Download Statistical Physics

Statistical Physics
Prepared by Sisay Shewamare.
NOTICE
TABLE OF CONTENTS
FOREWORD
This module has four major sections
A. The first one is the INTRODUCTORY section that consists of five parts vis:
1.
TITLE:- The title of the module is clearly described
2.
PRE-REQUISITE KNOWLEDGE: In this section you are provided with information regarding
the specific pre-requisite knowledge and skills you require to start the module. Carefully look
into the requirements as this will help you to decide whether you require some revision work or
not.
3.
TIME REQUIRED: It gives you the total time (in hours) you require to complete the module.
All self tests, activities and evaluations are to be finished in this specified time.
4.
MATERIALS REQUIRED: Here you will find the list of materials you require to complete the
module. Some of the materials are parts of the course package you will receive in a CD-Rom or
access through the internet. Materials recommended to conduct some experiments may be
obtained from your host institution (Partner institution of the AVU) or you may acquire /borrow
by some other means.
5.
MODULE RATIONALE: In this section you will get the answer to questions like “Why should
I study this module as pre-service teacher trainee? What is its relevance to my career?”
B. The second one is the CONTENT section that consists of three parts:
6.
OVERVIEW: The content of the module is briefly presented. In this section you will find a
video file (QuickTime, movie) where the author of this module is interviewed about this
module. The paragraph overview of the module is followed by an outline of the content
including the approximate time required to complete each section. A graphic organization of
the whole content is presented next to the outline. All these three will assist you to picture how
the content is organized in the module.
7.
GENERAL OBJECTIVE(S): Clear informative, concise and understandable objectives are
provided to give you what knowledge skills and attitudes you are expected to attain after
studying the module.
8.
SPECIFIC LEARNING OBJECTIVES (INSTRUCTIONAL OBJECTIVES): Each of the
specific objectives, stated in this section, is at the heart of a teaching learning activity. Units,
elements and themes of the module are meant to help you achieve the specific objectives and
any kind of assessment is based on the objectives intended to be achieved. You are urged to
pay maximum attention to the specific objectives as they are vital to organize your effort in the
study of the module.
C. The third section is the bulk of the module. It is the section where you will spend more time and is
referred to as the TEACHING LEARNING ACTIVITIES. The gist of the nine components is listed
below:
9.
PRE-ASSESSMENT: A set of questions, that will quantitatively evaluate your level of
preparedness to the specific objectives of this module, are presented in this section. The preassessment questions help you to identify what you know and what you need to know, so that
your level of concern will be raised and you can judge your level of mastery. Answer key is
provided for the set of questions and some pedagogical comments are provided at the end.
10.
KEY CONCEPTS: This section contains short, concise definitions of terms used in the module.
It helps you with terms which you might not be familiar in the module.
11.
COMPULSORY READINGS: A minimum of three compulsory reading materials are provided.
It is mandatory to read the documents.
12.
COMPULSORY RESOURCES: A minimum of two video, audio with an abstract in text form
is provided in this section.
13.
USEFUL LINKS: A list of at least ten websites is provided in this section. It will help you to
deal with the content in greater depth.
14.
TEACHING AND LEARNING ACTIVITIES: This is the heart of the module. You need to
follow the learning guidance in this section. Various types of activities are provided. Go
through each activity. At times you my not necessarily follow the order in which the activities
are presented. It is very important to note:

formative and summative evaluations are carried out thoroughly

all compulsory readings and resources are done

as many as possible useful links are visited

feedback is given to the author and communication is done
Enjoy your work on this module.
I.
Statistical Physics
BY SISAY SHEWAMARE GEBREMICHAEL JIMMA UNIVERSITY ETHIOPIA
II
PREREQUISITE COURSE OR KNOWLEDGE
In order to successfully study this module, it is recommended that you need to have either completed or
to concurrently study the AVU Thermal Physics, Mathematical Physics and Quantum Mechanics
Teachers’ Training modules.
III
TIME
This module can be completed in 120hrs.
IV
MATERIALS
The materials in this module are different books, and from the soft copy available on the internet.
V
MODULE RATIONALE
In this module we are focusing on the system of macroscopic particles and we study the statistical
description of systems in terms of probability and the behavior of the density of state which help to
measure the macroscopic parameters like heat, absolute temperature and entropy. As a result of this
discussion we will acquired some very powerful tools for calculating the macroscopic properties of any
system in equilibrium from knowledge of its microscopic constitutes then we shall illustrate their
usefulness by discussing the application of macroscopic thermodynamics and distribution of systems of
particles.
VI
OVERVIEW
The central concepts of this module are the macroscopic systems of particles and macroscopic
measurement. The module begins with the study of statistical description of systems with statistical
thermodynamics and measuring the macroscopic parameters and its application.
Activities are related to the interaction of the macroscopic systems of particle and analyzing the
distribution of macroscopic systems in terms of the mean energy, entropy and pressure. The inter relation
between the macroscopic parameter discussed in the application of macroscopic thermodynamics and in
the partition function.
6.1 OUTLINE
1 Unit 1 Statistical description of systems of particle









2 Macroscopic measurements




(30 hours)
Equilibrium condition and constraints,
Entropy of the combined system,
The approach to thermal equilibrium.
Heat reservoir.
Dependency of the density of states on the external parameters.
4 Some application of statistical and macroscopic thermodynamics









(25 hours)
Work and internal energy.
Absolut temperature,
Heat capacity and specific heats.
Entropy
3 Statistical thermodynamics





(25 hours)
Specification of the state of the system,
Statistical enesemble.
Probability,
Simple random walk problem in one dimenssion;
Binomial distribution.
Gaussian distribution.
Principles of equal priori probability
Relaxation time
The probability of the density of states.
(40 hours)
Thermodynamics potential and their relation with thermodynamical variables,
Enesembles systems,
Connection of canonical distribution with thermodynamics .
Partition function and their properties.
Gibs paradox.
Validity of the classical approximation.
The equi partition theorem
Kinetic theory of dilute gases in equilibrium
Distribution of systems of particles
6.2 GRAPHIC ORGANIZER
A. Statistical Description of
Systems of Particles
C. Statistical Thermodynamics
Statistical Theories,
Equilibrium condition and constraints,
Specification of the state of the system,
Entropy of a combined system,
phase space
The approach to thermal equilibrium.
Statistical enesemble.
Heat reservoir.
Accessible States,
Dependency of the density of states
on the external parameters.
Probability calculations,
Simple random walk problem in one dimenssion;
D. Some Applications
Thermodynamic potentials
and their relation with thermodynamical variables,
Enesembles systems,
Connection of canonical distribution with thermodynamics .
Partition function and their properties.
Gibs paradox,
Validity of the classical approximation.
The equi partition theorem
Kinetic theory of dilute gases in equilibrium
Distribution of systems of particles
Statistical
Physics
Binomial distribution.
Gaussian distribution.
Principles of equal priori probability
Relaxation time
The probability of the density of states.
B. Macroscopic Measurements:
Work and internal energy.
Absolut temperature,
Heat capacity and specific heats.
Entropy
VII. GENERAL OBJECTIVE(S)
After completing this module you will be able to
 Appreciate that the statistical distribution of systems of particle and their solution at equilibrium
 Understand the concept of temperature, heat and internal energy
 Understand the underlying basis and the total statistical thermodynamics law
 Understand the macroscopic parameters and their measurements
 Understand the basic generalized force and entropy
 Understand the application of statistical and macroscopic thermodynamics
 Understand the partition function
 Derive the macroscopic measurements using the partition function
 Derive the distribution of systems of particles
VIII. Specific Learning Objectives (Instructional Objectives)
Learning objectives
After Completing this section
you would be able to:
Content
1. Unit 1 Statistical description of systems of
particle (25 hours)









Specification of the state of the system,
Statistical enesemble.
Probability,
Simple random walk problem in one dimenssion;
Binomial distribution.
Gaussian distribution.
Principles of equal priori probability
Relaxation time
The probability of the density of states.
2. Macroscopic measurements




Work and internal energy.
Absolut temperature,
Heat capacity and specific heats
Entropy
3. Statistical thermodynamics





(25 hours)
(30 hours)
Equilibrium condition and constraints,
Entropy of the combined system,
The approach to thermal equilibrium.
Heat reservoir.
Dependency of the density of states on the
external parameters
.
4. Some application of statistical and
macroscopic thermodynamics (40 hours)
 Thermodynamics potential and their relation with
thermodynamical variables,
 Enesembles systems,
 Connection of canonical distribution with





thermodynamics .
Partition function and their properties.
Gibs paradox.
Validity of the classical approximation.
The equi partition theorem
Kinetic theory of dilute gases in equilibrium.
Deriving the statistical equation
Discussion on two state system to apply
random walk problem
Define the relaxation time
Deriving the binomial distribution
Derive the Gaussian equation
Define the work done and the internal
energy
Describe the absolute and entropy relation
State the heat capacity at constant V,P
Define and derive the entropy
 Write the equilibrium conditions
Derive the entropy equation for two systems
Solve problems related to entropy
Calculate the density of state at equilibrium
condition
Relates different thermodynamics equation
Find thermodynamics quantity relations
using the thermodynamics potentials
Define the ensemble systems
Derive the partition function
Show the thermodynamics quantities using
the partition function
Show the Gibbs paradox
Derive different distribution
IX.
PRE-ASSESSMENT: Are you ready for this module?
Figure 5: Ludwig Boltzmann (1844–1906)
Dear Learners:
In this section, you will find self-evaluation questions that will help you test your preparedness to
complete this module. You should judge yourself sincerely and do the recommended action after
completion of the self-test. We encourage you to take time and answer the questions.
Dear Instructors:
The Pre-assessment questions placed here guide learners to decide whether they are prepared to take the
content presented in this module. It is strongly suggested to abide by the recommendations made on the
basis of the mark obtained by the learner. As their instructor you should encourage learners to evaluate
themselves by answering all the questions provided below. Education research shows that this will help
learners be more prepared and help them articulate previous knowledge.
9.1 SELF EVALUATION ASSOCIATED WITH STATISTICAL PHYSICS
Evaluate your preparedness to take the module on thermal physics. If you score greater than or equal to
60 out of 75, you are ready to use this module. If you score something between 40 and 60 you may need
to revise your school physics on topics of heat. A score less than 40 out of 75 indicates you need to
physics.
1) How many calories of heat are required to raise the temperature of 3kg of aluminum from 200C
to 550C? Given specific heat capacity of aluminum C  910J kg-1K -1 and 4.2J=1 calorie
a. 13000
c. 35750
b. 22750
d. 95550
2) If 200g of water is contained in a 300g aluminum vessel at 100C and an additional 100g of water
at 1000C is poured into the container, what is the final equilibrium temperature of the system? In
degree Celsius
a. 77
c. 35
b. 45
d. 20

