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Physics Module Statistical Physics

Prepared by Sisay SHEWAMARE

African Virtual University �

Notice

This document is published under the conditions of the Creative Commons http://en.wikipedia.org/wiki/Creative_Commons Attribution http://creativecommons.org/licenses/by/2.5/ License (abbreviated “cc-by”), Version 2.5.


Table of ConTenTs

I. StatisticalPhysics__________________________________________ 5

II. PrerequisiteCourseorKnowledge_____________________________ 5

III. Time____________________________________________________ 5

IV. Materials_________________________________________________ 5

V. ModuleRationale __________________________________________ 5

VI. Content__________________________________________________ 5

6.1 Overview___________________________________________ 5 6.2 Outline_____________________________________________ 6 6.3 GraphicOrganizer_____________________________________ 7

VII. GeneralObjectives _________________________________________ 8

VIII. SpecificLearningObjective(s)_________________________________ 9

IX. Pre-assessment __________________________________________ 11

X. TeachingandLearningActivities______________________________ 18

XI. GlossaryofKeyConcepts__________________________________ 111

XII. ListofCompulsoryReadings_______________________________ 112

XIII. CompiledListof(Optional)MultimediaResources_______________ 113

XIV. CompiledlistofUsefullinks________________________________ 116

XV. SynthesisoftheModule___________________________________ 117

XVI. SummativeEvaluation_____________________________________ 118

XVII.References _____________________________________________ 124

XVIII.MainAuthoroftheModule________________________________ 125

XIX.FileStructure ___________________________________________ 125


Foreword

This module has four major sections

The first one is the Introductory section that consists of five parts viz:

Title:- The title of the module is clearly described

Pre-requisit Knowledge: In this section you are provided with information regarding the specific pre-requisite knowledge and skills you require starting the module. Ca-refully look into the requirements as this will help you to decide whether you require some revision work or not.

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.

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.

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?”

The second is the CONTENT section that consists of three parts:

Overview: The content of the module is briefly presented. In this section you will fined 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 content is organized in the module.

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.

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 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.

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:


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 pre-assessment 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.

Key Concepts (Glossary): This section contains short, concise definitions of terms used in the module. It helps you with terms which you might not be familiar with in the module.

Compulsory Readings: A minimum of three compulsory reading materials are provided. It is mandatory to read the documents.

Compulsory Resources: A minimum of two video, audio with an abstract in text form is provided in this section.

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.

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 PhysicsBy Sisay Shewamare Gebremichael Jimma University Ethiopia

II. Prerequisite Course or KnowledgeIn 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. TimeThis module can be completed in 120hrs.

IV. MaterialsThe materials in this module are different books, and from the soft copy available on the internet.

V. Module RationaleIn 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 useful-ness by discussing the application of macroscopic thermodynamics and distribution of systems of particles.

VI. Content

6.1 Overview

The central concepts of this module are the macroscopic systems of particles and macroscopic measurement. The module begins with the study of statistical descrip-tion 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.

6.2 Outline

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 (25 hours)

• Work and internal energy.• Absolut temperature,• Heat capacity and specific heats.• Entropy

3 Statistical thermodynamics (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• Distribution of systems of particles


6.3 Graphic Organizer

6

Statistica lPhysics

A. Statistical Description ofSystems of Particles

B. Macroscopic Measurements:

C. Statistical Thermodynamics

D. Some Applications

Statistical Theories,

Specification of the state of the system,

Statistical enesemble.

Probability calculations,

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.

phase space

Accessib le Sta te s,

Work and internal energy.

Absolut temperature,

Heat capacity and specific heats.

Entropy

Equilibrium condition and constraints,

Entropy of a combined system,

The approach to thermal equilibrium.

Heat reservoir.

Dependency of the density of stateson the external parameters.

Thermodynamic potentialsand 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

Statistica lPhysic


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)

Content Learning objectives AfterCompletingthissectionyouwouldbeableto:

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 pro-

bability• Relaxation time• The probability of the den-

sity of states.

• Deriving the statistical equation • Discussion on two state system to

apply random walk problem • Define the relaxation time • Deriving the binomial distribu-

tion• Derive the Gaussian equation

2. Macroscopic measurements (25 hours)

• Work and internal energy.• Absolut temperature,• Heat capacity and specific

heats • Entropy

• Define the work done and the internal energy

• Describe the absolute and en-tropy relation

• State the heat capacity at constant V,P

• Define and derive the entropy

African Virtual University �0

3. Statistical thermodynamics (30 hours)

• Equilibrium condition and constraints,

• Entropy of the combined system,• The approach to thermal equili-

brium.• Heat reservoir. • Dependency of the density of

states on the external parameters .

• Write the equilibrium conditions • Derive the entropy equation for

two systems• Solve problems related to entropy• Calculate the density of state at

equilibrium condition

4. Some application of statistical and macroscopic thermodynamics (40 hours)

• Thermodynamics potential and their relation with thermodyna-mical variables,

• Enesembles systems,• Connection of canonical distri-

bution with thermodynamics .• Partition function and their pro-

perties. • Gibs paradox.• Validity of the classical approxi-

mation.• The equi partition theorem• Kinetic theory of dilute gases in

equilibrium.

• Relates different thermodyna-mics equation

• Find thermodynamics quantity relations using the thermodyna-mics potentials

• Define the ensemble systems• Derive the partition function• Show the thermodynamics

quantities using the partition function

• Show the Gibbs paradox• Derive different distribution

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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. 13000b. 22750c. 35750d. 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. 77b. 45c. 35d. 20

3) Two moles of an ideal gas (γ =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. 1.4b. 3.4c. 3d. 1

4) An ideal gas (γ =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. 10b. 20c. 16d. 12


5) Following question 4 above, by what factor does its pressure change?

a. 1b. 1.2c. 0.02 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 tem-perature of the gas

a. 200Kb. 100Kc. 400Kd. 300K

7) Following question 6 above, calculate initial volume of the gas.

a. 20lb. 30lc. 22ld. 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. 30,000Jb. 40,000Jc. 50,000Jd. 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. 57Jb. 37Jc. 50Jd. 400J

10) Using question 9 above, what is the internal energy lost by the system

a. 500Jb. 600Jc. 456Jd. 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 ca-

sesc. 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 stan-

dard.

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 largeb. The molecules are in random motionc. The molecules do not exert any appreciable force on one another or on the

wallsd. 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 wallb. the average distance travelled by a molecule in one secondc. the root mean square velocityd. 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 compres-sion

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 largeb. The molecules are in random motionc. The molecules exert negligible force on one another or on the wallsd. 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

o K is

a. 1

3kT

b. 3

2kT

c. 1

2kT

d. 2

3kT


9.2 Answer Key:

1. b

2. c

3. a

4. c

5. c

6. d

7. c

8. d

9. a

10. c

11. a

12. c

13. c

14. d

15. .b

16. c

17. d

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 phe-nomena 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 mi-croscopic 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 thermodyna-mic 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 Me-chanics; Computational Stat Mech: Ising and Markov; Order Parameters, Broken Symmetry, and Topology; Deriving New Laws; Correlations, Response, and Dissi-pation; 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

Reference http://jersey.uoregon.edu/vlab/Piston/index.html Date Consulted:-Nov 2006

Reference:-: http://lectureonline.cl.msu.edu/~mmp/kap10/cd283.htm. Date Consulted:- August 2006 .

Reference http://en.wikipedia.org/wiki/Binomial_distribution Date Consulted:-Nov 2006

Reference:-: http://www.stat.yale.edu/Courses/1997-98/101/binom.htm. Date Consulted:- August 2006

Reference: http://en.wikipedia.org/wiki/Normal_distribution Date Consulted: Nov 2006 Complete Reference:- Computer calculation of Phase Diagrams.


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 one-dimensional Ising model; the mean field theory; Ising model on the Bethe Lattice; The Spherical Model; Duality and Star Triangle Trans-formations of Planar Ising Models; Square-Lattice Ising Model; Ice-Type Models; Alternative Way of Solving the Ice-Type Models; Squared Lattice Eight-Vertex Mo-del; Kagomé Lattice Eight-Vertex 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 met-als 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)

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.

Statistical Description of Systems of Particles

• Statistical Theories,• Ensemble• Accessible state• Probability calculation• Phase space

Specification 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 si-multaneously 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

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.


Probability

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 pro-blem

c) Magnetism: An atom has a spin

1

2 and a magnetic momentμ ; in accordance

with quantum mechanics, its 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?

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.

WN (n1 ) = N !n1 !n2 !

pn 1qn2


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...p

nq

1q

2...q

n= pn1 qn2

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 !n1 !n2 !

pn 1qn2


Binomial Distribution

Indeed, we recall that the binomial expansion is given by the formula

(p + q)N =n= 0

N

∑N !

n!(N − n)!pnqN − n

Read the binomial distribution in the fundamentals of thermodynamics book (Fede-rick Reif) pp.7-23

Group discussion

Given that n1 =12

(N + m), n2 =12

(N − m)

Show that

pN (m) = N !

[(N + m) / 2]![(N − m) / 2]!12

⎛⎝⎜

⎞⎠⎟

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 )

Mean Value

If f(u) is any function of u, then the mean value of f(u) is defined by

f (u) ≡p(u

i) f (u

i)

i =1

M

∑

P(ui)

i =1

M

∑ This expression can be simplified. Since P(u

i) is defined as a probability, the quan-

tity

African Virtual University 26

P (u1 ) + P (u2 ) + ...+ P (uM ) ≡ P (ui )i =1

M

∑

P (ui ) = 1i =1

M

∑ Thisistheso-called“normalizationcondition”

f (u) ≡ p(ui ) f (ui )

i =1

M

∑

Activity

Derivethesummationandtheproductofthemeanvalueofdifferentfunction

Solution

Iff(u)andg(u)areanytwofunctionsofu,then

f (u)_ g(u) = P (ui )[ f (uii =1

M

∑ ) + g(ui )] = P (ui ) f (ui ) + P (ui )g(ui )i =1

M

∑i =1

M

∑

Or

f (u) + g(u) = f (u) + g(u)

Ifcisanyconstant,itisclearthat

cf (u) = cf (u)

Deviation dispersion and standard deviation

Δu = u − u deviation

(Δu)2 ≡ P (ui )(uii −1

M

∑ − u)2 ≥ 0 secondmomentofuaboutitsmean,”ormore

simplythe“dispersionofu”since( (Δu)2 ≥ 0 canneverbenegative,

Thevarianceofuisproportionaltothesquareofthescatterofuarounditsmeanvalue.Amoreusefulmeasureofthescatterisgivenbythesquarerootofthevariance,

( ) ( )f u g u+


( )[ ]21

2* uu Δ=Δ which is usually called the standard deviation of u.

