MP203 Statistical and Thermal Physics

Jon-Ivar Skullerud and James Smith

December 20, 2017

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Contents

1 Introduction 4

1.1 Temperature and thermal equilibrium . . . . . . . . . . . . . . . . . . . . 5

1.1.1 The zeroth law of thermodynamics . . . . . . . . . . . . . . . . . 6

2 The ideal gas law 7

2.1 The ideal gas law and absolute temperature . . . . . . . . . . . . . . . . 7

2.1.1 What is an ideal gas? . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Microscopic model of an ideal gas . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Equipartition of energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.4 Mean Free Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 The First Law of Thermodynamics 15

3.1 Heat and Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.2 Compression Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3 Isothermal and Adiabatic Compression . . . . . . . . . . . . . . . . . . . 19

3.4 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

3.4.1 Molar Heat Capacities . . . . . . . . . . . . . . . . . . . . . . . . 23

3.4.2 Calculating Heat Capacities . . . . . . . . . . . . . . . . . . . . . 24

3.5 Phase Transitions and Latent Heat . . . . . . . . . . . . . . . . . . . . . 26

4 Heat Transport 28

4.1 Heat Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4.1.1 Fourier Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.1.2 Heat Conductivity of an Ideal Gas . . . . . . . . . . . . . . . . . 30

4.2 The Heat Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4.2.1 Solutions to the Heat Equation . . . . . . . . . . . . . . . . . . . 32

5 Entropy and the Second Law of Thermodynamics 36

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5.1 Two-State Systems and Combinatorics . . . . . . . . . . . . . . . . . . . 38

5.1.1 The Two-State Paramagnet . . . . . . . . . . . . . . . . . . . . . 40

5.1.2 Interacting Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 43

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Chapter 1

Introduction

We all think we know what temperature is, but it is notoriously difficult to define. Inthis course, we will arrive at a rigorous definition of temperature, but this definition doesnot appear to bear any direct or obvious relation to our commonsense understandingof the concept. This is one example of the paradoxical nature of the topic of statisticaland thermal physics: on the one hand, it is a very down-to-earth subject – we will forexample study the performance of heat engines and refrigerators, and heat loss throughwalls, ceilings and windows of houses; but on the other hand it is full of riddles, notleast how irreversible processes can arise from the motion of atoms and molecules whichis described by time-reversible laws.

Thermal physics has a history dating back to the seventeenth, with the discovery ofBoyle’s Law, providing a relation between the pressure and volume of a gas. It involvesfamiliar concepts such as temperature, energy, pressure, heat flow and heat capacity;but also less familiar concepts such as entropy and latent heat. Most of the quantitiescan be directly measured and are macroscopic.

Statistical physics dates back to the second half of the nineteenth century, withthe work of Maxwell, Boltzmann and Gibbs. It describes the average behaviour of anextremely large number of particles, and the fluctuations around these averages, andhow this gives rise to the thermal properties we can observe, including different statesof matter, heat and the ideal gas law.

Overview of topics

• Thermal equilibrium and temperature. The ideal gas law.

• Heat, work and the first law of thermodynamics

• Adiabatic, isothermal and cyclic processes

• Heat capacity; phase transitions and latent heat

• Heat transport: Fourier’s law of heat conduction

• Entropy and the second law of thermodynamics; heat engines and refrigerators

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• Microscopic description of thermal systems, derivation of the ideal gas law

• Counting of quantum states; paramagnetism

• The relation between entropy and temperature

• Systems in contact with a heat bath. The Boltzmann distribution.

Learning outcomes

• Define and distinguish the concepts of thermal equilibrium, temperature, heat,energy and entropy

• Calculate or estimate properties of an ideal gas from a particle model

• State the equipartition theorem and apply it to solids and gases

• Determine the heat exchanged and work performed in various thermal processes

• State the laws of thermodynamics and discuss some of their macroscopic implica-tions, eg heat engines

• Calculate thermal averages using the Boltzmann distribution

1.1 Temperature and thermal equilibrium

We start with some concepts and definitions:

Thermodynamic system: a certain amount of stuff limited by a closed surfaceThermodynamic variable: a macroscopic (measurable) quantity, such as energy or pressureIsolated system: no energy or matter exchange with the environmentIntensive quantity: independent of the total mass of the systemExtensive quantity: proportional to the total mass of the system

Examples of thermodynamic variables:Extensive:

• Internal energy U

• Entropy S

• Volume V

• Particle number N

Intensive:

• Temperature T

• Pressure p

• Chemical potential µ

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1.1.1 The zeroth law of thermodynamics

Let us start with a fundamental fact, which is often called the zeroth law of thermody-namics:

If two systems are both in thermal equilibrium with a third system, they are inthermal equilibrium with each other.

This fact is what allows us to define temperature as an objective property of a system,namely the property (whatever it is) that systems in thermal equilibrium with eachother have in common. We can therefore say:

If two systems are in thermal equilibrium with each other, they have thesame temperature.

Another fundamental fact of thermal equilibrium is:

If two systems have been in contact for long enough, they are in thermalequilibrium with each other.

Here, two objects or systems are deemed to be in contact if they are able to exchangeenergy. This allows us to achieve thermal equilibrium in practice. Combining thesefacts, we can come up with a provisional definition of temperature:

Temperature is the thing that is the same for two objects if they have been in contactfor long enough.

In general, a system with higher temperature will transfer energy to a system with lowertemperature until they achieve equilibrium. We will get back to this. For now, we willleave these rather abstract definitions and turn to some more concrete relations betweenthermodynamic variables.

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Chapter 2

The ideal gas law

2.1 The ideal gas law and absolute temperature

The first relation discovered between thermodynamic quantities (and, indeed, the firstlaw of physics written as an equation relating two quantities) was Boyle’s Law. It wasfirst discovered by Henry Power and Richard Towneley in the mid-17th century, andthen confirmed and published by the 14th child of the 1st Earl of Cork, Robert Boyle.It states that at constant temperature, the pressure of a quantity of gas is inverselyproportional to its volume,

pV = c , (2.1)

where c is a constant. This law holds for most gases at constant temperature, and canbe verified in air pumps.

Much later, Jacques Charles (in the 1780s), John Dalton (1801) and Joseph Louis Gay-Lussac (1802) discovered a relation between temperature and volume at fixed (atmo-spheric) pressure, namely that when a gas heats up, it expands in proportion to thetemperature,

V1 − V2 = kV1(T1 − T2) , (2.2)

where k is another constant. If we take T1 to be the freezing point of water, and measuretemperatures in degrees Celsius, we find that k = 1

273.15C

We can see from Charles’s Law (2.2) that if we lower the temperature, the volume ofthe gas will get smaller, and at some temperature T0 it will become zero. Below thattemperature, the volume will be negative, which probably does not make any physicalsense. Setting V2 = 0 in (2.2) we find

V1 = kV1(T1 − T0) =⇒ T0 = T1 −1

k. (2.3)

This was realised by William Thomson, who later became Lord Kelvin, in 1848.

Since the notion of a negative volume makes no sense, we can postulate that T0 is thelowest temperature that can be attained, and define this as absolute zero. We can thendefine absolute temperature relative to this, ie

Tabs ≡ T − T0 . (2.4)

From now on, temperature will always mean absolute temperature.

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Absolute temperature is usually measured in Kelvin (K) [not K]

Boyle’s Law (2.1) and Charles’s Law (2.2) are both special cases of

The ideal gas law pV = nRT ≡ NkBT (2.5)

In this equation,

n = the molar weight of the gas,

R = the gas constant = 8.31 J/(mol K) ,

N = the number of molecules in the gas,

kB = Boltzmann’s constant = 1.381 · 10−23J/K .