3) Two moles of an ideal gas ( =1.4) expand quasi-statically and adiabatic ally from pressure of 5
atm. and a volume of 12 liters to final volume of 30 liters a. What is the final pressure of the gas?
a. 1.4
c. 3
b. 3.4
d. 1
4) An ideal gas (  =1.4) expands quasi-statically and adiabatically. If the final temperature is one
third the initial temperature so by what factor does its volume change?
a. 10
c. 16
b. 20
d. 12
5) Following question 4 above, by what factor does its pressure change?
a. 1
c. 0.02
b. 1.2
d. 2
6) One mole of an ideal gas does 3000J of work on the surroundings as it expands isothermally to a
final pressure of 1atm. and volume of 25l. Determine the temperature of the gas
a. 200K
c. 400K
b. 100K
d. 300K
7) Following question 6 above, calculate initial volume of the gas.
a. 20l
c. 22l
b. 30l
d. 25l
8) Five moles of an ideal gas expands is isothermally at 1270C to four times its initial volume. Find
the work done by the gas
a. 30,000J
c. 50,000J
b. 40,000J
d. 32,012J
9) A gas is compressed at a constant pressure 0.8 atm from a volume of 9 liters to a volume of 2
liters. If in the process 400J of heat energy flows out of the gas what is the work done by the gas?
a. 57J
c. 50J
b. 37J
d. 400J
10) Using question 9 above, what is the internal energy lost by the system
a. 500J
c. 456J
b. 600J
d. 400J
11) There are two thermometers based on different thermometric properties of two different
materials. The two thermometers show identical readings because
a. each property changes uniformly with temperature.
b. the relation between the property and temperature is identical in the two cases
c. the property of one of increases with temperature and the property of the other decreases
at a uniform rate
d. the two thermometers have been calibrated with reference to a common standard.
12) In a Carnot cycle
a. work done during adiabatic expansion is less than work done during adiabatic
compression
b. work done by working substance during adiabatic expansion is greater than work done
during adiabatic compression.
c. work done during adiabatic expansion is equal to work done during adiabatic compression
d. work done during adiabatic expansion is equal to the heat absorbed from the source.
13) Which of the statements below is wrong about an ideal gas?
a. The total number of molecules is large
b. The molecules are in random motion
c. The molecules do not exert any appreciable force on one another or on the walls
d. The volume of the molecule is negligibly small compared with the volume occupied by
the gas.
14) The mean free path in a gas is
a. the distance travelled by a molecule before hitting a wall
b. the average distance travelled by a molecule in one second
c. the root mean square velocity
d. the average distance travelled by molecules between any two successive collisions
15) In adiabatic process work done
a. by working substance during adiabatic expansion is greater than work done during
adiabatic compression.
b. during adiabatic expansion is equal to work done during adiabatic compression
c. during adiabatic expansion is equal to the heat absorbed from the source.
d. by working substance during adiabatic expansion is equal to the heat that enters.
16) Which of the following statements is wrong about a real gas?
a. The total number of molecules is large
b. The molecules are in random motion
c. The molecules exert negligible force on one another or on the walls
d. The volume of the molecule is appreciable compared with the volume occupied by the
gas.
17) The root mean square velocity of a gas
a. does not depend on the temperature but on the pressure of the gas.
b. increases with the density of the gas.
c. decreases with the volume of the gas.
d. depends on both the pressure and temperature of the gas.
18) The average molecular kinetic energy at a temperature T K is
a. 13 kT
c. 12 kT
b. 32 kT
d. 23 kT
9.2 ANSWER KEY:
1. b
7. c
13. c
2. c
8. d
14. d
3. a
9. a
15. .b
4. c
10. c
16. c
5. c
11. a
17. d
6. d
12. c
18. b
9.3 PEDAGOGICAL COMMENT FOR THE LEARNER:
Physics, as a discipline that attempts to describe phenomena and processes in nature, has succeeded
in developing theoretical frameworks that describe processes and phenomena ranging from
subatomic particles to celestial bodies in galaxies. Theoretical framework, sufficient enough to
describe nonlinear systems such as the properties of granular media, earthquakes, friction and many
other systems, is still lacking.
Statistical physics gives a rational understanding of Thermodynamics in terms of microscopic
particles and their interactions. It allows calculation of macroscopic properties from microscopic
considerations. The tools and methods developed in statistical physics are extensively used in
frontier research areas to understand non-linear systems.
The material presented in this module is highly sequential. You need to follow the activities in the
order they are presented in the module. If you don't understand something go and refer to the
compulsory materials and visit the useful links there in; don't just write it down and hope that you'll
figure it out later.
Extensive research in recent years has shown that the students who do best in physics (and other
subjects) are those who involve themselves actively in the learning process. This involvement can
take many forms: writing many questions in the margins of the module; asking questions by email;
discussing physics in the AVU discussion fora doing exercises and self-assessments on schedule
etc.
X. TEACHING AND LEARNING ACTIVITIES
ACTIVITY 1: Statistical Description of System of Particles
You will require 25 hours to complete this activity. In this activity you are guided with a series of
readings, Multimedia clips, worked examples and self assessment questions and problems. You are
strongly advised to go through the activities and consult all the compulsory materials and as many
as possible of the useful links and references.
Specific Teaching and Learning Objectives
Deriving the statistical equation
Discussion on two state system to apply random walk problem
Define the relaxation time
Deriving the binomial Gaussian distribution
Summary of the Learning Activity
Description of a system of particles is an effort where theory is applied to a large numbers of
particles. We are not interested in all the details of the underlying microscopic dynamics of
individual particles that constitute a large number of particles like a given of mass of gas.
Instead, it is the systems’ macroscopic properties – among which are the thermodynamic functions
that we wish to understand or to deduce, and these are gross averages over the detailed dynamical
states. That is the reason for the word “statistical” in the name of our subject. Prominent feature in
the landscape of statistical mechanics is the Boltzmann distribution law, which tells us with what
frequency the individual microscopic states of a system of given temperature occur. An informal
statement of that law is given in the next section, where it is seen to be an obvious generalization of
two other well known distribution laws: the Maxwell velocity distribution and the “barometric”
distribution. We also need to note here that the exponential form of the Boltzmann distribution law
is consistent with – indeed, is required by – the rule that the probability of occurrence of
independent events is the product of the separate probabilities.
List of Required Readings
Reading #1:.
Complete reference : Statistical Mechanics
From Cornell Universit
URL : http://pages.physics.cornell.edu/sethna/StatMech
Accessed on the 23rd September 2007
Abstract :
Contents: Random Walks and Emergent Properties; Temperature and Equilibrium; Entropy; Free Energies
and Ensembles; Quantum Statistical Mechanics; Computational Stat Mech: Ising and Markov; Order
Parameters, Broken Symmetry, and Topology; Deriving New Laws; Correlations, Response, and
Dissipation; Abrupt Phase Transitions; Continuous Phase Transitions.
Rationale:
This chapter covers most of the topics in the second and third activities of the module...
List of Relevant MM Resources
1. Reference http://jersey.uoregon.edu/vlab/Piston/index.html
Date Consulted:-Nov 2006
Description:2. Reference:-: http://lectureonline.cl.msu.edu/~mmp/kap10/cd283.htm.
Date Consulted:- August 2006
Description:.
3. Reference http://en.wikipedia.org/wiki/Binomial_distribution
Date Consulted:-Nov 2006
Description:4. Reference:-: http://www.stat.yale.edu/Courses/1997-98/101/binom.htm.
Date Consulted:- August 2006
Description:5. Reference: http://en.wikipedia.org/wiki/Normal_distribution
Date Consulted: Nov 2006
Complete Reference:- Computer calculation of Phase Diagrams.
Rationale:
List of Relevant Useful Links
Useful Link #1
Title: Exactly Solved Models in Statistical Mechanics
URL: http://tpsrv.anu.edu.au/Members/baxter/book
Screen Capture:
Description: Rodney Baxter's classic book is officially out of print. Contents: basic statistical mechanics; the onedimensional Ising model; the mean field theory; Ising model on the Bethe Lattice; The Spherical Model; Duality
and Star Triangle Transformations of Planar Ising Models; Square-Lattice Ising Model; Ice-Type Models;
Alternative Way of Solving the Ice-Type Models; Squared Lattice Eight-Vertex Model; Kagomé Lattice EightVertex Model; Potts and Ashkin-Teller Models; Corner Transfer Matrices; Hard Hexagon and Related Models;
Elliptic Functions. .
Rationale: This book can be downloaded and used for personal and non-commercial use
Date Consulted: - Aug 2007
Useful Link #2
Title: STATISTICAL PHYSICS An Introductory Course
URL: http://www.worldscibooks.com/physics/3526.html
Screen Capture
By Daniel J Amit (Universita di Roma "La Sapienza" & The Hebrew University) & Yosef
Verbin (The Open University of Israel
Description: This invaluable textbook is an introduction to statistical physics that has been written
primarily for self-study. It provides a comprehensive approach to the main ideas of statistical
physics at the level of an introductory course, starting from the kinetic theory of gases and
proceeding all the way to Bose–Einstein and Fermi–Dirac statistics. Each idea is brought out with
ample motivation and clear, step-by-step, deductive exposition. The key points and methods are
presented and discussed on the basis of concrete representative systems, such as the paramagnet,
Einstein's solid, the diatomic gas, black body radiation, electric conductivity in metals and
superfluidity.
The book is written in a stimulating style and is accompanied by a large number of exercises
appropriately placed within the text and by self-assessment problems at the end of each
chapter. Detailed solutions of all the exercises are provided.
Introduction to the Activity
Detailed Description of the Activity (Main Theoretical Elements)
1: Statistical description of systems of particles
Consideration of non interactive systems of particles to analyze the probability with binomial and
Gaussian distribution by consideration of the statistical approach and with the density of systems of
particles.
1: Statistical Description of Systems of Particles:
 Statistical Theories,
 Ensemble
 Accessible state
 Probability calculation
 Phase space
1.1Specification of the state of the system
How do we determine the state of a many particle system? Well, let us, first of all, consider the
simplest possible many particle system, which consists of a single spinless particle moving classically
in one dimension. Assuming that we know the particle’s equation of motion, the state of the system is
fully specified once we simultaneously measure the particle’s position q and momentum p. In
principle, if we know q and p then we can calculate the state of the system at all subsequent times
using the equation of motion
1.2 Statistical ensemble
If we are informed about any of the initial conditions of a thrown up coin like its position, the height
of the throw and the corresponding velocity of the coin, we would indeed predict the out come of the
experiment by applying the law of classical mechanics.
In an experiment that describes the outcome in terms of the probability of a single coin, we consider
an ensemble consisting of many such single experiments.
1.3Probability
In this section we will discuss some of elementary aspect of probability theory. It is important to
keep in mind that whenever it is desired to described a situation from a statistical point of view
(i.e., in terms of probabilities), It is always necessary to consider an assembly ( ensemble) consists
of a very large number of similar prepared systems.
Group discussion
Give some example which can be described by two states of systems of particles
Answer
a)
In throwing a pair of dice, one gives a statistical description by considering a very large number.
b) In the basic probability concept, it will be useful to keep in mind a specific simple but important, illustrative
example the so called random walk problem
c)
Magnetism: An atom has a spin
1
and a magnetic moment  ; in accordance with quantum mechanics, its
2
spin can therefore point either “up” or “down” with respect to a given direction. If both these possibilities are
equally likely, what is the net total magnetic moment of N such atoms?
d) Diffusion of a molecule in a gas: A given molecule travels in three dimensions a mean distance
l between
collisions with other molecules. How far is it likely to have gone after N collisions?
1.4 The simple random walk problem in one dimension
For the sake of simplicity we shall discuss the random walk problem in one dimension. A particle
performing successive steps, or displacements, in one dimension after a total of N such steps, each
of length l , the particle is located at
x  ml
Where m is an integer lying between N mN
The probability PN (m) of finding the particle at the position x  ml after N such steps.
W N (n1 ) 
N ! n 1 n2
p q
n1!n2 !
Group discussion
Derive the probability WN (n1 ) for finding the particle at position x=ml after N steps
You can see the derivation as follow
The total number of steps N is simply
N  n1  n2
The net displacement where
m  n1  n2
m  n1  n2  n1  ( N  n1 )  2n1  N
Our fundamental assumption was that successive steps are statistically independent of each other.
Thus one can assert simply that, irrespective of past history, each step is characterized by the
respective probabilities
P = probability that the step is to the right
q =1 – p = probability that the step is to the left
Now, the probability of any one given sequence of n1 steps to the right and n2 steps to the left is
given simply by multiplying the respective probabilities, i.e., by
p1p 2 p3
p n q1q 2q 3
q n  p n1 q n 2
The number of distinct possibilities is given by
N!
n1!n2!
The probability W N (n1 ) of taking n1 steps to the right and n2 = N - n1 steps to the left, in any
order, is obtained by multiplying the probability of this sequence by the number of possible
sequences of such steps. This gives
WN (n1 ) 
N ! n 1 n2
p q
n1!n2 !
1.5 Binomial Distribution
Indeed, we recall that the binomial expansion is given by the formula
N
(p + q)N = 
n 0
N!
pnq N n
n!( N  n)!
Read the binomial distribution in the fundamentals of thermodynamics book (Federick Reif)
pp.7-23
Group discussion
Given that n1 
1
( N  m),
2
n2 
1
( N  m)
2
Show that
pN (m) 
N!
1
 
[( N  m) / 2]![( N  m) / 2]!  2 
N
In this discussion you may consider the probability PN (m) that the particle is found at position m
after N steps is the same as WN (n1 ) given by
PN (m) = WN (n1 )
1.5.1 Mean Value
If f(u) is any function of u, then the mean value of f(u) is defined by
M
f (u ) 
 p(u ) f (u )
i
i 1
i
M
 P(u )
i
i 1
This expression can be simplified. Since P(ui) is defined as a probability, the quantity
M
P(u1 )  P(u2 )  ...  P(uM )   P(ui )
i 1
M
 P(u )  1
i 1
i
This is the so-called “normalization condition”
M
f (u )   p (u i ) f (u i )
i 1
Activity
Derive the summation and the product of the mean value of different function
Solution
If f(u) and g(u) are any two functions of u, then
M
M
M
i 1
i 1
i 1
f (u ) _ g (u )   P(ui )[ f (ui )  g (ui )]   P(ui ) f (ui )   P(ui ) g (ui )
Or
f (u )  g (u )  f (u )  g (u )
If c is any constant, it is clear that
cf (u )  cf (u )
1.5.2 Deviation dispersion and standard deviation
u  u  u deviation
M
(u ) 2   P(ui )(ui  u ) 2  0 second moment of u about its mean,” or more simply the
i 1
“dispersion of u” since ( (u ) 2  0 can never be negative,
The variance of u is proportional to the square of the scatter of u around its mean value. A more
useful measure of the scatter is given by the square root of the variance,

 * u  u 

1
2 2
which is usually called the standard deviation of u.
1.6 The Gaussian Distribution
P ( n) 


 nn 2 
1

exp 
2
2  * n1
 2 * n1  
1
This is the famous Gaussian distribution function. The Gaussian distribution is only valid in the
limits N>>1 and n1 >>1
Activity
Using the Taylor expansion and derive the Gaussian distribution
Solution
Let us expand lnP around n = n~ . Note that we expand the slowly varying function lnP(n), instead of
the rapidly varying function P(n), because the Taylor expansion of P(n) does not converge
sufficiently rapidly in the vicinity of n = n~ to be useful. We can write

ln P(n~  n)  ln P(n~)  B1 
B2  ...
2
2
where
Bk 
d k ln P
dnk nn


 nn 2 
1

Pn   P n1 exp 
2
 2 * n1  
 
The constant P( n1 ) is most conveniently fixed by making use of the normalization condition
For discrete case
N
 P (n )  1
N
n1  0
!
For continues case
N
 P (n)dn  1
N
0
for a continuous distribution function. Since we only expect P (n) to be significant when n lies in the
relatively narrow range n1   * n1 , the limits of integration in the above expression can be replaced
by   with negligible error. Thus,


 nn 2 
1
 dn  P n1
P n  exp 
2


2

*
n



1



 



2 * n1  exp  x 2 dx  1

1.7 The principle of equal a priori probabilities
Activity
Take a bottle of gas which is isolated with the external environment.
Solution
In this situation, we would expect the probability of the system being found in one of its accessible
states to be independent of time.
This implies that the statistical ensemble does not evolve with time.
Individual systems in the ensemble will constantly change state; but the average number of systems
in any given state should remain constant.
Thus, all macroscopic parameters describing the system, such as the energy and the volume, should
also remain constant.
There is nothing in the laws of mechanics which would lead us to suppose that the system will be
found more often in one of its accessible states than in another. We assume, therefore, that the system
is equally likely to be found in any of its accessible states. This is called the assumption of equal a
priori probabilities, and lies at the very heart of statistical mechanics.
1.8 The relaxation time
Activity
Take an isolated many particle systems will eventually reach equilibrium, irrespective of its initial
state.
Number of particle
Solution
Relaxation Time Fluctuation Time
Time
The typical time-scale for this process is called the relaxation time, and depends in detail on the
nature of the inter-particle interactions.
The principle of equal a priori probabilities is only valid for equilibrium states.
The relaxation time for the air in a typical classroom is very much less than one second. This suggests
that such air is probably in equilibrium most of the time, and should, therefore, be governed by the
principle of equal a priori probabilities.
1.8 Behavior of the density of states
A macroscopic system is one which has many degrees of freedom denote the energy of the system by
E. We shall denote by E  the number of states whose energy lies between E and E+dE in a
system. Let  E  denote the total number of possible quantum states of the system which are
characterized by energies less than E. Clearly  E  increase when E increases. The number of states
E  in the range between E and E+dE is then
E    E  E    E  

E
E
Activity
Consider the case of a gas of N identical molecules enclosed in container of
volume V. The energy of the system can be written
E=K+U+Eint Where
K=K(p1,p2,….pN)=
1 N 2
 pi , U=U(r1,r2,…rN)
2m i 1
Considering the system for mono atomic ideal gas
U=0, Eint=0
Solution
The number of states  (E, V) lying between the energies E and E+  E is simply equal to the
number of cells in phase-space contained between these energies.
Py
R
Px
E
E+dE
In other words,  (E, V) is proportional to the volume of
phase-space between these two energies:
  E ,V  
E  E

d 3r1...d 3 rN d 3 p1...d 3 pN
E
Here, the integrand is the element of volume of phase-space, with
d3 r = dxi dyi dzi
d3 p = dpi x dpi y dpi z ,
the number of states E  lying spherical shell between energies E and E+dE is given
  BV N E
3N
2
In other words, the density of states varies like the extensive macroscopic parameters of the system
raised to the power of the number of degrees of freedom. An extensive parameter is one which
scales with the size of the system (e.g., the volume). Since thermodynamic systems generally
possess a very large number of degrees of freedom, this result implies that the density of states is an
exceptionally rapidly increasing function of the energy and volume. This result, which turns out to
be quite general, is very useful in statistical thermodynamics.
Problem
1. A penny is tossed 400 times. Find the probability of getting 215 heads. (Suggestion: use the
Gaussian approximation)
Solution
A penny is tossed 400 times. Find the probability of getting 215 heads is given by the Gaussian
approximation
P ( n) 


 nn 2 
1

exp 
2
2  * n1
 2 * n1  
1
where
N=400, n1=251, p=1/2, q=1/2
 * n1  Npq  400 x1 / 2 x1 / 2  100  10
n1  Np
 * n1 2  100 ,
n1  200
Substituting in the Gaussian equation
P(251,400) 
1
10 2
e

