Thermal Physic a Sample Material

8/21/2019 Thermal Physic a Sample Material

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A Students Handbook of Physics

Part III. Thermal Physics

February 2009


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Contents

1 The Kinetic Theory of Ideal Gases 5

1.1 Ideal gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 The Boltzmann distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 The Maxwell-Boltzmann distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3.1 Velocity distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3.2 Speed distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4 Properties of gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4.1 Particle energies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4.2 Heat capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4.3 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.4 Particle flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.4.5 Mean free path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.5 Establishing equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.5.1 Chemical equilibrium: particle transport . . . . . . . . . . . . . . . . . . . . . . 191.5.2 Mechanical equilibrium: momentum transport . . . . . . . . . . . . . . . . . . 231.5.3 Thermal equilibrium: heat transport . . . . . . . . . . . . . . . . . . . . . . . . 23

2 Classical Thermodynamics 272.1 Thermodynamic systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.2 The Laws of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.2.1 The Zeroth Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.2.2 The First Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.2.3 The Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2.4 The Third Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2.3 Thermodynamic potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.3.1 Internal energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.3.2 F, H and G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.3.3 The Maxwell relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2.3.4 Chemical potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3 Real gases and phase transitions 473.1 Real gas equation of state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

3.1.1 The van der Waals equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.1.2 The Boyle temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.1.3 The coexistence region . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.1.4 Isothermal compressibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.2 Phase transitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3.2.1 Conditions for phase equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . 513.2.2 Maxwell construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

3.2.3 Phase diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

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4 CONTENTS

3.2.4 Clausius Clapeyron equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.3 Irreversible gaseous expansions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

3.3.1 Joule expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.3.2 Joule-Thomson expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4 Statistical mechanics 63

4.1 Statistical ensembles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 634.2 The postulates of statistical mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.2.1 States and energy levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.2.2 Microstates and macrostates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 644.2.3 Properties of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

4.3 The density of states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.3.1 Derivation ink-space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.3.2 Converting to energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

4.4 Classical statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.4.1 The Boltzmann distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 734.4.2 The number density distribution . . . . . . . . . . . . . . . . . . . . . . . . . . 744.4.3 The partition function and thermodynamic properties . . . . . . . . . . . . . . 744.4.4 Agreement with kinetic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.5 Indistinguishable particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 794.5.1 Exchange symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 794.5.2 Bosons and fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.6 Quantum statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 824.6.1 Grand canonical partition function . . . . . . . . . . . . . . . . . . . . . . . . . 824.6.2 Quantum occupation functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 844.6.3 The classical approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 864.6.4 Black body radiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89


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Chapter 1

The Kinetic Theory of Ideal Gases

Kinetic theory describes the gross thermalproperties of ideal gases, from a consideration ofthe average behaviour of the individual particlesthat make up the gas. For example, by knowingthe probability distribution of particle energies or

velocities in an ideal gas, kinetic theory is able topredict large-scale properties of the gas, such asits internal energy, heat capacity, pressure andflux.

1.1 Ideal gases

Kinetic theory applies only to ideal gases, whichconstitute a simple model of real gases. In theideal gas model, we treat gas particles as hard

spheres that are moving randomly (Brownian mo-tion), and constantly colliding with each otherand with the walls of the container. The assump-tions of ideal gases and kinetic theory are:

1. Molecular size intermolecular separation(particle volumes are negligible)

2. Intermolecular forces are negligible

3. All collisions are perfectly elastic, whichmeans that momentum and energy are con-

served between the colliding objects

4. Energy exchange between particles is solelydue to elastic collisions

5. The molecular speed distribution is constantin time

All gases behave like ideal gases at very low pres-sures and high temperatures, since under theseconditions the particles relative volumes and in-termolecular forces are negligible. Inert gases,

such as neon and argon, behave like ideal gases

even at low temperatures, since they have com-plete outer electron shells and therefore weak in-termolecular forces.