Example 2.1

How many molecules are there in 1 m3 of air at room temperature and atmosphericpressure? Take the air to be an ideal gas.

Answer:

We must first define what we mean by room temperature and atmospheric pressure.The standard values for these are

Troom = 20C = 293 K ,

patm = 1013 hPa = 1013 · 102Pa = 1.013 · 105Pa .

With these values, we find

N =pV

kBT=

1.01013 · 105(N/m2) · 1.0 m3

1.381 · 10−23(J/K) · 293 K= 2.5 · 1025 . (2.6)

Example 2.2

A cylinder contains 12 litres of oxygen at 20C and 15 atmospheres pressure. Thetemperature is increased to 35C and the volume reduced to 8.5 litres. What is theresulting pressure?

Answer:

We have T1 = 20C = 293 K, T2 = 35C = 308 K. The ideal gas law gives us

p1V1 = NkBT1 =⇒ NkB =p1V1

T1

p2V2 = NkBT2

=⇒ p2 =NkBT2

V2

=p1V1T2

T1V2

= p1V1

V2

T2

T1

= 15 atm · 12

8.5

308

293= 22atm .

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2.1.1 What is an ideal gas?

As the name implies, the ideal gas law is an idealised description of real gases, and isnot exactly satisfied for all gases (not to mention liquids and solids). In practice, itholds for sufficiently dilute gases, and it is a good description for most normal gases atnormal conditions.

The theoretical definition of an ideal gas is

An ideal gas is a gas of point particles whose only interactions are perfectly elasticcollisions.

If the average distance between molecules in the gas is much larger than their size, thisis a good approximation. We will later express this in terms of the mean free path of amolecule.

2.2 Microscopic model of an ideal gas

Let us now change perspective and consider a gas with N ∼ 1023 molecules enclosed in avolume V , and try to work out the pressure it exerts on the container walls. Computingthe motion of all the molecules from Newton’s laws is clearly hopeless, even if they nevercollide with each other, only with the walls. Let us instead look at averages.

We start by considering the ‘pressure’ exerted on a single wall of the box by a singlemolecule.

L -

uPPPPPPq

~v

Figure 2.1: A single molecule bouncing off thewall of an enclosed box.

The molecule exerts a force on thewall only when it bounces off the wall.Let us assume that this collision iselastic, and that no momentum is im-parted to the wall in the y or z direc-tions — this will be true on average.

The average force is given by

Fx = −∆(mvx)

∆t, (2.7)

where ∆(mvx) is the change in themomentum of the molecule (in the x-direction) as it bounces off the wall,

and ∆t is the average time interval between collisions.

For an elastic collision with the wall, ∆(mvx) = −2mvx: the molecule just bounces backwith the same speed, but with vx reversed.

∆t is the time it takes for the molecule to get to the opposite wall, bounce off that wall,and come back again, ie

∆t =2L

vx. (2.8)

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Putting this together, we find that the average force is

Fx =2mvx2L/vx

=mv2

x

L. (2.9)

The pressure p is defined as the force divided by the area A of the wall, ie

p(x) =FxA

=mv2

x

AL=mv2

x

V, (2.10)

where V is the volume of the box.

We now assume that the gas is isotropic, ie that the average velocities and pressure arethe same in all directions. This means that

v2x = v2

y = v2z =

1

3(v2x + v2

y + v2z) =

1

3v2 , (2.11)

where X means the average of X. If we have N molecules, each one of them willcontribute to the pressure according to (2.10). Adding all this up, we bet

pV = Nmv2x =

1

3Nmv2 . (2.12)

But the ideal gas law tells us

pV = NkBT =⇒ kBT =1

3mv2 (2.13)

〈qq〉 1

2mv2 =

3

2kBT (2.14)

We see that the temperature is directly related to the average kinetic energy of themolecules in the gas.

This expression also gives us the root mean square (rms) average speed of a molecule inthe gas

vrms =

√3kBT

m. (2.15)

Note that this is not the same as the average, the most common, or the median speed,although all of these have similar values. The rms average is often used as the ‘best’measure of the typical magnitude of a quantity which can be both positive and negative.

Example 2.3

Find the rms average speed of a nitrogen molecule in air at room temperature.

Answer:

We need the mass of a nitrogen (N2) molecule. We can look this up in a table, orremember that

mN2 = 2mN = 2 · 14u = 28 · 1.661 · 10−27kg ,

where u is the atomic mass unit. Inserting this in (2.15) gives us

vrms =

√3 · 1.381 · 10−23J/K · 293 K

28 · 1.661 · 10−27kg= 511 m/s . (2.16)

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Comments

When we ‘derived’ this expression, we made several simplifying assumptions:

1. There are no forces between the molecules which would cause them to slow downor deviate from their paths.

2. The collisions between the molecules and the walls are elastic.

3. The gas is isotropic: no direction is preferred over any other.

The third condition is innocent: most gases (and liquids) are isotropic. If not, we wouldhave different pressures in different directions (which could happen in some circum-stances, but we will not deal with those here).

The second condition is not problematic either:

(a) The wall can be introduced as a ‘thought experiment’, allowing us to compute thepressure exerted by the gas on whatever it borders (including other bits of gas).

(b) If the collisions are inelastic, the kinetic energy that is lost will go into internalenergy (eg, vibrations) in the wall which can be released in subsequent collisitons.If the gas and wall are in thermal equilibrium, the collisions will on average beelastic, ie on average as much energy is gained as lost from the collisions.

The first condition is the crucial one, but it can be relaxed:

(a) Elastic collisions between molecules do not spoil the argument, since they do notchange the total (or average) kinetic energy.

(b) Inelastic collisions — where energy is lost to internal motion inside molecules,or even to molecules breaking up – do ruin the argument, but the result can berewritten by taking these types of motion into account, as we shall see later.

(c) The most important assumption is that molecules move in straight lines mostof the time. This breaks down if the gas is very dense (so that collisions hap-pen almost continuously), or if there are strong long-ranged forces between them.Electromagnetic forces might be examples of this.

2.3 Equipartition of energy

Our result for the average kinetic energy can be written

1

2mv2 =

1

2mv2

x +1

2mv2

y +1

2mv2

z = 3 · 1

2kBT (2.17)

and (since we assume the gas is isotropic)

1

2mv2

x =1

2mv2

y =1

2mv2

z =1

2kBT . (2.18)

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This is an example of equipartion of energy :Motion in the three directions (x, y, z) corresponds to three different degrees of freedom,where each degree of freedom receives 1

2kBT of energy.

For a monatomic gas — where the molecules are single atoms, such as helium (He), neon(Ne), argon (Ar) — this is the only form of motion that can exist. Also, in this case, atnormal temperatures, all collisions are elastic and the atoms are essentially spherical.

However, for most molecules, other forms of motion are also possible.

Rotation: A diatomic gas — where the molecules consists of two atoms, such asnitrogen (N2), oxygen (O2), hydrogen (H2) — can rotate about two axes. If we call theaxis joining the two molecules the z-axis, then the molecule can rotate about the x andy axes (but not the z axis). This is also the case for a linear molecule such as CO2. Wetherefore have

Krot =1

2Iω2

x +1

2Iω2

y = 2 degrees of freedom. (2.19)

Most multiatomic gases (eg, H2O, CH4) can rotate about all three axes, so

Krot =1

2Ixω

2x +

1

2Iyω

2y +

1

2Izω

2z = 3 degrees of freedom. (2.20)

Note that in general, the moments of inertia Ix, Iy, Iz about the three axes are different.