251 2002
200
P(251,400)  1.3x10 2
Problem
2. A particle of mass m is free to move in one dimension. Denote its position coordinate by x and its
momentum by p. Suppose that this particle is confined with a box so as to be located between x=0
and x=L, and suppose that its energy is known to lie between E and E+dE. Draw the classical phase
space of this particle, indicating the regions of this space which are accessible to the particle
Solution
Let us represent the particle motion in the coordinate of p, x
p
P+ dp
p
0
L
x
The particle with position x and momentum p position lies between x=0 and x=L, energy lies
between E and E+dE
The momentum of the particle is given by
E=p2 /2m
p  2mE
the accessible state in the phase space E  
energy E in phase space is given by
d E 
E the number of states which have an
dE
 E  = p  2mE
3. What is the probability of throwing a three or a six with one throw of die?
solution
the probability that the face exhibit either 3 or 6 is
1 1 1
 
6 6 3
ACTIVITY 2: Macroscopic Parameters and their Measurements
You will require 25 hours to complete this activity. In this activity you are guided with a series of
readings, Multimedia clips, worked examples and self assessment questions and problems. You are
strongly advised to go through the activities and consult all the compulsory materials and use as
many as possible useful links and references.
Specific Teaching and Learning Objectives




Define the work done and the internal energy
Describe the absolute and entropy relation
State the heat capacity at constant V,P
Define and derive the entropy
Summary of the Learning Activity
This activity defines the relation between work done and internal energy of a system. The concept
of Entropy is derived for a combined system and problems related to entropy and density of states
for the equilibrium are treated.
List of Required Readings
Reading #2:.
Complete reference : From Classical Mechanics to Statistical Mechanics
From Draft chapters of Thermal and Statistical Physics
URL : http://stp.clarku.edu/notes/chap1.pdf
Accessed on the 23rd September 2007
Abstract :
Thermal and Statistical Physics: From Classical Mechanics to Statistical Mechanics; thermodynamic
Concepts and Processes; Concepts of Probability;The Methodology of Statistical Mechanics; Magnetic
Systems; Noninteracting Particle Systems; Thermodynamic Relations and Processes; Theories of Gases
and Liquids; Critical Phenomena and the Renormalization Group; Introduction to Many-Body Perturbation
Theory...
Rationale:
This chapter covers most of the topics in the second and third activities of the module...
List of Relevant Resources
1. Reference http://en.wikipedia.org/wiki/Absolute_zero
Date Consulted:-Nov 2006
Description: - Absolute zero is the lowest possible temperature, occurring when no heat
energy remains in a substance. Absolute zero is the point at which particles have a
minimum energy, determined by quantum mechanical effects, which is called the zero-point
energy. By international agreement, absolute zero is defined as precisely 0 K on the Kelvin
scale, which is a thermodynamic (absolute) temperature scale, and -273.15°C on the Celsius
scale.[1] Absolute zero is also precisely equivalent to 0 °R on the Rankine scale (also a
thermodynamic temperature scale), and –459.67 °F on the Fahrenheit scale
2. Reference:-: http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Entropy/Entropy.html
Date Consulted:- February 1999
Description:The entropy is a measure of the probability of a particular result.
The entropy is a measure of the disorder of a system.
The entropy measures the heat divided by the absolute temperature of a body.
Introduction to the Activity
The laws that govern the relationships between heat and work are studied in thermal physics. Since
heat is a form of energy and work is the mechanism by which energy is transferred, these laws are
based on the basic principles that govern the behaviour of other types of energy such as the
principle of conservation of energy.
In this activity you will be guided through a series of tasks to understand heat as a form of energy
and define terms like heat capacity, heat of fusion and heat of vaporization.
Detailed Description of the Activity (Main Theoretical Elements)
Figure: compression of gas molecules
Macroscopic Measurements:
 Work and internal energy
 Absolute temperature
 Heat capacity and specific heat capacity
 Entropy
2.1 Work and internal energy
The macroscopic work done by a system is determined by the volume of a system if changed quasistatically from Vi to V f and throughout this process the mean pressure of the system has the
measurable value p V  .
Vf
W
 pdV
Vi
If the system is isothermally insulated so it can’t absorb any heat then Q=0
The internal energy  E  W
Activity
Consider a system that consists of the cylinder containing a gas. Supply the external energy to the
system by switching the circuit. What do you observe? Consider a standard macrostate i of volume
Vi and mean pressure pi , where E  Ei . How would one determined the mean energy E j of any
other macrostate j of volume V j and the mean pressure p j ?
Figure A system consists of cylinder containing gas.
The volume V of the gas is determined by the position of the piston. The resistance can brings
thermal contact to the system.
Solution
The microstate of the system can be specified by the two parameters, volume V and internal
energy E . Each macrostate can be represented by a point on pV diagram.
As the gas expand from 1 to its final volume 3 the mean pressure decrease to some value p3 and the
work done by the piston W13
To bring the pressure p3 without changing the volume, work is done by the electric resistance by an
amount WR and if the  amount of energy consumed by the resistance then the energy supplied by
the external system is WR   .
The total internal energy of the system in state in state 2 is then given by
E  E a  Wac  (WR   )
The amount of heat absorbed from a macrostate 1 to a macrostate 2 is given by
E 2  ( E 2  E1 )  W12
Heat
The heat Qab absorbed by the system in going from a macrostate a to another macrostate is given by
Qab   Eb  Ea   Wab
2.2 Absolute temperature
Properties of absolute temperature
1. The absolute temperature provides one with a temperature parameter which is completely
independent of the nature of the particular thermometer used to perform the temperature
measurement.
2. The absolute temperature T is a parameter of fundamental significance which enters all the
theoretical equations. Hence all the theoretical predictions will involve this particular
temperature.
Activity
From the equation of state p 
N
kT = nkT
V
2.3 Heat capacity and specific heat
Consider a macroscopic system whose macrostate can be specified by its absolute temperature T and
some other macroscopic parameter y (y might be volume or mean pressure)
Activity
 Take a macroscopic system at temperature T, an infinitesimal amount of heat dQ is added to
the system and the other parameters y kept fixed.
 The resulting change dT in temperature of the system depends on the nature of the system as
well as on the parameters T and y specifying the macrostate of the system
Result
The specific heat capacity at constant y is defined by
 dQ 
Cy  

 dT  y
The specific heat per mole or heat capacity per mole is thus defined by
cy 
1

Cy 
1  dQ 
  dT  y
Eventually the specific heat per gram is defined as
c 'y 
1
1  dQ 
Cy  

m
m  dT  y
Task
Take a gas or a liquid whose macrostate can be specified by two parameters say the temperature T
and volume. Calculate the heat capacity at constant volume C and at constant pressure C p
Figure Diagram illustrated specific heat measurement of a
gas kept at constant volume or at constant pressure
1. To determine C
We clamp the piston in position that the volume of the system is kept fixed.
In this case the system cannot do any work, and the heat dQ added to the system goes entirely to
increase the internal energy of the system
dQ  dE
2. To determine C p
The piston left completely free to move the weight of the piston being equal to the constant force per
unit area (mean pressure) on the system
In this case the piston will move when heat dQ is added to the system; as the result the system does
also mechanical work. Thus the heat dQ is used both to increase the internal energy of the system and
to do mechanical work on the piston
dQ  dE  pdV which is the fundamental law of thermodynamics
From the result we expected
i). dE is increase by small amount( and hence the temperature T will also increase by smaller
amount) in the second case compared to the first.
ii). C p  C
2.3.1 Heat capacity using the second law of thermodynamics
The second law of thermodynamics is given by dQ  TdS the heat capacity
 S 
Cy  T 

 T  y
If all external parameters of the system kept constant, then the system dose no macroscopic work,
dW  0 then the first law reduced to dQ  dE
 E 
 S 
CV  T 

 
 T V  T V
Example
Let us consider heat measurements by the method of mixtures in terms of the specific heats of the
substance involved. Consider that two substances A and B, of respective masses m A and mB , are
brought into thermal contact under condition where the pressure is kept constant. Assume that before
the substance are brought into thermal contact their respective equilibrium temperature are TA and
TB respectively. Compute the final temperature T f
Solution
2.4 Entropy
The entropy can readily be determined by using the second law dQ  TdS for an infinitesimal quasistatic process.
Given any macrostate b of the system, one can find the entropy difference between this state and
some standard state a to state b and calculating for this process
b
dQ
T
a
Sb  Sa  
Suppose that the macrostate of a body is specified by its temperature, since all its other parameters
are kept constant.
b
S Tb   S Ta   
a
dQ b C y T ' dT '

T Ta
T'
T
then
S Tb   S Ta   C y ln
Tb
Ta
Problem
Consider two system A and system B with constant specific heat C ' A and C 'B and originally at
respective temperature TA and TB , are brought into thermal contact with each other. After the system
come to equilibrium, they reach a come final temperature T f . What is the entropy change of the
entire system in this process?
System A,TA
B,TB System
Isolated system
Answer
To calculate the entropy change of system A, we can imagine that it is brought from its initial
temperature TA to its final temperature T f by a succession of infinitesimal heat additions.
dQ  mAC ' A dT
T
f
Tf
m C ' dT
dQ
dS 
 S A (T f )  S A (TA )   A A
 mAC ' A ln
T
T
TA
TA
Similarly for the system B
T
f
Tf
m C ' dT
dQ
dS 
 S B (T f )  S B (TB )   B B
 mB C 'B ln
T
T
TB
TB
The total entropy change
S A  S B  mAC ' A ln
Tf
TA
+ mB C 'B ln
Tf
TB
Problems
(a) One kilogram of water at 00C is brought into contact with a large heat reservoir at 1000C. When
the water has reached 1000C, what has been the change in entropy of the water? Of the heat
reservoir? Of the entire system consisting of both water and heat reservoir?
b) If the water had been heated from 00C to 1000C by first bringing it is contact with a reservoir
at 500C and then with a reservoir at 1000C, what would have been the change in entropy of the
entire system?
C) Show how the water might be heated from 00C to 1000C with no change in the entropy of the
entire system.
Answer
Entropy of water
dS 01000 C 
=
dQ
where dQ  mCdT
T
mCdT
T
373k
S  mC
dT
T
273k

S  mC ln
Tf
Ti
S water  mC ln
373
(where mass of water =1kg)
273
= 1310J/K
The entropy of reservoir
The amount of heat loss by the reservoir
Qwater  Qreservoir
Qreservoir  mC (T f  Ti )
S reservoir  
mC(T f  Ti ) water
T373
=-1126J/K
Total entropy
S total  S reservoir + S water
S total  
mC(T f  Ti ) water
Stotal  184J/K
T373
+ mC ln
373
273
ACTIVITY 3: Statistical Thermodynamics
You will require 30 hours to complete this activity. In this activity you are guided with a series of
readings, Multimedia clips, worked examples and self assessment questions and problems. You are
strongly advised to go through the activities and consult all the compulsory materials and use as
many as possible useful links and references.
Specific Teaching and Learning Objectives
Define the work done and the internal energy
Describe the absolute and entropy relation
State the heat capacity at constant V,P
Define and derive the entropy
Summary of the Learning Activity
In this activity you will investigate the relationship between pressure, temperature, volume, and the
amount of gas occupying an enclosed chamber. This activity consists of three sections. In section
one amount of gas and the importance of Avogadro’s number is discussed. In the second section
the relationship between pressure and volume will be covered. In part three the relationship
between pressure and volume as well the amount of gas present in a chamber will be determined.
The results learnt in these tasks will be used to derive the Ideal Gas Law.
List of Required Readings
Reading #2:.
Complete reference : From Classical Mechanics to Statistical Mechanics
From Draft chapters of Thermal and Statistical Physics
URL : http://stp.clarku.edu/notes/chap1.pdf
Accessed on the 23rd September 2007
Abstract :
Thermal and Statistical Physics: From Classical Mechanics to Statistical Mechanics; thermodynamic
Concepts and Processes; Concepts of Probability;The Methodology of Statistical Mechanics; Magnetic
Systems; Noninteracting Particle Systems; Thermodynamic Relations and Processes; Theories of Gases
and Liquids; Critical Phenomena and the Renormalization Group; Introduction to Many-Body Perturbation
Theory...
Rationale:
This chapter covers most of the topics in the second and third activities of the module...
Reading 5:
Complete reference: Introduction To Statistical Mechanics | Free eBooks Download!
From
URL:http:// www.ebookee.com/Introduction-To-Statistical-Mechanics_139834.html Accessed
Abstract
Rationale:
Reading 6:
r.
Complete reference: Molecular Driving Forces: Statistical Thermodynamics in
Chemistry ...
From
URL http:// www.ebookee.com/Molecular-Driving-Forces-Statistical-Thermodynamics-in-Chemistry-ampBiology_145376.html
Abstract:
Rationale:
.
List of Relevant Readings for all activities
1. Reference:- Kittel C. and Kroemer H., (1980) Thermal Physics, 2nd ed., W. H. Freeman
and Co., San Francisco, CA..
Abstract:
Rationale: This classic reference on thermal physics is recommended for a serious
student of Physics. The contents have been treated in detail with adequate mathematical
support.
2. Reference: Fundamentals of statistical and thermal physics: F. Reif (McGraw-Hill, New
York NY,1965).
Abstract:
Rationale: This reading provides easy sources of information. The contents have been treated
in lucid manner with adequate mathematical support.
List of Relevant Resources
1. Reference http://en.wikipedia.org/wiki/Entropy
Date Consulted:Description:2. Reference:-: http://lectureonline.cl.msu.edu/~mmp/kap10/cd283.htm.
Date Consulted: Description: - This Java applet helps you understand the effect of temperature and
volume on the number of collisions of the gas molecules with the walls. In the applet, you
can change the temperature and volume with the sliders on the left side. You can also
adjust the time for which the simulation runs. The applet counts all collisions and displays
the result after the run. By varying temperature and volume and keeping track of the
number of collisions, you can get a good feeling of what the main result of kinetic theory
will be.
3. Reference: video.google.com
Date Consulted: Nov 2006
Complete Reference: - Computer calculation of Phase Diagrams.
http://video.google.com/videoplay?docid=1397988176780135580&q=Thermodynamics&
hl=en
Rationale: Thermodynamic models of solutions can be used together with data to
calculate phase diagrams. These diagrams reveal, for a given set of all parameters (such as
temperature, pressure, and magnetic field), the phases which are thermodynamically
stable and in equilibrium, their volume fractions and their chemical compositions...
List of Relevant Useful Links
1. Title: The P-V Diagram and Engine Cycles
URL: http://www.antonineeducation.co.uk/Physics_A2/options/Module_7/Topic_4/topic_4.htm
Abstract: This site contains a good summary on Representation of processes on p – V
diagram, Estimation of work done in terms of area below the graph, Expressions for work
done are not required except for the constant pressure case, W = p  V , Extension to cyclic
processes: work done per cycle = area of loop
2. Title: Avogadro's Number
URL: http://njsas.org/projects/atoms/avogadro.php
Abstract: A historic as well as scientific of the origin of Avogadro’s number is presented
on this page
Introduction to the Activity
The Ideal Gas Law describes the relationship between pressure, volume, the number of atoms or
molecules in a gas, and the temperature of a gas. This law is an idealization because it assumes an
“ideal” gas. An ideal gas consists of atoms or molecules that do not interact and that occupy zero
volume.
A real gas consists of atoms or molecules (or both) that have finite volume and interact by forces of
attraction or repulsion due to the presence of charges. In many cases the behaviour of real gases can
be approximated quite well with the Ideal Gas Law. and this activity focuses on the description of
an ideal gas.
Detailed Description of the Activity (Main Theoretical Elements)
Introduction
Thermal relay switch and dispersion systems (Boltzmann and Gibbs factors, partition and connection
functions with thermodynamics
3.1 Equilibrium conditions and constraints
Consider an isolated system whose energy is specified to lie in a narrow range. As usually, we denote
by  then number of states accessible to this system. From the fundamental postulate we know that in
equilibrium such a system is equally likely to be found in any one of these states. If a system has a
constraint y1,y2,…yn then the accessible state given by    y1 , y 2 ,... y n  .
If some constraints of an isolated system are removed, the parameters of the system tend to readjust
themselves in such a way that    y1 , y 2 ,... y n  approaches a maximum  f   i
3.2 Thermal interaction between macroscopic systems
Activity
Consider a purely thermal interaction between two macroscopic systems, A and A’,
…………
…………
A
………………
………………
A’
Energy of the systems E and E’, the external parameters are constant, so that A and A’ cannot do
work on one another and the systems are thermally contact heat will exchange. Considering the
energy width  E
 Let us calculate the accessible state
 The temperature at equilibrium
 The entropy at equilibrium
Result
The number of microstates of A consistent with a macrostate in which the energy lies in the range E
to E +  E is denoted  (E). Likewise, the number of microstates of A’ consistent with a macrostate
in which the energy lies between E’ and E’ +  E is denoted  ’(E ).
The combined system A(0) = A + A’ is assumed to be isolated (i.e., it neither does work on nor
exchanges heat with its surroundings). The number of accessible to the entire system A0 let us denote
by  0 (E) when A has energy between E and E+dE.
The probability
P(E)=C  0 (E)
Total accessible state

