An Introduction to Thermal Physics - 1

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Contents

1 Schroeder Chapter 1 Introduction & Thermal Equilibrium 3

2 The ideal gas 18

3 Heat and Work 26

4 Compression Work 27

5 Heat Capacity 31

6 Schroeder Chapter 2 The second law 39

7 Two-state systems 44

8 Einstein Solid 53

9 Interacting Systems 55

10 Large systems and large numbers 63

11 Ideal gas 69

12 Entropy 81

13 Supplemental: Combinatorics 9413.1 Permutation with repetition . . . . . . . . . . . . . . . . . . . 9413.2 Permutation without repetition . . . . . . . . . . . . . . . . . 9413.3 Combination without repetition . . . . . . . . . . . . . . . . . 9513.4 Combination with repetition . . . . . . . . . . . . . . . . . . . 9513.5 Hypergeometrical . . . . . . . . . . . . . . . . . . . . . . . . . 97

14 Supplemental: NA=NB 9715 Schroeder Chapter 3 Interactions and Implications 100

16 Entropy and Heat 111

17 Paramagnetism 118

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18 Supplemental: Gospers approximation ofN! 132

19 Schroeder Chapter 4 Engines and Refrigerators 148

20 Heat Engines 148

21 Refrigerator 156

22 Real Heat Engines 158

23 Schroeder Chapter 5 Free energy and chemical thermody-namics 179

24 Free Energy 180

25 Free energy as a force towrads Equilibrium 189

26 Phase transformation of Pure Substances 203

27 Phase transition of Mixtures 216

28 Uses of thermodynamic potentials 219

29 Schroeder Chapter 6 Boltzmann Statistics 221

30 Quantum Statistics 23230.1 Degenerate Fermi Gas . . . . . . . . . . . . . . . . . . . . . . 233

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1 Schroeder Chapter 1 Introduction & Ther-

mal Equilibrium

Thermal physics is study of the behavior of many-body systems as afunction of temperature(hence thermal).

Statistical Mechanics is the microscopic theory that uses statisticalideas to analyze the macroscopicproperties of many-body systems.

Thermal physics DO NOThave to be derived from statistical mechan-ics. It is, like anything in physics, an empirical science based on a smallnumber of basic principles such as energy conservation.

One can measure pressure and energy of a particular system as afunction of temperature.

One can measure susceptibilities of a particular system withoutknowing any details of microscopic interactions inside.

Once these are known, the interaction of such systems with othermacroscopic systems can be readily calculated.

For instance, you dont have to know much about interactionsamong water molecules to calculate how much ice is needed tocool a cup of boiling water to a reasonable drinking temperature.

Just look up a table of latent heat from a book (more about thatlater).

Real fun thing is to see if we can actually calculate whats measuredin an experiment and also to predict the properties of physical systemsyet to be studied experimentally from Stat Mech.

We study thermal EquilibriumStatic And small deviations from itEveryday materials are made up of many molecules and atoms. For in-

stance, a mole of gas (thats about 22.4 liters under 1 atm and 0 C) containsabout 6.02 1023 molecules. Thats a huge number goes by the name ofAvogadros number = 6.02 1023 If we want to describe a system thatcontains that many number of particles, it is impossible to give detailed in-formation about the motion of each particle. First of all, we cant really do

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that. Not even a fastet computer in this day and age can possibly track the

motion of 10 to the 23 number of particles. Second, to solve Newtons equa-tions, we need to know the initial position and the velocity of all particles.Suppose each number takes 8 byte to specify. We need 6 such numbers. Sothe initial condition of each particle takes about 50 bytes. Therefore we need

M= 50 6 1023 = 3 1024bytes (1)

A good hard disk takes about 1011 bytes. So one would need about 1013

such hard disks to store the information about the initial condition alone.One hard disk takes up about 0.02 cubic meter. So the volume of hard disksalone (not to mention the computers) would be aboutV 1011 cubic meter.Thats kilometer by kilometer by 100 kilometer.Fortunately, we are not really interested in the details of such system.What we are interested in are

Macroscopic quantities Intensive quantities Dont depend on the size of the system

Temperature T Pressure P Chemical Potential

Densityn

... Extensive quantities Do depend on the size of the system

Volume V NumberN EnergyU EntropyS HeatQ ...

Responses Coefficient of expansion V/T Compressibilityt V /p Heat Capacity Q/T

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Magnetic susceptibility M/H ...

These are all average quantites which are averaged not only over theparticles in the system but also over all possible initial states. Thereforethe physical equations we are interested in are not the microscopic Newtonsequation

md2ridt2

=Fi(rj) (2)

but the equations that govern the behavior of pressure, energy, temperature,etc. In this regard, the large number actually helps us because it lets us to

use statistical ideas.There arent that many things in physics that are exactly solvable even if

you have the greatest computer ever built at your disposal. Usually solvablesystems are simple systems. For instance, any single particle or two-particleproblem in Mechanics in 1-D is ultimately solvable. But as soon as youincrease the number of particles or the dimensions, things get complicated.Again, systems that can be simplified due to symmetries, fundamental oraccidental, can be solved. An example is the Kepler problem, that is the mo-tion of a planet or an asteroid with respect to the Sun. With the introductionof computers, the calculation of orbits became so advanced any deviations

from the calculated orbits are taken as the sign of a new object such as the10-th planet.However, this kind of problems are few and far between. As soon as the

number of bodies (with similar sizes) becomes three, there isnt much thereti-cal physics can do about it. One has to resort to a computer calculation. Butthen when the number of bodies becomes realistic 1010, even the fastestcomputer available cant do much about that.

In the late 19-th century, physicists started to realize that there is an-other limit where analytic calculation is possible. This is the extremely largenumber limit. The reasoning is as follows. Suppose you have one mole ofa certain gas. You know that there are about 6

1023 gas molecules in the

container. It is not only impossible but absurd to keep track of the motion ofevery individual molecules 1.2 1024 microscopicdegrees of freedom.What one is interested in is just a few average macroscpic quantities such asthe pressure, energy density, number density, etc. The idea is then to usestatisticsto analyze the many-body system. From statistics, we know that

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the relative error in measuring the an average quantity behaves like 1/

N.

Now ifNis the Avogadros number, this is 1012

which is surely negligible.Therefore if we can formulate manybody problem in terms of averagequantities using concepts borrowed from statistics, we may be able to go farin solving for the characteristics of the system.

Let me give you a quick example.Suppose you have 3 particles interacting with a potential that attracts at

long distances but repulses at short distances. Put them in a large box andask yourself What is the density of this small box as a function of time?

My box

Figure 1: 3 bodies in a big box

Well, most of the time, it would be zero. But to know the density as a

function of time, we have to know the trajectories of the all three particlesand thats hard no matter how simple the interaction is.However, now suppose that instead of 3 particles, we have 6 times 10 to

the 23rd number of particles in the box.Actually there are only 10,000 dots in this figure. However, it is clear that

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Figure 2: 10,000 bodies in a box

unless clumping happens for some reason (it does. Condensation of waterdroplets, but that means changing temperature), the density of this small

box in the corner as a function of time is justn = N/Vno matter what howcomplicated the interaction among the molecules are as long as they remaingas.

The question is the, can clumping happen? That is, how is it likely thatthat a large deviation from n = N/V occur in this small volume? Well,it clearly depends on the size of the volume. If the size is too small tobe about the size of the molecular volume, then the answer could be veryfrequently. However, thats not what we are interested. We often talkabout macroscopically large but microscopically small volume. That is,we would like to think that our system is made up of a large enough numberof boxes so that calculus applies, but the box size is big enough to include

many particles. This, of course, is an approximation. The question is, howgood is this approximation?

Suppose we have Nparticles in a volume V. We divide the volume in B

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number of boxes. So on average, there are

NB = (N/V)B (3)

particles in each box. Now we ask: How likely is it for the number of particlesin a box to deviate from NB by-percent?

Since things are distributed almost randomly, we can use Binomial dis-tribution to approximate the real situation. For a singleparticle, the proba-bility that it is in this box is p = B/Vwhich we take to be a small number.Therefore, the probability that there are n particles in this box is given by

P(n) = N!

n!(N n)!pn(1

p)Nn (4)

Now since B V, typically nN. We know that the mean is NB = pNand the variance is

n2 n2 =N p(1 p) Np= NB (5)

IfNB is large enough, we can also approximate P(n) with a normal distri-bution with the mean ofNB and and the width of

NB:

P(n)dn dn 12NB

exp

(n NB)

2

2NB

= dx 1

2exp

x

2

2

(6)

where we definedx = (n NB)/

NB.Lets think about the probability that the number is within NB(10.01),

that is the probability that the actual number in the box is within1% ofNB.