The Gaussian Distribution

( )( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡

Δ

−−

Δ≅

21

2

1

1 *2exp

*2

1)(

n

nn

nnP

π

This is the famous Gaussian distribution function. The Gaussian distribution is only

valid in the limits N>>1 and 1n >>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) +ηβ

1+η2

2β

2+ ...

where

Bk =

dk ln Pdnk

n=n

P n( ) ≅ P n1( )exp −n− n1( )2

2 Δ * n1( )2

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥


The constant P(1n ) is most conveniently fixed by making use of the normalization

condition

For discrete case

PN (n

!)

n1 =0

N

∑ = 1

For continues case

PN (n)dn

0

N

∫ = 1

for a continuous distribution function. Since we only expect P (n) to be significant

when n lies in the relatively narrow range 11 * nn Δ± , the limits of integration in the above expression can be replaced by ∞± with negligible error. Thus,

P n( ) exp −n− n1( )2

2 Δ * n1( )2

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

dn = P n1( ) 2Δ * n1 exp −x2( )dx−∞

∞

∫−∞

∞

∫ ≅ 1

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.

The relaxation time

Activity


Take an isolated many particle systems will eventually reach equilibrium, irrespective of its initial state.

Time

Number of particle

Relaxation Time Fluctuation Time

Solution

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.

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 Ω E( ) the number of states whose energy

lies between E and E+dE in a system. Let φ E( ) denote the total number of possible quantum states of the system which are characterized by energies less than E. Clearly

φ E( ) increase when E increases. The number of states Ω E( ) in the range between E and E+dE is then

Ω E( ) = φ E + δE( ) + φ E( ) = ∂φ

∂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,p

2,….p

N)=

12m

pi2

i =1

N

∑ , U=U(r1,r

2,…r

N)

Considering the system for mono atomic ideal gas U=0, Eint

=0

Solution

The number of states Ω (E, V) lying between the energies E and E+ δ E is simply equal to the number of cells in phase-space contained between these energies.

E E+dE

R Px

Py

In other words, Ω (E, V) is proportional to the volume of phase-space between these two energies:

Ω E ,V( ) ∝ d3r

1...d3rN d3 p

1...d3 pN

E

E +δ E

∫

Here, the integrand is the element of volume of phase-space, with

d3 r = dxi dy

i dz

i

d3 p = dpi x

dpi y

dpi z

,

the number of states Ω E( ) lying spherical shell between energies E and E+dE is given

Ω = BV N E3 N

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. (Sugges-tion: use the Gaussian approximation)

Solution

A penny is tossed 400 times. Find the probability of getting 215 heads is given by the Gaussian approximation

( )( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡

Δ

−−

Δ≅

21

2

1

1 *2exp

*2

1)(

n

nn

nnP

πwhere

N=400, n1=251, p=1/2, q=1/2

n1 = Np Δ * n1 = Npq = 400x1 / 2x1 / 2 = 100 = 10

Δ * n1( )2= 100 , n1 = 200

Substituting in the Gaussian equation

P (251,400) = 1

10 2πe−

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 coor-dinate 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

x 0 L

p

P+ dp

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 Ω E( ) = dφ E( )dE

δE the number of states

which have an energy E in phase space is given by φ 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

16+

16=

13


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.


• 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


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.


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 Sta-tistical Mechanics; thermodynamic Concepts and Processes; Concepts of Probability;The Methodology of Statistical Mechanics; Magnetic Systems; Nonin-teracting Particle Systems; Thermodynamic Relations and Processes; Theories of Gases and Liquids; Critical Phenomena and the Renormalization Group; Introduc-tion to Many-Body Perturbation Theory...



List of Relevant Resources

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 parti-cles 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) temper-ature 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

Reference:-: http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Entropy/Entropy.htmlDate Consulted:- February 1999Description:- 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.



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 vapo-rization.


Figure: compression of gas molecules

Macroscopic Measurements:

• Work and internal energy• Absolute temperature• Heat capacity and specific heat capacity• Entropy

Work and internal energy

The macroscopic work done by a system is determined by the volume of a system if

changed quasi-statically from Vi to V f and throughout this process the mean pressure

of the system has the measurable value p V( ) .

W = pdV

Vi

V f

∫

If the system is isothermally insulated so it can’t absorb any heat then Q=0

The internal energy ΔE = −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 = E i . How would one determined the mean energy

E j of any other macrostate j of volume

V j and the mean pressure

pj ?

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 3p and the work done by the piston 13W

To bring the pressure 3p without changing the volume, work is done by the elec-

tric resistance by an amount RW and if the εΔ amount of energy consumed by the

resistance then the energy supplied by the external system is RW ε− Δ .

The total internal energy of the system in state in state 2 is then given by

( )a ac RE E W W ε= − + −Δ

The amount of heat absorbed from a macrostate 1 to a macrostate 2 is given by

2 2 1 12( )E E E W= − +

Heat

The heat abQ absorbed by the system in going from a macrostate a to another macrostate is given by

( )ab b a abQ E E W= − +

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 pV = NK T you can show equivalently where

,

a

pV R T

R N K

ν=

=


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

y

y

dQC

dT⎛ ⎞= ⎜ ⎟⎝ ⎠

The specific heat per mole or heat capacity per mole is thus defined by

1 1y y

y

dQc C

dTν ν⎛ ⎞= = ⎜ ⎟⎝ ⎠

Eventually the specific heat per gram is defined as

1 1

'y yy

dQc C

m m dT⎛ ⎞= = ⎜ ⎟⎝ ⎠

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 Cν and

at constant pressure pC

Figure : Diagram illustrated specific heat measurement of a gas kept at constant volume or at constant pressure


To determine Cν

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=

To determine pC

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

ν


Heat capacity using the second law of thermodynamics

The second law of thermodynamics is given by dQ = TdS the heat capacity

C y = T

∂S∂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

CV = T

∂S∂T

⎛

⎝⎜⎞

⎠⎟V

=∂E∂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 respec-

tive masses mA 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

Entropy

The entropy can readily be determined by using the second law dQ = TdS for an infinitesimal quasi-static 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

Sb − Sa =

dQTa

b

∫

Suppose that the macrostate of a body is specified by its temperature, since all its other parameters are kept constant.


S Tb( ) − S Ta( ) = dQ

Ta

b

∫ =C y T '( )dT '

T 'Ta

Tb

∫

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?

Isolated system

B,TB System System A,TA

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

dS =

dQT

= S A(T f ) − S A(TA ) =mAC A 'dT

TTA

T f

∫ = mAC 'A lnT f

TA


Similarly for the system B

dS =

dQT

= SB (T f ) − SB (TB ) =mBC B 'dT

TTB

T f

∫ = mBC 'B lnT f

TB

The total entropy change

ΔS A + ΔSB = mAC 'A ln

T f

TA

+ mBC 'B ln

T f

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→ 100 0 C=

dQT

where dQ = mCdT

=mCdT

T

ΔS = mC

dTT273 k

373 k

∫

ΔS = mC ln

T f

Ti


ΔSwater = mC ln 373

273(where mass of water =1kg)

= 1310J/K

The entropy of reservoir

The amount of heat loss by the reservoir

ΔQwater = −ΔQreservoir

ΔQreservoir = −mC (T f − Ti )

ΔSreservoir = −

mC (T f − Ti )water

T373 =-1126J/K

Total entropy

ΔStotal = ΔSreservoir +ΔSwater

ΔStotal = −mC (T f − Ti )water

T373

+ mC ln 373273

ΔStotal

= 184J/K


ACTIVITY 3: Statistical Thermodynamics



• 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


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.


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...



Reading 5

Complete reference: Introduction To Statistical Mechanics | Free eBooks Down-load! URL:http:// www.ebookee.com/Introduction-To-Statistical-Mechanics_139834.html Accessed

Reading 6

Complete reference: Molecular Driving Forces: Statistical Thermodynamics in Chemistry ...

URL http:// www.ebookee.com/Molecular-Driving-Forces-Statistical-Thermodyna-mics-in-Chemistry-amp-Biology_145376.html

List of Relevant Readings for all activities

Reference:- Kittel C. and Kroemer H., (1980) Thermal Physics, 2nd ed., W. H. Freeman and Co., San Francisco, CA.. Rationale: This classic reference on thermal physics is recommended for a serious student of Physics. The contents have been treated in detail with adequate math-ematical support.

Reference: Fundamentals of statistical and thermal physics: F. Reif (McGraw-Hill, New York NY,1965). Rationale: This reading provides easy sources of information. The contents have been treated in lucid manner with adequate mathematical support.


Reference http://en.wikipedia.org/wiki/Entropy

Reference:-: http://lectureonline.cl.msu.edu/~mmp/kap10/cd283.htm. 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 ap-plet, 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.


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 thermo-dynamically stable and in equilibrium, their volume fractions and their chemical compositions...


Title: The P-V Diagram and Engine Cycles URL: http://www.antonine-education.co.uk/Physics_A2/options/Module_7/Top-ic_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, Expres-sions for work done are not required except for the constant pressure case, W = pΔ V , Extension to cyclic processes: work done per cycle = area of loop

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



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.


