All 15 Chapters Covered
SOLUTION MANUAL
, Table of Contents
Chapter 1……………………………………………………..1
Chapter 2……………………………………………………14
Chapter 3…………………………………………………....47
Chapter 4…………………………………………………....72
Chapter 5…...……………………………………………….96
Chapter 6……………………………………………….….128
Chapter 7…………………………………………………..151
Chapter 8………………………………………………..…169
Chapter 9…………………………………………………..183
Chapter 10…………………………………………………203
Chapter 11…………………………………………..……..226
Chapter 12…………………………………………………249
Chapter 13………………………………………………....269
Chapter 14……………………………………………..…..288
Chapter 15…………………………………………..……..305
Sample Formula Sheet for Exams………………………….
viii
, Chapter 1
This chapter presents a review of some topics from classical physics. I have often
heard from instructors using the book that “my students have already studied a year of
introductory classical physics, so they don’t need the review.” This review chapter gives
the opportunity to present a number of concepts that I have found to cause difficulty for
students and to collect those concepts where they are available for easy reference. For
example, all students should know that kinetic energy is 12 mv 2 , but few are readily
familiar with kinetic energy as p m , which is used more often in the text. The
expression connecting potential energy difference with potential difference for an electric
charge q, ΔU = qΔV , zips by in the blink of an eye in the introductory course and is
rarely used there, while it is of fundamental importance to many experimental set-ups in
modern physics and is used implicitly in almost every chapter. Many introductory
courses do not cover thermodynamics or statistical mechanics, so it is useful to “review”
them in this introductory chapter.
I have observed students in my modern course occasionally struggling with
problems involving linear momentum conservation, another of those classical concepts
that resides in the introductory course. Although we physicists regard momentum
conservation as a fundamental law on the same plane as energy conservation, the latter is
frequently invoked throughout the introductory course while former appears and virtually
disappears after a brief analysis of 2-body collisions. Moreover, some introductory texts
present the equations for the final velocities in a one-dimensional elastic collision,
leaving the student with little to do except plus numbers into the equations. That is,
students in the introductory course are rarely called upon to begin momentum
conservation problems with pinitial = pfinal . This puts them at a disadvantage in the
application of momentum conservation to problems in modern physics, where many
different forms of momentum may need to be treated in a single situation (for example,
classical particles, relativistic particles, and photons). Chapter 1 therefore contains a
brief review of momentum conservation, including worked sample problems and end-of-
chapter exercises.
Placing classical statistical mechanics in Chapter 1 (as compared to its location in
Chapter 10 in the 2nd edition) offers a number of advantages. It permits the useful
expression K av = 23 kT to be used throughout the text without additional explanation. The
failure of classical statistical mechanics to account for the heat capacities of diatomic
gases (hydrogen in particular) lays the groundwork for quantum physics. It is especially
helpful to introduce the Maxwell-Boltzmann distribution function early in the text, thus
permitting applications such as the population of molecular rotational states in Chapter 9
and clarifying references to “population inversion” in the discussion of the laser in
Chapter 8. Distribution functions in general are new topics for most students. They may
look like ordinary mathematical functions, but they are handled and interpreted quite
differently. Absent this introduction to a classical distribution function in Chapter 1, the
students’ first exposure to a distribution function will be |ψ|2, which layers an additional
level of confusion on top of the mathematical complications. It is better to have a chance
to cover some of the mathematical details at an earlier stage with a distribution function
that is easier to interpret.