Vibration: A diatomic gas can also vibrate (oscillate). The energy in the oscillationis given by

Evib =1

2mv2

r +1

2kr2

r = 2 degrees of freedom, (2.21)

where rr is the distance (or displacement from equilibrium) between the two atoms, andvr is their relative velocity.

Inelastic collisions can transfer energy between linear motion, rotation and vibration.As a result, all active degrees of freedom receive 1

2kBT of energy on average.

There is a catch to this, which is that quantum mechanics sets a lower limit to allow-able vibration (and rotaion) energies, so usually vibrations do not feature at ‘normal’temperatures, since the lowest allowable vibration energy is usually larger than 1

2kBT .

This means that for a diatomic gas we have

1

2mv2

x =1

2mv2

y =1

2mv2

z =1

2Iω2

x =1

2Iω2

y =1

2kBT . (2.22)

The total energy in a gas of N such molecules is

U = N · 5 · (1

2kBT ) =

5

2NkBT . (2.23)

2.4 Mean Free Path

Let us now look at how far a molecule travels between collisions. The average value ofthis is called the mean free path. We will assume all molecules are perfect spheres with

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Figure 2.2: The motion of a single molecule with diameter 2d pictured as a cylinder.

diameter d. Two molecules collide when their centres come within a distance d of eachother.

Consider a single molecule (assume all others at rest). We can treat this molecule ashaving a diameter of 2d; all others are point particles. We can picture the motion of themolecule as a cylinder (series of cylinders), colliding with anything inside the cylinder.

The number of molecules inside a cylinder with length l is

Nc =N

VVcyl =

N

Vπd2l (2.24)

This is the average number of collisions the molecule experiences over the distance l(time t = l

v). The mean free path is

λ =l

Nc

=V

πd2N≡ 1

nσ(2.25)

where n = number density (not the molar number) and σ = cross-section.

Here we assumed all other molecules were at rest. If we take their motion into account,it turns out that we get a factor of

√2.

⇒ λmfp =1√2nσ

=V√

2πd2N(2.26)

We can compare this with the size d of the molecules. The ideal gas approximationholds if

λmfp d (2.27)

Then, by using the ideal gas law we get

V

N=kbT

p⇒ λmfp =

kbT√2πd2p

(2.28)

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Note: We assumed here that all molecules are perfect hard spheres, which is clearly nottrue for real atoms and molecules. However, the cross section σ can be defined (andmeasured/computed) for all real molecules - it is a measure of the strength and rangeof the interactions between particles. Therefore, we can still use these expressions todetermine the mean free path.

Example 2.4

What is

(a) the mean free path and

(b) the average collision rate

for nitrogen at 293K and 101kPa? A nitrogen molecule has an effective diameterd = 3.15 · 10−10m.

Answer:

(a)

λmfp =kbT√2πd2p

=(1.38 · 10−23J/K) · (293K)

(√

2π) · (3.15 · 10−10m)2 · (101 · 103N/m2)= 9.08 · 10−8m

This works out to be 300d, which means that the ideal gas condition is fulfilled.

(b) From example 2.3, we have that vrms = 511m/s (vrms is not exactly equal tovavg but it’s close). The collision rate is

1

time between collisions=

vavgλmfp

≈ 5.63 · 109s−1

which is billions of collisions per seconds!

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Chapter 3

The First Law of Thermodynamics

3.1 Heat and Work

Until the mid-19th century, heat was seen as a substance (caloric fluid) that was trans-ferred in the process of heating or cooling objects. It was measured in calories.

1cal = the amount of heat required to increase temperature of 1g of water from145C to 155C at 1atm pressure.

Note: 1 ’calorie’ in food is actually 1kcal = 1000cal.

James Joule showed that you could achieve the same temperature increase by doingmechanical work, without any heat. This established that heat is not a substance, buta form of energy, or rather a form of energy transfer (like work).

In thermodynamics, we define heat and work as

Heat: Spontaneous flow of energy from one abject to another, caused by a temperaturedifference between them.

Work: All other types of energy transferred to or from a system (mechanical, electrical,chemical, etc)

There are three types of heat:

1. Conduction - Heat transfer between nearby molecules

2. Convection - Bulk motion of a fluid

3. Radiation - Emission of electromagnetic waves

The first law of thermodynamics states that the total energy is conserved in any processand can be represented mathematically as

The First Law of Thermodynamics ∆U = Q+W (3.1)

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Figure 3.1: Work done by a piston compressing a gas.

In this equation,

∆U = change in internal energy of a system

Q = heat transferred to the system

W = work done on the system

Comments

(a) W is often taken to be the amount of work done by the system. In that case, thefirst 1st law reads ∆U = Q−W .

(b) According to the definitions above, it makes no sense to talk about the amount ofheat in a system. We can talk about the amount of energy in the system, but itis never stored at ’heat’ or ’work’. This is clearly not the same as the everydaymeaning of ’heat’, just as ’work’ also has a different meaning in physics.

3.2 Compression Work

The most important type of work done on a system(typically a gas) is compressing it,e.g. pushing a piston. Unless you slam the piston really fast (faster than the speed ofsound), the force will equal the pressure force of the gas.

F = pA (3.2)

Where A equals the area of the piston. The work done pushing the piston a smalldistance ∆x is

W = F∆X = pA∆x (3.3)

But A∆x is just the reduction in the volume of the gas,

∆V = −A∆x =⇒ W = −p∆V (3.4)

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In general, the pressure (and hence the force) can change as volume is reduced, and wehave

W =

∫ x2

x1

Fdx =

∫ x2

x1

pAdx =

∫ V2

V1

pdV (3.5)

=⇒ W = −∫ V2

V1

p(V )dV (3.6)

If we know the pressure as a function of volume, we can use this to calculate the work.

Example 3.1

1mol of ideal gas is compressed from an initial volume of 20.0L at atmosphericpressure, while keeping the temperature constant, to a final volume of 15.0L. Findthe amount of work done.

Answer:We know that

pV = nRT =⇒ p(V ) =nRT

Vfor constant T. (3.7)

Hence,

W = −∫ V2

V1

nRT

VdV = −nRT

∫ V2

V1

dV

V(3.8)

= −nRT [ln(V2)− ln(V1)] = nRT ln(V1

V2

)(3.9)

=⇒ W = p1V1 ln(V1

V2

)(3.10)

= (1.01 · 103N/m2) · (20.0 · 10−3m3) · ln(20.0m3

10.0m3

)(3.11)

= 2020J · 0.288= 581J (3.12)

This is an example of isothermal compression. For an ideal gas, we also know that thetotal energy is given by the equipartition theorem, U = f

2NkbT where f is the number

of degrees of freedom. So if N and T do not change, the gas loses 581J of heat to theenvironment.

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Example 3.2

An ideal gas is made to do undergo the cyclic process shown on the right. Foreach step A,B,C, determine whether the following are positive, negative or zero:

(i) the work done on the gas,

(ii) the change in the energy content of the gas,

(iii) the heat added to the gas. Finally, determine the sign of each of these for thewhole cycle.

Answer:

A. (i) Here we have expansion at constant pressure. The work done is then

W = −∫ V2

V1

pdV = −p(V2 − V1) < 0 (3.13)

since V2 > V1

(ii) From the Ideal Gas Law, pV = NkbT , if V increases while p,N areconstant, the temperature T must increase. Since the energy U ∝ NkbTfor an ideal gas, this implies ∆U > 0.

(iii) The gas loses energy through work, but gains energy overall, so heat mustbe added i.e Q > 0.