Ideal gas law

A defining characteristic of ideal gases is thatthey obey the ideal gas law. This law relatesthe macroscopic properties pressure, p [Pa], tem-perature, T [K], and volume, V [m3], for a gaswithn moles (Eq. 1.1). It applies only to systemsin equilibrium, meaning that their macroscopicproperties are not changing with time.

Ideal gas law:

pV = nRT (1.1)

= N kBT (1.2)

where the molar gas constant is:

R = NAkB

= N

nkB (1.3)

The ideal gas law comes from an amalgama-tion of three other laws, which were discoveredempirically:

1. Boyles law: V 1/p at constant T and n

2. Charless law: V T at constantp andn

3. Avogadros law: V n at constant T and p

Therefore, combining these we have

pV

nT= constant =R

5


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6 CHAPTER 1. THE KINETIC THEORY OF IDEAL GASES

The constant of proportionality is the molar gasconstant, R [8.31 JK1mol1]. This is relatedto the Boltzmann constant,kB [1.381023 JK1]by Eq. 1.3. Avogadros number, NA [6.0221023mol1], gives the number of gas particles, N, permole of gas:

NA=N

n

where n is the number of moles. Therefore, byeliminating R from the ideal gas law, we can ex-press it in terms of the Boltzmann constant, kB,for an ideal gas with N particles (Eq. 1.2).

1.2 The Boltzmann distribution

A postulate of kinetic theory is that particles in

an ideal gas obey the Boltzmann distribution ofenergy,1 f() (Eq. 1.4). This is a probability dis-tribution, wheref() dgives the probability thata particle has an energy between and + d.The Boltzmann distribution is an exponentiallydecaying function of energy, and depends only onthe temperature of the gas (Fig. 1.1). The expo-nential term in Eq. 1.4 is called the Boltzmannfactor, and the coefficient, A, is the normalisa-tion constant. The value ofA is determined fromthe requirement that the total probability, which

is the Boltzmann distribution integrated over allenergies, equals unity (Eqs. 1.51.6).

The Boltzmann distribution:

f()d = A e/kBT Boltzmann

factor

d (1.4)

where the normalisation condition is

0

f()d = 1 (1.5)

Therefore, the normalisation constant is

A = 1

0

e/kBTd(1.6)

1For a derivation of the Boltzmann distribution, see

Section 4.4.

f()

Thot

Tcold

d

Figure 1.1: Variation of the Boltzmann distri-

bution with temperature and energy (not nor-malised). The probability of a particle having anenergy between0and0 + dat Tcold is given bythe shaded area. Mark 0 on figure + plot usingmathematica.

Precision of statistical methods

Its not hard to convince yourself that using sta-tistical methods and probability distributions for

systems with many particles is very precise. Con-sider, for example, one mole of gas, which hasNA 1024 particles. If we split the Boltzmanndistribution into 106 bins of equal width d,then there are, on average, of order Nb 1018particles per bin. Therefore, the standard devia-tion for each bin is

1Nb

109

This means that statistical fluctuations in the

number of particles per bin are of order one partin a billion. So the difference between a real dis-tribution and a statistically-derived mean distri-bution is essentially negligible.

1.3 The Maxwell-Boltzmann

distribution

The Maxwell-Boltzmann distribution is a smallextension from the Boltzmann distribution;

whereas the latter gives the probability that a


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1.3. THE MAXWELL-BOLTZMANN DISTRIBUTION 7

particle has a given energy, the former gives theprobability that a particle has a given velocity.For an ideal gas, in the absence of external forcessuch as gravity, the total energy of a particleequals its kinetic energy, which means that