1
, Suggestions for Additional Reading
Some descriptive, historical, philosophical, and nonmathematical texts which give good
background material and are great fun to read:
A. Baker, Modern Physics and Anti-Physics (Addison-Wesley, 1970).
F. Capra, The Tao of Physics (Shambhala Publications, 1975).
K. Ford, Quantum Physics for Everyone (Harvard University Press, 2005).
G. Gamow, Thirty Years that Shook Physics (Doubleday, 1966).
R. March, Physics for Poets (McGraw-Hill, 1978).
E. Segre, From X-Rays to Quarks: Modern Physicists and their Discoveries (Freeman, 1980).
G. L. Trigg, Landmark Experiments in Twentieth Century Physics (Crane, Russak, 1975).
F. A. Wolf, Taking the Quantum Leap (Harper & Row, 1989).
G. Zukav, The Dancing Wu Li Masters, An Overview of the New Physics (Morrow, 1979).
Gamow, Segre, and Trigg contributed directly to the development of modern physics and
their books are written from a perspective that only those who were part of that
development can offer. The books by Capra, Wolf, and Zukav offer controversial
interpretations of quantum mechanics as connected to eastern mysticism, spiritualism, or
consciousness.
Materials for Active Engagement in the Classroom
A. Reading Quizzes
1. In an ideal gas at temperature T, the average speed of the molecules:
(1) increases as the square of the temperature.
(2) increases linearly with the temperature.
(3) increases as the square root of the temperature.
(4) is independent of the temperature.
2. The heat capacity of molecular hydrogen gas can take values of 3R/2, 5R/2, and 7R/2
at different temperatures. Which value is correct at low temperatures?
(1) 3R/2 (2) 5R/2 (3) 7R/2
Answers 1. 3 2. 1
B. Conceptual and Discussion Questions
1. Equal numbers of molecules of hydrogen gas (molecular mass = 2 u) and helium gas
(molecular mass = 4 u) are in equilibrium in a container.
(a) What is the ratio of the average kinetic energy of a hydrogen molecule to the
average kinetic energy of a helium molecule?
K H / K He = (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4
2
, (b) What is the ratio of the average speed of a hydrogen molecule to the average
speed of a helium molecule?
vH / vHe = (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4 (C)
(c) What is the ratio of the pressure exerted on the walls of the container by the
hydrogen gas to the pressure exerted on the walls by the helium gas?
PH / PHe = (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4
2. Containers 1 and 3 have volumes of 1 m3 and container 2 has a volume of 2 m3.
Containers 1 and 2 contain helium gas, and container 3 contains neon gas. All three
containers have a temperature of 300 K and a pressure of 1 atm.
He He Ne
1 2 3
(a) Rank the average speeds of the molecules in the containers in order from largest to
smallest.
(1) 1 > 2 > 3 (2) 1 = 2 > 3 (3) 1 = 2 = 3
(4) 3 > 1 > 2 (5) 3 > 1 = 2 (6) 2 > 1 > 3
(b) In which container is the average kinetic energy per molecule the largest?
(1) 1 (2) 2 (3) 3
(4) 1 and 2 (5) 1 and 3 (6) All the same
3. (a) Consider diatomic nitrogen gas at room temperature, in which only the
translational and rotational motions are possible. Suppose that 100 J of energy is
transferred to the gas at constant volume. How much of this energy goes into the
translational kinetic energy of the molecules?
(1) 20 J (2) 40 J (3) 50 J
(4) 60 J (5) 80 J (6) 100 J
(b) Now suppose that the gas is at a higher temperature, so that vibrational motion is
also possible. Compared with the situation at room temperature, is the fraction of the
added energy that goes into translational kinetic energy:
(1) smaller? (2) the same? (3) greater?
Answers 1. (a) 4 (b) 3 (c) 4 2. (a) 2 (b) 6 3. (a) 4 (b) 1
3
, Sample Exam Questions
A. Multiple Choice
1. A container holds gas molecules of mass m at a temperature T. A small probe
inserted into the container measures the value of the x component of the velocity of
the molecules. What is the average value of 12 mvx2 for these molecules?
(a) 32 kT (b) 12 kT (c) kT (d) 3kT
2. A container holds N molecules of a diatomic gas at temperature T. At this
temperature, rotational and vibrational motions of the gas molecules are allowed. A
quantity of energy E is transferred to the gas. What fraction of this added energy is
responsible for increasing the temperature of the gas?
(a) All of the added energy (b) 3/5 (c) 2/5 (d) 2/7 (e) 3/7
3. Two identical containers with fixed volumes hold equal amounts of Ne gas and N2 gas
at the same temperature of 1000 K. Equal amounts of heat energy are then transferred
to the two gases. How do the final temperatures of the two gases compare?