B. (i) We are now at constant volume (V3 = V2) so no work is done, i.e. W = 0.

(ii) For the same reason as in step 1., T must increase and therefore ∆U > 0.

(iii) ∆U = Q+W < 0 but W > 0, so Q > 0.

C. (i) Now we have compression along a line where p ∝ V , say p = bV . Then

W = −∫ V1

V3

pdV = −∫ V1

V3

bV dV = −[1

2bV 2

]V1V3

=1

2b(V 2

3 − V 21 ) > 0.

(3.14)

(ii) In this case, p and V both decrease, so pV = NkbT decreases, and ∆U >0.

(iii) ∆U = Q+W < 0 but W > 0, so Q < 0.

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Total: (i) We have

Wtot = WA +WC = −p1(V2 − V1) +1

2b(V 2

2 − V 21 ) (3.15)

since V3 = V2. But we also have p1 = bV1, so

W =1

2b(V 2

2 − V 21 )− bV1(V2 − V1) (3.16)

=1

2bV 2

2 − bV1V2 +1

2bV 2

1 (3.17)

=1

2(V2 − V1)2 > 0 (3.18)

(ii) Since we get back to where we started, U is the same, i.e. ∆U = 0

(iii) ∆U = Q+W = 0 =⇒ Q = −W , i.e. heat escapes from the gas.

The net result result of this process is that work is done on the gas andis converted into heat.

3.3 Isothermal and Adiabatic Compression

We will consider two important special cases of compression work:

Isothermal : The temperature does not change (slow)

Adiabatic: No heat flow in or out (faster)

We already looked at isothermal compression in Example 3.1. For an ideal gas, pV =NkbT , so if T (and N) is constant, we have p(V ) = NkbT

V.

W = −∫ V2

V1

p(V )dV = −NkbT∫ V2

V1

dV

V(3.19)

= NkbT[

lnV]V2V1

= NkbT ln(V1

V2

)(3.20)

Also, for an ideal gas, U ∝ NkbT = constant⇒ ∆U = 0. Therefore

Q = ∆U −W = −W = NkbT ln(V1

V2

)(3.21)

If V1 > V2 (compression), W > 0 (work done on the gas) and Q < 0 (heat leaves thegas)

Now consider adiabatic compression, i.e. Q = 0.

In this case, ∆U = Q+W = W . According to the equipartition theorem, U = f2NkbT

(f is the number of degrees of freedom).

19

Clearly, the temperature will have to change, as will the pressure.

Consider an infinitesimal compression dV , with temperature change dT . The change ininternal energy is thus

dU =f

2NkbdT = W = −pdV (3.22)

⇒ f

2NkbdT = −NkbTdV

V(3.23)

⇔ f

2

dT

T= −dV

V(3.24)

using the fact that p = NkbTV

. Integrating on both sides, we get

f

2

∫ T2

T1

dT

T=f

2ln(T2

T1

)= −

∫ V2

V1

dV

V= − ln

(V2

V1

)(3.25)

Now, exponentiate both sides, using eln(x) = x, and a ln(x) = ln(xa)

(T1

T2

)f/2=V2

V1

⇒ V1Tf/21 = V2T

f/22 or V T f/2 = constant (3.26)

For any given adiabatic compression, from V1 to V2, this gives us the change in temper-ature and therefore the energy, ∆U = f

2Nkb∆T = W . We may also find an expression

for the pressure

pV γ = constant, γ =f + 2

f= adiabatic exponent (3.27)

Deriving this is left as an exercise.

Example 3.3

1 mol of Oxygen gas (O2) has temperature 310K and volume 12l.

(a) What is the final temperature if the gas expands adiabatically?

(b) How much work is done by the gas in the process?

Oxygen gas at this temperature has rotations, but no oscillation/vibrations.

Answer:

(a) The number of degrees of freedom is f = 3(translation) + 2(rotation) = 5. Foradiabatic expansion, we have

V1Tf/21 = V2T

f/22 ⇒ T2 =

(V1

V2

)2/f

T1 =(12

19

)2/5

· 310K= 258K (3.28)

20

(b) Here we can use that W = ∆U for an adiabatic process, and

U =f

2NkbT =

f

2nmRT (3.29)

U1 =5

2nmRT1 =

5

2· 1mol · 8.31J/(mol K) · 310K = 6440J (3.30)

U2 =5

2nmRT1 =

5

2· 1mol · 8.31J/(mol K) · 258K = 5360J (3.31)

W = U2 − U1= −1080J (3.32)

This is the work done on the gas. The work done by the gas is 1080J .

We can also consider work done at constant pressure. In this case the expres-sion for the work is quite simple:

W = −∫pdV = −p

∫dV = −p(V2 − V1) = −p∆V (3.33)

Summary

First law of thermodynamics: the change in internal energy = heat transferred to thesystem + work done on the system.

∆U = Q+W

Compression work:

W = −∫pdV

Isothermal compression/expansion: (T=constant) ⇒ p(v) = NkbTV

W = NkbT ln(V1

V2

), ∆U = 0, Q = −W

Constant volume:

W = 0⇒ Q = ∆U =f

2Nkb∆T

Constant pressure:W = −pdV

Adiabatic expansion/compression: Q = 0

V T f/2 = constant, pV γ = constant, γ =f + 2

f

W = ∆U =f

2Nkb(T2 − T1)

Underlined expressions hold for ideal gases. but not in general.

21

3.4 Heat Capacity

The heat capacity of an object is the amount of heat required to increase its temperature,per unit of temperature increase.

C ≡ Q

∆T(3.34)

We will also be looking at the specific heat, which is heat capacity per unit mass

c ≡ C

m(3.35)

C is an extensive quantity. c is an intensive quantity - does not depend on the amountof stuff.

But out definitions are ambiguous!

We know that energy can be added as either heat or work, depending on the specificprocess. Even if the internal energy only depends on T , we may be doing work, so theexpression

C =∆U(T )−W

∆T

is ambiguous. There are two important special cases:

1. Constant Volume

Here, no work is done, so

Cv = lim∆T→0

Q

∆T= lim

∆T→0

∆U

∆T

∣∣∣V= const

=(∂U∂T

)V

(3.36)

Where Cv is heat capacity at constant volume.

2. Constant Pressure

This is more common in everyday life: objects expand and do work on the sur-rounding aire (for example), which remains at the same pressure.

Here we have

W = −p∆V (3.37)

Cp = lim∆T→0

Q

∆T= lim

∆T→0

∆U + p∆V

∆T

∣∣∣p= const

=(∂U∂T

)p

+ p(∂V∂T

)p

(3.38)

Where Cp is heat capacity at constant pressure

Note:

The partial derivatives (∂U∂T

)V 6= (∂U∂T

)p in general since U may depend on pressure orvolume. It matters which of them is kept constant!

Example 3.4

Water has a specific heat capacity cp = 1.00cal/(g K) = 4187J/(kg K). How muchenergy is needed to

22

(a) boil enough water to make a cup of tea; and

(b) heat the water for a bath?

Answer:

We will assume the mains water in (a) starts out with a temperature of 7C. Thewater in (b) might come from your attic tank with a temperature of 15C.

(a) A cup is approx 0.20l, so its mass is approx 0.20kg.

The temperature is increased by ∆T = 100C-7C = 93K

The heat required is

Q = Cp∆T = cpm∆T = 4187J/(kg K) = 0.20kg · 93K≈ 78kJ (3.39)

(b) Now we heat water from 15C to body temperature, 35C ⇒ ∆T = 22K

The volume of a bath tub is≈ 1.5m·0.5m·0.5m = 0.38m3 = 380l⇒ m ≈ 380kg

Q = cpm∆T = (4187 · 380 · 22)J= 35MJ (3.40)

If we like, we may convert this to kWh: 1W = 1J/s

1kWh = 1000W · 1hr = 1000W · 3600s = 3.6MJ (3.41)

Heating the bath water requires approx 10kWh.