= 12

mv2

Therefore, eliminating from the Eq. 1.4, we canwrite the probability distribution in terms of par-ticle velocity:

fv(v) =Aemv2/2kBT

In this case, the Boltzmann factor is a Gaussianfunction,2 whereas for the Boltzmann distribu-tion ofenergies, it is an exponentially decreasing

function. The subscript in fv(v) therefore indi-cates that this is not the same function as theBoltzmann distribution, which was denoted sim-ply byf(). It is important to be consistent withnotation, since there are several ways of express-ing the Maxwell-Boltzmann distribution:

3D velocity distribution: fv(v) d3v

1D velocity distribution: fv(vx) dvx

speed distribution: Fv(v) dv

The velocity distributions take into account theparticlesdirection, and therefore are functions ofthe velocityvector (vin 3D orvxin 1D), whereasthe speed distribution depends only on the mag-nitude,v, and is a function different from fv(thusdenoted by Fv). Below, we will look at each dis-tribution in turn.

1.3.1 Velocity distributions

Lets first look at the 1D velocity distribution,

fv(vx) dvx. Using = mv2x/2 in Eq. 1.4, the 1D

velocity distribution is

fv(vx)dvx = Aemv2x/2kBTdvx

This gives the probability of a particle havinga velocity in the x-direction between vx andvx+ dvx. (Similar distributions exist for velocityin either they or thez direction.) The normalisa-tion constant,A, is determined by integrating thedistribution over all 1D velocities ( to +),

2A Gaussian is a function of ex2

.

dvx

dvz

dvy

v

Figure 1.2: Element of velocity-space, d3v =dvx dvydvz, for a particle with velocity v.

and setting the integral equal to one, as shown inDerivation 1.1. This condition ensures that thetotal probability equals unity. The normalised

velocity distribution is a Gaussian distribution(Eq. 1.7).

Velocity distributions: (Derivations 1.1 and1.2)In 1D:

fv(vx) dvx=

m

2kBT emv

2x/2kBT

Boltzmann

factor

dvx (1.7)

In 3D:

fv(v) d3v=

m

2kBT

32

emv2/2kBT

Boltzmannfactor

d3v (1.8)

It is straightforward to generalise to 3D, bychanging vx to the total velocity, v. The mag-nitude of the total velocity is

v2 =v2x+ v2y+ v

2z

and the 3D element of velocity-space (Fig. 1.2) is

d3v = dvx dvydvz

The expression for the 3D distribution, f(v) d3v,is also a Gaussian distribution (Eq. 1.8 andDerivation 1.2). Eq. 1.8 gives the probability thata particles velocity lies in an element of velocity-

spaced3v from velocity v (Fig. 1.2).


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Table 1.1: Derivation of 1D velocity distribution (Eq. 1.7)

fv(vx)dvx = A

emv2x/2kBTdvx

= A

2kBT

m

Therefore, from the normalisation condition:

A = m2kBTfv(vx)dvx =

m

2kBTemv

2x/2kBTdvx

Starting point: The Maxwell-Boltzmanndistribution for velocity in 1D, fv(vx)dvxis given by substituting = mv2x/2 intoEq. 1.4. The normalisation condition re-quires that the integral over all velocitiesequals one:

fv(vx)dvx= 1

To integrate, use the standard integral:

ex2

dx=

Table 1.2: Derivation of 3D velocity distribution (Eq. 1.8)

fv(v) d3v = (fv(vx) dvx) (fv(vy) dvy) (fv(vz) dvz)

= m

2kBT

32em(v

2x+v2y+v2z)/2kBTd3v

=

m

2kBT

32

emv2/2kBTd3v

The 3D velocity distribution is theproduct of the three 1D componentdistributions, where:

d3v = dvx dvydvz

and v2 = v2x+ v2y+ v

2z

Use the normalised 1D distribu-tion from Eq. 1.7. (fv(vy) dvy andfv(vz) dvz are identical to fv(vx) dvx,but with different subscripts).


