substances. The internal energy is the sum of the kinetic and potential energy of the
molecules in a substance.
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