(a) T(Ne) = T(N2) (b) T(Ne) > T(N2) (c) T(Ne) < T(N2)
Answers 1. b 2. e 3. b
B. Conceptual
1. A container of volume V holds an equilibrium mixture of N molecules of oxygen gas
O2 (molecular mass = 32.0 u) and also 2N molecules of He gas (mass = 4.00 u). Is
the average molecular energy of O2 greater than, equal to, or less than the average
molecular energy of He? EXPLAIN YOUR ANSWER.
2. Consider two containers of identical volumes. Container 1 holds N molecules of He
at temperature T. Container 2 holds the same number N molecules of H2 at the same
temperature T. Is the average energy per molecule of He greater than, less than, or
the same as the average energy per molecule of H2? EXPLAIN YOUR ANSWER.
Answers 1. equal to 2. the same as
C. Problems
1. N molecules of a gas are confined in a container at temperature T. A measuring
device in the container can determine the number of molecules in a range of 0.002v at
any speed v, that is, the number of molecules with speeds between 0.999v and 1.001v.
When the device is set for molecules at the speed vrms, the result is N1. When it is set
for molecules at the speed 2vrms, the result is N2. Find the value of N1/N2. Your
answer should be a pure number, involving no symbols or variables.
2. A container holds 2.5 moles of helium (a gas with one atom per molecule; atomic mass =
4.00 u; molar mass = 0.00400 kg) at a temperature of 342 K. What fraction of the gas
molecules has translational kinetic energies between 0.01480 eV and 0.01520 eV?
4
,3. One mole of N2 gas (molecular mass = 28 u) is confined to a container at a
temperature of 387 K. At this temperature, you may assume that the molecules are
free to both rotate and vibrate.
(a) What fraction of the molecules has translational kinetic energies within ±1% of
the average translational kinetic energy?
(b) Find the total internal energy of the gas.
Answers: 1. 11.25 2. 0.0066 3. (a) 0.0093 (b) 11.2 kJ
5
, Problem Solutions
1. (a) Conservation of momentum gives px ,initial = px ,final , or
mH vH,initial + mHe vHe,initial = mH vH,final + mHe vHe,final
Solving for vHe,final with vHe,initial = 0 , we obtain
mH (vH,initial − vH,final )
vHe,final =
mHe
(1.674 × 10−27 kg)[1.1250 × 107 m/s − (−6.724 ×106 m/s)]
= −27
= 4.527 ×106 m/s
6.646 × 10 kg
(b) Kinetic energy is the only form of energy we need to consider in this elastic
collision. Conservation of energy then gives K initial = K final , or
1
2
2
mH vH,initial + 12 mHe vHe,initial
2
= 12 mH vH,final
2
+ 12 mHe vHe,final
2
Solving for vHe,final with vHe,initial = 0 , we obtain
2
mH (vH,initial − vH,final
2
)
vHe,final =
mHe
(1.674 ×10−27 kg)[(1.1250 ×107 m/s) 2 − (−6.724 × 106 m/s) 2 ]
= −27
= 4.527 ×106 m/s
6.646 ×10 kg
2. (a) Let the helium initially move in the x direction. Then conservation of momentum
gives:
px ,initial = px ,final : mHe vHe,initial = mHe vHe,final cos θ He + mO vO,final cos θ O
p y ,initial = p y ,final : 0 = mHe vHe,final sin θ He + mO vO,final sin θ O
From the second equation,
mHe vHe,final sin θ He (6.6465 ×10−27 kg)(6.636 × 106 m/s)(sin 84.7°)
vO,final = − =− = 2.551×106 m/s
mO sin θ O −26
(2.6560 ×10 kg)[sin(−40.4°)]
(b) From the first momentum equation,
6
, mHe vHe,final cosθ He + mO vO,final cos θ O