This is

P =NB(1+)

n=NB(1)

P(n)

xx

dx ex2/2 (7)

wherex= NB/

NB =

NB.

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Now suppose we divide 1 mole of gas in 1 m3 boxes. 1 mole of gas is

about 22.4 litres at room temperature so thats about 2.24 1016

boxes. Inthat case on average each box has NB 6.02 1023/(2.24 1016) 3 107molecules. Square-root of that is about 5 103. One percent of that is about50. The limits of the integral are therefore about50 to 50. This deviatesfrom 1 by about

e1250 10540 (8)which is practically never. The same goes for 0.1 % 0.01 % and so on. There-fore, as long as NB is large enough, we have practically no deviation fromthe average values.

What did we learn here? We learned that for some quantites in many

body system, the detailsof particle interactions dont matter much. In par-ticular, unless something dramatic happens (well get to that later), clumping(practically) never happens. In this sense, the problem of keeping track of1023 particles reduces to a much simpler problem of keeping track of only afew average quantities Thats the idea of Stat-Mech.

In this course, we are going to study thermal physics from the view pointof statistical mechanics. Stat Mech, however, is not the answer to all ques-tions. If you think about it, stat mech is the ultimate theory of matter. Allmacroscopic system can be dealt with using stat mech. However, systems likeliving cells are notoriously hard to describe in terms of stat mech or the mo-

tion of high speed wind passing a wing for that matter. This is because theseare dynamicproblems. In these problems, system properties change macro-scopically all the time, sometimes drastically. Stat Mech is hard pressed tosolve such problems, this time due to sheer complexity of the system itself.For instance, suppose that box of gas we were thinking about is actually apart of a wind which sometimes rotates or suddenly changes directions orencounters a brick wall, etc. Yes, the basic equations may be derived fromStat Mech, but the problem of solving for the properties as a function of timeis far from simple.

Now things get calmer if one thinks about static systems. These are thesystems which were left alone in an isolated box for a long time. In that case,

all the turbulances, gusts, vorticies etc have all calmed down and the systembecomes uniform. This is what we refer to as the Equilibrium State Ithas come to an equilibrium with its environment.

Studying equilibrium state is much simpler than the non-equilibriumstate. Of course, that doesnt mean that we can solve all problems in equi-

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librium. But we know a lot. Also, in many cases, the answer can be guessed

well before any actual calculation.In fact to know the answer beforehand, there are only a few things youreally needs to know. And I am going to tell you right now what they are.You can take it as a mini summary of what this course is about:

Extensive and Intensive quantitiesExtensive quantities are the ones that grows like system size. Theseare volume V, number of particles N, total energy U, entropyS, heatQ, Helmholtz free energy F, Gibbs free energy G, enthalpyH, etc.

Intensive quantities are the ones thats independent of system size.

These are temperature T, pressure P, chemical potential , densityn= N/V, in fact any ratio of two extensive quantities or derivative ofone w.r.t. another are intensive quantities.

Energy is conserved. The amount of energy thatenteredthe system throughthermalcon-

tact (in other words, temperature difference) is the heat Q.

The amount of energy thatenteredthe system throughnon-thermalcontact is the work W.

total energy change = U = Q +W. In many cases, this reduces to:T dS=dU+ P dV dN.

There is also this relationship: T S=U+ P V N Total entropy always increases. Equilibrium meansT, and p are the same. Temperature is proportional to the energy per particle E kBT.

Goes by the name of equi-partition theorem. kB = 1.38 1023J/K:Boltzmann constant. Hard to remember. Easier to remember

300K kB (1/40) eV (More precisely, 290 K kB = 1/(40.016) eV) 1 eV 12000K kB

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The amount of kinetic energy and the potential energy in the boundsystems are of the same order of magnitude.

Kinetic energy means pressure. If you have a large number, sum and integral dont differ that much. A particle can occupy a phase space volume ofd3xd3p/h3. The probability to have energy E is proportional to the Boltzmann

factorp eE/T

Fermions are like cats Fermi energy

Bosons are like dogs Bose-Einstein Condensate Stirlings formula N! 2N NN eN

hc 2000 eV A hc 200eV nm me= 0.511 MeV/c2 0.5MeV/c2

mN= 940 MeV/c2 1GeV/c2

Potential energy tries to organize Thermal energy tries to randomize

These arent that many and they are mostly qualitative. However, a largeamount of qualitative answers can be obtained from these facts. And gettingthe qualitative answer is just as important as getting the quantitative answerbecause getting the qualitative answer right shows that you understandtheproblem and what is actually going on.

For instance, if you know that the temperature is proportional to theenergy and the pressure is too, then you can easily guess

kBT P (9)Now to make up the dimensions, you need V

kBT P V (10)

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But the left hand side is intensive and the right hand side is extensive. Since

the dimensions match up, we should use dimensionless N to getNkBT =c P V (11)

wherec must be an order 1 number which in our case turns out to be just 1or

P V =N kBT (12)

Thats the ideal gas law.Another system we are going to study later is the system of spin 1/2

particles which acts like tiny magnets. The question to ask is, suppose you

have Nspin 1/2 particles in a magnetic field B. What is the average mag-netization? First of all, we need to know the energy of each particle. Letssuppose that each particle has the magnetic moment . Then if the spin linesup with the magnetic field, its energy isB. If the spin is anti-parallel toB, its energy is B. So naturally, when left alone, each particle would liketo align itself with the magnetic field. In that case, the magnetization wouldbe simplyNin the direction ofB.

But if the system is at finite temperature, then what thermal energy doesis to randomize the orientation of the spin. The magnets are colliding witheach other and the other particles in the system and getting agitated allthe time. Now the typical thermal energy scale is kT. So we can make the

following guess.

If the particles are left alone, that is atT= 0, the magnetization wouldbe simply M = N since all particles line up with the magnetic fieldatT = 0.

If the temperature is very high so that the thermal energy is muchgreater than the magnetic energy B, Mwill be very small becausethe orientation of the spins will be practically random.

In between, the magnetization would be a function of the ratio B/kTso that

M=N f(B/kT) (13)

This functionf(x) should go to 0 when x goes to zero (tinyB or largeT limit) and go to 1 when x becomes large (tiny T or large B limit).

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Furthermore, if the direction ofB is reversed, M should also reverse.

That is, f(x) should be an odd function ofx.

There are a few elementary functions that exhibit such behavior: One isarctangent and the other is hyperbolic tangent

-1

-0.5

0

0.5

1

-4 -2 0 2 4

atan(x)/(3.1416/2)tanh(x)

Figure 3: Arctangent and Hyperbolic Tangent

As you can see here, arctangent approaches 1 in a polynomial way andhyperbolic tangent approaches 1 in an exponential way. Now if you go backto the list we made, you will see that the energy and the temperature combi-nation naturally occurs in an exponetial way thats the Boltzmann factor.So we would guess that the magnetization should behave like

M N tanh(cB/kT) (14)

where c is again as yet unknown number of order 1. Again in this case, cturns out to be 1 and in fact,

M=N tanh(B/kT) (15)

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Similar analogy can be made about melting and boiling. Ordinary materials

such as iron are held together by molecular bonding. As you heat up thematerial, the atoms in the crystal becomes more and more agitated. Thismeans that first of all, each one needs more room. Think of a harmonicoscillator. The more the energy of an SHO, the bigger the amplitude. Sodoes the atoms in a crystal. Now, the atomic potential is not really a simpleharmonic potential. So unlike the SHO case, when the amplitude becomestoo large (kinetic energy is too large), the bonding will break down. That is,as the atoms agitate more and more, the thermal kinetic energy overcomesthe binding potential energy and the solid melts or water boils.

There you have it. What we are going to do from now on are how tomake more quantitative calculations of these quantities and many related

ones. But the spirit is the same.TemperatureWe are going to study thermal physics. Naturally then the most im-

portant concept is the Temperature. We kind of intuitively know whattemperature is. For instance we know that boiling water is much hotter thanthe ice. But what exactly is the temperature? How do we define it the waywe can define other physically measurable quantities such as the mass or thevolume of an object?

There exists a precise definition of what a temperature is. However,to talk about that we need to introduce the concept of entropy first and

that can wait. For now, lets think about how we measure the temperaturepractically. Well, we use thermometers, of course. But what exactly are thethermometers? Whats happening when you stick a thermometer in a boilingwater and say that the temperature is 100 C?

To begin with the thermometer would be at the room temperature. Thatis about 20 C. When you stick it in a boiling water, it starts to heat up.That is, the temperature of the thermometer gradually becomes the same asthe temperature of the boliling water and it will show up in the scales. Thisis operational defintion of temperature.