 0 E   E ' E 0  E

Temperature at equilibrium
The probability of system A having the energy an energy near E is given by

P(E)=C E ' E 0  E

~
To locate the maximum position of P(E) at E= E
 ln P ( E ) 1 P

=0
E
P E
ln P( E)  ln C  ln E   ln ' E'
 ln P ( E )  ln E   ln ' E '


=0
E
E
E
where E0=E+E’ which is dE=-dE’ then
 ln E   ln ' E '

=0
E
E '
~
~ 
 E   ' E'
Entropy of the combined system
Activity
~
~
where E and E ' denote the corresponding energies of A and A’ at the maximum, and where we have
introduced the definition
 E  
 ln 
1
kT  where k is some positive constant having the dimension of energy and whose
E

magnitude in some convenient arbitrary way.
The parameter T is then defined as kT 
S
E
Solution
Where we have introduced the definition S  k ln  this quantity S is given the name of entropy


Total accessible state  0 E   E ' E 0  E and taking the logarithm

ln 0  E   ln   E   ln  ' E 0  E

S  0  S  S '
The condition of maximum probability is expressible as the condition that the total entropy
S  S ' max imum entropy occurs when T=T’
3.3 The approach to thermal equilibrium
If the two systems are subsequently placed in thermal contact, so that they are free to exchange heat
energy until the two systems attain final mean energies E f and E f '
which are
f  f'
It follows from energy conservation that
E f  E ' f  E i  Ei
'
The mean energy change in each system is simply the net heat absorbed, so that
Q  E f  Ei ;
Q '  E' f E ' i
The conservation of energy then reduces to
Q+Q’=0:
It is clear, that the parameter  , defined

 ln 
E
Temperature
1. If two systems separately in equilibrium are characterized by the same value of the
parameter, then the systems will remain in equilibrium when brought into thermal contact
with each other.
2. If the systems are characterized by different values of the parameter, then they will not
remain in equilibrium when brought into thermal contact with each other.
If two systems are n thermal equilibrium with a third system, then they must be in thermal
equilibrium with each other
3.4 Heat reservoir
A
A’
If A’ is sufficiently large compared to A so A’ is a reservoir.
Suppose the macroscopic system A’ has ' E' accessible states and absorbs heat Q'  E ' using
Expanding ln '  E ' , Q '  at E’=Q
1   2 ln '  2
  ln ' 
Q' ...
ln ' E ' , Q'  ln ' E '  
Q' 
2  E ' 2 
 E ' 
using approximation
Q'
  ln ' 
the higher order becomes zero

Q ' =
kT '
 E ' 
ln ' E' , Q'  ln ' E' 
Q'
kT '
k (ln  '  E ', Q '  ln  '  E ') =
S ' 
3.5
Q '
T'
Q '
For a heat reservoir
T'
Dependence of the density of states on the external parameter
Activity
Now that we have examined in detailed the thermal interaction between systems, let us turn to the
general case where mechanical interaction can also take place, i.e. where the external parameters of
the systems are also free to exchange. We begin, therefore, by investigating how the density of states
depends on the external parameters.
Solution
E+
E
Figure shaded area indicate the energy range
occurred by states with a value of whose energy
changes from E to E+when the external
parameter is changed from x to x+dx
The number of states accessible to the system microstates accessible to the system when the overall
energy lies between E and E +  E depends on the particular value of x, so we can write
  E, x .
The number of states  (E, x) whose energy is changed from a value less than E to a value greater
than E when the parameter changes from x to x + dx is given by the number of microstates per unit
energy range multiplied by the average shift in energy of the microstates, Hence
 E , x  
E , x  E r
dx
E x
where the mean value of  Er/  x is taken over all accessible microstates (i.e., all states where the
energy lies between E and E +  E and the external parameter takes the value x). The above equation
can also be written
 E , x   
 E , x 
X dx
E
where
X E , x   
E r
is the mean generalized force conjugate to the external parameter x.
x
Consider the total number of microstates between E and E +  E. When the external parameter
  
changes from x to x + dx, the number of states in this energy range changes by 
dx . In symbols
 x 
E , x 

dx   E    E  E  
E
x
E
which yields
 
   X

x
E


   X

X


X 
x
E
E
E
then
 ln   ln 
X

X
x
E
E
 ln   ln 

X X
x
E
Thus,
 ln 
  X
x
where X  is the mean generalized force conjugate to the parameter x
Infinitesimal quasi static process
Activity
Consider a quasi static process in which the system A, by virtue of its interaction with systems A',
is brought from an equilibrium state describe by E and x   1, 2,...n  to an infinitesimally
different, equilibrium state described by E  dE and x  dx .
What is the resultant change in the number of states  accessible to A?
Solution
The accessible state
    E; x1 ,..., xn 
d ln  
n
 ln 
 ln 
dE  
dx
E
 1 x
Substituting the in the above equation  
 ln 
,
E
 ln 
  X
x


d ln     dE   X  dx 



dW   X  dx

Then d ln     dE  dW    dQ
The fundamental relation valid for any quasi-static infinitesimal process
dQ  TdS   dE  dW  or equivalently
dS 
dQ
T
Adiabatic process
dQ  0 which asserts
dS  0
Equilibrium
Consider the equilibrium between the systems A and A’ in the simple case the external parameters
are the volumes V and V’ of the two systems. The number of state available to the combined
system A0 is given by the simple product.
0  E,V     E,V   '  E ',V '
Activity
Using the accessible state given for the combined system derive the equation that guarantee for
thermal and mechanical equilibrium.
Solution
For the combined system the accessible state given as 0  E,V     E,V   '  E ',V '
Taking the logarithm
ln 0  E,V   ln   E,V   ln  '  E ',V '
The total entropy of the system given by
S0  S  S '
At the maximum value the total accessible state d ln  0  0
d ln 0  E,V   d ln   E,V   d ln  '  E ',V '  0
d ln  
 ln   E ,V 
V
dV 
 ln   E,V 
E
dE +
 ln  '  E ',V '
V '
where
p

 ln   E ,V 
V
 ln   E , V 
E
similarly  ' p ' 
similarly  ' 
 ln  '  E ',V '
V '
 ln  '  E ',V '
E '
Substituting in the above equation
d ln    pdV   dE   ' p ' dV '  ' dE ' =0
from the combined system
E  E '  E0
V V '  V 0
Then dE  dE ',
dV  dV '
Substituting in the above equation
 pdV   dE   ' p ' dV   ' dE =0
Collecting terms
 pdV   ' p ' dV =0
 dE   ' dE =0
 dE   ' dE
Then at thermal equilibrium
 '
 pdV =  ' p ' dV
Then mechanical equilibrium
p = p'
dV '
 ln  '  E ',V ' 
E '
dE ' =0
Thermodynamics laws and basic statistics relation
Summery of thermodynamic laws
 Zero law: If two systems are in thermal in equilibrium with a third system, they must be in
thermal equilibrium with each other.
 First law: an equilibrium macrostate of a system can be characterized by a quantity E
(called internal energy) which has the property that for an isolated E =constant. If the
system is allowed to interact and thus goes from one macrostate to another, the resulting
change in E can be written in the form E  W  Q
 Second law: an equilibrium macrostate of a system can be characterized by a quantity S
(called entropy ) which has the property that

In any process in which a thermally isolated system goes from one macrostate to
another, the entropy tends to increase S  0

If the system is not isolated and under goes a quasi-static infinitesimal process in
dQ
which it absorbs heat dQ, then dS 
T
 Third law: The entropy S of a system has the limiting property that T  0 , S  S0 where
S 0 is a constant independent of all parameters of the particular system
ACTIVITY 4: Some Application of Statistical and
Macroscopic Thermodynamics
You will require 40 hours to complete this activity. In this activity you are guided with a series of
readings, Multimedia clips, worked examples and self assessment questions and problems. You are
strongly advised to go through the activities and consult all the compulsory materials and use as
many as possible useful links and references.
Specific Teaching and Learning Objectives

Calculate the thermodynamics relations

Derive the equation for the canonical distribution and kinetic theory of dilute gasses in
equilibrium
Summary of the Learning Activity
List of Required Readings (for the Learning Activity).
Useful Link # 1
Title MACROSCOPIC AND STATISTICAL THERMODYNAMICS
URL: http://www.worldscibooks.com/physics/6031.html
Screen Capture
By Yi-Chen Cheng (National Taiwan University, Taiwan)
Description
This textbook addresses the key questions in both classical thermodynamics and statistical
thermodynamics: Why are the thermodynamic properties of a nano-sized system different from
those of a macroscopic system of the same substance? Why and how is entropy defined in
thermodynamics, and how is the entropy change calculated when dissipative heat is involved?
What is an ensemble and why is its theory so successful?
They include the introduction of the grand canonical ensemble, the grand partition function
and its application to ideal quantum gases, a discussion of the mean field theory of the Ising
model and the phenomenon of ferromagnetism, as well as a more detailed discussion of ideal
quantum gases near T = 0, for both Fermi and Bose gases.
Reading 2:.
Complete reference: Fundamentals Of Statistical And Thermal physics
From
URL: http://www.ebookee.com/Reif-Fundamentals-Of-Statistical-And-Thermal-Physics_
Abstract:
Rationale:
Reading 3
Complete reference: Maxwell Velocity Distribution simulation
URL: http://www.kfki.hu/physics/physedu/kinetic_gas_model/exp/veldistr.html
Abstract: A graph which show the Maxwell velocity distribution
Reading 4
http://colos1.fri.unilj.si/~colos/COLOS/TUTORIALS/JAVA/THERMODYNAMICS/THERMO_UK/HTML/Vel_
Distri.html
Abstract: Distribution of particles in their energy leel
Detailed Description of the Activity (Main Theoretical Elements)
Partition function and their properties Ideal gas, validity of classical approximation, equipartition
theory, harmonic oscillator at high temperature Distribution of particles Maxwell Boltzmann, Bose
Einstein and Fermi-Dirac statistics
Introduction to the Activity
The gas laws described in activity 3 were found by experimental observation, but Boyle’s law and
Charles’ law are not obeyed precisely at all pressures. A gas which obeys the above laws perfectly
at all pressures would be a “perfect” or “ideal” gas, and the kinetic theory resulted from an attempt
to devise a mechanical model of such a gas based on Newton’s laws of motion.
First Law of thermodynamics
dQ  dE  dW
If the process is quasi-static, the second law gives
dQ  TdS
The work done by the system when the volume is changed by an amount dV in the process is given
by
dW  pdV
Then the fundamental thermodynamics
TdS  dE  pdV
The equation of state of an ideal gas
Macroscopically, an ideal gas is described by the equation of state relating its pressure p, volume V,
and the absolute temperature T. For v moles of gas, this equation of state is given by
pV  vRT
The internal energy of an ideal gas depends only on the temperature of the gas, and is independent of
the volume
E = E (T) independent of V.
Entropy
The entropy of an ideal gas can readily be computed from the fundamental thermodynamic relation
TdS  dE  pdV
ds  vCV
dT vR

dV
T
V
Adiabatic expression or compression
pV   cons tan t
V  1T  cons
Thermodynamic potentials and their relation with thermodynamic variables
The thermodynamic state of a homogeneous system may be represented by means of certain selected
variables, such as pressure p, volume v, temperature T, and entropy S. Out of these four variables ,
any two may vary independently and when known enable the others to be determined. Thus there are
only two independent variables and the others may be considered as their function.
The first and the second law of thermodynamics give the four thermodynamic variables
dQ  dE  pdV the first law of thermodynamics
dQ  TdS the second law of thermodynamics
dE  TdS  pdV combined the two laws
Activity
For two independent variables S and V using the fundamental thermodynamics derive the
thermodynamics state of a homogeneous system.
Answer
The independent thermodynamic function
E  E  S ,V  the internal energy
Differentiating the function
 E 
 E 
dE  
 dS  
 dV
 S V
 V S
From the fundamental thermodynamic equation
dE  TdS  pdV
Comparing the two equations we can get
 E 
T 