Introduction

Thermal relay switch and dispersion systems (Boltzmann and Gibbs factors, partition and connection functions with thermodynamics

Equilibrium conditions and constraints

Consider an isolated system whose energy is specified to lie in a narrow range. As usually, we denote by Ω 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 y

1,y

2,…y

n then the

accessible state given byΩ = Ω y1 , y2 ,...yn( ) .

If some constraints of an isolated system are removed, the parameters of the system

tend to readjust themselves in such a way that Ω = Ω y1 , y2 ,...yn( ) approaches a

maximum Ω f ≥ Ωi

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 δ 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 + δ E is denoted Ω (E). Likewise, the number of microstates of

A’ consistent with a macrostate in which the energy lies between E’ and E’ + δ E is denoted Ω ’(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 Ω0 (E) when A has energy between E and E+dE.

The probability

P(E)=CΩ0 (E)

Total accessible state

Ω0 E( ) = Ω E( )Ω ' E 0 − E( )

Temperature at equilibrium

The probability of system A having the energy an energy near E is given by

P(E)=CΩ E( )Ω ' E 0 − E( )To locate the maximum position of P(E) at E=E

∂ ln P (E )∂E

=1P∂P∂E

=0

ln P (E ) = lnC + lnΩ E( ) + lnΩ ' E '( )

∂ ln P (E )∂E

=∂ lnΩ E( )

∂E+∂ lnΩ ' E '( )

∂E=0

～


where E0=E+E’ which is dE=-dE’ then

∂ lnΩ E( )∂E

−∂ lnΩ ' E '( )

∂E ' =0

Entropy of the combined system

Activity

where and denote the corresponding energies of A and A’ at the maximum, and where we have introduced the definition

β E( ) = ∂ lnΩ∂E

kT ≡1β

where k is some positive constant having the dimension of

energy and whose 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 lnΩ this quantity S is given the name of entropy

Total accessible state Ω0 E( ) = Ω E( )Ω ' E 0 − E( ) and taking the logarithm

lnΩ0 E( ) = lnΩ E( ) + lnΩ ' E 0 − E( ) S

0( ) = S + S '

The condition of maximum probability is expressible as the condition that the total entropy

S + S ' = max imumentropy occurs when T=T’


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

β f = β f'

It follows from energy conservation that

E f + E ' f = E i +E i

'

The mean energy change in each system is simply the net heat absorbed, so that

Q ≡ E f − E i ;Q ' ≡ E ' f − E 'i

The conservation of energy then reduces to

Q+Q’=0:

It is clear, that the parameterβ , defined

β =

∂ lnΩ∂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


Heat reservoir

A’ A

If A’ is sufficiently large compared to A so A’ is a reservoir.

Suppose the macroscopic system A’ has Ω ' E '( ) accessible states and absorbs heat

Q ' = ΔE ' using Expanding lnΩ' E ' ,Q '( ) at E’=Q

lnΩ ' E ',Q '( ) − lnΩ ' E '( ) = ∂ lnΩ '

∂E '⎛⎝⎜

⎞⎠⎟

Q '+ 12

∂2 lnΩ '∂E '2

⎛

⎝⎜⎞

⎠⎟Q '2 + ...

using approximation

∂ lnΩ '∂E '

⎛⎝⎜

⎞⎠⎟

Q ' =Q 'kT '

the higher order becomes zero

lnΩ ' E ',Q '( ) − lnΩ ' E '( ) = Q 'kT '

kΔ(lnΩ ' E ',Q '( ) − lnΩ ' E '( )) =

ΔQ 'T '

ΔS ' =

ΔQ 'T '

For a heat reservoir


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+

F igur e 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 + δ E depends on the particular value of x, so we can write

Ω ≡ Ω E , x( ) .

The number of states σ (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

σ E , x( ) = Ω E , x( )

δE∂E r

∂xdx

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 + δ E and the external parameter takes the value x). The above equation can also be written

σ E , x( ) = −

Ω E , x( )δE

Xdx

where

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 chan-ged from x to x+dx


X E , x( ) = −∂E r

∂xis the mean generalized force conjugate to the external parameter x.

Consider the total number of microstates between E and E + δ E. When the external parameter changes from x to x + dx, the number of states in this energy range changes

by ∂Ω

∂x⎛⎝⎜

⎞⎠⎟

dx . In symbols

∂Ω E , x( )∂x

dx = σ E( ) − σ E + δE( ) ≅ ∂σ

∂EδE

which yields

∂Ω

∂x=∂ ΩX( )∂E

∂Ω∂x

=∂ ΩX( )∂E

=∂Ω∂E

X + Ω∂X∂E

then

∂ lnΩ∂x

=∂ lnΩ∂E

X +∂X∂E

∂ lnΩ∂x

=∂ lnΩ∂E

X = βX

Thus,

∂ lnΩ∂xα

= βXα


where Xα is the mean generalized force conjugate to the parameter xα

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 xα α = 1,2,...n( )

to an infinitesimally different, equilibrium state described by E + dE and xα + dxα

.

What is the resultant change in the number of states Ω accessible to A?

Solution

The accessible state

Ω = Ω E ; x

1,..., xn( )

d lnΩ =

∂ lnΩ∂E

dE +∂ lnΩ∂x

αα =1

n

∑ dxα

Substituting the in the above equationβ =∂ lnΩ∂E

, ∂ lnΩ∂xα

= βXα

d lnΩ = β dE + X

αdx

αα∑

⎛

⎝⎜⎞

⎠⎟

dW = X

αdx

αα∑

Then d lnΩ = β dE + dW( ) = βdQ

The fundamental relation valid for any quasi-static infinitesimal process


dQ = TdS = dE + dW( )or equivalently

dS =

dQT

Adiabatic process

dQ = 0 which asserts

dS = 0

Equilibrium

Consider the equilibrium between the systems A and A’ in the simple case the exter-nal 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.

Ω0 E ,V( ) = Ω E ,V( )Ω ' 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

Ω0 E ,V( ) = Ω E ,V( )Ω ' E ',V '( )

Taking the logarithm

lnΩ0 E ,V( ) = lnΩ E ,V( ) + lnΩ ' E ',V '( )

The total entropy of the system given by

S0 = S + S '

At the maximum value the total accessible state d lnΩ0 = 0

d lnΩ0 E ,V( ) = d lnΩ E ,V( ) + d lnΩ ' E ',V '( ) = 0


d lnΩ =

∂ lnΩ E ,V( )∂V

dV +∂ lnΩ E ,V( )

∂EdE +

∂ lnΩ ' E ',V '( )∂V '

dV '+∂ lnΩ ' E ',V '( )

∂E 'dE ' =0

where

β p =

∂ lnΩ E ,V( )∂V

similarly β ' p' =

∂ lnΩ ' E ',V '( )∂V '

β =

∂ lnΩ E ,V( )∂E

similarly β ' =

∂ lnΩ ' E ',V '( )∂E '

Substituting in the above equation

d lnΩ = β pdV + βdE + β ' p'dV '+ β 'dE ' =0

from the combined system

E + E ' = E 0

V +V ' = V 0

Then dE = −dE ', dV = −dV '

Substituting in the above equation

β pdV + βdE −β ' p'dV − β 'dE =0

Collecting terms

β pdV −β ' p'dV =0

βdE −β 'dE =0

βdE = β 'dE


Then at thermal equilibrium

β = β '

β pdV = β ' p'dV

Then mechanical equilibrium

p = p'

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 isola-

ted 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

ΔE = −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 ΔS ≥ 0

-If the system is not isolated and under goes a quasi-static infinitesi

mal process in which it absorbs heat dQ,

then dS =

dQT

• Third law: The entropy S of a system has the limiting property that T → 0+

,

S → S0

where S0is a constant independent of all parameters of the particular

system


ACTIVITY 4: Some Application of Statistical and Macroscopic Thermodynamics



• Calculate the thermodynamics relations • Derive the equation for the canonical distribution and kinetic theory of dilute

gasses in equilibrium

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 thermo-dynamics 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 URL: http://www.ebookee.com/Reif-Fundamentals-Of-Statistical-And-Thermal-Physics_

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.uni-lj.si/~colos/COLOS/TUTORIALS/JAVA/THERMODYNA-MICS/THERMO_UK/HTML/Vel_Distri.html

Abstract: Distribution of particles in their energy leel



Partition function and their properties Ideal gas, validity of classical approxima-tion, equipartition theory, harmonic oscillator at high temperature Distribution of particles Maxwell Boltzmann, Bose Einstein and Fermi-Dirac statistics


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 thermo-dynamic relation

TdS = dE + pdV

ds = vCV

dTT

+vRV

dV

Adiabatic expression or compression

pV γ = cons tan t

V γ −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 va-riables

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

dE =

∂E∂S

⎛

⎝⎜⎞

⎠⎟V

dS +∂E∂V

⎛

⎝⎜⎞

⎠⎟ S

dV

From the fundamental thermodynamic equation

dE = TdS − pdV

Comparing the two equations we can get

T =∂E∂S

⎛⎝⎜

⎞⎠⎟

V

p = −∂E∂V

⎛⎝⎜

⎞⎠⎟

SUsing the second order differential and dE is a perfect differential. E must be inde-pendent of the order of differentiation.

∂

∂V⎛⎝⎜

⎞⎠⎟

S

∂E∂S

⎛⎝⎜

⎞⎠⎟

V

=∂T∂V

⎛⎝⎜

⎞⎠⎟

S

∂

∂S⎛⎝⎜

⎞⎠⎟

V

∂E∂V

⎛⎝⎜

⎞⎠⎟

S

= −∂p∂S

⎛⎝⎜

⎞⎠⎟

V Then

∂T∂V

⎛⎝⎜

⎞⎠⎟

S

= −∂p∂S

⎛⎝⎜

⎞⎠⎟

V

Activity

For two independent variables S and p using the fundamental thermodynamics derive the thermodynamics state of a homogeneous system.


Answer


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

dH =

∂H∂S

⎛

⎝⎜⎞

⎠⎟ p

dS +∂H∂p

⎛

⎝⎜⎞

⎠⎟ S

dp

From the thermodynamic equation

dH = TdS +Vdp


T =

∂H∂S

⎛

⎝⎜⎞

⎠⎟ p

V =

∂H∂p

⎛

⎝⎜⎞

⎠⎟ S

Using the second order differential and dH is a perfect differential. H must be in-dependent of the order of differentiation.