Boiling a cup of tea requires approx 0.02kWh

Note:

For liquids and solids, ’the heat capacity’ nearly always means constant pressure (pres-sure of the surrounding medium). That means we allow the object to expand thermally(preventing this by applying pressure would be difficult). The difference between cp andcv is very small in practice.

3.4.1 Molar Heat Capacities

Another useful quantity is the molar heat capacity [molar specific heat], which is heatcapacity per mol:

cm =C

nm

. Here are the specific heats and molar specific heats of some substances:

23

Substance cv [J/kg K] cm [J/mol K]Lead, Pb 128 26.5Silver, Ag 236 24.8Copper, Cu 386 25.5Aluminium, Al 900 24.5Tungsten, W 134 24.4Is there a pattern here?Helium, He 3123 12.5Argon, Ar 317 12.6Oxygen, O2 647 20.7Nitrogen, N2 742 20.8Hydrogen, H2 10160 20.5Ammonia, NH2 1703 29.0Carbon Dioxide, CO2 675 29.7Methane, CH4 1700 27.3

3.4.2 Calculating Heat Capacities

Assume the equipartition theorem holds where f is the number of active degrees offreedom.

Then, U = f2NkbT

Cv =(∂U∂T

)v

=∂

∂T

(f2NkbT

)=f

2Nkb =

f

2nmR (3.42)

assuming f does not depend on T .

The molar specific heat is

Cmv =

Cvnm

=f

2R (3.43)

For monatomic gases: f = 3⇒ cmv = 32R = 12.47J/mol K

diatomic gases: f = 5⇒ cmv = 52R = 20.79J/mol K

polyatomic gases: f = 6⇒ cmv = 3R = 24.95J/mol Kmetals/solids: f = 6⇒ cmv = 3R = 24.95J/mol K

The actual values for polyatomic gases such as NH4, CH4, CO2 are closer to 72R =

29.10J/mol K suggesting that a vibrational degree of freedom may be active.

The number of active degress of freedom depends on the temperature - at low temper-ature all degrees of freedom freeze out (so cv → 0 as T → 0 for solids), while rotationsand then vibrations (in gases) become active at higher temperatures. Figure 3.2 showsa typical plot of cmv vs T for a diatomic gas.

Note that at very low T the substance will become liquid, while at very high T theatoms will dissociate. In both cases the ideal gas description is no longer valid.

What about cp (constant pressure)?

We found that

Cp =(∂U∂T

)p

+ p(∂V∂T

)p

(3.44)

24

Figure 3.2: A plot showing how the specific heat of a diatomic gas increases withtemperature.

For an ideal gas:

U =f

2NkbT ⇒ Cv =

(∂U∂T

)p

=(∂U∂T

)V

=f

2Nkb (3.45)

pV = NkbT ⇒ V =NkbT

p⇒(∂V∂T

)p

=Nkbp

(3.46)

Putting this together:

Cp =f

2Nkb + p

(Nkbp

)= Cv +Nkb (3.47)

Cp − CV = Nkb = nR (3.48)

For specific heat capacities we would have

cp = cv +Nkbm

(3.49)

It is important to remember that these equation hold for ideal gases only.

The relationship between cv and cp for solids and liquids is much more complicatedthan for ideal gases and is beyond the scope of this course. But since liquids and solidscan practically be assumed to be incompressible, cp and cv have almost the same value.Hence, it is common to just have one value for their heat capacity.

Example 3.5

We transfer 1000J as heat to a diatomic gas, allowing the gas to expand with p helpconstant. The molecules can rotate but not vibrate.

(a) How much of the 1000J goes into increasing the internal energy of the gas?

(b) How much of this goes into translational kinetic energy, and how much intorotational motion?

(c) If there is 7.60mol of the gas, by how much does its temperature increase?

25

Answer:

(a) Since the pressure is kept constant we can use

Cp = Cv + nmR =5

2nmR + nmR =

7

2nmR (3.50)

⇒ ∆T =Q

Cp=

Q72nmR

(3.51)

But U =5

2nmRT (3.52)

⇒ ∆U =5

2nmR∆T =

5

2nmR

( Q72nmR

)=

5

7Q= 714J (3.53)

(b) We have ktrans = 32nmRT ; krot = 2

2nmRT = 2

5U

⇒ ∆ktrans =3

2∆U= 429J; ∆krot =

2

5∆U= 286J (3.54)

(c) The temperature increase is

∆T =Q

72nmR

=1000J

3.5 · 7.60mol · 8.31J/mol K= 4.52K (3.55)

Note that (a) and (b) could be answered without knowing nm and ∆T .

3.5 Phase Transitions and Latent Heat

When a system undergoes a phase transition - e.g. ice melting to become water, orwater boiling to become vapour - it can absorb heat without increasing its temperature.

Essentially, this energy goes into breaking the bonds between the molecules in the lowertemperature phase.

This energy is called latent heat, defined as the heat per unit mass required to com-pletely melt/boil a substance.

L ≡ Q

m(3.56)

By convention, this is always taken to be at constant pressure.

Example 3.6

The latent heat for melting ice is 333J/g.

How much heat is required to melt 6.0 litres of ice at 0C and heat it to 45C towash dishes?

26

Answer:

Qmelt = L ·m = 333 · 103J/kg · 6.0kg = 2.0 · 106J (3.57)

Qheat = cwater ·m ·∆T = 4.186J/gK · 6000g · 45K = 1.13 · 106J (3.58)

Qtot = 3.1MJ = 0.87kWh (3.59)

27

Chapter 4

Heat Transport

4.1 Heat Conduction

So far we have only considered systems in thermal equilibrium or the beginning andend of the equilibrium porcess (e.g. two bodies with different temperatures are broughtinto contact). But we may also want to know how fast the equilibrium state is reached- this can be the difference between mild discomfort and a visit to A&E, or between asmall heating bill and fuel poverty.

Specifically, we will look at how fast heat flows from a hot object to a cold object whenthey are put in contact: heat conduction

(Heat can also be transferred through radiation and convection. The radiation lawrequires more advanced physics, while convection is a lot more complicated so we willnot deal with it here.)

Figure 4.1:

Imagine a slab of material with thickness ∆x and area A.On the left side of the slab, the temperature is T = T2 andon the right side T = T1. If T2 > T1, heat will flow fromleft to right. On general grounds, we can expect that

1. Q ∝ A: the bigger the area, the higher the heat flow.

2. Q ∝ t: if T1, T2 are kept constant, the heat flows ata constant rate per time unit.

3. The thicker the slab, the smaller the heat flow.

4. The bigger the temperature difference ∆T = T2−T1,the bigger the flow. In particular, Q = 0 if T1 = T2.

The simplest expression consistent with 1-4 is

Q

∆t∝ A∆T

∆xor

Q

∆t= −κAdT

dx(4.1)

Note the - sign: dTdx< 0 if T2 > T1

28

We can write this as

JQ =Q

A∆T= −κdT

dxFouriers Law of Heat Conduction (4.2)

The quantity JQ is called the heat flux = heat per area per time.

κ [kappa] is the thermal conductivity - it is a property of the material.

Example 4.1

Calculate the rate at which heat would be lost on a cold winters day through a 62mx 38m brick wall 32cm thick, if the temperature inside is 22C and outside is −2C.The thermal conductivity of the brick is 0.74W/(m K).