More theoretical defintion would be

Temperature is the thing thats the same for two objects, aftertheyve been in contact long enough.

Thats intuitive. But what do all these word mean exactly? What doesin contact mean? In the context of temperature, this means that the twoobjects can exchange energy in some form. What about long enough?

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Well, this is different from system to system. This depends on the rate of

heat transfer or heat conductivity. For instance, steel conduct heat fairlyquickly. So if you build a house out of steel, your house will become coldvery quickly when winter comes and heating it will take a lot of energy. Inthis case, we say that the relaxation time is short.

Figure 4: You start with this

On the other hand, if you put styrofoam between you and the steel wall,it will take a long time for the air inside of your house to be as cold as theoutside air once it was heated up. But eventually, without additional heatsource, it will become as cold. It just takes much longer than the steel wallalone. In this case we say the relaxation time is long.

The short and long of it, however, is relative term. The above examplesmeasure time in human scale. But thats good enough. All we want to getout of this is that there is a characteristic time for each system to becomeacclamatized with its surroundings. Long enough means that longer thanthis characteristic relaxation time.

Now when two system are in contact for long enough, theyll come to the

state ofThermal Equilibrium. This is the state whenon averagethere isno energy exchange between two systems. That is on average, things becomestatic or independent of time.

Remember our example of 3 particles and 10,000 particles in a box? Evenif the particles are still actively moving around, the density of the system

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Figure 5: end up with this.

remains (practically) the same for all time. If you think of each particlecarrying a certain amount of energy, then you can say that the temperatureof any small box is the same as the temperature of the whole box. That is,they have come to the state of thermal equilibrium.

In this example, there is another quantity that remains the same. That is,the average number of particles in the box. You can start with an initial state

wherer all particles are in the right half of the box but quickly the systemwill become homogenized and never can go back to the initial state. This isanother kind of equilibrium calleddiffusive equilibrium. This time, thereis no exchange of the number of particles between the systems. Now if thereis a movable wall between two systems, depending on the pressure the wallcan move around changing volumes of the two systems in contact. When thetwo pressures become the same, then the forces on the wall balances and thewall stops moving.

This is called mechanical equilibriumand in this case, whats ceasingto be exchanged is the volume.

In all these examples of equilibrium, something is flowing such as energyor number of particles. When two objects are brought in contact with eachother, usually one has more tendency to give up the energy than the others.This has nothing to do with the absolute amount of energy each system has.The atmosphere has a lot more energy than a hot piece of steel. But still it

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P P

F = P AF = P A

Figure 6: Mechanical Equilibrium

is the hot steel that gives up the energy.Therefore something makes the energy flow from one system to another.

Looking ahead, this is ultimately the role entropy plays. However, we willjust say here that

Temperature is a measure of the tendency of an object tospontaneously give up energy to its surroundings.

Now that we have a fair bit of idea what temperature means, we needa unit. In everyday life, we use Celsius (Centigrade) or Fahrenheit. The

official SI unit, however, iskelvin(notdegrees Kelvin). 1 kelvin difference isthe same is 1 C difference. But the 0 point is different. In Celsius, 0 degreeis defined by the freezing point of water. In kelvin, 0 degree (often calledabsoulte zero) is defined by the point to which the pressure of low densitygas goes to zero. In Celsius, zero kelvin is273.15 C. Please note thatotherwise stated,all formulas in thermodynamics work with temperature inkelvin. (C.F. Triple point of water: 273.16K = 0.01 C).

Operationally, we use the fact that certain properties of materials are wellknown as a function of temperature such as the expansion of mecury oralcohol, also see Fig.1.3 of the textbook to measure the temperature.More sophisticated instrument that measure extremely cold or hot temper-

ature may use the change in the resistance as a function of temperature orthe spectrum of infrared radiation generated by the surface.

Standard temperature and pressure (STPThis is 0 C and 1 atm (= 1.013 105 P a). 1 mole of gas occupies 22.4

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litre at STP. At room temperature, it occupies

V300 = VSTP(300/273) = 24.6litre (16)

2 The ideal gas

Summary of Lecture 1

Temperature energy flows from higher to lower Relaxation time Characteristic time to achieve equilibrium

Thermal equilibrium no net exchange of energy

Diffusive equilibrium no net exchange of particles Mechanical equilibrium no net exchange of volume Unit of temperature: kelvin 0 K = 273C Low density gas has zero pressure.

Low density gas and Ideal gas lawEmpirically, we know that properties of low density gases can be well

described by the ideal gas law

P V =nRT (17)

where

P: pressure measured in pascal: Pa = N/m2. V: volume measured in m3. n: number of moles of gas R: A universal constant: 8.31 J/molK

T: measure in kelvin

1 mole is defined to contain one Avogadros number of molecules

NA= 6.02 1023 (18)Other measures of pressure includes

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bar = 105 Pa atm= 1.013 105 Pa = 1013 mbar

This is the form often used in chemistry. In physics, it is more useful torewrite it as

P V = (nNA)(R/NA)T =N kT (19)

whereNis the total number of particles (molecules) in the system and

k (R/NA) = 1.381 1023J/K (20)is the Boltzmann constant.

This constant is one of the most important ones in physics because it pro-

vides connection between macroscopic physics and the microscopic physics.Notice the unit ofk. It is joule per kelvin or energy per temperature. There-fore the existence of this constant indicates that energy can be converted intotemperature and temperature can be converted into energy. An analogy isthe speed of light c which is another constant. It provides a way to converttime to length and vice versa and ultimately the existence of the constantc gave birth to Einsteins relativity. In the case of Boltzmann constant, itgave birth to the statistical mechanics.

The above value ofk in joule and kelvin is, however, often inconvenientwhen considering microscopic physics. Joule is simply to big. The energyunit most often used in atomic and subatomic physics is electron-volt. This

is defined to be the potential energy gained by an electron when it is traversesa potential difference of 1 volt. In terms of eV, this is easier to remember:

k (300 K) 140

eV (21)

or

1 eV k (12, 000 K) (22)or if you have memorized the surface temperature of sun 6, 000K,

k (6, 000K) 0.5 eV (23)

These values are fine for rough estimates but for more quantitative values,you may memorize:

k 290K 140.02

eV (24)

Now, when we started this section, we said

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Ideal gas law is valid for low density gas It is an approximation.

What do we mean by that? What does low density mean? Lets thinkabout what really happens when the temperature becomes very small. Theideal gas law dictates that in this limit, the product P V is zero. Suppose wekeep the pressure constant. Now we know that real molecules and atoms havea finite size. Therefore the volume, however small the temperature is, cantshrink further than N vmolecule where vmolecule is the volume of the moleculeitself. In other words, there is a maximum density that a gas can reachthat is

maximum= 1/vmolecule (25)

This happens when there is no room what-so-ever between each molecule.Now let me right the ideal gas law in this way:

P

kT = (26)

where= N/V is the density. IfP is constant, then as T becomes smallerand smaller, the left hand side becomes larger and larger and eventually willexceed maximum. But that cant happen. Therefore we have this conditionfor the validity of the ideal gas:

maximum= 1/vmolecule (27)

That is, the average space between each molecules must be much larger thanthe size of the molecule. Another way of saying it is that thepoint particleapproximation is a good approximation. At a constant temperature, thisalso means that the temperature must be high enough. This makes sense. Ifthe temperature becomes low enough, any gas liquifies and the ideal gas lawof course breaks down.

Microscopic Model of Ideal Gas and Equipartion of Energy

Now lets see if we can get any more information out of the ideal gas law.

The ideal gas law itself is an emperical law that has been verified manytimes in laboratory experiements with low density gases. To get any moreinformation, we need to add some more physical intuition/ingredients. Inthis case what we add is our knowledge that all rarified gases are made upofweakly interacting molecules.

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By Newtons third law (the action is equal to the reaction), the wall experi-

ences a force

Fwall= Fptcl (33)

when a particle bounces off of it.Then the average force on the wall due to N such particles is

Ftotal = N Fwall= 2Nm(vx, 0, 0)/(2L/vx) =N m(v2x/L, 0, 0) (34)

Pressure is the perpendicular force per unit area:

P =

|F

|/A= N mv2x/(AL) =N m

v2x

/V (35)

or

P V =N mv2x (36)

This is what I meant when I said that pressure is kinetic energy.Comparing this with the ideal gas law

P V =N kT (37)

we conclude

mv2x =kT (38)

Now there is nothing special about the x direction. Therefore

mv2x =mv2y =mv2z =kT (39)

The average kinetic energy of a molecule is then

K =12

mv2x + v2y + v2z

=

3

2kT (40)

This is what I meant when I said that energy is temperature.This is a remarkable formula. We started with an emperical ideal gaslaw, threw in a basic microscopic physics and got a profound result eachmomentum degree of freedom contributes kT /2 to the total energy of aparticle.