∂∂p

⎛

⎝⎜⎞

⎠⎟ S

∂H∂S

⎛

⎝⎜⎞

⎠⎟ p

=∂T∂p

⎛

⎝⎜⎞

⎠⎟ S

∂∂S

⎛

⎝⎜⎞

⎠⎟ p

∂H∂p

⎛

⎝⎜⎞

⎠⎟ s

=∂V∂S

⎛

⎝⎜⎞

⎠⎟ p Then

∂T∂p

⎛

⎝⎜⎞

⎠⎟ S

=∂V∂S

⎛

⎝⎜⎞

⎠⎟ p

Activity

For two independent variables T and V using the fundamental thermodynamics derive the thermodynamics state of a homogeneous system.

Answer


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( )

dF =

∂F∂T

⎛

⎝⎜⎞

⎠⎟V

dT +∂F∂V

⎛

⎝⎜⎞

⎠⎟ T

dV


dF = −SdT − pdV



S = −

∂F∂T

⎛

⎝⎜⎞

⎠⎟V

p = −

∂F∂V

⎛

⎝⎜⎞

⎠⎟ TUsing the second order differential and dH is a perfect differential. H must be in-dependent of the order of differentiation.

∂∂T

⎛

⎝⎜⎞

⎠⎟V

∂F∂V

⎛

⎝⎜⎞

⎠⎟ T

= −∂p∂T

⎛

⎝⎜⎞

⎠⎟V

∂∂V

⎛

⎝⎜⎞

⎠⎟ T

∂F∂T

⎛

⎝⎜⎞

⎠⎟V

= −∂S∂V

⎛

⎝⎜⎞

⎠⎟ T Then

∂p∂T

⎛

⎝⎜⎞

⎠⎟V

=∂S∂V

⎛

⎝⎜⎞

⎠⎟ T Activity

For two independent variables T and pusing the fundamental thermodynamics derive the thermodynamics state of a homogeneous system.

Answer


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( )

dG =

∂G∂T

⎛

⎝⎜⎞

⎠⎟ p

dT +∂G∂p

⎛

⎝⎜⎞

⎠⎟ T

dp


dG = −SdT +Vdp


S = −

∂G∂T

⎛

⎝⎜⎞

⎠⎟ p

V =

∂F∂p

⎛

⎝⎜⎞

⎠⎟ T Using the second order differential and dH is a perfect differential. H must be in-dependent of the order of differentiation.

∂∂T

⎛

⎝⎜⎞

⎠⎟ p

∂G∂p

⎛

⎝⎜⎞

⎠⎟ T

=∂V∂T

⎛

⎝⎜⎞

⎠⎟ p

∂∂p

⎛

⎝⎜⎞

⎠⎟ T

∂G∂T

⎛

⎝⎜⎞

⎠⎟ p

= −∂S∂p

⎛

⎝⎜⎞

⎠⎟ T Then

∂V∂T

⎛

⎝⎜⎞

⎠⎟ p

= −∂S∂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∂V

⎛⎝⎜

⎞⎠⎟

s

= −∂P∂S

⎛⎝⎜

⎞⎠⎟

v

∂T∂p

⎛

⎝⎜⎞

⎠⎟s

=∂V∂S

⎛⎝⎜

⎞⎠⎟

p

∂S∂V

⎛⎝⎜

⎞⎠⎟

T

=∂p∂T

⎛⎝⎜

⎞⎠⎟

v

∂S∂p

⎛

⎝⎜⎞

⎠⎟T

= −∂V∂T

⎛⎝⎜

⎞⎠⎟

p

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 func-tion

dE = TdS − pdVdH = Tds +VdpdF = −SdT − pdVdG = −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


Solution for b

Considering the independent variable S = S T , p( ) and second law of thermo

dynamics

dQ = TdS = T∂S∂T

⎛⎝⎜

⎞⎠⎟

p

dT +∂S∂p

⎛

⎝⎜⎞

⎠⎟T

dp⎡

⎣⎢⎢

⎤

⎦⎥⎥

it is possible to express dp in terms

of dT and dV

dQ = T∂S∂T

⎛⎝⎜

⎞⎠⎟

p

dT +∂S∂p

⎛

⎝⎜⎞

⎠⎟T

∂p∂T

⎛⎝⎜

⎞⎠⎟

V

dT +∂p∂V

⎛⎝⎜

⎞⎠⎟

T

dV⎡

⎣⎢

⎤

⎦⎥ where at

V=constant

dV=0

dQ = T∂S∂T

⎛⎝⎜

⎞⎠⎟

p

dT +∂S∂p

⎛

⎝⎜⎞

⎠⎟T

∂p∂T

⎛⎝⎜

⎞⎠⎟

V

dT then

∂Q∂T

⎛⎝⎜

⎞⎠⎟

p

= T∂S∂T

⎛⎝⎜

⎞⎠⎟

p

+∂S∂p

⎛

⎝⎜⎞

⎠⎟T

∂p∂T

⎛⎝⎜

⎞⎠⎟

V

C p = CV +

∂S∂p

⎛

⎝⎜⎞

⎠⎟T

∂p∂T

⎛⎝⎜

⎞⎠⎟

V

from the Maxwell relation ∂S∂p

⎛

⎝⎜⎞

⎠⎟T

= −∂V∂T

⎛⎝⎜

⎞⎠⎟

p

The volume coefficient of expansion of the substance

α =

1V

∂V∂T

⎛⎝⎜

⎞⎠⎟

p

=-1V

∂S∂p

⎛

⎝⎜⎞

⎠⎟T

∂S∂p

⎛

⎝⎜⎞

⎠⎟T

=-Vα

we can express V in terms of T and P

dV =∂V∂T

⎛⎝⎜

⎞⎠⎟

p

dT +∂V∂p

⎛

⎝⎜⎞

⎠⎟T

dp=0 since V= constant


∂p∂T

⎛⎝⎜

⎞⎠⎟

V

= −

∂V∂T

⎛⎝⎜

⎞⎠⎟

p

∂V∂p

⎛⎝⎜

⎞⎠⎟

T

from the isothermal compressibility of the substance

k = −1V

∂V∂p

⎛

⎝⎜⎞

⎠⎟T

, −kV =∂V∂p

⎛

⎝⎜⎞

⎠⎟T

∂S∂p

⎛

⎝⎜⎞

⎠⎟T

=-Vα

∂p∂T

⎛⎝⎜

⎞⎠⎟

V

=α

kSubstituting in the above equation which yields

C p = CV +

∂S∂p

⎛

⎝⎜⎞

⎠⎟T

∂p∂T

⎛⎝⎜

⎞⎠⎟

V

= CV + -Vαα

k

= CV −α 2V

k


Ensembles system -Canonical distribution

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 E

r, the probability P

r of finding the system in state r

is given by

Pr = C If E<Er<E+δE

Pr = 0 Other wise

Pr∑ = 1 Normalized

An ensemble representing an isolated system in equilibrium consists then of sys-tem distributed in the above expression. It is some times called a microcanonical ensemble.

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 P

r 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-E

r.

The number of state Ω' (E 0 − E r ) 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 'Ω' (E ' )

Pr

r

∑ = 1

Using


lnΩ' (E 0 − E r ) ≈ lnΩ' E 0( ) − ∂ lnΩ'

∂E '

⎡

⎣⎢

⎤

⎦⎥

E ' = E 0

E r + ....

lnΩ' (E 0 − E r ) ≈ lnΩ' E 0( ) − βE r

Ω ' E '( ) ≈ Ω ' E 0( )e− β E r

then

Pr = C 'Ω ' E 0( )e− β E r

Pr = C ' Ω '(E 0 )e− β E r∑ = 1∑

C ' = 1Ω ' E 0( )e− β E r

r

∑

Pr =e− β E r

e− β E r

r

∑The probability of the canonical distribution

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 μ

State Magnetic moment Energy+ μ

E += −μH

_ -μ E −

= +μH


The particles has two states + or – the probability

P+ = Ce− β E + = Ceβμ H

P− = Ce− β E − = Ce− βμ H from the normalization condition

P++P

-=1 then we get

C =

1eβμ H + e− βμ H

P+=

eβμH

eβμH + e−βμH

P−=

e−βμH

eβμH + e−βμH

Molecule in ideal gas

Activity

Consider a monatomic gas at absolute temperature T confined in a container of vo-lume 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

E=

12

mV 2 =P 2

2m• If the molecule’s position lies in the range between r and r+dr and momen-

tum lies between P and P+dP then the volume in phase space is given by d3rd3P=(dxdydz)dp

xdp

ydp

z)

• 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

P(r,p)d3rd3p

∝

d3rd3 ph

03

⎛

⎝⎜

⎞

⎠⎟ e

−βp2

2m


• The probability that P(p)d3p that a molecule has momentum lying in the range between p and p+dp

P p( )d3 p = P r , p( )d3rd3 p = Ce− β p2

2 m⎛

⎝⎜⎞

⎠⎟ d3 pr( )∫

where we have p=mv d3p=md3v

Then

P ' V( ) = P p( )d3 p = Ce−βmV 2 / 2

Generalized force

Activity

Using the canonical distribution write the generalized force

Solution

If the a system depends on the external parameter x, then Er=E

r(x) and from the defi-

nition of the generalized force we have that

Xr = −

∂E r

∂xthe mean value of the generalized force we can write as

X =e− β E r −

∂E r

∂x⎛⎝⎜

⎞⎠⎟

r

∑e− β E r

r

∑ then

X =

1β

∂ ln Z∂x

the average work done

dW = Xdx


where the external parameter is V

dW =

1β

∂ ln Z∂V

dV

p =

1β

∂ 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−βE r x( ) so it can be represented in terms of β,x since E

r=E

r(x)

Z=Z (β , x) considering a small change

d ln z =

∂ ln zdx

dx +∂ ln Z∂β

dβ

d ln Z = βdW − E dβ

The last term can be written inn terms of the change in E rather than the change in

β . Thus

d ln Z = βdW − d Eβ( ) + βdE

d ln Z + βE( ) = β dW + dE( ) ≡ βdQ

using the second law of thermodynamics

dS =dQT

therefore

( )EZkS β+≡ ln

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−βE r

r∑ partition function

If a system can be treated in the classical approximation then its energy

E = E q

1,...qn, p

1,..pn( )depends on some f generalized coordinates and f

momenta.