vHe,initial =
mHe
(6.6465 × 10 kg)(6.636 × 106 m/s)(cos 84.7°) + (2.6560 × 10−26 kg)(2.551 × 106 m/s)[cos(−40.4°)]
−27
=
6.6465 × 10−27 kg
= 8.376 × 106 m/s
3. (a) Using conservation of momentum for this one-dimensional situation, we have
px ,initial = px ,final , or
mHe vHe + mN vN = mD vD + mO vO
Solving for vO with vN = 0 , we obtain
mHe vHe − mD vD (3.016 u)(6.346 × 106 m/s) − (2.014 u)(1.531× 107 m/s)
vO = = = −7.79 × 105 m/s
mO 15.003 u
(b) The kinetic energies are:
K initial = 12 mHe vHe
2
+ 12 mN vN2 = 12 (3.016 u)(1.6605 ×10−27 kg/u)(6.346 × 106 m/s) 2 = 1.008 × 10−13 J
K final = 12 mD vD2 + 12 mO vO2 = 12 (2.014 u)(1.6605 × 10−27 kg/u)(1.531×107 m/s) 2
+ 12 (15.003 u)(1.6605 × 10−27 kg/u)(7.79 ×105 m/s) 2 = 3.995 ×10−13 J
As in Example 1.2, this is also a case in which nuclear energy turns into kinetic
energy. The gain in kinetic energy is exactly equal to the loss in nuclear energy.
4. Let the two helium atoms move in opposite directions along the x axis with speeds
v1 and v2 . Conservation of momentum along the x direction ( p x ,initial = px ,final ) gives
0 = m1v1 − m2 v2 or v1 = v2
The energy released is in the form of the total kinetic energy of the two helium atoms:
K1 + K 2 = 92.2 keV
Because v1 = v2 , it follows that K1 = K 2 = 46.1 keV , so
2 K1 2(46.1 ×103 eV)(1.602 ×10−19 J/eV)
v = = −27
= 1.49 ×106 m/s
m1 (4.00 u)(1.6605 ×10 kg/u)
v2 = v1 = 1.49 × 106 m/s
7
, 5. (a) The kinetic energy of the electrons is
K i = 12 mvi2 = 12 (9.11× 10−31 kg)(1.76 × 106 m/s) = 14.11× 10−19 J
In passing through a potential difference of ΔV = Vf − Vi = +4.15 volts , the potential
energy of the electrons changes by
ΔU = qΔV = (−1.602 × 10−19 C)( + 4.15 V) = −6.65 × 10−19 J
Conservation of energy gives K i + U i = K f + U f , so
K f = K i + (U i − U f ) = K i − ΔU = 14.11× 10−19 J + 6.65 × 10−19 J = 20.76 × 10−19 J
2Kf 2(20.76 ×10−19 J)
vf = = −31
= 2.13 × 106 m/s
m 9.11×10 kg
(b) In this case ΔV = −4.15 volts, so ΔU = +6.65 × 10−19 J and thus
K f = K i − ΔU = 14.11× 10−19 J − 6.65 × 10−19 J = 7.46 × 10−19 J
2Kf 2(7.46 ×10−19 J)
vf = = −31
= 1.28 ×106 m/s
m 9.11×10 kg
6. (a) Δx A = vΔt A = (0.624)(2.997 × 108 m/s)(124 × 10−9 s) = 23.2 m
(b) ΔxB = vΔt B = (0.624)(2.997 × 108 m/s)(159 × 10−9 s) = 29.7 m
7. With T = 35°C = 308 K and P = 1.22 atm = 1.23 × 105 Pa ,
N P 1.23 ×105 Pa
= = = 2.89 ×1025 atoms/m3
V kT (1.38 ×10 J/K)(308 K)
-23
so the volume available to each atom is (2.89 × 1025/m3)−1 = 3.46 × 10−26 m3. For a
spherical atom, the volume would be
4
3 π R 3 = 43 π (0.710 ×10−10 m)3 = 1.50 × 10−30 m3
The fraction is then
1.50 × 10−30
−26
= 4.34 × 10−5
3.46 × 10
8
SOLUTION MANUAL
, Table of Contents
Chapter 1……………………………………………………..1
Chapter 2……………………………………………………14
Chapter 3…………………………………………………....47
Chapter 4…………………………………………………....72
Chapter 5…...……………………………………………….96
Chapter 6……………………………………………….….128
Chapter 7…………………………………………………..151
Chapter 8………………………………………………..…169
Chapter 9…………………………………………………..183
Chapter 10…………………………………………………203
Chapter 11…………………………………………..……..226
Chapter 12…………………………………………………249
Chapter 13………………………………………………....269
Chapter 14……………………………………………..…..288
Chapter 15…………………………………………..……..305
Sample Formula Sheet for Exams………………………….