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From above formula, we can also get the average root-mean-square speed

of a molecule at a temperature ofT:mv2

2 =

3kT

2 (41)

or

vrms=

3kT

m (42)

Lets plug in some numbers. At room temperature, we know that kT1/40 eV. The air is mostly made up of Nitrogen molecules which are in turn

made up of 2 nitrogen atoms. Each nitrogen atom carries 14 nucleons. Eachnucleons weight about 940 MeV/c2 or roughtly 109 eV/c2. Here we are usingenergy as a unit of mass using the Einsteins famous E=mc2. Therefore

m= 28 109 eV/c2 3 1010 eV/c2 (43)then

vrms =

3 1

40eV/(3 1010eV/c2)

= 3 1

40eV/(3

1010eV/c2)

1.6 106 c 480m/s (44)Thats slightly larger than the speed of sound.

This division of energy in equal amount among degrees of freedom goes bythe name ofEquipartition of energyor simplyEquipartition theorem.Well get to the theorem part but what it states is that if any quadratic termin the energy, be it kinetic, rotational or potential, contribute kT /2 to thetotal energy. This includes the translational kinetic energy

Ktr

= p2

2m (45)

for any value ofm and the rotational kinetic energy

Krot =L2

2I (46)

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where L is the angular momentum and I is the moment of inertia and any

simple harmonic potential energy

VSHO=2x2

2 (47)

or vibration energy. Often times, when the whole system is in a structurallystable configuration (such as in a crystal), the potential energy near theequilibrium point of each molecule or atom can be approximated by a SHOpotential. So this is not as artificial as it first looks.

If a molecule hasfsuch degrees of freedom, then the total energy of thesystem is

U=N fkT2

(48)

However, not all degrees of freedom contributes at all temperatures. Thetranslational kinetic energy is always there so f is at least 3. For rotationalenergy, quantum mechanics dictates that there is a minimum energy. Sounless kT reaches this minimum energy, this degree of freedom does notcontribute. This is calledfreeze out. When it does however, it very quicklyeach rotational degree of freedom contribute kT /2 to the energy. For thevibrational energy, again, there is a minimum energy dictated by quantummechanics (zero point energy, if you remember) thats required to excite this

sort of motion. So again unless kTis above the minimum energy, vibrationsdo not contribute to the total energy. But again once they do, they quicklycontributekT /2 per d.o.f.

Note that we are already talking about quantum mechanics here. Manyphenomena easily observed in nature is impossible to explain without quan-tum mechanics. Now, we arenot going touse any heavy machinery of QM.But as the opportunities arise, we wont shy away from it either. Having saidthat, lets have consider a simple example where classical consideration andquantum consideration gives very differentsimple results.

Monatomic gas: f= 3

Diatomic gas with two identical atoms O2, N2,...: 3 translational d.o.f.

2 rotational d.o.f. Rotation around the symmetry axis doesnt

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2 vibrational d.o.f. kinetic and potential

Total f= 7

Polyatomic molecule without axial symmetry: 3 translational d.o.f.

3 rotational d.o.f.

Sub Total f= 6

Many different kinds of vibrational mode stretching, bending, ...

Crystal lattice:

3 translational d.o.f. 3 quadratic potential energies

Total f= 6

Kinetic Energy Rotational Energy

P^2/(2m) L^2/(2I)

Vibrational Energy

p^2/(2m) + w x^2/2

Figure 7: Different energies of diatomic molecule.

Again, some of these can be frozen out at low temperatures.For instance, the air molecules around room temperature only exhibits 5

degrees of freedom, missing the vibrational ones.

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3 Heat and Work

There are a few fundamental principles of physics which are never vilolatedso far as we know. One of them is the conservation of total energy. Oth-ers are the conservation of total momentum, conservation of electric charge.If you are only concerned about non-relativistic physics (chemistry for in-stance), then you may add the conservation of mass to the list. Since theselaws are obeyed by most fundamental particles and their interactions, macro-scopic systems must also obey the same law. Trouble is, unlike electric charge,energy can assume many different forms kinetic energy, potential energy,rotational energy, vibrational energy, ...

If you are concerned about a system of gas in static or near static situ-

ation, you dont really care about all these forms of energies. Most of thetimes, what you are concerned about are

How much energy did I put into the system? Conversely, how muchenergy is spent by the system?

Whats the accompanying temperature change? How muchmechanical work did the system do?

For instance, if you are designing a refrigerator, the temperature is whatyou most care about. But as we will soon learn, to make the temperature go

down, you need to make a volume of gas to do work. And if you are designingan engine, what you really care about is the amount of energy put in versusthe amount of mechanical work the system has done.

In equation, we express this as

U=Q + W (49)

where Uis the total change of energy for the system. Q is the amount ofenergy that entered the system from thermal contacts with other systemsand W is the amount of the energy that entered the system from non-thermal contacts (e.g. mechanical, electrical, etc). Negative Q or W

means the energy was taken out of the system thru thermal contacts andnon-thermal contacts respectively. This is referred to as The first law ofthermodynamics. But thats just another way of saying that total energyis conserved.

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Now in the textbook, the change in the energy is denoted with symbol

while the heat and the work do not carry such a symbol. Mathematicallythis is becausedUis a perfect differential whose integral does not depend onthe path of integration. In other words, for energy, if you are at a certainpoint in the phase space, it doesnt matter how you get there. The energy isdetermined by the point you occupy. However things like mechanical workcan and will depend of the path that lead to the final point.

This is nothing mysterious. In geometrical term, perfect differentials suchas the energy is like the vector displacement. It doesnt matter how yougot to the final position. The displacement is always

x= xfinal

xinit

dx= xfinal

xinitial (50)

However thelengthof your journey is a totally different matter. The lengthof your journey

L= tfti

dxdt dt (51)

depends on the path you take even when you are in 1-D. So dx is a perfectdifferential while the line element dL =

dxdt

dtis not.Note that for the thermal equilibrium to be established, heat Q must

be exchanged between two systems brought into contact. For mechanicalequilibrium, W is the relevant quantity. There are different ways heat canbe transferred between the systems.

Conduction: In contact. Kinetic energy is exchanged. Convection: Circulation of gas and liquid driven by temperature dif-

ference and the density changes.

Radiation: Emission of photons.

4 Compression WorkIn the Mechanics, a work is defined by

W = F dr (52)

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If the force is conservative, that is if a potential energy can be found so that

F = V (53)

then the change inWwhen a particle moves from one point to another doesnot depend on the path it took. However, if no such potential exists, thenthe change in W does depend on the path. Thats why the book doesntwrite W.

PF

dx

Figure 8: Compression work

Now suppose you have a cylinder full of air with a piston at one end. Ifyou push the piston in, you know from everyday experience that you need acertain amount of force to do so especially as the piston goes deeper into thecylinder. Now from the defintion of pressure, we know that

Fn= P A (54)

where Fn is the component of the force normal to the surface and A is thearea of the surface. Surface in our case, of course, refers to the surface of thepiston. Plugging this into the first equation gives

W =Fndr= P Adr = P dV (55)where dV is the amount of volume displaced by the piston moving a smalldistancedr (the distance in the normal direction). The minus sign indicatesthat the system got smaller by this amount.

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Now for this formula to apply, the movement of the piston has to be

slow so that the system always has the time to adjust to the new volumeand establish an equilibrium accordingly. This sort of slow movement iscalled quasi-static movement. Usually, this is a good approximation foran everyday object (translation: size of O(1 m). For this to be not a goodapproximation, the piston has to move close to the speed of sound (330m/s).

Now before we plunge into some calculations, lets stop here and thinkaboutwhypressing the piston needs force. Not only that, why it gets harderas the volume becomes smaller. To see this, we go back to our simple pictureof lots of balls bouncing around the room. Now remember that when a ball

V

V

L

A

Figure 9: Microscopic view of pressure

bounces off of a wall, the momentum changes by

|p| = 2m|vx| (56)and the rate of bounce is

t= 2L|vx| (57)so that the average force exerted by a single particle is

fx = |p|t

=m v2x

L (58)

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Average force due to N such particle is

Fx =mNv2x

L (59)

Pressure due toN such particles is therefore

P =NFx/A= mNv2x

L A (60)

Now suppose the average speed of particles, or the average kinetic energy ofparticles does not change during the course of volume change. In other words,suppose the cylinder is in contact with a big system with a temperature T.

Since we are talking about quasi-static change, the temperature in the systemis maintained. This sort of change is calledisothermal. Iso in latin meaningthe same. In that case, we can see from the force expression that if Lgets reduced by 1/2, then the force doubles because the rate of collisionsdoubles.