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1.4. PROPERTIES OF GASES 9

f(v)

v0

= kT/m

Figure 1.3: Maxwell-Boltzmann 1D velocity dis-tribution. The distribution is a Gaussian cen-tred about the mean velocity,vx, which is zerosince there is no preferred direction. The stan-dard deviation is

kBT /m, and increases with

increased temperature. The shaded area repre-

sents the probability of a particle having a veloc-ity betweenv0 andv0+ dv. The total area underthe curve is equal to unity (total probability = 1).Plot using mathematica. Show Thot and Tcoldandv0 andv+ dv0.

Average values

Both the 1D and 3D velocity distributions areGaussian distributions, centred about the origin.

The standard deviation for a Gaussian of theform ex

2is = 1/

2. Therefore, the stan-

dard deviations for the velocity distributions aretemperature-dependent:

1D =

kBT

m (1.9)

and 3D = 3 1D=

3kBT

m (1.10)

The centre of the distribution corresponds to theaverage velocity,vx, and the standard devia-tion corresponds to the rms velocity,

v2x. Inworked example 1.4, these average values are cal-culated explicitly using Eq. 1.7. The 1D velocitydistribution is shown in Fig. 1.3.

1.3.2 Speed distribution

The speed distribution, Fv(v) dv, gives the prob-ability of a particle having aspeedbetweenv andv+dv, and thus is independent of the particles

direction of travel. If we express the 3D velocity

distribution in spherical polar coordinates, thenthe 3D element of velocity-space is:

d3v = v2 sin dv d d (1.11)

Therefore, by integrating over angular coordi-nates and , we can remove the directional-

dependence of Eq. 1.8 (Derivation 1.3). Thespeed distribution is therefore given in Eq. 1.12.This distribution has the same Gaussian Boltz-mann factor as the velocity distributions, but theentire distribution is no longer Gaussian since theBoltzmann factor now is multiplied by the fac-tor of v2, which appeared from Eq. 1.11. Thismeans that the probability distribution is skewedtowards higher speeds, as shown in Fig. 1.5. Theaverage speed, v, the rms speed,

v2, and the

modal speed, vm, which are labelled in Fig. 1.5,

are calculated in worked example 1.6.

Speed distribution: (Derivation 1.3)

Fv(v) dv= 4v2

m

2kBT

32

emv2/2kBT

Boltzmannfactor

dv

(1.12)

1.4 Properties of gases

The Maxwell-Boltzmann probability distribu-tions can be used to determine average velocitiesand speeds, as shown in worked examples 1.4 and1.6. We can then use these averages to explainthe macroscopic properties of systems in equilib-rium, as shown below.

1.4.1 Particle energies

Using the expression for the average squaredspeed,

v2

, from worked example 1.6, the av-erage translational kinetic energy per particle is:

KE3D = 1

2m

v2

= 3

2kBT (1.13)

Or, for translational motion in one-dimension:

KE1D = 1

2m

v2x

= 1

2 kBT (1.14)


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Figure 1.4: Worked example: Using Eq. 1.7 to calculate average values of particle velocity

1. The average velocity,

vx

, equals the aver-

age of the Gaussian distribution, x. Theaverage velocity equals zero since there isno preferred direction of travel.

2. The root mean square velocity,v2x, does

not depend on direction, therefore we wouldexpect it to be non-zero. We see that

v2xequals the standard deviation, x, of theGaussian distribution.

3. Since the three components vx, vy and vzare independent, the square of the standarddeviation of the total velocity distributionis equal to the sum of the square of the com-ponents:

2 =2x+ 2y +

2z

Therefore, the root mean square of the totalvelocity is:

v2 =

v2x

+

v2y

+

v2z

=

3kBTm

1. vx =

vxfv(vx)dvx

=

m

2kBT

vxemv2x/2kBTdvx

= 0

x

2.

v2x

=

v2xfv(vx)dvx

= m2kBT

v2xemv2x/2kBTdvx

= kBT

m 2x

To integrate, use the standard integrals:

xex2

dx = 0

x2ex2

dx = 1

2

Table 1.3: Derivation of the speed distribution (Eq. 1.12)

Fv(v)dv =

20

0

fv(v)v2 sin dvdd

= v2

m

2kBT

32

emv2/2kBTdv

20

d

0

sin d

= 4v2

m

2kBT

32

emv2/2kBTdv

Starting point: The speed distribu-tion is found by integrating the 3Dvelocity distribution (Eq. 1.8) over allangles and , to remove the direc-tional dependence of the velocity.