 S V
 E 
p  

 V  S
Using the second order differential and dE is a perfect differential. E must be independent of the
order of differentiation.
    E 
 T 

   

 V  S  S V  V  S
    E 
 p 
  
   
 S V  V  S
 S V
Then
 T 
 p 

   
 V  S
 S V
Activity
For two independent variables S and p using the fundamental thermodynamics derive the
thermodynamics state of a homogeneous system.
Answer
The independent thermodynamic function
dE  TdS  pdV
dE  TdS  d  pV   Vdp
d  E  pV   TdS  Vdp
let H  E  pV which we call it enthalpy
H  H  S, p
dH  TdS  Vdp
Differentiating the function
 H 
 H 
dH  
 dp
 dS  
 S  p
 p  S
From the thermodynamic equation
dH  TdS  Vdp
Comparing the two equations we can get
 H 
T 

 S  p
 H 
V 

 p  S
Using the second order differential and dH is a perfect differential. H must be independent of the
order of differentiation.
    H 
 T 
  

 
 p  S  S  p  p  S
    H   V 
 
  

 S  p  p  s  S  p
Then
 T   V 

 

 p  S  S  p
Activity
For two independent variables T and V using the fundamental thermodynamics derive the
thermodynamics state of a homogeneous system.
Answer
The independent thermodynamic function
dE  TdS  pdV
dE  d TS   SdT  pdV
d  E  TS   SdT  pdV
let F  E  TS which we call it Helmholtz free energy
F  F T ,V 
dF   SdT  pdV
Differentiating the function F  F T ,V 
 F 
 F 
dF  
 dT  
 dV
 T V
 V T
From the thermodynamic equation
dF   SdT  pdV
Comparing the two equations we can get
 F 
S  

 T V
 F 
p  

 V T
Using the second order differential and dH is a perfect differential. H must be independent of the
order of differentiation.
    F 
 p 

 
  

 T V  V T
 T V
    F 
 S 

 
  

 V T  T V
 V T
Then
 p   S 

 

 T V  V T
Activity
For two independent variables T and p using the fundamental thermodynamics derive the
thermodynamics state of a homogeneous system.
Answer
The independent thermodynamic function
dE  TdS  pdV
dE  d TS   SdT d  pV   Vdp
d  E  TS  pV   SdT  Vdp
let G  E  TS  pV which we call it Gibbs free energy
G  G T , P 
dG   SdT  Vdp
Differentiating the function G  G T , p 
 G 
 G 
dG  
 dp
 dT  
 T  p
 p T
From the thermodynamic equation
dG   SdT  Vdp
Comparing the two equations we can get
 G 
S  

 T  p
 F 
V 

 p T
Using the second order differential and dH is a perfect differential. H must be independent of the
order of differentiation.
    G   V 
 

 

 T  p  p T  T  p
    G 
 S 
  
   
 p T  T  p
 p T
Then
 S 
 V 

   
 T  p
 p T
Summary for the thermodynamics function
Maxwell relations
The entire discussion of the preceding section was based upon the fundamental thermodynamics
relation
dE  TdS  pdV
 T 
 P 

   
 V  s
 S  v
 T 
 V 

  

 p  s  S  p
 S 
 p 

  
 V  T  T  v
 S 
 V 
   

 T  p
 p  T
Thermodynamics functions
E.............................E  E ( S , V )
H  E  pV ............H  H ( S , p )
F  E  TS ..............F  F (T , V )
G  E  TS  pV ......G  G (T , p)
Next we summarize the thermodynamic relations satisfied by each of these function
dE  TdS  pdV
dH  Tds  Vdp
dF   SdT  pdV
dG   SdT  Vdp
Specific heats
Consider any homogeneous substance whose volume V is the only relevant external parameter.
The heat capacity at constant volume is given by
 dQ 
 S 
CV  
  T 
 dT V
 T V
The heat capacity at constant pressure is similarly given by
 dQ 
 S 
Cp  
  T

 dT  p
 T  p
Activity
a) For an infinitesimal process of a system the molar specific heat at constant volume and at
constant pressure is given by CV and C p respectively. Show that C p  C v  R which shows
C p  Cv
b) Using the heat capacity and thermodynamics function relation show that the heat capacity at
constant volume and at constant pressure related by C p  CV 
Solution for a
At constant volume dV  0
Then first law of thermodynamics reduced to dQ  dE
 2V
k
Using the molar heat capacity
1  dQ 
1  E 
Cv  
   
v  dT  v v  T  v
We have E which is depend on T and independent of V
 E 
dE    dT
 T  v
The change of energy depends only on the temperature change of the gas
dE  vCv dT
Substituting in the fundamental equation
dQ  vCv dT  pdV
Using the ideal gas equation
pV  vRT
pdV  vRdT
The heat absorbed at constant pressure
dQ  vCv dT  vRdT
From the definition we have
1  dQ 
Cp  

v  dT  p
Then
1  dQ 

  Cv  R
v  dT  p
C p  C v  R Which shows C p  C v
Solution for b
Considering the independent variable S  S T , p  and second law of thermo dynamics
 S 

 S 
dQ  TdS  T 
 dT    dp  it is possible to express dp in terms of dT and dV
 p  T 
 T  p

 S   p 
 S 
 p 
dQ  T   dT      dT  
 dV  where at V=constant dV=0
 T  p
 V T 
 p T  T V
 S   p 
 S 
dQ  T 
 dT    
 dT then
 T  p
 p  T  T V
 S   p 
 S 
 Q 
    


  T
 T  p  p  T  T V
 T  p
 S   p 
C p  CV    
 from the Maxwell relation
 p  T  T V
The volume coefficient of expansion of the substance

1  V 
1

 =V  T  p V
 S 
 
 p  T
 S 
  =- V
 p  T
we can express V in terms of T and P
 V 
 V 
 dp =0 since V= constant
dV  
 dT  
 T  p
 p  T
 V

 T
 p 

 
 V
 T V

 p


p


T
from the isothermal compressibility of the substance
 S 
 V 
   

 T  p
 p  T
k
1  V

V  p

 V 
 ,  kV  


p
T

T
 S 
  =- V
 p  T

 p 
   Substituting in the above equation which yields
 T V k
 S   p 
C p  CV    

 p  T  T V
= CV  - V
= CV 

k
 2V
k
Ensembles system -Canonical distribution
1) Isolated system
An isolated system consists of N number of particles in a specified volume v, the energy of the
system being known to lie in some range between E and E + dE. The fundamental statistical postulate
asserts that in an equilibrium situation the system is equally likely to be found in any of its accessible
states. Thus, if the energy of the system in state r is denoted by Er, the probability Pr of finding the
system in state r is given by
Pr  C
If E<Er<E+ E
Pr  0
Other wise
P
Normalized
r
1
An ensemble representing an isolated system in equilibrium consists then of system distributed in the
above expression. It is some times called a microcanonical ensemble.
2) In contact with reservoir
A
A’ T
We consider the case of a small system A in thermal interaction with a heat reservoir A’. What is the
probability Pr of finding the system A in any one particular microstate r of energy Er?
The combined system A0=A+A’ and from the conservation of energy E0=Er+E’
When A has an energy Er, the reservoir A’ must then have an energy near E’=E0-Er.
The number of state  ' ( E 0  Er ) accessible to A’
The probability of occurrence in the ensemble of a situation where A in state r is simply proportional
the number of state accessible to A0
Pr  C '  ' ( E ' )
P
r
1
r
Using
  ln  ' 
ln  ' ( E 0  Er )  ln  ' E 0  
Er  ....
' 
 E  E '  E 0
 
 
ln  ' ( E 0  Er )  ln  ' E 0  Er
 
' E '  ' E 0 e  Er
then
 
Pr  C ' ' E 0 e  Er
P
r
C' 
 C'  ' ( E 0 )e  Er  1
1
 ' E 0 e  Er
 
r
e  Er
The probability of the canonical distribution
Pr 
 e  Er
r
Application of canonical ensemble
Activity
Spin system: paramagnetic particles which has N atoms in a system with spin ½
Answer
Considering a system which contains N atoms, spin ½ particles interact with external magnetic field
H with the magnetic moment 
state
Magnetic moment

+
-
_
The particles has two states + or – the probability
P  Ce  E  Ce H
P  Ce  E  Ce  H from the normalization condition
P++P-=1 then we get
C
e
H
1
 e  H
P 
e  H
e  H  e  H
P 
e  H
e  H  e  H
Energy
E   H
E    H
Molecule in ideal gas
Activity
Consider a monatomic gas at absolute temperature T confined in a container of volume V. The
molecule can only be located somewhere inside the container. Derive the canonical distribution for a
monatomic non interacting gas
Solution

The energy of the monatomic gas in a system is given by purely kinetic
1
P2
E= mV 2 =
2
2m

If the molecule’s position lies in the range between r and r+dr and momentum lies between P
and P+dP then the volume in phase space is given by d3rd3P=(dxdydz)dpxdpydpz)

The probability that the molecule has position lying in the range between r and r+dr and
momentum in the range between p and p+dp
 d 3rd 3 p    2pm
P(r,p)d rd p  
e
3
 h0 
2
3

3
The probability that P(p)d3p that a molecule has momentum lying in the range between p and
p+dp
P p d p   Pr , p d rd p  Ce
3
3
3
 2

   p

 2m 
r 
where we have p=mv d3p=md3v
Then
P ' V   P  p  d 3 p  Ce   mV
2
/2
Generalized force
Activity
Using the canonical distribution write the generalized force
d3p
Solution
If the a system depends on the external parameter x, then Er=Er(x) and from the definition of the
generalized force we have that
Xr  
Er
x
the mean value of the generalized force we can write as
 E r 

x 
 e   
X 
e 
 Er
r
 Er
r
then
X 
1  ln Z
 x
the average work done
dW  X dx
where the external parameter is V
dW 
p
1  ln Z
dV
 V
1  ln Z
 V
Connection of canonical distribution with thermodynamics
Activity
One can write the thermodynamics function in terms of the partition function derive the equation
Solution
The partition function given by Z  e  Er  x  so it can be represented in terms of  , x since Er=Er(x)
Z=Z (  , x) considering a small change
d ln z 
 ln z
 ln Z
dx 
d
dx

d ln Z  dW  Ed
The last term can be written inn terms of the change in E rather than the change in  . Thus
 
d ln Z  dW  d E  d E

 

d ln Z   E   dW  d E  dQ
using the second law of thermodynamics
dQ
therefore
T
dS 

S  k ln Z   E

TS  kT ln Z  E
From Helmholtz free energy F= E  TS
Thus ln Z is very simply related to Helmholtz free energy F
F= E  TS =-kT ln Z
Partition function and their properties
Z   e  Er partition function
r
If
a
system
can
be
treated
in
the
classical
approximation
then
E  E  q1 ,...qn , p1 ,.. pn  depends on some f generalized coordinates and f momenta.
The partition function in the phase space given by
Z   ... e  E ( q1 ,...qn , p1 ,... pn )
dq1 ,...dqn , dp1 ,...dpn
hf
its
energy
Activity
Consider the energy of the system is only defined by a function to which is an arbitrary additive
constant. If one changes by a constant amount  0 the standard state r the energy state becomes
E r  Er   0 using the partition function
a. Show the corresponding mean energy shifting by the amount of  0
b. Show the entropy of the combined system will not change S   S
Solution
a. The mean value of the energy when shifting the system energy by  0
Partition function
Z    e  ( Er  0 )
= e  0  e  Er = e  0 Z
r
r
ln Z   ln Z   0
from the definition E  

 ln Z
 ln Z 
and E   


 ln Z 
 ln Z

 0


E  E  0
The mean energy also shifted
b. The entropy
let the partition function in terms of the variables Z   Z  (  , x)
 ln Z 
 ln Z 
 ln Z 
 ln Z 

and  dW 
dx
d ln Z 
d 
dx where E  
x

x


Then we can find
d ln Z    Ed    dW
using the relation Ed   d   E    dE
d ln Z   d   E    dE +  dW
d ln Z   d   E  =  dE +  dW =  dQ
d (ln Z    E ) =  dQ =

S   k ln Z *   E 
dQ
kT

Since we can write
 E    E   0 and ln Z   ln Z   0 substituting in the above equation


S   k ln Z *   E  = k ( ln Z   0   E   0 ) =k ( ln Z +  E )=S
S   S the entropy keeping constant
Activity
The second remark concerns the decomposition of partition function for a system A which consists
of two parts A’ and A’’ which interact weakly with each other, if the states of A’ and A’’ are
labelled respectively by r and s find the partition function for the total system
Solution
Part A’ state r corresponding energy Er
Part A’’ state s corresponding energy Es
System A state r,s corresponding energy Ers
The partition function for the system A is given by Z
Z   e   ( Er  Es ) where Er , s  Er  Es
r ,s
then
Z   e   ( Er  Es ) =  e  ( Er )
r ,s
e 
 ( Es )
r
s
Z  Z ' Z ''
ln Z  ln Z ' ln Z ''
Calculation of Thermodynamics quantities with partition function
Activity
Consider a gas consisting of N identical monatomic molecules of mass m enclosed in a container of
volume V. The position vector of the ith molecule denoted by ri
Pi 2
, its momentum by pi the total energy given by E  
 U  r1 , r2 ,...rN  where for non-interacting
i 1 2m
N
monatomic ideal gas U=0 and write the partition function in phase space
Solution
Taking a gas consisting of N identical monatomic molecules of mass m enclosed in a container of
volume V. The position vector of the ith molecule denoted by ri
N
, its momentum by pi the total energy given by E  
i 1
Pi 2
 U  r1 , r2 ,...rN  where for non-interacting
2m
monatomic ideal gas U=0 therefore the partition function in phase space can be given as follows
3
3
3
3
  1
  d r ...d rN dp1 ...dp N
Z   exp   
p12  ...  pN 2  U  r1 ,...rN   1
h03 N

  2m

Z
1
h0
3N


 1
 exp   2m  p
2
1
 exp  U  r ,...r  d
1
Z
N
3
1

 ...  pN 2   dp13 ...dpN 3  exp  U  r1 ,...rN  d 31...d 3 N



...d 3 N = V N
  1
VN

exp   
p12  ...  pN 2   dp13 ...dpN 3
3N 
h0

  2m


where p21  p1x 2  p1 y 2  p1z 2 , dp13  dp1x dp1 y dp1z so for the ith particle

  1
V

exp   
p 2   dp
3 
h0

  2m
 

Z N

e
  p1 x 2
2m
2m
dp1x 



e
,
  p2
2m

3
 2m 
V  2m  2
 3 
=V  2 

h0   
 h 0 
3
 
2

2
m

Z   N = V  2  
  h 0  


3
 2m 
dp  

  
N
the thermodynamics quantities with the partition function
3  2m

ln Z  N ln V  ln  2
2  h0


 3
  ln  
 2

Activity
With the given partition function, find
i) The value for the mean pressure,
iii) The heat capacity,
iV) The entropy
Solution
i) The mean pressure
p
1  ln Z NkT