viii
, Chapter 1
This chapter presents a review of some topics from classical physics. I have often
heard from instructors using the book that “my students have already studied a year of
introductory classical physics, so they don’t need the review.” This review chapter gives
the opportunity to present a number of concepts that I have found to cause difficulty for
students and to collect those concepts where they are available for easy reference. For
example, all students should know that kinetic energy is 12 mv 2 , but few are readily
familiar with kinetic energy as p m , which is used more often in the text. The
expression connecting potential energy difference with potential difference for an electric
charge q, ΔU = qΔV , zips by in the blink of an eye in the introductory course and is
rarely used there, while it is of fundamental importance to many experimental set-ups in
modern physics and is used implicitly in almost every chapter. Many introductory
courses do not cover thermodynamics or statistical mechanics, so it is useful to “review”
them in this introductory chapter.
I have observed students in my modern course occasionally struggling with
problems involving linear momentum conservation, another of those classical concepts
that resides in the introductory course. Although we physicists regard momentum
conservation as a fundamental law on the same plane as energy conservation, the latter is
frequently invoked throughout the introductory course while former appears and virtually
disappears after a brief analysis of 2-body collisions. Moreover, some introductory texts
present the equations for the final velocities in a one-dimensional elastic collision,
leaving the student with little to do except plus numbers into the equations. That is,
students in the introductory course are rarely called upon to begin momentum
conservation problems with pinitial = pfinal . This puts them at a disadvantage in the
application of momentum conservation to problems in modern physics, where many
different forms of momentum may need to be treated in a single situation (for example,
classical particles, relativistic particles, and photons). Chapter 1 therefore contains a
brief review of momentum conservation, including worked sample problems and end-of-
chapter exercises.
Placing classical statistical mechanics in Chapter 1 (as compared to its location in
Chapter 10 in the 2nd edition) offers a number of advantages. It permits the useful
expression K av = 23 kT to be used throughout the text without additional explanation. The
failure of classical statistical mechanics to account for the heat capacities of diatomic
gases (hydrogen in particular) lays the groundwork for quantum physics. It is especially
helpful to introduce the Maxwell-Boltzmann distribution function early in the text, thus
permitting applications such as the population of molecular rotational states in Chapter 9
and clarifying references to “population inversion” in the discussion of the laser in
Chapter 8. Distribution functions in general are new topics for most students. They may
look like ordinary mathematical functions, but they are handled and interpreted quite
differently. Absent this introduction to a classical distribution function in Chapter 1, the
students’ first exposure to a distribution function will be |ψ|2, which layers an additional
level of confusion on top of the mathematical complications. It is better to have a chance
to cover some of the mathematical details at an earlier stage with a distribution function
that is easier to interpret.
1
, Suggestions for Additional Reading
Some descriptive, historical, philosophical, and nonmathematical texts which give good
background material and are great fun to read:
A. Baker, Modern Physics and Anti-Physics (Addison-Wesley, 1970).
F. Capra, The Tao of Physics (Shambhala Publications, 1975).
K. Ford, Quantum Physics for Everyone (Harvard University Press, 2005).
G. Gamow, Thirty Years that Shook Physics (Doubleday, 1966).
R. March, Physics for Poets (McGraw-Hill, 1978).
E. Segre, From X-Rays to Quarks: Modern Physicists and their Discoveries (Freeman, 1980).
G. L. Trigg, Landmark Experiments in Twentieth Century Physics (Crane, Russak, 1975).
F. A. Wolf, Taking the Quantum Leap (Harper & Row, 1989).
G. Zukav, The Dancing Wu Li Masters, An Overview of the New Physics (Morrow, 1979).
Gamow, Segre, and Trigg contributed directly to the development of modern physics and
their books are written from a perspective that only those who were part of that
development can offer. The books by Capra, Wolf, and Zukav offer controversial
interpretations of quantum mechanics as connected to eastern mysticism, spiritualism, or
consciousness.