The element of 3D velocity space inspherical polar coordinates is

d3v= v2 sin dvdd


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Figure 1.6: Worked example: Using Eq. 1.12 to calculate average particle speeds

1. The average speed, v, is given by the aver-age of the speed distribution,Fv(v). The in-tegration limits are 0 to since we are nottaking direction into account, and thereforecannot have negative speeds.

2. The root mean square speed,v2, equals

the root mean square velocity calculated

from the 3D distribution (see worked ex-ample 1.4). This follows since the square ofthe velocity of a particle is identical to thesquare of its speed.

3. The modal speed is the turning point ofFv(v), and therefore is the speed at which

dFv(v)

dv

vm

= 0

4. The ratio of average speeds is therefore

vm: v :

v2 =

2 :

8

:

3

1 : 1.13 : 1.22

(refer to Fig. 1.5)

To integrate, use the standard integrals:

0

x3ex2

dx = 1

22

0

x4ex2

dx = 3

82

1. v =0

vFv(v)dv

= 4 m2kBT3/2

0

v3

emv2/2kBT

dv

=

8kBT

m

2.

v2

=

0

v2Fv(v)dv

= 4 m

2kBT3/2

0v4emv

2/2kBTdv

= 3kBT

m

3. Differentiate Fv(v) and set derivative equalto zero:

vm = 2kBTm


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Tcold

Thot

F(v)

v0 vm

Figure 1.5: Maxwell-Boltzmann speed distribu-tion. The shaded area represents the proba-bility of a particle having a speed between v0and v0+ dv. The total area under the curve istherefore equal to unity (total probability = 1).For hot systems, the distribution is skewed to-

wards higher particle speeds. The ratio of aver-age speeds is vm :v :

v2 = 1 : 1.13 : 1.22,and is independent of both T and m. Plot usingmathematica. Show Thot and Tcold and v0 andv+ dv0.

These expressions show that the kinetic energyof particles in a gas depends only on its temper-ature. There are three independent degrees offreedom3 for translational kinetic energy, corre-sponding to motion in the x, y or z directions,respectively. Therefore, Eqs. 1.13 and 1.14 showthat there is a kinetic energy of kBT /2 associ-ated with each translational degree of freedom.Since the average kinetic energy per particle de-pends only on the temperature of the gas, it isoften referred to as the thermal energyof the sys-tem. Atoms are said to be in thermal motionas aresult of their being at a particular temperature,and their thermal motion increases with increasedtemperature. Owing to collisions, the thermalmotion of atoms appears as a random walk. Inlarger particles, the random walk due atomic andmolecular collisions is known as Brownian mo-tion.

Internal energy

The internal energy, U[J], of a gas is the sum ofthe kinetic and potential energies of its individual

3Adegree of freedomin this context is a parameter that

defines an independent direction of movement.

particles. Thepotentialenergy includes chemicalenergy stored in atomic bonds or intermolecularforces. Since an ideal gas has by definition nointermolecular forces, the potential energy of itsparticles is assumed to be zero. Therefore the in-ternal energy of an ideal gas results only from thekinetic energy of its particles. For a monatomicideal gas ofNparticles, with three translationaldegrees of freedom (x, y and z), the internal en-ergy of the gas is therefore:

U1D = NKE3D=

3

2N kBT

32

nRT (1.15)

where we have used Eq. 1.3 to eliminate nkB.Similarly, for translational motion in one-dimension:

U1D = NT1D=

1

2N kBT

= 1

2nRT (1.16)

Therefore the internal energy of an ideal gas de-

pends only on its temperature. (For real gases,intermolecular forces are not negligible, thereforetheir internal energy is a function of the gass vol-ume as well as its temperature.)