Answer:

The heat loss is

dQ

dt=κ · A ·∆T

∆x=

(0.74W/(m K) · 62m · 3.8m · 24K

0.32m≈ 13000W (4.3)

4.1.1 Fourier Relaxation

Fourier’s law of heat conduction, for a system where the temperature T varies only inthe x-direction, is

JQ =dQ

Adt= −κ∂T

∂x. (4.4)

Here, JQ is the heat flux, which is the heat flow per unit time and area, A is the areaof the surface through which the heat flows and κ is the thermal conductivity of thesystem.

Imagine now that we have an amount of matter with heat capacity CV in a containerwith area A, thickness d and thermal conductivity κ, and that the temperature outsidethe container is kept constant at To. The temperature Tin on the inside of the containerwill now vary with time as heat flows out of (or into) the container. Across the walls ofthe container itself, we have

∂T

∂x=Tin − To

d. (4.5)

Substituting this into (4.4) and using dQ = CV dT , we get for the temperature Tin = T (t)of the matter inside the container,

dQ

Adt=CV dTiAdt

=CV d(T − To)

Adt= −κ

d(T − To) (4.6)

⇐⇒ d(T − To)T − To

= − κA

dCVdt (4.7)

⇐⇒ ln(T − To) + C = − κA

dCVdt (4.8)

⇐⇒ T (t) = To +Be− κAdCV

t. (4.9)

29

The constant B = eC can be determined if we know the temperature Ti at the start ofthe process, t = 0:

T (0) = To +B = Ti ⇐⇒ B = Ti − To . (4.10)

Note that we made some assumptions here which may not be very well satisfied in reallife cases. In particular, we assume that the temperature does not vary across the pieceof material we are considering, which would imply that the heat transport inside thematerial is perfect (or that the piece of matter is small enough that it always remains ininternal thermal equilibrium). We also assume that there are no temperature gradientsin the medium outside, so that any heat coming out from the container is immediatelycarried away.

4.1.2 Heat Conductivity of an Ideal Gas

Figure 4.2:

Let us consider two ’boxes’ of ideal gas, each ofwidth l, but with slightly different temperaturesT and ∆T . This means that the molecules in boxon the right are moving slightly faster on average.

1

2m〈v2

x〉left =1

2kbT (4.11)

1

2m〈v2

x〉right =1

2kb(T + ∆T ) (4.12)

In a time ∆t, some of the molecules on the rightwill have moved into the left box, and visa versa.

Specifically, take l ≈ λmfp (mean free path).

∆t = average time between collisions =λmfp

〈|v|〉

In this time, the molecules will (on average) havemoved freely (without collisions).

Half the molecules on the left (those with vx > 0)will have moved into the right hand box, and half

of those on the right will have moved to the left. [We are cheating a bit here. A carefulcalculation shows that molecules move l = 2

3λmfp in the x-direction].

This implies a transfer of heat from right to left,

−Q =1

2UR −

1

2UL =

1

2

(f2NRkb(T + ∆T )− f

2NLkbT

)(4.13)

=1

2(f

2Nkb)∆T =

1

2Cboxv ∆T (4.14)

[We take the density to be the same on both sides]

If we write Cboxv = V cvol = l · A · cvol, where cvol = Cv

V= f

2nkb is the heat capacity per

unit volume.

30

We can confirm Fouriers law,

JQ =Q

A∆T=

12lcvol∆T

l/〈|v|〉=

1

2〈|v|〉lcvol ·

∆T

l= κ

∆T

l, (4.15)

with κ =1

2〈|v|〉lcvol ≈

1

2

√〈v2〉λmfpcvol (4.16)

We can now substitute the expressions for 〈v2〉, λmfp, cvol;

κ ≈ 1

2

√〈v2〉λmfpcvol =

1

2

√3kBT

2m· 1√

2nσ· f

2nkb (4.17)

=

√3

4σ· f

2kb ·(kbTm

)1/2

=

√3

4σcmoleculev

(kbTm

)1/2

(4.18)

The precise prefactor here is not important since we have made a lot of approximations.The important lessons are:

1. κ ∝√T - This is experimentally well confirmed in practice.

2. κ is independent of the pressure and density.

3. At constant T , κ ∝ 1√mσ

- experimentally well confirmed

Note that σ (cross-section) is a measure of the strength of the interaction betweenthe molecules.

So if there are no interactions, the thermal conductivity is infinite, while it is smallif the interactions are strong. This observation holds also for other ’transportproperties’ such as viscosity.

- the stronger the interaction, the smaller the viscosity/conductivity etc.

4. κ ∝ cmoleculev ∝ f = the number of degrees of freedom.

4.2 The Heat Equation

Figure 4.3:

Look at a uniform rod, with T varying alongthe x-direction.

Look at heat flowing into a segment oflength ∆x.

The heat flowing in from the left in a time∆t is

∆QL = A∆t · κ∂T (x)

∂x(4.19)

The heat flowing in from the right is

−∆QR = −A∆t · κ∂T (x+ ∆x)

∂x(4.20)

31

The net heat flow equals the change in energy,

∆QL + ∆QR = A∆tκ[∂T (x)

∂x− ∂T (x+Deltax

∂x

](4.21)

= ∆U = cm∆x∆T = cρA∆x∆T, (4.22)

where ∆T is the change in temperature in time ∆t.

This gives us∆T

∆t=

κ

cρ

[∂T (x)/∂x− ∂T (x+ ∆x)/∂x

∆x

](4.23)

or in the limit ∆t→ 0, ∆x→ 0,

∂T

∂t= K

∂2T

∂x2with K =

κ

cρThe Heat Equation (4.24)

4.2.1 Solutions to the Heat Equation

The heat equation is a partial differential equation (depending on x and t).

Solving such equations is in general very complicated, and you do not have the mathe-matical or numerical tools for this.

We can looks at some special cases though:

1. Stationary (time-independent) solution

Assume the temperatures T1, T2 at either end of a heat conducting material oflength L are help constant in time. In that case, there is no reason why thetemperature within the material should change with time. Specifically,

∂T

∂t

∣∣∣x=0

=∂T

∂t

∣∣∣x=L

= 0 ; we can take∂T

∂t= 0∀x (4.25)

Then∂T

∂t= 0 = K

∂2T

∂x2⇒ ∂2T

∂x2= 0 (4.26)

This now looks like an ordinary differential equation, T ′′(x) = 0

Which has the solutionT (x) = Ax+B (4.27)

The constants A,B are determined from the boundary conditions

T (x = 0) = T1, T (x = L) = T2 ⇒ B = T1, A =T2 − T1

L(4.28)

2. Heat flow from a small region

Let us assume that at some time t0 the temperature profile of a thin rod is

T (x, t0) = T1e−x2/a + T0 (4.29)

32

If a is small this will describe a system with temperature T1 − T0 in a very smallregion, and T0 elsewhere.

Consider the following expression:

T (x, t) = T0 +A√te−x

2/4Kt (4.30)

This gives

∂T

∂x=

A√t

(− 2x

4Kt

)e−x

2/4Kt = − Ax

2Kt3/2e−x

2/4Kt (4.31)

∂2T

∂x2= − A

2Kt3/2e−x

2/4Kt − Ax

2Kt3/2

(− 2x

4Kt

)e−x

2/4Kt (4.32)

= − A

2Kt3/2

(1− x2

2Kt

)e−x

2/4Kt (4.33)

∂T

∂t= − A

2t3/2e−x

2/4Kt + e−x2/4Kt +

A√t

(+

x2

4Kt2

)e−x

2/4Kt (4.34)

= − A

2t3/2

(1− x2

2Kt

)e−x

2/4Kt (4.35)

We can see that this satisfies the heat equation.