 V
V
2
2
Taking the logarithm
ii) The mean energy,
3
pV  NkT
ii) The total mean energy
E
E
 ln Z

3N 3
 NkT
2 2
3
2
  kT
E  N
iii) The heat capacity at constant volume
 E 
3
CV  
  R
 T V 2
iV)The entropy


S  k ln Z   E , where
E 
3
N
2

3  2m
ln Z  N ln V  ln  2
2  h0

3  2m

S  Nk ln V  ln  2
2  h0


 3
  ln  

 2
 3
3 
  ln   
2 
 2


3
3  2m k 
3   2m k  
S  Nk ln V  ln T  (ln  2   1)  where   ln  2   1
2
2  h0 
2   h 0  


3


S  Nk ln V  ln T   
2


Then the Mean Energy
   i  E '
i
e

 
e
i
 e  dp  e



 e dp  e

 i dp1 ,...dp f
i E '

dp1 ,...dp f
  i
 E '
i

i
 E'
i
i
 e  dp


 e dp
dp1 ,...dp f
dp1 ,...dp f
  i
i
i

i
considering that
i
i
i 
p2
 bp 2 then
2m
p2
 i

 i   ln  e 2 m i dpi let

i  
i 
4 mi

ln


kT
2
The Harmonic Oscillator at high thermal energy
Summery of harmonic oscillator
For a1D-harmonic oscillator which is in equilibrium with a heat reservoir at absolute temperature
T.
P2 1 2
 kx the energy of the oscillator
2m 2

E

1

E  n  
2

Is the energy of the oscillator in quantum mechanics the angular frequency  
k
m
Activity
Using the partition function of the harmonic oscillator derive the mean energy of the oscillator for
   1 and    1
Solution
The mean energy for the harmonic oscillator given by

E
e 
 En
n 0

En
e 
 En
n 0
E
 ln Z

where

Z  e
  En
n 0
Z e

 e
 1
   n  
 2
n 0
 
2

e 
 n 
n 0
Z e
Z e



2

2
1  e
 
 e2   .....

1  e
 

1
  
E
ln e
 


E


ln(e


1  e
2

2
 

1




)  ln 1  e 




e   
E

2 1  e  

i) Considering the case
   1
From the Taylor expansion
e

 1   
1
2
     ...
2
substituting in the equation
1 
1
E     

 2 e 1 
1
1 
E   

2  
   1 ,
E
1

1
1
1


2    
= kT
ii) Considering
   1
1 
1
then E      

 2 e 1 
neglecting the higher order since    1
1
E     e 
2
E


 which shows T  0 the ground state energy given by

1

2
Kinetic theory of dilute gasses in equilibrium
Maxwell velocity distribution
Summery for Maxwell velocity distribution
Consider a molecule of mass m in a dilute gas the energy  of the molecule is equal to

P2
  int
2m
P2
due to the kinetic energy of the centre of mass motion
2m
 int the molecule is not monatomic the internal energy due to rotation and vibration of the atom
with respect to the molecular centre of mass
The probability Ps  r, p  d 3rd 3 p of finding the molecule with centre-of –mass variables in the
ranges (r,dr) and (p,dp) and with internal state specified by s the result
Ps  r , p  d rd p  e
3
3
 p 2 int 
  
 
 2m

where e  contributes for the constant proportionality
int

p2
2m

V2
2m
Ps  r , p  d rd p  e
3
3
f  r ,V  d rd V  Ce
3
3
d 3rd 3 p
d 3rd 3V
d 3rd 3 p
Activity
Using the normalization condition for N number of molecules in a system derive the value of C and
write the Maxwell velocity distribution
Solution

f  r,V  d 3rd 3V  N
r V

Ce

V2
2m
d 3rd 3V  N
r V
3
    mV x

C  d r   e 2 m dVx   N
 

r


2
3
3
 2 
CV 
 N
 m 
3
N
C
V
N
 m2
 2  , n  V total number of molecule per unit volume


3
  m  2   2m 3 3
f  r ,V  d rd V  n 
d rd V Maxwell velocity distribution
 e
 2 
3
V2
3
Activity
Derive the velocity distribution component
Solution
Let the number of molecule per unit volume with x-component of velocity in the range between Vx
and Vx+dVx, irrespective of the values of their other velocity is given by
g (Vx )dVx  
Vx
 f V d V
3
Vy
 m 
g (Vx )dVx  n 

 2 kT 
3
2
e

 m
V dV e m 2 kT V d 3V
y
z
2
2 kT
2
y
Vy
3
z
Vz


 m
V2
 m
V2
 m  2  m 2 kT Vx2
2 kT  y
2 kT  z
g (Vx )dVx  n 
e
dV
e
dVz
y
 e


 2 kT 
3
1
 m  2  m   m 2 kT Vx2
g (Vx )dVx  n 
dVx
 
 e
 2 kT   2kT 
The graph g (Vx ) versus Vx
Problem
Solve the value for
Vx and Vx 2
Formulation of the statistical
Problems
Consider a gas of identical
particles in a volume V in
equilibrium at the temperature T.
We shall use the following
notation
 Label
the
possible
quantum states of a single
particle by r or s
 Denote
particles in state r by  r
the
energy
of
 Denote the number of particles in state r by nr
 Label the possible quantum states of the whole gas by R
The total energy of the gas when it is in some state R where there are n1 particle r=1, n2 particles in
state r=2 etc.,
ER  n11  n2 2  ...   nr  r
r
The total number of the gas N is given by
n
r
N
r
In order to calculate the thermodynamic function of the gas it is necessary to calculate its partition
function
Z   e   ER
R
Z   e    n11  n2 2 ...
R
Activity
Derive the mean number of the particles in state s
Solution
 n e  

e  

n1 1  n2 2 ...
s
ns
R

n1 1  n2 2 ...
R
ns  
1  ln Z
  s
Problem
Calculate the dispersion
Solution
One can similarly write down an expression for the dispersion of the number of particles in state s.
One can use the general relation.
(ns ) 2  (ns  ns ) 2  ns 2  ns
2
For the case ns 2
 n e  

e  
n1 1  n2 2 ...

2
s
ns 2
R
n1 1  n2 2 ...

R
n2 s  
1  2 ln Z
 2 Z  2 s
2
1    1 Z  1  Z  
ns 2



 
   s  Z  s  Z 2   s  


2
n2s 
 ns 
1  

 2   s
2

 1 Z 
2
2

   ns 
 Z  s 


1    1 Z  



 2   s  Z  s  
=
1  2 ln Z
 2  s 2
 ns 
2
=
1  ns
the dispersion of the distribution of particles
  s
Photon Statistics
The average numbers of particles in state s in case of photon statistics
n e  

e  
 ns
s
s
ns
 ns

ns 
s
1 
e  ns s

  s
 e  ns s
ns  

1 
ln  e   ns s Using the geometric series
  s
e 
 ns s
 1  e  s  e2  s  ... 
ns 0
ns  
ns 
ns 
1 
1
ln
  s 1  e   s
1 
ln 1  e   s
  s

1
e
1
1  e  s
  s
1

The average number of particles in Plank’s distribution
Fermi-Dirac Statistics
Activities
Consider particles in a system where the total number N of particles is fixed n1 , n2,.... such that
nr  0 and nr  1 for each r, but these numbers must always satisfy  nr  N , let us derive the
r
average number of particles in a given system
Solution
Considering the above mentioned condition where the total number N of particles is fixed
n1 , n2,.... such that nr  0 and nr  1 for each r, but these numbers must always satisfy  nr  N , to
r
derive the average number of particles in a given system for Fermi-Dirac Statistics we consider
the partition function
zs  N  
s
 e 
n1 , n2,...

n11  n2 2 ...
then
  n
N
s
r
s state omitted
r
n e 

 ns  s
s
ns 
ns
e
s
n1 , n2 ,..
  ns  s
ns

s
e    n11  n2 2 ...
e    n11  n2 2 ...
n1 , n2 ,..
since ns=0 and 1
0  e  s
ns 

s
e   n11  n2 2 ...
n1 , n2 ,..

s
e
   n11  n2 2 ...
e
  s
n1 , n2 ,..
ns 
ns 
s
 e 

n11  n2 2 ...
n1 , n2
0  e  s Z s  N  1
Z s  N   e  s Z s  N  1
taking the ratio of the equation
1
Zs  N 
  s
e 1

Z s  N  1 

taking the Taylor expansion of ln Z s  N  N  for N  N
ln Z s  N  N   ln Z s  N  
 ln Z s  Z 
N
N
 Z  N  1 
 ln Z s  N 
ln  s
=- N where  

 Z N 
N
 s

Zs  N  N   Zs  N  eN if we approximate N  1
Zs  N 1  Z s  N  e
since we have
ns 
ns 
1
Zs  N 
  s
e 1

Z
N

1

s

1
   s
e
1
and substituting
which is Fermi-Dirac Distribution
Bose-Einstein Statistics
Activity
Derive the distribution of the particles in a system considering the case where the total number N of
particles is fixed n1 , n2,.... such that nr  0 ,1,2,….but these numbers must always satisfy
n
r
r
Solution
zs  N  
s
 e 

n11  n2 2 ...
n1 , n2,...
n e 
 ns  s
s
ns 
ns
e
s
n1 , n2 ,..
  ns  s
ns
ns 


s
e    n11  n2 2 ...
e    n11  n2 2 ...
n1 , n2 ,..
0  e  s Z s  N  1  2e2  s Z s  N  2   ...
Z s  N   e  s Z s  N  1  e2  s Z s  N  2   ...
where
Zs  N 1  Z s  N  e and
Zs  N  2  Z s  N  e2
ns 
ns 
Z s  N  0  e  s e  2e2  s e2  ....
Z s  N 1  e  s e  e2  s e2  ....
 0  e  e  2e2  e2  ....
1  e  e  e2  e2  ....
s
s
s
s
N

ne 


 e  

   s  ns
s
ns
s 

 ns
considering
 n e  

s 
 ns
s

ne 


 e  



 e  s  ns

   s  ns
s
ns

s 
 ns
=


   n
ln  e  s  s
 ns 0
taking the expansion

 e  

s 
 ns
 1 e
  s  
s 
 ns
 1 e
e
2  s  
ns 0

 e  


ns  0


  
ln 1  e  s 

ns  


e
1 e


  s  

1
1
1
  s  
  s  
1
e
  s  

 ...  1  e
  s  
1
Bose-Einstein Distribution
Maxwell-Boltzmann statistics
Activity
  n   n  ...
With the help of the partition function is z   e  1 1 2 2  compute the Maxwell-Boltzmann
R
distribution distribution
Solution
Hence, the partition function is z   e    n11  n2 2 ...
R
For N number of molecules there are, for given values of (n1 ,n2,…)
N!
possible ways in which the particle can be put into the given single- particle states, so that
n1 !n2 !..
there are n1 particles in state 1, n2 particles in state 2, etc. By virtue of the distinguishability of
particles, each of these possible arrangements corresponds then to a distinct state for the whole gas.
Hence the partition function can be written
z
N!
e   n11  n2 2 ...
n1 , n2 ,.. n1 ! n2 !...

where the sum overall values nr  0 ,1,2,….for each r, subject to the restriction
n
r
r
z
N!
e  1 n1 e  2 n2 ...
n1 , n2 ,.. n1 ! n2 !...

expanding the polynomial
z

N!
e  1 n1 e  2 n2 ... = e  1  e   2  ...
n1 , n2 ,.. n1 ! n2 !...


N


ln Z  N ln   e  r 
 r

from the mean values of the distribution of the particle we have defined as
ns  
1  ln Z
1   e  s
 N
  s
  e  r
where
 e 

r
s
r
 e   s  e   r
r 1
r
e  s
this is called the Maxwell-Boltzmann distribution
ns  N
  r
e

r
N
List of Relevant Resources
1.
Reference http://jersey.uoregon.edu/vlab/Balloon/
Date Consulted:Description: This experiment is designed to further demonstrate the properties of the
ideal gas law. In addition, our balloon will also serve as a planetary atmosphere for the
second part of the experiment
2.
Reference:-: http://lectureonline.cl.msu.edu/~mmp/kap10/cd283.htm.
Date Consulted: - August 2006
Description: - This Java applet helps you understand the effect of temperature and
volume on the number of collisions of the gas molecules with the walls. In the applet,
you can change the temperature and volume with the sliders on the left side. You can
also adjust the time for which the simulation runs. The applet counts all collisions and
displays the result after the run. By varying temperature and volume and keeping track
of the number of collisions, you can get a good feeling of what the main result of kinetic
theory will be.
3.
Reference: video.google.com
Date Consulted: Nov 2006
Complete Reference: - Computer calculation of Phase Diagrams.
http://video.google.com/videoplay?docid=1397988176780135580&q=Thermodynamics
&hl=en
Rationale: Thermodynamic models of solutions can be used together with data to
calculate phase diagrams. These diagrams reveal, for a given set of all parameters (such
as temperature, pressure, and magnetic field), the phases which are thermodynamically
stable and in equilibrium, their volume fractions and their chemical compositions...
List of Relevant Useful Links
1. Title: Heat Engines
URL: http://en.wikipedia.org/wiki/Heat_engines
Abstract: - The article in wikipedia presents an overview of heat engines, everyday
examples, examples of heat engines, efficiency of heat engines etc. A good number of
external links are also provided
2. Title: Heat Engines and Refrigerators
URL: http://theory.phy.umist.ac.uk/~judith/stat_therm/node15.html
Abstract: In any heat engine, heat is extracted from a hot source (e.g. hot combustion
products in a car engine). The engine does work on its surroundings and waste heat is
rejected to a cool reservoir (such as the outside air). It is an experimental fact that the waste
heat cannot be eliminated, however desirable that might be. Indeed in practical engines,
more of the energy extracted from the hot source is wasted than is converted into work.
This web page presents a good comparison of different web pages.
3. Title: Second law of thermodynamics
URL: http://en.wikipedia.org/wiki/Second_law_of_thermodynamics
Abstract:
4. Title: Second law of thermodynamics
URL: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html
Abstract: The second law of thermodynamics is a general principle which places
constraints upon the direction of heat transfer and the attainable efficiencies of heat engines.
In so doing, it goes beyond the limitations imposed by the first law of thermodynamics.
This webpage presents a visualization in terms of the waterfall analogy.
Formative Evaluation 5
Optional Formative Evaluation 3
1. What is the probability of throwing three dice to obtain a total of score of 6 or less?
Solution
Each dice have the numbers 1, 2, 3,4,5,6
When we throwing the dice the accessible state of the total sum 6 or less will be
1+1+1, 1+1+2, 1+1+3, 1+1+4, 1+2+1, 1+2+2, 1+2+3, 1+3+1,1+3+2 , 1+4+1, 2+1+1, 2+1+2, 2+1+3,
2+2+1, 2+2+2, 2+3+1, 3+1+1, 3+2+1, 3+1+2, 4+1+1
then state of the sum 6 or less is 20
the total number of accessible states is 63=216
Then the probability of throwing three dice to obtain 6 points or less is
Px  
 x  20
=
=0.093
 total 216
2. A penny is tossed 400 times. Find the probability of getting 215 heads. (Suggestion: use the
Gaussian approximation)
Solution
A penny is tossed 400 times. Find the probability of getting 215 heads is given by the Gaussian
approximation
P ( n) 


 nn 2 
1

exp 
2
2  * n1
 2 * n1  
1
where
N=400, n1=251, p=1/2, q=1/2
n1  Np
 * n1  Npq  400 x1 / 2 x1 / 2  100  10
 * n1 2  100 ,
n1  200
Substituting in the Gaussian equation
P(251,400) 
1
10 2
e