Eqs. 1.15 and 1.16 show that there is an inter-nal energy ofnRT/2 associated with each degreeof freedom. Diatomic ideal gases also possess ro-tational and vibrational kinetic and potential en-ergies, which means they have additional degreesof freedom. Linear diatomic gases have two ad-ditional rotational degrees of freedom, and two

additional vibrational degrees of freedom, whichare illustrated in Fig. 1.7. (Need to draw this bet-ter.) For an ideal gas withf independent degreesof freedom, it is found that its internal energy is:

U = f

2N kBT

= f

2nRT (1.17)

where f = 3 for a monatomic ideal gas (trans-lational motion in x, y or z directions). For a

linear diatomic gas undergoing rotation with no


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Figure 1.7: Degrees of freedom for a linear di-atomic ideal gas. Rotation and vibration eachintroduce two additional degrees of freedom.

vibrations, f = 5, and if the gas also has radialvibrations, thenf= 7.

Using Eq. 1.17 together with the ideal gas law,we find that the relationship between pressure,

p [Pa], and internal energy per unit volume, u[Jm3], is:

p = 2

fu (1.18)

Therefore, for an ideal monatomic gas, where f=3, the pressure is

p = 2

3u

The equipartition theorem

Equation 1.17 forms the basis of theequipartitiontheorem, which states that the internal energyof a system is shared equally among its differ-ent contributions. Said another way, with eachdegree of freedom, there is an energy factor ofRT /2 [J/mole] or kBT /2 [J/molecule] (Eqs. 1.19and 1.20).

Equipartition theorem: The average energyassociated with each quadratic degree of free-dom is

E = 12

RT per mole (1.19)

12

kBT per molecule (1.20)

The equipartition theorem holds true providedthat the energy has a quadratic dependence onthe particular degree of freedom.4 For example,the dependence of kinetic energy on v or p isquadratic:

KE= 12

mv2 p22m

Alternatively, the dependence of the potential en-ergy of a harmonic oscillator is quadratic in x:

VSHO=1

2kx2

where k [Nm1] is the spring constant. Refer toclassical mechanics section?

1.4.2 Heat capacity

We saw above that the temperature of an idealgas is a measure of its internal energy. Heat flowinto or out of an ideal gas changes its internalenergy, and hence its temperature. The heat ca-pacity, C [JK1], of a system is the amount ofheat,Q, required to raise its temperature by 1 K.It is often expressed as a partial derivative, andis given by Eq. 1.21. The variable defines theconstraint on the system. For example, if heat isadded at constant pressure, then = p, whereasif heat is added at constant volume, then = V.The specific heat capacity, c [JK1kg1], is theheat capacity per unit mass, and the molar heatcapacity, cm [JK

1mol1], is the heat capacityper unit mole (Eqs. 1.22 and 1.23 respectively).

Heat capacity:

C =

Q

T

(1.21)

c = 1

m

QT

(1.22)

cm = 1

n

Q

T

(1.23)

The actual rise in temperature with increasingenergy for a given substance depends on the na-ture of its particles. For example, for monatomic

4For a statistical mechanics treatment, see Sec-

tion 4.4.4.


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gases, the extra energy only goes into increas-ing their translational kinetic energy, whereas fordiatomic gases, the energy also increases theirrotational or vibrational energies. Therefore, inqualitative terms, for the same amount of inputenergy, the temperature of a monatomic gas in-creases more than for a diatomic gas.