The partition function in the phase space given by

Z = ... e−βE (q1 ,...qn , p1 ,... pn )

∫∫dq

1,...dqn,dp

1,...dpn

h f

Activity

Consider the energy of the system is only defined by a function to which is an arbi-

trary additive constant. If one changes by a constant amount ε0 the standard state r

the energy state becomes E∗

r = E r + ε0using the partition function

a. Show the corresponding mean energy shifting by the amount of ε0

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 ε0

Partition function

Z ∗ = e−β ( E r +ε0 )

r∑ =

e−βε0 e−βE r

r∑ = e

−βε0 Z

ln Z ∗ = ln Z − βε0


from the definition E = −

∂ ln Z∂β

and E ∗ = −

∂ ln Z ∗

∂β

−∂ ln Z ∗

∂β= −

∂ ln Z∂β

+ ε0

E∗ = E + ε

0The mean energy also shifted

b. The entropy

let the partition function in terms of the variables Z∗ = Z ∗(β, x)

d ln Z ∗ =

∂ ln Z ∗

∂βdβ +

∂ ln Z ∗

∂xdx where

E ∗ = −

∂ ln Z ∗

∂β and

βdW =

∂ ln Z ∗

∂xdx

Then we can find

d ln Z ∗ = −E dβ + βdW

using the relation E dβ = d βE( ) − βdE

d ln Z ∗ = −d βE( ) + βdE + βdW

d ln Z ∗ + d βE( ) = βdE + βdW =β dQ

d(ln Z ∗ + βE ) =β dQ =

dQkT

S ∗ = k ln Z * + βE ∗( )

Since we can write

βE ∗ = βE + βε0and ln Z ∗ = ln Z − βε

0 substituting in the above equation


S ∗ = k ln Z * + βE ∗( )= k ( ln Z − βε

0+ βE + βε

0) =k ( ln Z + βE )=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 E r

Part A’’ state s corresponding energy E s

System A state r,s corresponding energy E rs

The partition function for the system A is given by Z

Z = e−β ( E r + E s )

r ,s∑ where

E r ,s = E r + E s

then

Z = e−β ( E r + E s )

r ,s∑ =

e−β ( E r )

r∑

e−β ( E s )

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 r

i

, its momentum by pi the total energy given by

E =

Pi2

2mi=1

N

∑ +U r1,r

2,...rN( )where

for

non-interacting 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 r

i

, its momentum by pi the total energy given by

E =

Pi2

2mi=1

N

∑ +U r1,r

2,...rN( )where

for

non-interacting monatomic ideal gas U=0 therefore the partition function in phase space can be given as follows

Z = exp −β

12m

p12 + ...+ pN

2( ) +U r1,...rN( )⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

d3r1...d3rN dp

13...dp3

N

h0

3N∫

Z =

1

h0

3Nexp −β

12m

p12 + ...+ pN

2( )⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪dp

13...dpN

3∫

exp −β U r

1,...rN( )⎡⎣ ⎤⎦{ }d3

1...d3

N∫

exp −β U r

1,...rN( )⎡⎣ ⎤⎦{ }d3

1...d3

N∫ = VN

Z =

V N

h0

3Nexp −β

12m

p12 + ...+ pN

2( )⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪dp

13...dpN

3∫


where p2

1= p

1x2 + p

1y2 + p

1z2 ,

dp

13 = dp

1xdp1ydp

1z so for the ith particle

ξ =

Vh

03

exp −β1

2mp2( )⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪dp∫

Z = ξN

e−β p1x

2

2m dp1x

−∞

∞

∫ =2mπβ

,

e−β p2

2m dp−∞

∞

∫ =2mπβ

⎛

⎝⎜⎞

⎠⎟

32

ξ =

Vh

03

2mπβ

⎛

⎝⎜⎞

⎠⎟

32

= V

2mπβh2

0

⎛

⎝⎜

⎞

⎠⎟

32

Z = ξN =

V2mπβh2

0

⎛

⎝⎜

⎞

⎠⎟

32⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

N

the thermodynamics quantities with the partition function

Taking the logarithm

2

0

3 2 3ln ln ln ln

2 2

mZ N V

hπ

β⎧ ⎫⎛ ⎞⎪ ⎪

= + −⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭


Activity

With the given partition function, find

i) The value for the mean pressure, ii) The mean energy, iii) The heat capacity, iv) The entropy

Solution

i) The mean pressure

p =

1β∂ ln Z∂V

=NkT

V

pV = NkT

ii) The total mean energy

E = −

∂ ln Z∂β

E =

32

Nβ

=32

NkT

ε =

32

kT

E = Nε

iii) The heat capacity at constant volume

CV =

∂E∂T

⎛

⎝⎜⎞

⎠⎟V

=32

R


iv)The entropy

( )EZkS β+≡ ln , where

βE =

32

N

2

0

3 2 3ln ln ln ln

2 2

mZ N V

hπ

β⎧ ⎫⎛ ⎞⎪ ⎪

= + −⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

S = Nk lnV +

32

ln2mπh2

0

⎛

⎝⎜

⎞

⎠⎟ −

32

lnβ +32

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

2

0

3 3 2ln ln (ln 1)

2 2

m kS Nk V T

hπ⎧ ⎫⎛ ⎞⎪ ⎪

= + + +⎨ ⎬⎜ ⎟⎪ ⎪⎝ ⎠⎩ ⎭

where

δ =

32

ln2mπk

h20

⎛

⎝⎜

⎞

⎠⎟ +1

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

S = Nk lnV +

32

lnT + δ⎧⎨⎩

⎫⎬⎭

Then the Mean Energy

ε i =e−β ε i + E '( )ε i dp

1,...dpf∫

e−β ε i + E '( )dp1,...dpf∫

ε i =e−βε i∫ ε i dpi e−βE 'dp

1,...dpf∫

e−βε i dpi∫ e−βE 'dp1,...dpf∫

ε i =e−βε i∫ ε i dpi

e−βε i dpi∫considering that

ε i =

p2

2m= bp2 then


ε i = −

∂∂β

ln e−β

pi2

2mi dpi∫ let

ε i = −

∂∂β

ln4πmi

β

,2

i

K Tε =

for one degree of freedom

For three degree of freedom 3

,2i K Tε =

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.

Ø E =

P 2

2m+

12

kx2 the energy of the oscillator

Ø E = n+

12

⎛

⎝⎜⎞

⎠⎟hω

Is the energy of the oscillator in quantum mechanics the angular frequency

ω =

km


Activity

Using the partition function of the harmonic oscillator derive the mean energy of the

oscillator for βhω << 1 and

βhω >> 1

Solution

The mean energy for the harmonic oscillator given by

E =e−βE n E n

n=0

∞

∑

e−βE n

n=0

∞

∑

E = −

∂ ln Z∂β

where

Z = e−βE n

n=0

∞

∑ = e−β n+

1

2

⎛

⎝⎜⎞

⎠⎟hω

n=0

∞

∑

Z = e

−βhω

2 e−βnhω

n=0

∞

∑

Z = e

−βhω

2 1+ e−βhω + e−2βhω .....( )

Z = e

−βhω

2 1− e−βhω( )−1

E = −

∂∂β

ln e−β

hω

2 1− e−βhω( )−1⎡

⎣⎢⎢

⎤

⎦⎥⎥

E = −

∂∂β

ln(e−β

hω

2 ) − ln 1− e−βhω( )⎡

⎣⎢⎢

⎤

⎦⎥⎥


E =

hω2

+e−βhω

hω

1− e−βhω

i) Considering the case βhω << 1

From the Taylor expansion

eβhω = 1+ βhω +

12

βhω( )2+ ... neglecting the higher order since

βhω << 1

substituting in the equation

E = hω

12+

1

eβhω −1

⎛

⎝⎜⎞

⎠⎟

E ≈ hω

12+

1βhω

⎛

⎝⎜⎞

⎠⎟

βhω << 1,

12+

1βhω

≈1

βhω

E ≈

1β

= kT

ii) Considering βhω >> 1

then E = hω

12+

1

eβhω −1

⎛

⎝⎜⎞

⎠⎟

E ≈ hω

12+ e−βhω⎛

⎝⎜⎞

⎠⎟which shows T → 0 the ground state energy given by

E =

12

hω


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 ε of the molecule is equal to

ε =

P 2

2m+ ε int

P 2

2mdue to the kinetic energy of the centre of mass motion

εint 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( )d3rd3 p of finding the molecule with centre-of –mass varia-

bles in the ranges (r,dr) and (p,dp) and with internal state specified by s the result