On the other hand, lets consider another extreme case when the systemis totally isolated from the outside. That is, put some big chunk of insulator(styrofoam will do) around the cylinder so that no heat can escape from it.What happens then?

In purely macroscopic terms, we can get the result in the following way.If the process is adiabatic, there is no heat enetering or leaving the system.

So the first law says

U=W = PV (61)

Note again the sign. However, we also know that

U=f

2NkBT (62)

and hence

U=f

2Nk

BT (63)

Equating the two, we get

f

2NkBT = PV (64)

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If the gas obeys the ideal gas law, we then get

f

2N kBT = NkBTV

V (65)

or

f2

T

T =

V

V (66)

Since

d ln x=dx

x (67)

we get

ln

V Tf/2

= Const. (68)

or

V Tf/2 = Const. (69)

From

P V =N kT (70)

we also get

P V

T P V1+f/2 = Const. (71)

5 Heat Capacity

O.K. So compressing or expanding gas can do raise or lower the temperatureof the gas by pumping the energy into the system or out of the system bymechanical work. Another way of changing temperature of the system is,

of course, make it in thermal contact with another system with differenttemperature.

Now experience shows that some system can soak up a lot of energybefore its temperature is substantially raised and for some other systems, it

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doesnt take much energy to raise/lower the temperature. This property of

the system/material under study is called heat capacity. This is defined as

C= Q

T (72)

In other words, the amont of heat needed to raise the temperature by 1kelvin.

Now before we do any calculation, lets see if we can guess whatC should be. What should it depend on? First of all, considerone litre of water and 10 litres of water. Which takes more energybefore the temperature can be raised by 1 degree? The 10 litres

of water, of course. And you would expect that the more thewater, the more energy you need to raise its temperature. Inother words,

C V N (73)

Now think about a gas of a monatomic molecules and a diatomicmolecules. Monatomic molecules can have only 3 degrees of free-dom. But we saw that a diatomic molecules can have 7 degrees offreedom. Now theequipartition theorem states that the en-ergy is equally shared among these degrees of freedoms.

Since each degree of freedom takes kBT /2 amount of energy nomatter what the situation, we can guess that it takes more energyto raise temperature if there are more degrees of freedom. Hence

C f (74)

What should its unit be? Well, sinceQ is energy Cmust havethe unit of energy/temperature. But this is precisely the unit ofthe Boltzmann constant. Therefore, we can guess that

C= const. kBNf (75)

where const.should be a order O(1) number.

A more fundamental quantity is the specific heat capacity defined by

c Cm

(76)

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wherem is the mass of the molecular unit in the system.

One thing to notice is that the above definition is ambiguous for theprecisely the same reason that we dont write Q. That is, the heat is aprocess dependent quantity. In other words, it is a function of how theenergy entered the system. Since there are many different ways for theheat to enter the system, this is not a well defined quantity. In mathematicalterm, again Q is not a perfect differential and therefore its integral is path-dependent.

But put that aside for a while and lets think about this thing for awhile. Before doing any analytic work, what can we say about the specificheat? Think about a monatomic gas and diatomic gas and remember theequipartion theorem. Any amount of energy entering the system will be

shared equally among the degrees of freedom. In a monatomic gas, theenergy will be shared by 3 translational degrees of freedom. But in for thediatomic gas, the energy must be shared by up to 7 degrees of freedom. Sogiven equal amount of heat and all else being equal, it is easier to heat upmonatomic gas than a diatomic gas. In other words, we needsmaller heatto raise temperature for the monatomic gas. That is, the heat capacity formonatomic gas must besmallerthan that of the diatomic gas. In fact, moredegrees of freedom to excite means that the heat capacity will be larger.That is, there are more sponges for each molecule to soak up the heat.

To see all this more explicity, use

Q= U W (77)and write

C=U W

T (78)

In case of compressional work,

C=U+ PV

T (79)

Just as in the consideration of the compressional work, it is the P dV termthat is the source of this trouble.

We can consider two extreme cases. First consider that the volume didntchange. In that case, there is no mechanical work and

CV =

U

T

V

(80)

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where the subscript V is there to remind that the volume is held fixed.

Naturally, this is calledheat capacity at constant volume.On the other hand, we can consider fixing the pressure but not the volume.In this case,

CP =

U

T

P

+ P

V

T

P

(81)

again the subscript P is there to remind that the pressure is held fixed.Naturally, this is calledheat capacity at constant pressure.

Which one should be larger? If you just look at the formulas, it lookslike that CPmust be larger than CVdue to the extra term. But is it true?

Is the sign of (V/T)P positive? Well, yes. Higher temperature meansbigger volume to have the same pressure. If you keep the same volume, thenthe pressure is going to be raised as the temperature goes up (rememberP kT). So to let the steam out, the volume must increase.

O.K. But the question still remains. Why is it reasonable to expect thatCP is larger than CV? This is simply a consequence of energy conservation.If the volume is held fixed, all energy goes into rasing the temperature ofthe system. On the other hand if the pressure is held fixed, some energymustbe spent in enlarging the system volume against the external pressure.Therefore it takes more energy to raise the temperature of the system atconstantPthan the system at a constant V.

How much more then? This depends on the detailed properties of the gasmolecules. For ideal gas with fdegrees of freedom,

CV =

U

T

V

= d

dT

NfkT

2 =

Nf k

2 (82)

and

CP =

U

T

P

+ P

V

T

P

= d

dT

NfkT

2

+ d

dT

(NkT) =CV + Nk (83)

Latent Heat

For some system, it is possible to pump in or out heat and not changethe temperature. It may sounds odd, but this is everyday phenomenon. If

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you let a glass of ice and water on the table, the temperature of the ice-

water system remain at 273 kelvin until all ice is dissolved. After that thewater temperature will rise some more to eventually equilibrated with theatmospheric temperature of the room. But this does not mean that no energywas pumped into the ice-water system while the ice was dissolving. Ice wasdissolving after all.

This example teaches us the following:

This sort of thing happens during phase transition. Well get tophase transitions later. For now it is suffice to have an intuition aboutthat. That is, you know that H2O has three phases: ice, water andvapor depending on the temperature andpressure.

The amount of energy put into the system must have been spent tochange one phase of matter to another. In the above example, the heatfrom the atmosphere was used to break up the bond between watermolecules in the ice and make them runny thats water.

To quantify the amount of energy used in such phase transition, we definethe latent heat

L Qm

(84)

where m is the unit mass of the ingredients of the system. In the aboveexample, the water molecule.Note again that since this definition involves heat, it is again ambiguous.

One must specify the exact circumstance in which L is measured. Thetables in textbooks usally list L values atP= 1atm. For ice

L= 333J/g (85)

For boling water

L= 2260J/g (86)

Where do these numbers come from? Are they natural? Well, we knowthat a typical atomic energy scale is

1 eV = 1.6 1019J (87)

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A water molecule has 2 hydrogen and 1 oxygen. Therefore

mH2O 30 1024 g (88)So the ratio is

1 eV

mH2O 5 103 J/g (89)

We are in the right ball park. The above numbers for the water means thatthe energies involved in breaking the ice into water and the water into vapormust be in the range of about 0.1 eV to 1 eV. This is, of course,very roughestimate. But we got it about right within an order of magintude and that

means that means that we are that much closer to actually understandwhat goes on at the molecular level.

Enthalpy

The energy conservation in the first law form is

U=Q + W (90)

This is the law of nature. You cant argue with that. In some situations,however, it is convenient to rewrite it. One such situation is when the systemis under a constant pressure. In that case, the compressional work done on

the systemwhile its volume changes by V is simply

Wcompressional= PV = (P V) (91)Again, note the minus sign. If the volume of the system decreases, a workwas doneon the system. The inclusion ofPunder sign is possible herebecause P is constant. Otherwise the last step is in general not permissive.In this case, one can rewrite the first law as

(U+ P V) =Q + Wothers (92)

whereWothers represents work done again on the system by contacts other

than thermal and mechanical. This could be magnetic, electric, graviational,etc.

Lets define Enthalpy

H=U+ P V (93)

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and rewrite

H=Q + Wothers (94)

Up to now, all we have done is to take Pconstant and rewrite the energyconservation law. The question is, why are we doing this? Why is thisdefintion useful?

First of all, a lot of everyday phenomena happens under approximatelyconstant pressure, i.e. 1 atm. Second, if there areno other works done onthe system, then the above equation simplifies to

H=Q (95)

By writing it this way, we eliminated pressure and volume dependence fromour consideration. This means that if we can measure the enthalpy justlike we measure energy, then all we need to know about energy flow underconstant pressureis, just that, the enthalpy. In other words, in the absenceof other types of work, enthalpy is heat.