CV and Cp

The heat capacity depends on which variable isheld constant while heat is added to the system.The two most common heat capacities are theheat capacity at constant volume, CV, and heatcapacity at constant pressure, Cp. From the con-servation of energy, we find that CV is equal to

the change of the systems internal energy withtemperature:

CV = dU

dT (1.24)

(Derivation 1.4). Using the expression for U fromEq. 1.17 in this, we find that the heat capacity atconstant volume is proportional to the number ofdegrees of freedom, f, of the system (Eq. 1.25).Therefore, since a monatomic gas has three trans-lational degrees of freedom, its heat capacity at

constant volume for one mole of gas (n= 1) is:

CV = 3

2R

Cp is greater than CV since the energy goes intoincreasing the volume of the gas as well as itsinternal energy (its temperature). Using the con-servation of energy, it is found that the differ-ence in heat capacities, Cp CV, is equal to nR(Eq. 1.27 and Derivation 1.4). Therefore, the heat

capacity at constant pressure is given by Eq. 1.26.For one mole of monatomic gas, this gives:

Cp=5

2R

The ratio of heat capacities is a constant and isreferred to as the adiabatic index, (Eq. 1.28).For a monatomic ideal gas, the adiabatic index istherefore:

=5

3

CV and Cp for ideal gases:(Derivation 1.4)Heat capacities for fdegrees of freedom

CV

= f

2nR (1.25)

Cp = f+ 2

2 nR (1.26)

Difference in heat capacities:

Cp CV = nR (1.27)

Ratio of heat capacities

CpCV

= (1.28)

= f+ 2f

(1.29)

Aside on latent heat: In this section, we haveonly considered ideal gases. However, when asubstance changes state, for example solid to liq-uid or liquid to gas, then heat input does notincrease the bodys temperature, but instead isabsorbed by the bonds between particles, whichtherefore increases the potential energy of theparticles. Since the kinetic energy of the parti-cles remains constant, the temperature of the sub-stance does not change during a change of state.The amount of heat absorbed or released per unitmass during such a transition is called the specificlatent heat,5 L[Jkg1] (Eq. 1.30). The latent heatoffusionis the energy required to melt unit massof solid to a liquid, whereas the latent heat ofva-porisationis the energy required to vaporise unitmass of liquid to a gas. Melting or boiling areendothermic processes, which means that latentheat is absorbed, whereas condensation or freez-ing are exothermic processes, meaning that latentheat is released.

Latent heat:

L = Q

m (1.30)

5The word latent is derived from the Latin latere, to

lie hidden.


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Table 1.4: Derivation of expressions forCV and Cp. (Eqs. 1.24 and 1.27)

dU = dQ + dW= dQ pdV

Constant volume:

U

T

V

=

Q

T

V

CV

CV = dU

dT

Constant pressure:

dQ = dU dW= CVdT+pdV

Q

T

p

= CV +p

V

T

p

Cp = CV + nR

Starting point: From the conservation of energy,the element of increase in internal energy of a sys-tem, dU, equals the element of heat added, dQ,plus the element of work done by the system, dW:

dU=dQ + dW

(This is the First Law of Thermodynamics; seeSection 2.2.2)

The work done equals the energy needed to ex-pand the gas against pressure, which results fromthe molecules impulse (see Section 2.2.2)

dW = pdV

Constant volume: Divide the expression for dUby dT, holding V constant. From the defini-tion of heat capacity (Eq. 1.21), CV results fromthe change in internal energy with temperature.Since U is a function ofT only, for an ideal gas,then CV can be written as the total derivative,dU/dT, instead of a partial derivative.

Constant pressure: Divide the expression for dQbydT, holdingp constant. Use the ideal gas law,

pV =nRT, to determine the volume-temperature

derivative at constant pressure.


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1.4.3 Pressure

From worked example 1.6, the average squaredspeed is:

v2

=3kBT

m

Using this to eliminate kBT from the ideal gaslaw, we can express the pressure in terms of

v2

:

p=1

3mnp

v2

wherenp [m3] is the particle number density:

np=N

VWe reach the same result if we assume that thepressure results from the force per unit area ex-erted by molecular collisions with the container

walls (Eq. 1.31 and Derivation 1.5). This con-firms that pressure is due to molecular impulse,and not due to intermolecular repulsion as wasoriginally thought by Newton.