Ps r , p( )d3rd3 p ∝ e

−βp2

2m+ε int

⎛

⎝⎜

⎞

⎠⎟

d3rd3 p

where e−βε int

contributes for the constant proportionality

Ps r , p( )d3rd3 p ∝ e

−βp2

2md3rd3 p

f r ,V( )d3rd3V = Ce

−βV 2

2md3rd3V


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

V∫

r∫

f r ,V( )d3rd3V = N

V∫

r∫ Ce

−βV 2

2md3rd3V = N

C d3r e−β

mV 2x

2m

−∞

∞

∫ dVx

⎛

⎝⎜⎜

⎞

⎠⎟⎟

3

= Nr∫

CV

2πβm

⎡

⎣⎢⎢

⎤

⎦⎥⎥

3

= N

C =

NV

βm2π

⎡

⎣⎢

⎤

⎦⎥

3

2

,n =NV

total number of molecule per unit volume

f r ,V( )d3rd3V = n

βm2π

⎛

⎝⎜⎞

⎠⎟

32

e−β

V 2

2md3rd3V Maxwell velocity distribution

Activity

Derive the velocity distribution component

Solution

Let the number of molecule per unit volume with x-component of velocity in the range between V

x and V

x+dV

x, irrespective of the values of their other velocity is

given by

g(Vx )dVx =Vx

∫ f V( )V y

∫ d3V


g(Vx )dVx = nm

2πkT⎛

⎝⎜⎞

⎠⎟

32

e− m

2kT( )V 2y

V y

∫ dV y e− m

2kT( )Vz2

Vz

∫ d3Vz

g(Vx )dVx = n

m2πkT

⎛

⎝⎜⎞

⎠⎟

32

e− m

2kT( )Vx2

e− m

2kT( )V y2

−∞

∞

∫ dV y e− m

2kT( )Vz2

−∞

∞

∫ dVz

g(Vx )dVx = n

m2πkT

⎛

⎝⎜⎞

⎠⎟

32 m

2kT⎛

⎝⎜⎞

⎠⎟

−1

e− m

2kT( )Vx2

dVx

The graph g(Vx ) versus Vx

Problem

Solve the value for

Vx and Vx2

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 the energy of particles in state r by ε r

• 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,

n2particles in state r=2 etc.,

E R = n

1ε

1+ n

2ε

2+ ... = nrε r

r∑

The total number of the gas N is given by

nrr∑ = N

In order to calculate the thermodynamic function of the gas it is necessary to calculate its partition function

Z = e−βE R

R∑

Z = e−β n1ε1 +n2ε2 +...( )

R∑

Activity

Derive the mean number of the particles in state s

Solution

ns =nse

−β n1ε1 +n2ε2 +...( )

R∑

e−β n1ε1 +n2ε2 +...( )

R∑

ns = −

1β∂ ln Z∂εs

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.

(Δns)2 = (n

s− n

s)2 = n

s

2 − ns

2


For the case ns

2

ns2 =

n2se

−β n1ε1 +n2ε2 +...( )

R∑

e−β n1ε1 +n2ε2 +...( )

R∑

n2

s = −1

β 2Z∂2 ln Z∂ε 2

s

n2

s=

1

β 2

∂

∂εs

1

Z∂Z∂ε

s

⎛

⎝⎜

⎞

⎠⎟ +

1

Z 2

∂Z∂ε

s

⎛

⎝⎜

⎞

⎠⎟

2⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

n2

s=

1

β 2

∂

∂εs

1

Z∂Z∂ε

s

⎛

⎝⎜

⎞

⎠⎟ + β 2 n

s

2⎡

⎣⎢⎢

⎤

⎦⎥⎥

Δn

s( )2

=1

β 2

∂

∂εs

1

Z∂Z∂ε

s

⎛

⎝⎜

⎞

⎠⎟

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=

1

β 2

∂2 ln Z∂ε s

2

Δn

s( )2

=

−

1

β

∂ns

∂εs

the dispersion of the distribution of particles

Photon Statistics

The average numbers of particles in state s in case of photon statistics

ns =nse

−βnsεs∑e−βnsεs∑

ns =

−1β

∂∂εs

e−βnsεs∑

e−βnsεs∑


ns = −

1β

∂∂εs

ln e−βnsεs∑Using the geometric series

e−βnsεs

ns =0

∞

∑ = 1+ e−βεs + e−2βεs + ... =1

1− e−βεs

ns = −

1β

∂∂εs

ln

1

1− e−βεs

ns =

1β

∂∂εs

ln 1− e−βεs( )

ns =

1

e−βεs −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 1 2,....,n n

such that 0rn = and 1rn = for each r, but these numbers must always satisfy,

nr

r∑ = N

let us derive the average number of particles in a given system

Solution Considering the above mentioned condition where the total number N of particles is

fixed 1 2,....,n n such that 0rn = and 1rn = for each r, but these numbers must always

satisfy

nrr∑ = N

, to derive the average number of particles in a given system for


Fermi-Dirac Statistics we consider the partition function

zs N( ) = e−β n1ε1 +n2ε2 +...( )

n1 ,n2,...

s( )

∑

then

s( )nrr∑ = N

s state omitted

ns =

nse−βnsεs

ns

∑ s( )e−β n1ε1 +n2ε2 +...( )

n1 ,n2 ,..∑

e−βnsεs

ns

∑ s( )e−β n1ε1 +n2ε2 +...( )

n1 ,n2 ,..∑

since ns=0 and 1

ns =

0 + e−βεss( )e−β n1ε1 +n2ε2 +...( )

n1 ,n2 ,..∑

s( )e−β n1ε1 +n2ε2 +...( ) + e−βεs e−β n1ε1 +n2ε2 +...( )

n1 ,n2

s( )

∑n1 ,n2 ,..∑

ns =

0 + e−βεs Zs N −1( )Zs N( ) + e−βεs Zs N −1( )

taking the ratio of the equation

ns =1

Zs N( )Zs N −1( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

eβεs +1

taking the Taylor expansion of ln Zs N − ΔN( ) for ΔN << N

ln Zs N − ΔN( ) = ln Zs N( ) −

∂ ln Zs Z( )∂N

ΔN


ln

Zs N −1( )Zs N( )

⎛

⎝⎜

⎞

⎠⎟

=- αΔN where α =

∂ ln Zs N( )∂N

Zs N − ΔN( ) = Zs N( )e−αΔN if we approximate ΔN ≈ 1

Zs N −1( ) = Zs N( )e−α

since we have

ns =1

Zs N( )Zs N −1( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥

eβεs +1 and substituting

ns =

1

eα +βεs +1 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 1 2,....,n n such that 0rn = ,1,2,….but these numbers

must always satisfy

nrr∑ = N

Solution

zs N( ) = e−β n1ε1 +n2ε2 +...( )

n1 ,n2,...

s( )

∑

ns =

nse−βnsεs

ns

∑ s( )e−β n1ε1 +n2ε2 +...( )

n1 ,n2 ,..∑

e−βnsεs

ns

∑ s( )e−β n1ε1 +n2ε2 +...( )

n1 ,n2 ,..∑


ns =

0 + e−βεs Zs N −1( ) + 2e−2βεs Zs N − 2( ) + ...

Zs N( ) + e−βεs Zs N −1( ) + e−2βεs Zs N − 2( ) + ...

where

Zs N −1( ) = Zs N( )e−α and

Zs N − 2( ) = Zs N( )e−2α

ns =

Zs N( ) 0( + e−βεs e−α + 2e−2βεs e−2α + ....)Zs N( ) 1( + e−βεs e−α + e−2βεs e−2α + ....)

ns =

0( + e−βεs e−α + 2e−2βεs e−2α + ....)1( + e−βεs e−α + e−2βεs e−2α + ....)

ns =ns∑ e− α +( βεs )ns

e− βεs +α( )ns∑

considering

nse

− βεs +α( )ns = −∂∂α∑ e− βεs +α( )ns∑

ns =ns∑ e− α +( βεs )ns

e− βεs +α( )ns∑ −

∂∂α

ln e− βεs +α( )ns

ns =0

∞

∑ =

taking the expansion

e− βεs +α( )ns

ns =0

∞

∑ = 1+ e− βεs +α( ) + e−2 βεs +α( ) + ... = 1− e− βεs +α( )( )−1

e− βεs +α( )ns

ns =0

∞

∑ = 1− e− βεs +α( )( )−1

ns = −

∂∂α

ln

1− e− βεs +α( )( )−1


=

e−βεs+α( )

1− e− βεs +α( )

=

1

e βεs +α( ) −1 Bose-Einstein Distribution

Maxwell-Boltzmann statistics

Activity

With the help of the partition function is z = e−β n1ε1 +n2ε2 +...( )

R∑ compute the Maxwell-

Boltzmann distribution distribution

Solution

Hence, the partition function is z = e−β n1ε1 +n2ε2 +...( )

R∑

For N number of molecules there are, for given values of (n1 ,n

2,…)

1 2

!

! !..

N

n n possible ways in which the particle can be put into the given single- particle

states, so that there are n

1 particles in state 1, n

2 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 !n

1!n

2!...

e−β n1ε1 +n2ε2 +...( )

n1 ,n2 ,..∑

where the sum overall values 0rn = ,1,2,….for each r, subject to the restriction

nr

r∑ = N


z =

N !n

1!n

2!...

e− β( ε1 )n1 e− βε2( )n2 ...

n1 ,n2 ,..∑

expanding the polynomial

z =

N !n

1!n

2!...

e− β( ε1 )n1 e− βε2( )n2 ...

n1 ,n2 ,..∑

=

e−βε1 + e−βε2 + ...( )N

ln Z = N ln e−βεr

r∑

⎛

⎝⎜⎞

⎠⎟

from the mean values of the distribution of the particle we have defined as

ns = −

1β∂ ln Z∂εs

= −1β

N−βe−βεs

e−βεr

r∑

where

e−βεr

r∑ = e−βεs e−βεr

r =1

s

∑

ns = Ne−βεs

e−βεr

r∑

this is called the Maxwell-Boltzmann distribution



Reference http://jersey.uoregon.edu/vlab/Balloon/ 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

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 ap-plet, 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.

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 thermo-dynamically stable and in equilibrium, their volume fractions and their chemical compositions...


Title: Heat Engines U R L : h t t p : / / e n . w i k i p e d i a . o r g / w i k i / H e a t _ e n g i n e s 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

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.