Materials for Active Engagement in the Classroom
A. Reading Quizzes
1. In an ideal gas at temperature T, the average speed of the molecules:
(1) increases as the square of the temperature.
(2) increases linearly with the temperature.
(3) increases as the square root of the temperature.
(4) is independent of the temperature.
2. The heat capacity of molecular hydrogen gas can take values of 3R/2, 5R/2, and 7R/2
at different temperatures. Which value is correct at low temperatures?
(1) 3R/2 (2) 5R/2 (3) 7R/2
Answers 1. 3 2. 1
B. Conceptual and Discussion Questions
1. Equal numbers of molecules of hydrogen gas (molecular mass = 2 u) and helium gas
(molecular mass = 4 u) are in equilibrium in a container.
(a) What is the ratio of the average kinetic energy of a hydrogen molecule to the
average kinetic energy of a helium molecule?
K H / K He = (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4
2
, (b) What is the ratio of the average speed of a hydrogen molecule to the average
speed of a helium molecule?
vH / vHe = (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4 (C)
(c) What is the ratio of the pressure exerted on the walls of the container by the
hydrogen gas to the pressure exerted on the walls by the helium gas?
PH / PHe = (1) 4 (2) 2 (3) 2 (4) 1 (5) 1/ 2 (6) 1/2 (7) 1/4
2. Containers 1 and 3 have volumes of 1 m3 and container 2 has a volume of 2 m3.
Containers 1 and 2 contain helium gas, and container 3 contains neon gas. All three
containers have a temperature of 300 K and a pressure of 1 atm.
He He Ne
1 2 3
(a) Rank the average speeds of the molecules in the containers in order from largest to
smallest.
(1) 1 > 2 > 3 (2) 1 = 2 > 3 (3) 1 = 2 = 3
(4) 3 > 1 > 2 (5) 3 > 1 = 2 (6) 2 > 1 > 3
(b) In which container is the average kinetic energy per molecule the largest?
(1) 1 (2) 2 (3) 3
(4) 1 and 2 (5) 1 and 3 (6) All the same
3. (a) Consider diatomic nitrogen gas at room temperature, in which only the
translational and rotational motions are possible. Suppose that 100 J of energy is
transferred to the gas at constant volume. How much of this energy goes into the
translational kinetic energy of the molecules?
(1) 20 J (2) 40 J (3) 50 J
(4) 60 J (5) 80 J (6) 100 J
(b) Now suppose that the gas is at a higher temperature, so that vibrational motion is
also possible. Compared with the situation at room temperature, is the fraction of the
added energy that goes into translational kinetic energy:
(1) smaller? (2) the same? (3) greater?
Answers 1. (a) 4 (b) 3 (c) 4 2. (a) 2 (b) 6 3. (a) 4 (b) 1
3
, Sample Exam Questions
A. Multiple Choice
1. A container holds gas molecules of mass m at a temperature T. A small probe
inserted into the container measures the value of the x component of the velocity of
the molecules. What is the average value of 12 mvx2 for these molecules?
(a) 32 kT (b) 12 kT (c) kT (d) 3kT
2. A container holds N molecules of a diatomic gas at temperature T. At this
temperature, rotational and vibrational motions of the gas molecules are allowed. A
quantity of energy E is transferred to the gas. What fraction of this added energy is
responsible for increasing the temperature of the gas?