This solution describes a very narrow, very tall peak at small t, and a broad, flatdistribution at late t. This is typical of how temperatures get evened out withtime.

Example 4.2

A certain solid material has heat capacity C = 150J/(kgK), thermal conductivityκ = 300W/(Km) and density 10kg/m3.

We consider a rod of this material with length 1m and cross-section 1cm2. Weassume the temperature only varies along the length direction of the rod.

(a) Assume the rod is attached to a tank of ice water at 0C on one end and atank of boiling water at 100circC at the other end.If the temperature of therod does not change with time, calculate the amount of heat flowing throughthe rod, from the boiling water to the ice water, per second.

(b) How long does it take for the heat flowing through the rod to melt a singlegram of ice in the ice water tank?

Answer:

(a) if T does not change with time, the heat equation tells us that

T (x) = T0 +T1 − T0

Lx (4.36)

33

Where x = 0 at the ice end, x = L = 1m at the boiling end, T0 = 0C andT1 = 100C Therefore,

∂T

∂x=T1 − T0

L, and (4.37)

dQ

dt= −κ · A · ∂T

∂x= 300W/Km · (10−2m)2 · 100K

1m= −3.0W (4.38)

The minus sign is because heat flows from right to left, into the ice water.

(b) The latent heat of melting ice is 333J/g. Therefore, to melt 1g of ice requiresQ = 333J.

This takes t = QdQ/dt

= 333J3.0J/s

= 111s (= 1min 51s) to achieve.

Summary

Fouriers law of heat conduction:

JQ =Q

A∆t= −κdT

dx

Heat conductivity for ideal gas:

κ ≈ 1

2< |v| > λmfpcvol ≈

cmoleculev

σ

(kbTm

)1/2

=f

2

kbσ

(kbTm

)1/2

Heat Equation:∂T

∂t= κ

∂2T

∂x2; K =

κ

cρ

Stationary Heat Flow (∂T/∂t = 0):

T (x) = Ax+B = T0 +T1 − T0

Lx

Figure 4.4: Astationary solu-tion

34

Figure 4.5: Heat flow fram a smallregion

35

Chapter 5

Entropy and the Second Law ofThermodynamics

Recall the first law,∆U = Q+W (5.1)

Basically this says that you cannot get more energy out of a system than you putin. Specifically for a cyclical process (which can repeat itself over and over, ∆U = 0,Qnet +Wnet = 0, or

Qin +Win = Qout +Wout

This means that you cannot make an engine that does more work than the energy youput into it (perpetual motion of the first kind)

But there is more!

You could imagine a ”perfect engine” (perpetual motion of the second kind) that takesheat in and produces the same amount of work. THIS IS NOT POSSIBLE There are

also processes that are irreversible - they only proceed in one direction, and not in theother:

• A stone falls to the ground and stays there. It does not jump back up on its ownaccord.

• A cup of coffee cools down in a cold room. it does not spontaneously heat up,taking heat from the room.

• Red wine is spilled on a table cloth. It does not by itself gather together and”unstain” the cloth.

From the point of view of conservation of energy (or momentum, or angular momentum,ect) there is nothing wrong with these processes going in reverse, or with a perfect engine.

These facts of life are related to the second law of thermodynamics.

There are several formulations of this law:

36

1. Kelvin Postulate

No series of processes is possible whose sole result is the absorption of heat froma thermal reservoir and the complete conservation of this energy to work.

2. Clausius Postulate

No process is possible whose sole result is the transfer of heat from a reservoir atone temperature to another reservoir at a higher temperature.

3. Entropy Formulation

In a closed system, the entropy of the system will either increase or remain con-stant. It will never decrease

The third formulation is the most general, and may be shown to entail the two others,but to understand it, we need to know what entropy is.

Clausius introduced entropy S as

∆S =

∫ f

i

dQ

T

We will come back to this later. A more fundamental definition was found by Boltzmann,

S = kbln(Ω) (5.2)

where Ω is the multiplicity of states.

To understand what this means, we need to take a step back and introduce ideas ofmicrostates, macrostates and combinatorics.

• A microstate is a complete enumeration of the states (eg. position, momentum, en-ergy levels, spin/angular momentum) of each and every one of the atoms/moleculesin a system.

• A macrostate is described and distinguished by the values of bulk quantities suchas total energy, pressure, magnetisation, ect.)

The fundamental assumption of statistical mechanics:

In an isolated system in thermal equilibrium, all microstates are equally probable

Let us consider a pair of dice. We throw the dice once.

The microstates are given by the values shown on each die.

There are 36 of them:

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The macrostates are given by the total value of the dice. There are 11 of them (thenumbers 2, 3, ..., 12).

The multiplicity is how many microstates there are for each macrostate.

Here we have:

Macrostate 2 3 4 5 6 7 8 9 10 11 12Multiplicity 1 2 3 4 5 6 5 4 3 2 1

Using the fundamental assumption of statistical mechanics (which here is the same assaying the dice are not loaded) we find that 7 is six times more likely to occur thaneither 2 or 12. According to the definition of entropy, we find that the ’state’ 7 has thehighest entropy, specifically

S(2) = S(12) = kb ln(1) = 0 (5.3)

S(7) = kb ln(6) ≈ 1.79kb (5.4)

What this means is that for 2 and 12, the microstate is uniquely determined, whereasthere are many ways of achieving a 7.

5.1 Two-State Systems and Combinatorics

Let us now move from dice to coins, and consider a sequence of N coin tosses, which caneach result in heads (↑) or tails (↓). This is actually a reasonable model of magnetisation,as we shall see soon.

The microstate now records the sequence of individual tosses, eg. ↑↑↓↑↑↓↓↓↑

The macrostate counts the number of heads and tails, eg N↑ = 5, N↓ = 4 The question

is now:

How many ways are there of choosing N↑ heads out of N (= N↑ + N↓) tosses? Thiswill give us the multiplicity Ω(N↑, N↓) [or Ω(N↑, Ntot)]. Let us proceed by induction,assuming we do N coin tosses.

N↑ = 0: this means all tosses were tails, which is a unique microstate, i.e.

Ω(0) = 1

N↑ = 1: any one of the N tosses could be head, i.e.

Ω(1) = N

N↑ = 2: we can choose the first head among among N tosses, and the second among theremaining N − 1 ⇒ N(N − 1). However, it does not matter which one we call’first’ or ’second’, so the number of pairs is

Ω(2) =N(N − 1)

2

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N↑ = 3: we can choose the first N ways: the second N − 1 ways and the third N − 2 ways⇒ N(N − 1)(N − 2) ordered triplets. But all orderings are equivalent, and thereare 3 · 2 ways or ordering 3 items, so

Ω(3) =N(N − 1)(N − 2)

2 · 3

N↑ = 4: we now have N(N − 1)(N − 2)(N − 3) ordered quadruplets, and 4 · 3 · 2 ways orordering each, i.e.

Ω(4) =N(N − 1)(N − 2)(N − 3)

2 · 3 · 4

In general, for N coins and n heads:

Ω(N, n) =N(N − 1)...(N − n+ 1)

n(n− 1)...1

=N(N − 1)...1

[(N − n)(N − n− 1)...1]n!=

N !

(N − n)!n!≡(N

n

)binomial coefficient

This result,

Ω(N, n) =N !

n!(N − n)!≡(N

n

), (5.5)

holds not only for coin tosses, but for any system of N items which can be in one of twostates (and not just in those cases either!).