251 2002
200
P(251,400)  1.3x10 2
3. A battery of total emf V is connected to a resistance R; as a result an amount of power P=V2 /R is
dissipated in this resistor. The battery itself consists of N individual cells connected in series so that V
is just equal to the sum of the emf’s of all these cells. The battery is old, however, so that not all cells
are imperfect condition. Thus there is only a probability p that the emf of any individual cell has its
normal value v; and a probability 1-p that the emf of any individual cell is zero because the cell has
become internally short. The individual cells are statistically independent of each other. Under these
condition Calculate the mean power P dissipated in the resistor, express the result in terms of N, v,
and p
Solution
The total potential of the connection is given by V
And the total power is given by P=V2/R
From the connection n1 number of the cells has emf each values v
The total potential is given by V=n1v
The mean value of the power is given by
P
P
V2
Where V=n1v ,v=constant
R
n1v 2
R
v 2 n1
=
R
2
Using the binomial distribution equation we can solve n1
2
N
N!
2
p n q N n n1 
n1 2 = 


n
!
N

n
!
n 0 1
1
1
1
1
and using from equation1.38 and equation 1.39 and rearranging the solution
 1 p
2
the substituting in the above equation
n1 = N 2 p 2 1 
Np 

P
n1v 2
=
R
=
v 2 n1
R
2
N 2v 2 2  1  p 
p 1 
R
Np 

4Consider the random walk problem with p=q and let m=n1 - n2 denote the net displacement to the
right. after a total of n steps, calculate the following mean values:
m, m2 , m3 , and m4 .
Solution
Where
a) m  n1  n2
then m  n1  n2
where
n1 =
WN n1  = 
n1  p
n1  p
N ! n1 n1 1
p n1
p q n1 using the relation n1 p n1  p
n1!n2 !
p

N ! n1 n2
p q and using the binomial distribution

p
n1!n2 !
N!
 n !n ! p
1

 p  q N
p
= pN ( p  q) N 1 where p  q  1 then
n1  Np Similarly you can find for n 2 = Nq
Then from the above equation
m  N ( p  q)
b) m 2  n1  n2 
2
= n1  n2  2n1 n2 using the following relation
2
2
N ! n1 n2 2   
n1  
p q n1   p 
n1 n1 ! n 2 !
 p 
2
2
N!
 n !n ! p
n1
1
n1
q n2
2
2
  
N
=  p   p  q 
 p 
  
=  p  pN ( p  q) N 1
 p 


= pN ( p  q) N 1  p 2 N ( N  1)( p  q) N 2 consider (p+q)=1 then
n1 = pN  p 2 N ( N  1) similarly for n 2 can be calculated as
2
2
n2  qN  q 2 N ( N  1) Substituting in the equation given below
2
2
n1
q n2 =  p  q 
N
m 2 = n1  n2  2n1 n2 = qN  q 2 N ( N  1) + pN  p 2 N ( N  1) - Np . Nq
2
2
5. An ideal gas has a temperature – independent molar specific heat cv at constant volume. Let
  c p / cv denote the ratio of its specific heats. The gas is thermally insulated and is allowed to
expand quasi-statically from an initial volume V, at temperature Tf to a final volume Vf
a) Use the relation pV  = constant to find the final temperature Tf of this gas.
b) Use the fact that the entropy remains constant in this process to find the final
temperature Tf .
Answer
We have given that pV  =cont.
from the ideal gas equation pV  nRT
then p 
nRT f
nRTi
nRT
which is pi 
, pf 
Vf
V
Vi
substituting in the above equation

piVi  p f V f


nRT f V f
nRTiVi

Vi
Vf

which is
TiVi
 1
 Tf Vf
V
T f  Ti  i
V
 f




 1
 1
b) Consider the entropy as a function of T, V
S=S(T,V)
 S 
 S 
dS    dT  
 dV  0
 T V
 V T
If we evaluate for the value of
dT
dV
 S 


 T 
 V  T

 
 S 
 V  S


 T V
Using the second law of thermodynamics
dQ  TdS
 S 
 dQ 
T   
 = CV
 T V  dT V
For monatomic deal gas the internal energy and molar heat capacity is given
 3
 E 
 3R
CV    
 RT  =
 T V T  2
V 2
Using one of the Maxwell relation
RT
p R
 S 
 p 

then

    the pressure from the ideal gas equation p 
V
T V
 V T  T V
using the above equation
 S 


 T 
 V  T

 
 S 
 V  S


 T V
R
 T 
V

 
3
R

V

S
2T
V  T 
2

 
T  V  S
3
 ln T
2

 ln V
3
which gives
2
TV 3  cons tan t
which is given as
2
5
   1   1 for the ideal gas
3
3
6. The Molar specific heat at constant volume of a monatomic ideal gas is known to be
3
R.
2
suppose that one mole of such a gas is subjected to a cyclic quasi-static process
which appears as a circle on the diagram of pressure p versus volume V shown in the figure below
106dynecm-2
P
B
3
2
C
A
1
D
1
2
3
103cm3
V
Find the following quantities.
a) The net work (in joules) done by the gas in one cycle.
b) The internal energy difference (in joules) of the gas between state C and state A.
c) The heat absorbed (in joules) by the gas in going from A to C via the path ABC of the
cycle.
Answer
a) The work done in one cycle
w   pdV
from the figure we can write the value of v and p
V  (2  cos  )cm 310 3
p  (2  sin  )dyn / cm 2 10 6
dV   sin cm 310 3 d
0
w    (2  sin  ) sin 10 9 d10 7 J
2
2
w   (2  sin  ) sin d10 2 J
0
w  314J
b) The internal energy of the ideal gas is given by
E  CV T
E
3
nRT
2
for ideal gas pV=nRT
then substituting in the above equation we will get
E
3
pV
2
The internal energy of the gas along the path of ac
Ec  Ea 
3
( p cV c  p a V a )
2

3
(2 x3  2 x1)10 2 J
2
Ec  Ea  600J
c) The heat energy from a to c along abc is given by
E  Q  w
compute for each value
w   pdV
0
w   (2  sin  ) sin d10 2 J


w  100 (2 sin   sin 2  )d J
0
w  (4 

2
)100 J
w  557.08
Ec  E a  600 J
Q  ( Ec  Ea )  w
Q  600 J  557 J
Q  1157 J
7. Compute the mean values of the magnetic moment
Answer
   Pr  r
r
  P    P_  _

e H  e  H
e H  e  H
   tanh
H
kT
8. Compute the mean energy of the canonical distribution of mean energy
Answer
The system in the representatives statistical ensemble are distributed over their accessible states in
according with the canonical distribution
Pr 
e  Er
 e  Er
r
the mean energy given by
e  E
E
e 
 Er
r
r
 Er
r
where
e 
 Er
 
r
r


  Er

e

Z


where Z=  e  Er the quantity Z defined the sum over state or partition function
r
E 
1 Z
 ln Z
=
Z 

9. Using the canonical distribution compute the dispersion of the energy
Answer
The canonical distribution implies a distribution of systems over possible energies; the resulting
dispersion of the energy is also readily computed
E 2  E  E 2
 E2  E
e  E

e 
 Er
E2
here
2
2
r
r
 Er
r
e
but
2
 Er
 
    

Er =
   e  Er 
  e  Er E r    
  r
     r

2
then
1 2Z
E 
Z  2
2
using the following


 1 Z 
1

   2
Z
 Z  
2
 Z  1  2 Z

 +
2
   Z 
-


E   E 2 - E

E 2   E

=
2
 2 ln Z
 2
10. The internal energy of the ideal gas is given by E=E(T) show that for the ideal gas its internal
energy does not depend on its volume
Answer
Let E=E(T,V)
Then we can write mathematically
 E 
 E 
dE    dT  
 dV
 T  v
 V  T
from the first law
TdS  dQ  dE  dW
dS 
1
vR
dE 
dV using the above equation for dE
T
V
dS 
 1  E 
1  E 
vR 
  dT   
   dV
T  T V
 T  V T V 
the entropy as a function of T and V
S=S(T,V)
 S 
 S 
dS    dT  
 dV Comparing the equation
 T V
 V  T
1  E 
 S 
    
 T V T  T V
1  E 
vR
 S 
with the second order differential equation

  
 
 V T T  V T V
2S
2S

VT TV
2
    S 
    1  E    1  E 

   
      

 V T  T V  V T  T  T V  T VT 
vR 
    S 
    1  E 
  
     
  
 T V  V T  T V  T  V  T V 
 1
=  2
 T
2
 E  1   E

  
 V  T  TV



comparing the two equation
 1  E  1   2 E
  
 2 

V
T

 T  TV

  1  2 E 
 = 
 the right and the left equations are equal when
  T VT 
 1  E 
 T 2  V   0 which implies E is independent of V