(a) All of the added energy (b) 3/5 (c) 2/5 (d) 2/7 (e) 3/7
3. Two identical containers with fixed volumes hold equal amounts of Ne gas and N2 gas
at the same temperature of 1000 K. Equal amounts of heat energy are then transferred
to the two gases. How do the final temperatures of the two gases compare?
(a) T(Ne) = T(N2) (b) T(Ne) > T(N2) (c) T(Ne) < T(N2)
Answers 1. b 2. e 3. b
B. Conceptual
1. A container of volume V holds an equilibrium mixture of N molecules of oxygen gas
O2 (molecular mass = 32.0 u) and also 2N molecules of He gas (mass = 4.00 u). Is
the average molecular energy of O2 greater than, equal to, or less than the average
molecular energy of He? EXPLAIN YOUR ANSWER.
2. Consider two containers of identical volumes. Container 1 holds N molecules of He
at temperature T. Container 2 holds the same number N molecules of H2 at the same
temperature T. Is the average energy per molecule of He greater than, less than, or
the same as the average energy per molecule of H2? EXPLAIN YOUR ANSWER.
Answers 1. equal to 2. the same as
C. Problems
1. N molecules of a gas are confined in a container at temperature T. A measuring
device in the container can determine the number of molecules in a range of 0.002v at
any speed v, that is, the number of molecules with speeds between 0.999v and 1.001v.
When the device is set for molecules at the speed vrms, the result is N1. When it is set
for molecules at the speed 2vrms, the result is N2. Find the value of N1/N2. Your
answer should be a pure number, involving no symbols or variables.
2. A container holds 2.5 moles of helium (a gas with one atom per molecule; atomic mass =
4.00 u; molar mass = 0.00400 kg) at a temperature of 342 K. What fraction of the gas
molecules has translational kinetic energies between 0.01480 eV and 0.01520 eV?
4
,3. One mole of N2 gas (molecular mass = 28 u) is confined to a container at a
temperature of 387 K. At this temperature, you may assume that the molecules are
free to both rotate and vibrate.
(a) What fraction of the molecules has translational kinetic energies within ±1% of
the average translational kinetic energy?
(b) Find the total internal energy of the gas.
Answers: 1. 11.25 2. 0.0066 3. (a) 0.0093 (b) 11.2 kJ
5
, Problem Solutions
1. (a) Conservation of momentum gives px ,initial = px ,final , or
mH vH,initial + mHe vHe,initial = mH vH,final + mHe vHe,final
Solving for vHe,final with vHe,initial = 0 , we obtain
mH (vH,initial − vH,final )
vHe,final =
mHe
(1.674 × 10−27 kg)[1.1250 × 107 m/s − (−6.724 ×106 m/s)]
= −27
= 4.527 ×106 m/s
6.646 × 10 kg
(b) Kinetic energy is the only form of energy we need to consider in this elastic
collision. Conservation of energy then gives K initial = K final , or
1
2
2
mH vH,initial + 12 mHe vHe,initial
2
= 12 mH vH,final
2
+ 12 mHe vHe,final
2
Solving for vHe,final with vHe,initial = 0 , we obtain
2
mH (vH,initial − vH,final
2
)
vHe,final =
mHe
(1.674 ×10−27 kg)[(1.1250 ×107 m/s) 2 − (−6.724 × 106 m/s) 2 ]
= −27
= 4.527 ×106 m/s
6.646 ×10 kg
2. (a) Let the helium initially move in the x direction. Then conservation of momentum