For instance, suppose you are boiling some water. To calculate how muchheat you need, you can do two things. You can explicitly use

U+ PV =Q (96)

and look up the needed energy change and the change of volume when, say,

a mole of liquid water becomes a mole of water vapor at 100 C.On the other hand, if you just know enthalpy of liquid water and the

water vapor, you can just subtract the two and come up with the answer.This is, of course, much easier. Chemistry books are full of tables of enthalpyfor different materials. The reason is exactly that it makes a chemists lifethat much easier.

O.K. Thats fine. But what is this mysterious quantity called enthalpy?What is the meaning of it? Well, what is P V anyway? We had

Wcompressional= PV = (P V) (97)

Remember that this is work done on the system. Now think of theatmo-sphere as the system. ThenPV is the amount of work done on theatmosphere systemto reduce its volume by|V|. In other words, in thiscase, something or somebody must do this amount of work on the atmo-sphereto create something other than air with a volume |V|. Or one may

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say that P V (note that V itself is positive while the change V can be of

either sign) is the amount of work somthing or somebody must do to pushthe atmosphere away to make a way for something else in its place, watervapor for instance.

In other words, in the expression

H=U+ P V (98)

U is analogous to the mechanical kinetic energy and P V is analogous tothe mechanical potential energy and H is analogous to the total energy.Think of pressing against a plate attached to a spring. To make a roomfor yourself, you push the wall away. By doing so, you have increased your

potential energy bykx2

/2. Water molecules from boiling water in a way needto do the same thing. They have to push away the air molecules to makeroom for themselves. However, CAUTION: Dont take this analogy toofar. Although there is some truth to it, its for illustration only. Uin generalcontains both the kinetic and the potential energy for the molecules.

Now just as the absolute amount of total energy has no meaning (youcan always define what you mean by zero by adding a constant), absoluteamount ofH has no meaning. The only thing that matters the differencein enthalpy when somethings change into somethings else. For instance, thechange in enthalpy when liquid water changes into water vapor is

HH2O= 40, 660J (99)

per mole of water. Now a mole of water is about 18 grams. That means theenthalpy change per gram of water is

HH2O/m= 2260J/g (100)

which is the same as the latent heat. In 40, 660J

P V =N kT =RT = (8.31 J/K)(373 K) = 3100 J (101)

is spent working against the atmospheric pressure. Thats about 8 %. Therest of it spent in breaking up the molecular bonds between water molecules.Another example is burning hydrogen

H2+1

2O2 H2O (102)

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For each mole of water produced,

H= 286 kJ (103)Huh? Negative enthalpy? Well, this has two explanations. One, you burnedapproximately one and half units of gas (1 for hydrogen and 1/2 for oxygen)and got one unit of gas (water vapor) that reduced the volume. Thereforethe second term in

H= U+ PV (104)

is negative. The potential energy between hydrogen and oxygen is reducedwhen they bind together to form water. This energy is then released as kinetic

energy thereby raising the temperature. Thats burninig. In terms of energyput into the system, U < 0. Therefore overall, heat is released fromthe system. In terms of heat enteringthe system, thats negative quantity.This is a good thing. Othewise, Montreal winter would be unbearable.

6 Schroeder Chapter 2 The second law

In this chapter, we are going to study the second law of thermodynamics.The first law

U=Q + W (105)

is an absolute law of nature. The equality is the equality. The second law isa bit different although in the end it doesnt really matter. The second lawof thermodynamics states:

The entropy always increases.

Stated in this way, it sounds mysterious. But this is not so strange. Ineveryday language, it sounds something like this.

Suppose you have a system of many particles, say a boxful. Theparticles inside the box flies around more or less randomly. There-

fore, you can consider the probability that the particles in the boxare in the phase space volume

N=Ni=1

xipi (106)

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around a particular configuration

N = {(x1, p1), (x2, p2), ..., (xN, pN)} (107)The second law of thermodyanmics states that a very small frac-

p

x

xp

x

p1

1

Figure 10: Phase space volume

tion of such configurations is overwhelmingly more likely thanall others. The system practically never leaves vicinity of thosemost likely configurations. And if the system started out at apoint far from those configuration, given enough time (usuallyvery short) the system will always end up near the most probableconfigurations.

Now notice here that we are starting to talk about probability. Thisis the key concept in Statistical mechanics. When do you need probabil-ity? Well, if you know exactly how a single particle behaves, for instance,the movement of a pendulum, then you dont need probability. You knowthe position and the momentum of that pendulum absolutely. There is no

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uncertainty. However, if you are watching a fly darting aroud the room with

no detectable pattern, you cant be absolutely sure where the fly will be 2minutes later. But by observing the motion of fly long enough, you can guesswhere it probably will be, i.e. at the garbage can. But you cant be certainbecause you dont know what the fly is thinking.

Thats it. When you know something about the system but not all, allyou can have is the probability. This could be correlated (since the fly is atthe garbage can right now, it will most likely be still there 2 minutes later)or uncorrelated (since there is no garbage can in the room, the probabilitythat it will stay at this corner is just as likely as it will be at another cornertwo minutes later) but in any case, you must consider the probability.

Now consider a typical example of thermodynamic system a box full

of gas molecules. You cant know the exact position and the momemtum ofeach 1024 particles in the box and frankly you dont want to know. But thismeans that you cant absolutely predict whats gonna happen to the systemtwo minutes later. The question is, can we then talk about the probability?

O.K. Suppose we want to do that. Then the next question is

How do you define probability anyway?

In this case, we proceed as follows. First we specify the global conditions ofthe system. Usually, we specify the total energy of the system and the totalnumber of the particles. Suppose we do that. Now thats only 2 constraints

among 10

23

degrees of freedom. This means thata lot of different configu-ration (state) of those 1023 degrees of freedom can have the same U and N.Now suppose we prepare many, many systems with the same U and N butdont specify anything else. The whole is called ensemble.

The probability to have any particular configuration (states) C (for in-stance configurations with 1/4 of particles having momentum smaller than,say,U /N) is then given by

P(C) = Number of systems satisfyingC

Total number of systems in the Ensemble (108)

In the limit of the large total number of systems, the total number of systems

can be thought of as the number of all possible states. And the numer-ator can be thought of as the number of states satisfying the condition C.Therefore one of the most important problem in stat-mech is the countingproblem. You need to know how to count, first of all all possible states, andthen need to know how to count all possible states under certain conditions.

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Now in Classical mechanics, the state of a particle at any given instance is

completely specified by its phase space coordinates and the energy {x, p, E}.All these variables are continuous variables and there are 7 of them.However, in reality we know that microscopic world is governed not by

Classical mechanics but by quantum mechanics. The most important factin quantum mechanics is the particle-wave duality. Fundamenetally, aparticle obeys wave equation. Only in the macroscopic limits, one can ap-proximate it with classical equation of motion. You will learn more about itin quantum mechanics course. Here well just briefly state the facts we needto proceed with the rest of the course.

If particles are fundamentally waves, there are many non-trivial conse-quences. For us, the followings are needed:

A wave cannot have zero size as particles can. It must have a finite ex-tend inphase space. The consequence is the Heisenberg Uncertaintyprinciple:

x p h/2 (109)

which is to say that one cannot measure the position and the momen-tum of a particle simultaneously. That is to say that there is a minimumphase space volume that a particle must occupy. In contrast, a classi-cal particle occupies a pointin the phase space which by defintion has

zero volume.

If we specify p, then the uncertainty principle tells us that we have noidea what-so-ever where the particle is actually located. So there is nopoint in worrying about the position of the particle. All one has tospecify is eitherx or p. In our applications, it is much more convenientto specifyp.

When confined either in a box or in a potential, we can have standingwaves. These are only stable form of waves in confined condition. Re-member that to form standing waves, certain relationship between the

size of the box (or potential) and the wavelength has to be satisfied.The consequence is that the energy levels are discrete. In classicalmechanics, x,p can have any real value so that the energy Ecan haveany real value even if the particle is confined in a potential. That is, theenergy levels are continous. In quantum mechanics, this is no longer

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true when particles are confined in some way either in a box or in a

potential. It is possible that different states (labeled by some other quantities

such as momentum or angular momentum) can have the same energy.The number of such states for a given energy level is referred to asthe multiplicity or the degeneracy of the energy level. In classicalmechanics, this number is infinite because any finite interval of realnumber line contains as many points as the whole real line. In thiscase, one would speak ofvolume instead of multiplicity. In quantummechanics, this is of course not true. Not only the energy levels arediscrete but other quantities are discrete as well. Therefore, we can

count how many different states share the same energy.