For example,

• distributing items (balls - or molecules!) between two containers

• a system of magnetic dipoles with two possible directions

The second example is called the two-state paramagnet, and is a good description ofmany real materials.

We will come back to it soon.

We can generalise our results for 2-state systems to systems with more possible states.The mathematics of the is called combinatorics.

For example, if we have a 3-state system (let us call the three possible outcomes/statesR,G,B with the number in each state labelled as nR, nG, nB), we can show that for asystem of size N (N items where nR + nG + nB = N),

The total number of microstates = 3N

The total number of macrostates = (N+1)(N+2)2

The multiplicity of the macrostate Ω(nR, nG, nB) = N !nR!nG!,nB !

Exercise: Check this for youself!

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5.1.1 The Two-State Paramagnet

Consider a material consisting of N ’elementary magnets’ (magnetic dipoles).

These can be electrons, or protons or nuclei or even protein molecules! In most solidsthey are usually unpaired electrons which carry a magnetic moment related to theirspin (internal angular momentum).

The magnetic moment µ characterises the strength of a magnet, defined by

~τ = ~µ× ~B (torque created by external magnetic field) (5.6)

A magnetic dipole in an external magnetic field ~B has energy V = −~µ · ~B.

For a single electron, we have µ2 = −geµBs2 where ge ≈ 2, µB = eh2me

= Bohr magneton,

and s2 = ±12

is the spin quantum number.

We see that for a single electron there are two possible values (spin up, spin down)for the magnetic moment, and if we have N atoms with one such electron each, wehave a 2-state system similar to our system of coin tosses! Paramagnetism is when

’elementary magnets’ tend to align in the external magnetic field (like a compass needle)

since this minimises the potential energy V = ~µ · ~B. This creates an increased magneticfield (magnetisation)

Ferromagnetism is when the magnets/spins interact with each other so that they lineup in the same direction spontaneously.

Diamagnetism is when the material creates a magnetic response opposite to the ex-ternal field - this happens because the field induces a ’rotating current’ - most materialsare diamagnetic to some extent.

Consider now a two-state paramagnet with elementary magnetic moment µ, in a mag-netic field ~B = B~z. Each particle has energy Ui = ±µB.

The total energy of a system of size N , with N↑ spin-up dipoles and N↓ spin-downdipoles is

Utot = −µB(N↑ −N↓) = µB(N − 2N↑) [ since N↑ + N↓ ] (5.7)

The total magnetisation is

M = µ(N↑ −N↓) = −UB

(5.8)

A macrostate is any state with a given (N↑, N↓), which gives a definite value for themacroscopic observables U , M .

We have already found the multiplicity of the macrostates:

Ω(N↑!, N) =N !

N↑!(N −N↑)!=

N !

N↑!N↓!(5.9)

If N (and N↑, N↓) is not very large, we can calculate this exactly, and use this to findthe entropy as a function of energy or magnetism.

But Ω(N↑, N) quickly becomes huge:

40

N N↑ N↓ Ω(N↑, N)20 10 10 184,75625 10 15 3,268,76030 10 20 30,045,01530 15 15 155,117,52040 15 25 4.22·1010

50 20 30 4.71·1013

100 35 65 1.10·1027

100 45 55 6.14·1028

200 112 88 2.15·1058

In real macroscopic pieces of matter we are considering not 50 or 200, but 1023 atoms.You will find that your calculator or computer is probably not able to compute 1023! -it is larger than the largest number the computer is able to store!

Thankfully, we only need to know the logarithm of Ω - but we still need to make someapproximations.

Stirling’s Approximation

A very useful approximation for the factorial of a large number is

N ! ≈ NNe−N√

2πN Stirlings’s Approximation (5.10)

[Try testing this with some not-very-large numbers like 50 or 200 and see how good itis]

Using this, we can easily compute lnN !:

ln(N !) ≈ ln(NN) + ln(e−N) + ln(√

2πN) = N ln(N)−N +1

2ln(2πN) (5.11)

Note that if N is really large, we can ignore the last term:

ln(N !) ≈ N ln(N)−N +1

2ln(2π) +

1

2ln(N) =

(N +

1

2

)ln(N)−N +

1

2ln(2π) (5.12)

⇒ ln(N !) ≈ N ln(N)−N (5.13)

since the difference between N + 12

and N is irrelevant, and the final term is a reallysmall number [1

2ln(2π) ≈ 0.919]

For example, if N = 1.000 · 108, then N + 12

is also 1.000 · 108, as is N + 1, N + 12

ln(2π),or N + 57.

Note that this effectively means that the term√

2πN can be ignored in N !, although itis a large number. This is because N ! is such an enormous number that large factors donot matter.

We can see how this can happen,

1010100 × 10100 = 10(10100+100) = 1010100 , [ax · ay = ax+y] (5.14)

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i.e. a googol-plex times a googol is still a googol-plex!

Back to the 2-State Paramagnet

S

kb= ln(Ω) = ln

( N !

N↑!(N −N↑)!

)= lnN !− lnN↑!− ln(N −N↑)! (5.15)

≈ N lnN −N −N↑ lnN↑ +N↑ − (N −N↑) ln(N −N↑) + (N −N↑) (5.16)

= N lnN −N↑ lnN↑ −N ln(N −N↑) +N↑ ln(N −N↑) (5.17)

= N ln( N

N −N↑

)−N↑ ln

( N

N −N↑

)(5.18)

We can write this in terms of y ≡ N↑N

= proportion of up-spins:

S

kb= −N ln

(N −N↑N

)+N

N↑N

ln(N −N↑

N↑

)(5.19)

= −N ln(1− y) +Ny ln(1− y

y

)[ln

a

b= ln a− ln b] (5.20)

= N [− ln(1− y) + y ln(1− y)− y ln y] (5.21)

(5.22)

⇒ S = kbN [(1− y) ln(1− y) + y ln y] (5.23)

Figure 5.1:

Note: 0 ≤ y ≤ 1⇒ ln y < 0, ln(1− y) < 0

We see that (as expected) the entropy is highestwhem roughly half the spins are pointing up androughly half are pointing down.

We can also determine the entropy in terms ofenergy/magnetisation:

M = −UB

= µ(N↑ −N↓) = µN(1− 2y)

⇒ y =1

2

(1− M

µN

)=

1

2

(1 +

U

µBN

)1− y =

1

2

(1 +

M

µN

)=

1

2

(1− U

µBN

)

S = −kbN[1

2

(1 +

M

µN

)ln

1

2

(1 +

M

µN

)+

1

2

(1− M

µN

)ln

1

2

(1− M

µN

)]= kbN

[ln 2− 1

2

(1 +

M

µN

)ln(

1 +M

µN

)− 1

2

(1− M

µN

)ln(

1 +M

µN

)]Later on, we will want to know the derivativew.r.t. U .

∂S

∂U= − 1

B

∂S

∂M=

1

2µBN

∂S

∂y(5.24)

42

where

∂S

∂y= kbN

[− ln(1− y)− 1− y

1− y+ ln y +

y

y

]= kbN ln

1− yy

(zero ay y =

1

2

)(5.25)

5.1.2 Interacting Systems

We can now look at two paramagnets which are in thermal contact so that they canexchange energy. We want to find the most likely macrostate for each of them, giventhe constraint that the total energy is constant.

If we call the two subsystems A and B, we have

U = UA + UB = µB(NA↑ −NA

↓ )− µB(NB↑ −NB

↓ ) (5.26)

= µB[NA(1− 2yA) +NB(1− 2yB)] = constant (5.27)

since NAandNB = constant⇒ N = NA +NB = constant (5.28)

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