gives:
px ,initial = px ,final : mHe vHe,initial = mHe vHe,final cos θ He + mO vO,final cos θ O
p y ,initial = p y ,final : 0 = mHe vHe,final sin θ He + mO vO,final sin θ O
From the second equation,
mHe vHe,final sin θ He (6.6465 ×10−27 kg)(6.636 × 106 m/s)(sin 84.7°)
vO,final = − =− = 2.551×106 m/s
mO sin θ O −26
(2.6560 ×10 kg)[sin(−40.4°)]
(b) From the first momentum equation,
6
, mHe vHe,final cosθ He + mO vO,final cos θ O
vHe,initial =
mHe
(6.6465 × 10 kg)(6.636 × 106 m/s)(cos 84.7°) + (2.6560 × 10−26 kg)(2.551 × 106 m/s)[cos(−40.4°)]
−27
=
6.6465 × 10−27 kg
= 8.376 × 106 m/s
3. (a) Using conservation of momentum for this one-dimensional situation, we have
px ,initial = px ,final , or
mHe vHe + mN vN = mD vD + mO vO
Solving for vO with vN = 0 , we obtain
mHe vHe − mD vD (3.016 u)(6.346 × 106 m/s) − (2.014 u)(1.531× 107 m/s)
vO = = = −7.79 × 105 m/s
mO 15.003 u
(b) The kinetic energies are:
K initial = 12 mHe vHe
2
+ 12 mN vN2 = 12 (3.016 u)(1.6605 ×10−27 kg/u)(6.346 × 106 m/s) 2 = 1.008 × 10−13 J
K final = 12 mD vD2 + 12 mO vO2 = 12 (2.014 u)(1.6605 × 10−27 kg/u)(1.531×107 m/s) 2
+ 12 (15.003 u)(1.6605 × 10−27 kg/u)(7.79 ×105 m/s) 2 = 3.995 ×10−13 J
As in Example 1.2, this is also a case in which nuclear energy turns into kinetic
energy. The gain in kinetic energy is exactly equal to the loss in nuclear energy.
4. Let the two helium atoms move in opposite directions along the x axis with speeds
v1 and v2 . Conservation of momentum along the x direction ( p x ,initial = px ,final ) gives
0 = m1v1 − m2 v2 or v1 = v2
The energy released is in the form of the total kinetic energy of the two helium atoms:
K1 + K 2 = 92.2 keV
Because v1 = v2 , it follows that K1 = K 2 = 46.1 keV , so
2 K1 2(46.1 ×103 eV)(1.602 ×10−19 J/eV)
v = = −27
= 1.49 ×106 m/s
m1 (4.00 u)(1.6605 ×10 kg/u)
v2 = v1 = 1.49 × 106 m/s
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, 5. (a) The kinetic energy of the electrons is
K i = 12 mvi2 = 12 (9.11× 10−31 kg)(1.76 × 106 m/s) = 14.11× 10−19 J
In passing through a potential difference of ΔV = Vf − Vi = +4.15 volts , the potential
energy of the electrons changes by
ΔU = qΔV = (−1.602 × 10−19 C)( + 4.15 V) = −6.65 × 10−19 J
Conservation of energy gives K i + U i = K f + U f , so
K f = K i + (U i − U f ) = K i − ΔU = 14.11× 10−19 J + 6.65 × 10−19 J = 20.76 × 10−19 J
2Kf 2(20.76 ×10−19 J)
vf = = −31
= 2.13 × 106 m/s
m 9.11×10 kg
(b) In this case ΔV = −4.15 volts, so ΔU = +6.65 × 10−19 J and thus
K f = K i − ΔU = 14.11× 10−19 J − 6.65 × 10−19 J = 7.46 × 10−19 J
2Kf 2(7.46 ×10−19 J)
vf = = −31
= 1.28 ×106 m/s
m 9.11×10 kg
6. (a) Δx A = vΔt A = (0.624)(2.997 × 108 m/s)(124 × 10−9 s) = 23.2 m
(b) ΔxB = vΔt B = (0.624)(2.997 × 108 m/s)(159 × 10−9 s) = 29.7 m
7. With T = 35°C = 308 K and P = 1.22 atm = 1.23 × 105 Pa ,
N P 1.23 ×105 Pa
= = = 2.89 ×1025 atoms/m3
V kT (1.38 ×10 J/K)(308 K)
-23
so the volume available to each atom is (2.89 × 1025/m3)−1 = 3.46 × 10−26 m3. For a
spherical atom, the volume would be
4
3 π R 3 = 43 π (0.710 ×10−10 m)3 = 1.50 × 10−30 m3
The fraction is then
1.50 × 10−30
−26
= 4.34 × 10−5
3.46 × 10
8