If two particles are identical, there is no distinction between the statewhere one particle has energy E1 and the other one E2. This maysound trivial. But it is not. Remember we have a counting problem. Ifthese are classical particles, we can distingish two particles even if theyhave identical properties. We can always mark them with a marker.If you exchange the position and the momentum of two particles, weend up with different state. In other words, ordering of particlesis important. The list (1, 2) and the list (2, 1) are different and hencerepresents two different states. On the other hand in quantum mechan-

ics, identical means identical. You cant label two identical particles inany way. So the list (1, 2) and (2, 1) are the same. That is, quantummechanics corresponds to orderless sets.

Actually, modern quantum mechanics grew out of a crisis in classicalstatistical mechanics. One of the reason quantum mechanics was discoveredwas 19th century physics inability to explain the momentum spectrum ofblack body radiation. Plancks briliant contribution was to assume that theenergy of photons were quantized. This lead to completely different countingfor the low enery photons. The crisis was averted and quantum mechanicswas born.

In summary:

1. There is a minimum phase space volume that a particlemust occupy.

2. Energy levels of a confined particle is discrete.

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3. Multiplicity or degeneracy of each energy level is countable.

4. Identical particles are absolutely identical. There cant be any order-ing for them.

7 Two-state systems

Now lets first think about classical counting. In this case, we can labeleach particle even if they are identical in all other properties. I.e. they aredistinguishable.

A prototype of classical counting problem is the coin toss. The questionis:

If you toss a coin Ntimes, what is the probability to have nnumber of heads?

To answer this question, we need to know that total number of possiblities

and the number of possibilities where there aren heads.The textbook has an example where you toss, a penny, a nickel and a dime

in that order. The reason for having different coins is to get away from theissue of identical particles well, coins. We dont have touse different coins.We might as well toss a single coin three times but remembering that the

orderis important. Here is the reproduction of table 2.1 from the textbook:

Penny Nickel Dime MacrostateH H H 3 heads

H H TH T H 2 heads (= 1 tail)T H T

H T T

T H T 1 head (= 2 tails)T T H

T T T 0 head (= 3 tails)

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There are a total of 8 possibilities according to this table. Intuitively

then we have 1/8 chance of getting either no tail or no head and 3 /8 chaceof getting either one tail or one head.This looks like an quite artificial example in that there is no physics

analogy. This is not so. The binary problem happens in physics all the time.This is because since we the number of states are countable, the simplestnon-trivial problem one can think of involves 2 states. Often enough, atlow temperatures, the most important energy levels are the ground energyand the first excited states.

In physics, each one of the above line corresponds to a microstate. Amicrostate is specified if you know all the details about the system. Onthe other hand, if you are only concerned about how many tails you have

but not when and how they appeared, these corresponds to macrostates.In our case these are states with 3 heads, 2 heads, 1 head and no head. Nowsince we are ignoring details in macrostates, each macrostate corresponds tomay microstate. The number of microstate put under a macrostate is calledthe multiplicity or the degeneracy of the macrostate. In stat mech, wemainly use the term multiplicity. The name degeneracy is usually reservedfor the multiple state with the same energy in quantum mechanical sense.However, this is not a rule. You need to be able to distinguish what meanswhat from the context. This, however, is usually quite clear. We will ingeneral denote the multiplicity with the greek letter Omega . For instance,

the multiplicity for 2 head macrostate in the above example will be(2) = 3 (110)

Note that the way we defined probability, the probability for this state canbe written as

P = (2)

(all)=

3

8 (111)

Now it is tedious but quite easy to enumerate all the possibilities of 3 cointoss. But what if we want to toss a coin many many times, say, 1023 times?

Writing down all the possibilities and counting them are out of question.Luckily there is a branch of mathematics that deals precisely this sort ofthings. This is called combinatorics. At the end of this note for chapter 2,you will find a summary of often used counting rules.

Let me quickly summarize it

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Permutation with Repetition : You are picking out numbers for a lot-

tery. To win, you not only need to pick the right numbers ( sof them)but also in the right order. There areNnumbers to choose from.However, repetition is allowed. That is, you can pick 1, 1, 1,...if youwant to. There are a total of

N

s= N

s (112)

possibilities.

......

s slots

Each slot can be filled withN number of symbols.

For instance if these are alphabets,there are 26 possibilities to fill each

slot. So multiplicity = N^s.

Figure 11: Permutation with repetition

Permutation without Repetition : You are picking out numbers for a

lottery. To win, you not only need to pick the right numbers (s ofthem), but also the right order. There are N numbers to choosefrom. However, this time, no repetition is allowed. There are atotal of

NPs= N(N 1)(N 2) (N (s 1)) = N!(N s)! (113)

possibilities.

Combination without repetition : You are picking out numbers for alottery. This time the rule is more lenient. You only have to pick the

right numbers (again s of them) regardless of the order. Again thereareNnumbers to choose from. No repetition is allowed. There area total of

Permutation ofs without repetition out ofN

Permutation ofs with repetition out ofs (114)

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NCs = N(N 1)(N 2) (N (s 1))

s!=

N!

(N s)!s!

Ns

(115)

possibilities.

Combination with repetition : You are picking out numbers for a lot-tery. This time the rule changed again. You only have to pick the rightnumbers (again s of them) regardless of the order. Again there are Nnumbers to choose from and this time, repetition is allowed. Thereare a total of

NHs=

N+ s 1s

(116)

For the problem at hand, we can reformuate it this way:Suppose you have Nslots to write Hor T.

......

......

......

N slots

......

......

H H H H H H H H H H H H

H H H H H H H H H H HT

T T T T T T T T T T T T

Figure 12: Number of words of length Nin 2 letter alphabet.

That is, you want to write down all possible words of length N in a 2letter alphabet. The alphabet in this world consists of only two letters H

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and T. Systematically, you would start out with all-heads configuration

Cfirst= HHHH HHH (117)and end with all-tail configuration

Clast= T T T T T T T T (118)How many such states are there? Well, for each slot, you have 2 possibilitiesand you have N slots. Therefore

(all) = 2N (119)

This is an example of permutation with repetition.If you want to know how many words have only one T orH, thats just

N. Because you can choose any one ofN slots to put T in and fill all theother slots with Hand vice versa or

C1 =

THHH HHHTHH HHHHTH HH...HHHH T HHHHH

HT

(120)

Now if you want to know how many words contain two and only twoT, you need to able to count the number of different ways of picking out 2different slots out ofN.

C2=

T T H H HHHT H T H HHHTHHT HHH...HHHH HT T

(121)

How many different possiblities are there? Well, for the first slot, you haveNchoices. For the second slot you have N 1 different choices because oneslot is already occupied. So you have N(N 1)ordered choices for 2 slotsin which, say, (1, 2) and (2, 1) are counted as different choices. This is anexample ofpermutation without repetition. But thats not right. These

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result in an identical word. So you must divide this by 2. Therefore the

multiplicity associated with 2-tail macrostate is

(2 tails) =N(N 1)

2 (122)

That is, the counting problem becomes combination without repetition.CAUTION: These identical states have nothing to do with identical

particles. The particles, or coins, here are still distinguishable. This ispurely a matter of counting different words.

You can go on like this. For three heads, you have N(N 1)(N 2)ordered choices. But states like (1, 2, 3) and (3, 1, 2) leads to the sameword. Now if you have 3 different objects there are 3! = 6 different ways ofordering it.

So in general, the number of possible events with s heads is

(s) =

N

s

=

N!

(N s)! s! (123)

This formula is good for s= 0 and Nwith the definition 0! = 1. In bothcases

(0) = (N) = 1 (124)

For large N, this is a huge number. Probability for such state is

P(s) = (s)

(all)=

N!(Ns)! s!

2N (125)

Since this a very typical and also important problem in counting, let medo it once more. Each state can be a head or a tail. So if you have 2such particles, all possible combination can appear in the expression

( + )( + ) = + + + (126)

If order is important, each state gets multiplicity of 1. If order is not im-portant, and are two microstates under the same macrostate.Therefore we should write the above as

( + )( + ) = +2 + (127)

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and read off the multiplicity of each macrostate as 1 for the two-heads

macrostate, 2 for one-head-one-tail macrostate and 1 for the two-tailsmacrostate. Likewise, if you have three such particles,

( + )( + )( + ) = + + + + + + + = +3 +3 + (128)

and read off the multiplicity of each macrostate as 1 for the three-heads, 3for the two-heads, 3 for the two-tails and 1 for the three tails.