Figure : Maxwell-Boltzmann distribution
It is possible to demonstrate that the partitioning we have found is not just the most probable but by
far the most probable one. In other words, any noticeable deviation from this distribution of
particle velocities is extremely improbable (see above: multinomial distribution.) This makes for
the great practical importance of the MB distribution: it is simply the distribution of velocities in a
many particle system which we may assume to hold, neglecting all other possible but improbable
distributions.
Course material with JAVA applets
Franz J. Vesely
Computational Physics Group
Institute of Experimental Physics, University of Vienna,
Boltzmanngasse 5, A-1090 Vienna, Austria, Europe
Copyright F. Vesely 1996-2005
XI
COMPILED LIST OF ALL KEY CONCEPTS (GLOSSARY)
1. System of Particles:- .
Source: http://www.answers.com/topic/coulomb-scattering
2.
Boltzmann’s Distribution:-.
Source: http://hep.uchicago.edu/cdf/cdfglossary.html
Scattering cross section - The area of a circle of radius b, the impact parameter.
3. Phase Space:Source: http://en.wikipedia.org/wiki/Cross_section_(physics)
4. Ensemble:Source : http://en.wikipedia.org/wiki/Statistical_ensemble
: http://srikant.org/core/node11.html
5. Macroscopic irreversibility from microscopically reversible laws of motion
Source: http://comp.uark.edu/~jgeabana/mol_dyn/
6. A University of Pennsylvania physical chemistry look at the Maxwell-Boltzmann distribution,
including applets..
Source: http://oobleck.chem.upenn.edu/~rappe/MB/MBmain.htm
XII
COMPILED LIST OF COMPULSORY READINGS
Reading 1:
Complete reference : Statistical physics
From
Wikipedia, the free encyclopedia .
URL : http://en.wikipedia.org/wiki/Statistical_physics
Accessed on the 20th April 2007
Abstract :
Rationale:
Reading 2: Complete reference :
From Statistical Physics
URL : http://www.oberlin.edu/physics/dstyer
Abstract :
Rationale: This book begins with the properties of matter in bulk that introduces statistical mechanics and shows why it
is so fascinating.
XIII COMPILED LIST OF (OPTIONAL) MM RESOURCES
Resource #1
Title: Motion of Centre of Mass
URL: http://surendranath.tripod.com/Applets/Dynamics/CM/CMApplet.html
Screen Capture:
Description: Applet shows the motion of the centre of mass of a dumbbell shaped object.
The red and blue dots represent two masses and they are connected by a mass
less rod. The dumbbell’s projection velocity can be varied by using the velocity and
angle sliders. The mass ratio slider allows shifting of centre of mass. Here m1 is the
mass of the blue object and m2 is the mass of red object. Check boxes for path1
and path2 can be used to display or turn off the paths of the two masses .
Rationale: This applet depicts the motion of centre of mass of two balls (shown in red and blue
colour). The applets speed and angle of projection can be varied...
Resource #2 Rotating Stool
url:- http://hyperphysics.phy-astr.gsu.edu/hbase/rstoo.html#sm
Complete Reference:- Good animation graphics and applet to visualize the dependence of moment
of inertia on distribution of matter on an object..
Rationale: Strengthens what is already discussed in Activity 2.
Resource #3;Hyper Physics
url:-: http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html
Date Consulted:-April 2007
Description:- This Java applet helps you to do a series of virtual experiments, . you can determine
the escape and orbital velocities by varying different parameters of the projectile.
Boltzmann's Transport Equation
With his ``Kinetic Theory of Gases'' Boltzmann undertook to explain the properties of dilute gases
by analysing the elementary collision processes between pairs of molecules.
Applet BM: Start
We may understand this prescription as the rule of a game of fortune, and with the aid of a
computer we may actually play that game!
Applet LBRoulette: Start
Resource #1
Title: Statistical Mechanics (Advanced Texts in Physics)
by Franz Schwabl (Author), W.D. Brewer (Translator
URL: http://www.ebookee.com
Screen Cupture
Description
The completely revised new edition of the classical book on Statistical Mechanics covers the basic concepts
of equilibrium and non-equilibrium statistical physics. In addition to a deductive approach to equilibrium
statistics and thermodynamics based on a single hypothesis - the form of the microcanonical density matrix this book treats the most important elements of non-equilibrium phenomena. Intermediate calculations are
presented in complete detail. Problems at the end of each chapter help students to consolidate their
understanding of the material. Beyond the fundamentals, this text demonstrates the breadth of the field and
its great variety of applications. Modern areas such as renormalization group theory, percolation, stochastic
equations of motion and their applications to critical dynamics, kinetic theories, as well as fundamental
considerations of irreversibility, are discussed. The text will be useful for advanced students of physics and
other natural sciences; a basic knowledge of quantum mechanics is presumed.
Resource #2
Title: MACROSCOPIC AND STATISTICAL THERMODYNAMICS
by Yi-Chen Cheng (National Taiwan University, Taiwan)
URL: http://www.worldscibooks.com
Screen capture
Description: This textbook addresses the key questions in both classical thermodynamics and
statistical thermodynamics: Why are the thermodynamic properties of a nano-sized system
different from those of a macroscopic system of the same substance? Why and how is entropy
defined in thermodynamics, and how is the entropy change calculated when dissipative heat is
involved? What is an ensemble and why is its theory so successful?
XIV COMPILED LIST OF USEFUL LINKS
Useful Link #1 Classical Mechanics
Title: Classical Mechanics
URL: http://farside.ph.utexas.edu/teaching/301/lectures/
Screen Capture:
Description: Advanced description of the topics discussed in mechanics I and II of the AVU
Physics module.
Rationale: This site has comprehensive coverage of most of physics, in the mechanics
courses. The learner can consult chapters 7, 8 and 9 of the book. The PDF version
is also available.
Useful Link #1 Classical Mechanics
Title: Classical Mechanics
URL: http://farside.ph.utexas.edu/teaching/301/lectures/
Screen Capture:
Description: Advanced description of the topics discussed in mechanics I and II of the AVU
Physics module.
Rationale: This site has comprehensive coverage of most of physics, in the mechanics
courses. The learner can consult chapters 7, 8 and 9 of the book. The PDF version
is also available.
Useful Link #1 Classical Mechanics
Title: Classical Mechanics
URL: http://farside.ph.utexas.edu/teaching/301/lectures/
Screen Capture:
Description: Advanced description of the topics discussed in mechanics I and II of the AVU
Physics module.
Rationale: This site has comprehensive coverage of most of physics, in the mechanics
courses. The learner can consult chapters 7, 8 and 9 of the book. The PDF version
is also available.
Useful Link #1 Classical Mechanics
Title: Classical Mechanics
URL: http://farside.ph.utexas.edu/teaching/301/lectures/
Screen Capture:
Description: Advanced description of the topics discussed in mechanics I and II of the AVU
Physics module.
Rationale: This site has comprehensive coverage of most of physics, in the mechanics
courses. The learner can consult chapters 7, 8 and 9 of the book. The PDF version
is also available.
Useful Link #1 Classical Mechanics
Title: Classical Mechanics
URL: http://farside.ph.utexas.edu/teaching/301/lectures/
Screen Capture:
Description: Advanced description of the topics discussed in mechanics I and II of the AVU
Physics module.
Rationale: This site has comprehensive coverage of most of physics, in the mechanics
courses. The learner can consult chapters 7, 8 and 9 of the book. The PDF version
is also available.
Useful Link #1 Classical Mechanics
Title: Classical Mechanics
URL: http://farside.ph.utexas.edu/teaching/301/lectures/
Screen Capture:
Description: Advanced description of the topics discussed in mechanics I and II of the AVU
Physics module.
Rationale: This site has comprehensive coverage of most of physics, in the mechanics
courses. The learner can consult chapters 7, 8 and 9 of the book. The PDF version
is also available.
Useful Link #1 Classical Mechanics
Title: Classical Mechanics
URL: http://farside.ph.utexas.edu/teaching/301/lectures/
Screen Capture:
Description: Advanced description of the topics discussed in mechanics I and II of the AVU
Physics module.
Rationale: This site has comprehensive coverage of most of physics, in the mechanics
courses. The learner can consult chapters 7, 8 and 9 of the book. The PDF version
is also available.
Statistical and thermal physics sites
http://oobleck.chem.upenn.edu/~rappe/MB/MBmain.html - A University of Pennsylvania
The Maxwell-Boltzmann distribution, including applets
http://csep10.phys.utk.edu/guidry/java/wien/wien.html
Some applets related to black body radiation
http://history.hyperjeff.net/statmech.html
A statistical physics timeline, for history buffs
http://www.cstl.nist.gov/div836/836.05/thermometry/home.htm
The thermometry research group at NIST, actively trying to improve our understanding and
standards of temperature, particularly below 1 K
http://comp.uark.edu/~jgeabana/mol_dyn/
An applet that shows an example of macroscopic irreversibility from microscopically reversible
laws of motion in the presence of infinitesimal perturbation
http://www.physics.buffalo.edu/gonsalves/Java/Percolation.html
An applet that shows the percolation phase transition
http://webphysics.davidson.edu/Applets/ising/intro.html
Another applet, this one showing a numerical approach to the 2d ising model
Statistics websites
http://www.ruf.rice.edu/~lane/rvls.html The Rice University virtual statistics laboratory
http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html
An applet for demonstrating the "Monty Hall" problem
http://www.stat.sc.edu/~west/javahtml/CLT.html
An applet by the same author for demonstrating the Central Limit Theorem
http://www.math.uah.edu/stat/index.xhtml
A large number of statistics demo applets
XV SYNTHESIS OF THE MODULE
Statistical Physics:
In this module you have learnt an important branch of physics namely Statistical Physics. i.e. this
module offered an introduction to probability, statistical mechanics, and thermodynamics. Specific
topics in probability include random variables, joint and conditional probability densities, and
functions of a random variable. Topics in statistical mechanics include macroscopic variables,
thermodynamic equilibrium, fundamental assumptions of statistical mechanics, and microcanonical
and canonical ensembles. Topics in thermodynamics include the first, second, and third laws of
thermodynamics.
Prerequisites
Physics III: Vibrations and Waves (8.03), Differential Equations (18.03) and Concurrent
Enrollment in Quantum Physics I (8.04) is recommended. In Learning Activity 2 of this module
you have been guided through In Learning Activity 3, you have been guided through the evolution
of In Learning Activity 4, you have been guided through the applications of
XVI. Summative Evaluation
Short answer questions
1. The heat absorbed by a mole of ideal gas in a quasi-static process in which its temperature T
changes by dT and its volume V by dV is given by dQ  cdT  pdV where c is its constant molar
RT
. Find an expression for the
V
change of entropy of this gas in quasi-static process which takes it from initial volume of
temperature Ti and volume Vi to final values Tf and Vf.
specific heat at constant volume and p is its mean pressure, p 
2. A 0.5kg of water had been heated from 70c to 870 c by first bringing it in contact with a reservoir
at 340c and then with a reservoir at 870c. When the water has reach 870c
i) What has been the change in entropy of water?
ii) What has been the change in entropy of the heat reservoir?
iii) What has been the change in entropy of the entire system consisting of both water and heat
reservoir?
3. Starting from the fundamental thermodynamic equation derive the general relation which
represent a necessary connection between the parameters T, S, P, V,
 T 
 V 
  

 P  S  S  P
3
R.
2
Suppose that one mole of such a gas is taken quasi-statically from state A to state B along straight
line on the diagram of pressure P versus volume V shown in the figure. Find the following
quantities:
i) The internal energy difference (in joule) of the gas between state A and state B
ii) The net work done (joule) by the gas between state A and state B
iii) The heat absorbed (joule) by the gas between state A and state B
4. The molar specific heat at constant volume of a monatomic ideal gas is known to be
In 106 dynes cm-2
P
B
2
1
A
2
4
103 cm3
V
5. The ideal gas is thermally insulated and is allowed to expand quasi-statically from an initial
volume Vi at temperature Ti to a final volume Vf . Using the relation PV  =const to find the final
temperature Tf of this gas.
6. Derive the mean energy equation using the canonical distribution.
7. The mean energy E and the work dW are expressible in terms of lnZ considering Z=Z (  , x)
Consider Using the canonical distribution show that the Helmholtz free energy equation given by
F  E  TS  kT ln Z
8. Consider an ideal gas of N molecules which is in equilibrium within a container of volume V0.
Denote by n the number of molecules located within any sub volume V of this container.
a) What is the mean number n of molecules located within V? Express your answer in terms
of N,V0, and V
b) i) Find the standard deviation
 n 
2
in the number of molecules located within the sub
volume V
ii) Calculate
n
, expressing your answer in terms of N,V0 and V
n
9. Consider a system A consisting of 2 spins ½ each having magnetic moment 20 , and another
system A’ consisting of 4 spins ½ each magnetic moment 0 . Both systems are located in the same
magnetic field B. The systems are located in contact with each other so that they are free to contact
with each other so that they are free to exchange energy. Suppose the total magnetic moment for
the combined system is M + M’= 40
.
a) Count the total number of states accessible number to the combined system A + A’
b) Calculate the ratio of
p ( M  0)
p ( M  4 0 )
c) Calculate the mean value
i)M
ii)M '
1
10. A simple harmonic one dimensional oscillator has energy level given by En  ( n  )  , where
2
 is the characteristic (angular) frequency of the oscillator and where the quantum number n can
assume the possible integral value n=0,1,2,….. Suppose that such an oscillator is in thermal contact
kT
1
with a heat reservoir at temperature T low enough so that

d) Find the ratio of the probability of the oscillator being in the 3rd excited state to the
probability of its being in the 2nd excited state.
e) Assuming that only the ground state and first excited state are appreciably occupied;
find the mean energy of the oscillator as a function of the temperature T.
11. The heat absorbed in an infinitesimal process is given by the first law islative probability
dQ  dE  pdV
Considering te ideal gas equation at constant pressure. Show that
  1
Cp
R
, where 
CV
CV
12. Two states with energy difference4.83 4.83 1021 joule occur with relative probability e 2 .
Calculate the temperature. Given k  1.38 1023 joule/K
13. A system can take only three different energy state 1  0,  2  1.38 1021 joules,
 3  2.76 1021 joules. These three states can occur in 2, 5 and 4 different ways respectively.
Find the probability that at temperature 100K the system may be
i)
In one of the microstate of energy  3
ii)
In the ground state 1
14. Let Vx , Vy , Vz represent the three Cartesian components of velocity of a molecule in a gas. Using
symmetry consideration and equipartition theorem deduced expressions for the following mean
values in terms of k, T and m
i) Vx 
ii) VxVz 
iii) V 2 x  iv) (Vx  bVz )2 
Answer key:
1. S  C ln
Tf
Ti
 R ln
Vf
Vi
2. i) 528J/K ii) -184J/K iii) 344J/K
4. i) 900J
V
5. T f  Ti  i
V
 f
6. E  
ii) 300J



iii) 1200J
 1
 ln Z

7. Consider Z=Z (  , x)

S  k ln Z   E

TS  kT ln Z  E
From Helmholtz free energy F= E  TS
Thus ln Z is very simply related to Helmholtz free energy F
F= E  TS =-kT ln Z
1
NV  V0  2
V
 1 , c)
8. a) n  N , b)
V0  V
V0

NV
V0
 V0 
 V  1


1
2
9. a) 8 b) 2/3 C) i) 3 0 ii) 1
10. a)
p3
1
 1  3e   
   b)
p2 e
2  1  e   
11. use PV  RT at constant pressure PdV  RdT ,
dE  CvdT
Hence dQ  dE  PdV Substituting the above equation we can write
 dQ 

  Cv  R  C p
 dT  p
E2  E3  4.83x1021
p3

   E2  E3 
2

e

e

12.
Comparing the two equations 2 
, T=175
p2
kT
1.38 x1023 T
13. p1  Ce
p3  Ce
 1 / kT
 3 / kT
 Ce  C
 Ce
0
2.76 x1021
 2 / kT
p2  Ce
1.38 x1023 x100
1.38 x1023 x100
 Ce1
 Ce2 and keeping in mind that 2,5,4 microstates can occur in
the three energy state, the probabilities are
C
 Ce
1.38 x1021
 p  1 then
p1  p2  p3  1 , 2C  5Ce1  4Ce2  1
1
=0.23
4.38
i)
The probability for the system to be in one of the microstates
energy  3 is p3 
ii)
14. i) 0 ii) 0 iii)
4C
4
=0.12

2
e
4.38 x(2.72)2
The probability p1  2C  0.45
kT
kT
iV) 1  b 2 
m
m
XVII. References:
1. Reif, F. Fundamentals of Statistical and Thermal Physics. New York, NY: McGraw-Hill, June
1, 1965. ISBN: 0070518009.
Abstract: This standard textbook is an excellent treatment of Statistical Physics. The chapter
end exercises and the summary correlate very well with the contents of the module.
Rationale: This reference on Fundamentals of statistical and thermal physics is recommended
for undergraduate text book. The contents have been treated in detail with adequate
mathematical support.
2. Gupta and Kumar Elementary statistical mechanics 21 edition 2006. ISBN 81-7556-988Abstract:
Rationale: This reading provides basic concept methods of ensemble Distribution law
3. Zemansky, M., and R. Dittman. Heat and thermodynamics: an intermediate textbook. 7th Ed.
New York, NY: McGraw-Hill Companies, 1997. ISBN: 0070170592
Abstract:
Rationale: This reference may serve as an optional reading for this module.
4. Joel Keizer “Statistical Thermodynamics of Nonequilibrium Processes” (Springer-Verlag)
1987.
Abstract:
Rationale: This reference may serve as an optional reading for this module.
5. Frank E. Beichelt, L. Paul Fatti “Stochastic Processes and Their Applications” (Taylor &
Francis) 1997.
Abstract:
Rationale: This reference may serve as an optional reading for this module.
6. V.G. Morozov, “On the Langevin Formalism for NonLinear & NonEquilibrium Hydrodynamic
Fluctuations” Physica 126A (1984) 443-460.
Abstract:
Rationale: This reference may serve as an optional reading for this module.
7. Ming Chen Wang, G.E. Uhlenbeck, “On the Theory of the Brownian Motion II” Reviews of
Modern Physics, Volume 17; 1945.
Abstract:
Rationale: This reference may serve as an optional reading for this module.
8. Walter Greiner, Ludwig Neise and Horst St¨ocker, Thermodynamics and Statistical Mechanics,
English edition, translated from the German by Dirk Rischke (Springer, New York, 2000) ISBN
0 387 94299 8
9. L. D. Landau and E. M. Lifshitz, Statistical Physics, 3rd Edition, Part I (Landau and Lifshitz
Course of Theoretical Physics, Volume 5)(Butterworth-Heinemann, Oxford, 1980)
ISBN 0 7506 3372 7
10. Chandler, D. 1987 Introduction to Modern Statistical Mechanics Oxford: Oxford University
Press.
XVIII. Main Author of the Module
About the author of this module:
Name: Title:
Address:
Sisay Shewamare
Lecturer of Physics
Jimma University
P.O.Box 378
Jimma
Ethiopia.,
E-mail: [email protected]
Breif Biography: I am a Graduate from Addis Ababa University, Ethiopia where I did M.Sc in
Physics in the Area of statistical Physics.
Currently I’m lecturer in physics at Jimma University Ethiopia.
You are always welcome to communicate with the author regarding any question,
opinion, suggestions, etc this module.
IXX. File Structure
Name of the module (WORD) file :
 Statistical Physics.doc
Name of all other files (WORD, PDF, PPT, etc.) for the module.

Compulsory readings Statistical Physics.pdf