Title: Second law of thermodynamics URL : http://en.wikipedia.org/wiki/Second_law_of_thermodynamics

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 thermo-dynamics. 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( ) = Ω x( )Ωtotal

= 20216

=0.093

2. A penny is tossed 400 times. Find the probability of getting 215 heads. (Sugges-tion: use the Gaussian approximation)

Solution

A penny is tossed 400 times. Find the probability of getting 215 heads is given by the Gaussian approximation

( )( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡

Δ

−−

Δ≅

21

2

1

1 *2exp

*2

1)(

n

nn

nnP

π

where

N=400, n1=251, p=1/2, q=1/2

n1 = Np Δ * n1 = Npq = 400x1 / 2x1 / 2 = 100 = 10

Δ * n1( )2= 100 , 2001 =n

Substituting in the Gaussian equation


P (251,400) = 1

10 2πe−

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

R

VP

2

= Where V=n1v ,v=constant

( )R

vnP

21= =

R

nv 21

2

Using the binomial distribution equation we can solve 2

1n

( )21n =

( )

( )∑=

−

−

N

n

nNn nqpnNn

N

0

21

111

11

!!

!

and using from equation1.38 and equation 1.39 and rearranging the solution

21n = N 2 p2 1+ 1− p

Np⎡

⎣⎢

⎤

⎦⎥

the substituting in the above equation


( )R

vnP

21= =

R

nv 21

2

=

N 2 v2

Rp2 1+ 1− p

Np⎡

⎣⎢

⎤

⎦⎥

4. Consider the random walk problem with p=q and let m=n1 - n

2 denote the net

displacement to the right. after a total of n steps, calculate the following mean values:

,m .,, 432 mandmm

Solution

Where

a) m= n1 − n2

then m= n1 − n2

where

1n = WN n1( )∑ = N !

n1 !n2 !pn1 qn1 −1n1∑ using the relation n1 pn1 = p

∂pn1

∂p

n1 = p

∂

∂pN !

n1 !n2 !pn1 qn2∑ a n d u s i n g t h e b i n o m i a l d i s t r i b u t i o n

N !

n1 !n2 !pn1 qn2∑

= ( )Nqp +

n1 = p∂

∂pp+ q( )N

= pN ( p+ q)N −1

where 1=+ qp then

n1 = Np Similarly you can find for 2n = Nq

Then from the above equation

m= N ( p− q)


b) m2 = n1 − n2( )2

= n12 + n2

2 − 2n1 n2 using the following relation

= p∂

∂p⎛

⎝⎜⎞

⎠⎟

2

p+ q( )N

= p∂

∂p⎛

⎝⎜⎞

⎠⎟pN ( p+ q)N −1( )

= pN ( p+ q)N −1 + p2 N (N −1)( p+ q)N − 2

consider (p+q)=1 then

2

1n = pN + p2 N (N −1) similarly for 2

2n can be calculated as

n22 = qN + q2 N (N −1) Substituting in the equation given below

2m = n12 + n2

2 − 2n1 n2 = qN + q2 N (N −1) + pN + p2 N (N −1) - Np Nq.

5. An ideal gas has a temperature – independent molar specific heat vc at constant

volume. Let γ ≡ cp / 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 T

f to a final volume V

f

a) Use the relation pV γ

= 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 T

f .

Answer

We have given that pV γ =cont.

from the ideal gas equation pV = nRT

then V

nRTp = which is

i

ii V

nRTp = ,

f

ff V

nRTp =

substituting in the above equation


piVi

γ = pf V fγ

nRTiVi

γ

Vi

=nRT f V f

γ

V f

which is

TiVi

γ −1 = T f V fγ −1

T f = Ti

Vi

V f

⎛

⎝⎜

⎞

⎠⎟

γ −1

b) Consider the entropy as a function of T, V

S=S(T,V)

dS =∂S∂T

⎛⎝⎜

⎞⎠⎟

V

dT +∂S∂V

⎛⎝⎜

⎞⎠⎟

T

dV = 0

If we evaluate for the value of dTdV

∂T∂V

⎛⎝⎜

⎞⎠⎟

S

= −

∂S∂V

⎛⎝⎜

⎞⎠⎟

T

∂S∂T

⎛⎝⎜

⎞⎠⎟

V

Using the second law of thermodynamics

dQ = TdS

T∂S∂T

⎛⎝⎜

⎞⎠⎟

V

=dQdT

⎛⎝⎜

⎞⎠⎟

V = VC

For monatomic deal gas the internal energy and molar heat capacity is given

CV =∂E∂T

⎛⎝⎜

⎞⎠⎟

V

=∂

∂T32

RT⎛⎝⎜

⎞⎠⎟

V =

2

3R


Using one of the Maxwell relation

∂S∂V

⎛⎝⎜

⎞⎠⎟

T

=∂p∂T

⎛⎝⎜

⎞⎠⎟

V the pressure from the ideal gas equation p =

RTV

then

∂p∂T

=RV

using the above equation

∂T∂V

⎛⎝⎜

⎞⎠⎟

S

= −

∂S∂V

⎛⎝⎜

⎞⎠⎟

T

∂S∂T

⎛⎝⎜

⎞⎠⎟

V

∂T∂V

⎛⎝⎜

⎞⎠⎟

S

= −

RV3R2T

3

2−=⎟

⎠

⎞⎜⎝

⎛

∂

∂

SV

T

T

V

∂ lnT∂ lnV

= −23

which gives

TV23 = cons tan t

which is given as 23= γ −1 = 5

3−1 for the ideal gas

6. The Molar specific heat at constant volume of a monatomic ideal gas is known

to be 2

3R. 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


B

A C

D

1 2

2

3

1

3

106dynecm-2 P

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

Δw = pdV∫

from the figure we can write the value of v and p

V = (2 + cosθ)cm3103

p = (2 + sinθ)dyn / cm2106

dV = − sinθcm3103 dθ

Δw = − (2 + sinθ)sinθ109

2π

0

∫ dθ10−7 J

Δw = (2 + sinθ)sinθdθ102 J

0

2π

∫

Δw = 314J


b) The internal energy of the ideal gas is given by

TCE V=

nRTE2

3=

for ideal gas pV=nRT

then substituting in the above equation we will get

E =

32

pV

The internal energy of the gas along the path of ac

E c − E a =

32

( pcVc − paVa )

=32

(2x3− 2x1)102 J

E c − E a J600=

c) The heat energy from a to c along abc is given by

ΔE = ΔQ + Δwcompute for each value

Δw = pdV∫

Δw = − (2 + sinθ)sinθdθ102 Jπ

0

∫

Δw = 100 (2sinθ + sin2 θ)dθ0

π

∫ J

Δw = (4 +

π

2)100J

Δw = 557.08


E c − E a = 600J

ΔQ = (E c − E a ) + Δw

ΔQ = 600J + 557JΔQ = 1157J

7. Compute the mean values of the magnetic moment

Answer

μ = Prμr

r

∑ μ = P+μ+ + P_μ_

μ =μeβμ H − μe− βμ H

eβμ H + e− βμ H

μ = μ tanh μH

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 acces-sible states in according with the canonical distribution

Pr =e− β E r

e− β E r

r

∑the mean energy given by

E =e− β E r E r

r

∑e− β E r

r

∑where


e− β E r

r

∑ = −∂

∂βe− β E r( ) = −

∂

∂βZ

r

∑

where Z= e− β E r

r

∑ the quantity Z defined the sum over state or partition function

E = −1Z∂Z∂β

= −∂ ln Z∂β

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

ΔE( )2

= E − E( )2= E 2 − E

2

here E 2 =e− β E r E r

2

r

∑e− β E r

r

∑

but e− β E r E r2∑

=

∂

∂βe− β E r E r

r

∑⎛

⎝⎜⎞

⎠⎟= −

∂

∂β

⎛

⎝⎜⎞

⎠⎟

2

e− β E r

r

∑⎛

⎝⎜⎞

⎠⎟

then

E 2 =1Z∂2 Z∂β 2

using the following

∂

∂β

1Z∂Z∂β

⎛

⎝⎜⎞

⎠⎟=

−

1Z 2

∂Z∂β

⎛

⎝⎜⎞

⎠⎟

2

+

1Z∂2 Z∂β 2

∂

∂βE( ) = - 2E - E( )2

ΔE( )2= −

∂E∂β

=

∂2 ln Z∂β 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

dE =

∂E∂T

⎛⎝⎜

⎞⎠⎟

v

dT +∂E∂V

⎛⎝⎜

⎞⎠⎟

T

dV

from the first law

TdS = dQ = dE + dW

dS =1T

dE +vRV

dV using the above equation for dE

dS =1T

∂E∂T

⎛⎝⎜

⎞⎠⎟

V

dT +1T

∂E∂V

⎛⎝⎜

⎞⎠⎟

T

+vRV

⎡

⎣⎢

⎤

⎦⎥dV

the entropy as a function of T and V

S=S(T,V)

dS =∂S∂T

⎛⎝⎜

⎞⎠⎟

V

dT +∂S∂V

⎛⎝⎜

⎞⎠⎟

T

dV Comparing the equation

∂S∂T

⎛⎝⎜

⎞⎠⎟

V

=1T

∂E∂T

⎛⎝⎜

⎞⎠⎟

V

∂S∂V

⎛⎝⎜

⎞⎠⎟

T

=1T

∂E∂V

⎛⎝⎜

⎞⎠⎟

T

+vRV

with the second order differential equation

∂2 S∂V ∂T

=∂2 S∂T ∂V

∂

∂V⎛⎝⎜

⎞⎠⎟

T

∂S∂T

⎛⎝⎜

⎞⎠⎟

V

=∂

∂V⎛⎝⎜

⎞⎠⎟

T

1T

∂E∂T

⎛⎝⎜

⎞⎠⎟

V

⎛

⎝⎜⎞

⎠⎟=

1T

∂2 E∂V ∂T

⎡

⎣⎢

⎤

⎦⎥


∂

∂T⎛⎝⎜

⎞⎠⎟

V

∂S∂V

⎛⎝⎜

⎞⎠⎟

T

=∂

∂T⎛⎝⎜

⎞⎠⎟

V

1T

∂E∂V

⎛⎝⎜

⎞⎠⎟

T

+vRV

⎛

⎝⎜⎞

⎠⎟

= −1

T 2

∂E∂V

⎛⎝⎜

⎞⎠⎟+

1T

∂2 E∂T ∂V

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢

⎤

⎦⎥

comparing the two equation

−1

T 2

∂E∂V

⎛⎝⎜

⎞⎠⎟+

1T

∂2 E∂T ∂V

⎛

⎝⎜⎞

⎠⎟⎡

⎣⎢

⎤

⎦⎥ =

1T

∂2 E∂V ∂T

⎡

⎣⎢

⎤

⎦⎥

the right and the left equations are equal when

−1

T 2

∂E∂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 devia-tion 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.CompiledListofAllKeyConcepts(Glossary)

System of Particles Source: http://www.answers.com/topic/coulomb-scattering

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.