We can continue like this indefinitely. In general if you have Nparticleswhich can occupy binary states, all possible states appear in the expansionof

Ni=1

( + ) = ( + )( + ) ( + ) (129)

where there areN( + ) factors. This is nothing but a well known binomialexpansion. Therefore, if order is not important, we can write this as

Ni=1

( + ) =N

N=0

NN

NN (130)

and read off the multiplicity as

(N) =

NN

(131)

You can easily extend this to multinomial expansion. Suppose that par-ticles in the system can have 3 states labelled a,b,c. If you have N suchparticles in the system, then all possible states of the system itself appearsin

(a+ b+ c)N =Ni=0

Nj=0

TN:na,nb,ncana bnb cnc (132)

wherena+ nb+ nc= N and

TN:na,nb,nc = N!

na! nb! nc! (133)

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is the trinomial coefficient which gives the multiplicity of a macrostate with

na particles in the astate andnb particles in bstate.The justification of this formula is as follows. First, think ofb and c asthe same. Then the multiplicity for the macrostate withna particles in theastate is

N!

na!(N na)! (134)

Now consider the b and c. There are N na of them. Now if I want aparticular state with nb particles in the b state, there are

(N

na)!

nb! (N na nb)! (135)

possibilities. So the total multiplicity for a macrostate with na, nb, ncparticlesin the a, b, c states is

(na, nb) = N!

na!(N na)! (N na)!

nb! (N na nb)!= N!

na! nb! nc! (136)

using N na= nb+ nc.You can continue on. In general, if you have N particles and k states,

(n1, n2, , nk1) = N!

n1! n2! , nk1! nk! (137)withn1+ n2+ + nk1+ nk =N.

The Two-State Paramagnet

O.K. So what is this good for? Is there any physical situation that thiscoin-flipping is relevant? One very practical problem is that of a magnet.Magnetism stems from spin of the constituents. A subatomic particle witha non-zero spin acts like a tiny magnet. If all these tiny magnets tend to lineup with the applied magnetic field, we call the material paramagnet. If the

line up persists even if we turned of the external magnetic field, we call suchmaterialferromagnet.

You know that magnets always come in dipole. That is, there is noknown (to human anyway) particle or material in the universe that has onlyS pole or N pole. Each magnet always come with both poles. Hence, the

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name dipole. Now if quantum mechanics allow the dipoles of the constituent

to have onlytwostates parallel or anti-parallel to the magnetic field , thenwe have two-state paramagnet. This happens, for instance, the relevantdegrees of freedom is electron spin. An electron has a half spin that meansyou can have only two states: Up or down. You will learn a lot more aboutit in QM course. For now, lets accept that as fact. The problem is, what isthe multiplicity of a state where N number of dipoles are parallel to the Hfield? If we have a total ofNparticles, the answer is

(N) =

NN

=

N!

N! N(138)

whereN=N+ N.When a magnetic field is applied, the energy of being parallel to it is

E= B (139)where is the magnetic moment of the particle and

E = B (140)

for the anti-parallel orientation. The total energy is therefore

U = NE+ NE

= (N N)B= (N 2N)B (141)

So specifying the number of up spins is the same as specifying the totalenergy but nothing else. So one can also say that (N) is the multiplicityof the macrostate with energy E = (N 2N)B. Well learn more aboutparamagnetic material later.

What about identical particles?

One thing you should be careful about is the question of identical parti-

cles. Suppose the paramagnetism here is caused by an electron. An electronis an electron. Any electron thats pointing up is as good as any others. Theyare identical. But the formula we used came from coin tossing where all thecoins were distinguishable! Whats going on here? Why isnt there only asinglestate when there are N up-spins?

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In this case, we are allowed to distinguish the electrons because the

paramagetic materials are usually in a crystallin structure. That is, eachelectron has an assigned lattice site. In that sense, we can say that thiselectron belongs to the site (0, 0, 0), this belongs to (1, 0, 0) and so on. Thisis possible because the lattice sites are well separated and the electrons welllocalized. If the atoms/electrons are not well separated, for instance we aredealing with a dense liquid of something, then this is not strictly true. Wehave to use full machinary of many-body quantum mechanics with built inidentical particle consideration. For now, unless I say otherwise (or the booksays otherwise) we deal with well separate paramagnetic material.

8 Einstein SolidEinstein Solid

k

k

Figure 13: Ziggling atoms in a crystal

Now lets think about somewhat more elaborate counting. This is theproblem of counting the multiplicity of a particular macrostate with a fixedenergyUfor a crystall with L cubic lattice sites. Thats a mouthful. Let me

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do this again. Suppose you have a crystal that contains L atoms. Further

suppose that these atoms are arranged in a way that they form a regularcubic lattice. Thats what you get if you draw bars (parallel to the axis)between integer points in a Cartesian space or if you build a bigg cube outof many small cubes.

Now at a finite temperature, the atoms dont stay at the same place.Thermal energy makes them jiggle around the equilibrium positions. If theamplitude of the oscillation is small (so that the crystal doesnt melt), thenit is always possible to approximate and consider each link as a spring, orsimple harmonic oscillator.

A simple harmonic osciilator is the most importantquantum mechani-cal system. This is because often times, small amplitude motions around

equilibrium position can be approximated by simple harmonic oscillator.Also, SHO problem isexactly solvable. Another important property is thatfor quantum mechanical SHO, the energy levels are equally spaced. That

is, the first excited states has the energy of h = hf ( =

ksm

, = 2f)from the ground state, the second excited states has the energy of 2 h fromthe ground state, and so on. Hereh is called the Placks constant. This is afundamental constant of nature. When we talked about quantum mechanicsa little bit we said that due to the wave-particle duality, each particlemustoccupy a phase-space volume larger than h in each dimension. On the otherhand, a classical particle can occupy a point in the phase space. That is,

volume 0. Hence, if h is zero, we wouldnt have quantum mechanics. Onthe other hand, if h is too big, then we would see all kinds of weird stuffin everyday life (well, we wouldnt think them as weird, just natural). Fornow, just think of it as a conversion constant between frequency and theenergy just as you can think of the Boltzmann constant k as the conversionconstant between temperature and energy. In your regular QM course, allthese will be extensively discussed. For now, lets accept this as a fact.

Now consider a crystal withLatoms. Since we live in a 3-D world, eachatom can oscillate in 3 different directions (x,y,z). Therefore, each atomcorresponds to 3 distinct oscillators. The total number of oscillators istherefore

N= 3L (142)

Suppose that each of these oscillators have the same . The question we askis:

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What is the multiplicity of a macrostate that has the total energy

ofU=qh?

In another words, we want to distribute qunits of energy to N differentlocations. A single site can have any unit of energy. In another words, wewant to know the number of ways to pick qnumbers out ofNpossible num-bers including repetition. Is the order important? No. So the countingproblem reduces to that of combination with repetition. The answer is

(q) =

N+ q 1

q

(143)

9 Interacting Systems

When we started this course, I told you that thermodynamics is study ofequilibrium. Intuitively, equilibrium is a state a system reaches if it left alonefor a very long time. For instance, suppose you pour a boiling water in cupand let it sit for awhile. While the water cools, the temperature constantlychanges. After something like an hour, the water temperature becomes thesame as the room temperature. From then on, it doesnt matter when youmeasure the temperature of the water. Itll be always be the same. That is,the water has reached thermal equilibrium with the air around it. (Mostlikely it didnt reach diffusive equilibrium with the air and eventually willdry up.)

Question is, why? Why does a system tend to reach equilibrium withthe surrounding?

In this section, we start to answer that question. Full answer will comelater, but already we can have a pretty good basic understanding as to whythis happens. This is all about probability and the laws of large numbers.

Intuitively, what happens is like this: A proto-typical example of equilib-rium is a box half-full of particles at time t = 0. So initially, then density ofthe other half of the box is 0.

However, very quickly, the particles fill up the whole box and the density

becomes homogeneous. That is, after a very short time, you wouldnt knowif this system started out as half filled.

SHOW THE MOVIE

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Figure 14:

Figure 15:

Now, Newtons law is time reversible. That is, you shouldnt be able totell whether a movie is run forward or backward, but if you run the moviebackwards, you know.

SHOW THE MOVIE

So somehow, nature consipres in a way that even if the underlying dy-namics is time reversible, things that started out in a non-equilibrium statewill evolve into the equilibrium state and stay there losing all memory ofinitial state except a few quantities such as the total energy.

So whats so special about the equilibrium state so that even if you startout with quite different states they are all driven to the same equilibriumstate? Well the answer is that the multiplicity of the equilibrium state is

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overwhelmingly largerthan any other state. For large number of particles

say 1023

particles, overwhelming means not 10 times or 100 times, or even1000 times, but more like at least ten billion times more than any others.Therefore once a system gets there, all it stays inside. Very rarely, the systemis outside the equilibrium state and this happens for very small fluctuations.Large fluctuations are very, very rare. Practically never.

Thats fin

An Introduction to Thermal Physics - 1

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