Phase Space Source: http://en.wikipedia.org/wiki/Cross_section_(physics)

Ensemble Source : http://en.wikipedia.org/wiki/Statistical_ensemble http://srikant.org/core/node11.html

Macroscopic irreversibility from microscopically reversible laws of motion Source: http://comp.uark.edu/~jgeabana/mol_dyn/

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.CompiledListofCompulsoryReadings

Reading 1

Complete reference : Statistical MechanicsFrom Cornell UniversitURL : 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 Me-chanics; Computational Stat Mech: Ising and Markov; Order Parameters, Broken Symmetry, and Topology; Deriving New Laws; Correlations, Response, and Dissi-pation; Abrupt Phase Transitions; Continuous Phase Transitions.

Rationale: This chapter covers most of the topics in the second and third activities of the module.

Reading 2

Complete reference : From Classical Mechanics to Statistical MechanicsFrom Draft chapters of Thermal and Statistical PhysicsURL : http://stp.clarku.edu/notes/chap1.pdf Accessed on the 23rd September 2007

Abstract : Thermal and Statistical Physics: From Classical Mechanics to Sta-tistical Mechanics; thermodynamic Concepts and Processes; Concepts of Probability;The Methodology of Statistical Mechanics; Magnetic Systems; Nonin-teracting Particle Systems; Thermodynamic Relations and Processes; Theories of Gases and Liquids; Critical Phenomena and the Renormalization Group; Introduc-tion to Many-Body Perturbation Theory...

Rationale: This chapter covers most of the topics in the second and third activities of the module.

Reading #3

Complete reference : Lecture Notes in Statistical MechanicsURL : http://www-f1.ijs.si/~vilfan/SM/ Accessed on the 23rd September 2007

Abstract : This document has advanced theory of statistical mechanics..

Rationale: These Lecture Notes in Statistical Mechanics were written for the students of the ICTP Diploma Course in Condensed Matter Physics at the Abdus Salam ICTP in Trieste, Italy.

The lectures cover classical and quantum statistical mechanics, however, some emphasis is put on classical spin systems. The author also gave an introduction to Bose condensation and superfluidity but I do not discuss phenomena specific to Fermi particles, being covered by other lecturers.


XIII. CompiledListof(Optional) MmResources

Resource #1

Title: Motion of Centre of Mass

URL: http://surendranath.tripod.com/Applets/Dynamics/CM/CMApplet.html

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 velo-city 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

Title : Rotating Stool

URL:- http://hyperphysics.phy-astr.gsu.edu/hbase/rstoo.html#sm

Complete Reference:- Good animation graphics and applet to visualize the de-pendence of moment of inertia on distribution of matter on an object..

Rationale: Strengthens what is already discussed in Activity 2.

Resource 3

Title : 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.


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 thermody-namics based on a single hypothesis - the form of the microcanonical density ma-trix - 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

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?


Boltzmann’s Transport Equation

With his ``Kinetic Theory of Gases’’ Boltzmann undertook to explain the proper-ties 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

XIV. CompiledListofUsefulLinks

Useful Link #1

Title: Classical Mechanics URL: http://farside.ph.utexas.edu/teaching/301/lectures/ 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 me-chanics 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. SynthesisOfTheModule

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 mecha-nics, 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.SummativeEvaluation

Short answer questions

1. The heat absorbed by a mole of ideal gas in a quasi-static process in which its tem-

perature T changes by dT and its volume V by dV is given by dQ = cdT + pdV

where c is its constant molar specific heat at constant volume and p is

its mean pressure, p =RTV

. Find an expression for the change of entropy

of this gas in quasi-static process which takes it from initial volume of temperature T

i and volume V

i to final values T

f and V

f.

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∂P

⎛⎝⎜

⎞⎠⎟

S

=∂V∂S

⎛⎝⎜

⎞⎠⎟

P

4. The molar specific heat at constant volume of a monatomic ideal gas is known to

be32

R . 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


2 4

1

2

V

P In 10 6 dynes cm-2

103 cm3

A

B

5. The ideal gas is thermally insulated and is allowed to expand quasi-statically from an initial volume V

i at temperature T

i to a final volume V

f . Using the relation

PV γ =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 (β, 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 V

0.

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,V

0, and V

b) i) Find the standard deviation

Δn( )2 in the number of molecules located

within the sub volume V

ii) Calculate

Δn

n, expressing your answer in terms of N,V

0 and V


9. Consider a system A consisting of 2 spins ½ each having magnetic moment 2μ0

, and another system A’ consisting of 4 spins ½ each magnetic moment μ0. 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’= 4μ0 .

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 = 4μ

0)

c) Calculate the mean value i) M ii) M '

10. A simple harmonic one dimensional oscillator has energy level given by

E n = (n+

12

)hω , where ω 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 with a

heat reservoir at temperature T low enough so that

kThω

=1

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 tempera-ture 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

γ = 1+

RCV

, whereγ =C p

CV


12. Two states with energy difference4.83 4.83×10−21 joule occur with relative

probability e2 . Calculate the temperature. Given k = 1.38 ×10−23 joule/K

13. A system can take only three different energy state ε1= 0,ε

2= 1.38 ×10−21 jou-

les, ε3= 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 ε3

ii) In the ground state ε1

14. Let Vx ,V y ,Vz represent the three Cartesian components of velocity of a mole-

cule 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) 〈V2

x 〉 iv) 〈(Vx + bVz )2 〉

Answer key

1.

ΔS = C ln

Tf

Ti

+ R lnV

f

Vi

2. i) 528J/K ii) -184J/K iii) 344J/K

4. i) 900J ii) 300J iii) 1200J

5.

T

f= T

i

Vi

Vf

⎛

⎝⎜

⎞

⎠⎟

γ −1

6. E = −

∂ ln Z∂β


7. Consider Z=Z (β , x)

( )EZkS β+≡ ln

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

8. a)

n = N

VV

0

, b)

NVV

0

V0

V−1

⎛

⎝⎜⎞

⎠⎟

12

, c)

NVV

0

V0

V−1

⎛

⎝⎜⎞

⎠⎟

12

9. a) 8 b) 2/3 C) i) 3μ0ii) 1

10. a)

p3

p2

=1

eβ hωb)

hω

2

1+ 3e− β hω

1+ e− β hω

⎡

⎣⎢

⎤

⎦⎥

11. use PV = RT at constant pressure PdV = RdT , dE = CvdT

Hence dQ = dE + PdV Substituting the above equation we can write

dQdT

⎛

⎝⎜⎞

⎠⎟p

= Cv+ R = C

p

12.

p3

p2

= e2 = e− β E 2 + E 3( ) Comparing the two equations

2 =

E2− E

3( )kT

=4.83x10−21

1.38x10−23T, T=175

13. p1= Ce− ε1 / kT = Ce0 = C p2

= Ceε 2 / kT = Ce−1.38 x10−21

1.38 x10−23 x100 = Ce−1

p3= Ce− ε3 / kT = Ce

−2.76 x10−21

1.38 x10−23 x100 = Ce−2 and keeping in mind that 2,5,4

microstates can occur in the three energy state, the probabilities are p∑ = 1 then


p1+ p

2+ p

3= 1 , 2C + 5Ce−1 + 4Ce−2 = 1

C =

1

4.38=0.23

i) The probability for the system to be in one of the microstates energy ε3 is

p3

=

4Ce2

=4

4.38x(2.72)2=0.12

ii) The probability p1= 2C = 0.45

14. i) 0 ii) 0 iii)

kTm

iV) 1+ b2( ) kT

m


XVII. ReferencesReif, F. Fundamentals of Statistical and Thermal Physics. New York, NY: Mc-

Graw-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 phys-ics is recommended for undergraduate text book. The contents have been treated in detail with adequate mathematical support.

Gupta and Kumar Elementary statistical mechanics 21 edition 2006. ISBN 81-7556-988- Rationale: This reading provides basic concept methods of ensemble Distribution law

Zemansky, M., and R. Dittman. Heat and thermodynamics: an intermediate textbook. 7th Ed. New York, NY: McGraw-Hill Companies, 1997. ISBN: 0070170592 Rationale: This reference may serve as an optional reading for this module.

Joel Keizer “Statistical Thermodynamics of Nonequilibrium Processes” (Springer-Verlag) 1987. Rationale: This reference may serve as an optional reading for this module.

Frank E. Beichelt, L. Paul Fatti “Stochastic Processes and Their Applications” (Taylor & Francis) 1997. Rationale: This reference may serve as an optional reading for this module.

V.G. Morozov, “On the Langevin Formalism for NonLinear & NonEquilibrium Hydrodynamic Fluctuations” Physica 126A (1984) 443-460 Rationale: This reference may serve as an optional reading for this mod-ule.

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.

Walter Greiner, Ludwig Neise and Horst St¨ocker, Thermodynamics and Sta-tistical Mechanics, English edition, translated from the German by Dirk Rischke (Springer, New York, 2000) ISBN 0 387 94299 8

L. D. Landau and E. M. Lifshitz, Statistical Physics, 3rd Edition, Part I (Landau and Lifshitz Course of Theoretical Physics, Volume 5)(Butterworth-Heine-mann, Oxford, 1980) ISBN 0 7506 3372 7

Chandler, D. 1987 Introduction to Modern Statistical Mechanics Oxford: Ox-ford University Press.


XVIII.MainAuthoroftheModuleAbout the author of this module:

Sisay Shewamare

Title: 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 Phy-sics 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.FileStructureName 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

statistical - [email protected] - african virtual university

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