Ideal gas pressure: (Derivation 1.5)

p = 1

3mnp

v2

(1.31)

1.4.4 Particle flux

We can use a similar treatment to determine anexpression for the particle flux, [m2s1], whichis the number of particles striking a surface perunit area and per unit time. The particle flux isfound to depend on the particle number density,np, and the average particle speed,v (Eq. 1.32and Derivation 1.6). These, in turn, depend onthe gass pressure and temperature, therefore theparticle flux can be expressed in terms ofp andT (Eq. 1.33). This shows that particle flux re-lates the microscopic motion of particles to thegass macroscopic properties, pressure and tem-perature.

Particle flux: (Derivation 1.6)

= 1

4np v (1.32)

p2mkBT

(1.33)

Effusion

Effusion is the process by which particles emergethrough a small hole. The rate of effusing parti-cles per unit area, RE [m

2s1], is equal to theparticle flux, , striking the hole if it were closed

off:

RE=1

4np v

Expressing the average velocity in terms of theMaxwell-Boltzmann speed distribution,

v =

vFv(v)dv

where Fv(v)dv is provided in Eq. 1.12, we findthat

RE = 14

np

vFv(v)dv

FE(v)dv

Therefore, we can define the effusion rate perunit area in terms of the velocity distributionof effusing particles, FE(v). Since the Maxwell-Boltzmann speed distribution,Fv(v), depends onv2, the effusion distribution, FE(v), is found todepend on v3:

FE(v)dv = 1

4npvFv(v)dv

v3emv2/2kBTdv (1.34)

Therefore, high velocity particles have a greaterprobability of effusing since the probability dis-tribution is skewed towards higherv. This meansthat both the pressure and temperature of a gasin an effusing vessel decrease, since high energyparticles are leaving. Effusion has practical ap-

plications in isotope separation, since the effusionrate depends on the mass of the effusing particle:

RE v 1m

In consequence, the heavier isotope is progres-sively enriched in the effusing vessel with time.

1.4.5 Mean free path

The mean free path, [m], is the average dis-

tance travelled by a particle before it undergoes


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Table 1.5: Derivation of the pressure of an ideal gas (Eq. 1.31)

vdt

dS

Cylinder volume = vdt dScos

vv

v = 2v cos

v cos

dp = F

dSdN

= F

dSNCN(Fv(v)dv)

= 2mv cos

dS dt

(npvdtdScos )(sin dd)(Fv(v)dv)

4=

1

2mnpv

2Fv(v)dv cos2 sin dd

p = 1

2mnp

v2Fv(v)dv

=v2

/20

cos2 sin d

20

d

= mnp

v2

13

cos3

/20

=

1

3 mnp v2

Starting point: An element of pres-sure, dp, is given by the force F, perunit area dS, multiplied by the num-ber of particles dN, striking the con-tainer wall in a time dt.

For a particle density ofnp[m3],dNis given by:

dN = NCN(Fv(v)dv)

where NC = npvdtdScos

N = d

4 =

sin dd

4

NC is the number of particles withinthe cylinder (refer to diagram);Fv(v)dv is the fraction of those withthe correct velocity (between v andv+ dv); N is the fraction of thosewithin the correct solid angle, d.

The force exerted on dS due to achange in particle momentum is

F = dp

dt

= mdv

dt

= m2v cos

dt

(refer to diagram)

Integrating over all space, we find thetotal pressure exerted by all particlesin a gas, in terms ofm, np and

v2

.


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Table 1.6: Derivation of particle flux (Eq. 1.32)

vdt

dS

Cylinder volume = vdt dScos

d = dN

dSdt

= NCN(Fv(v)dv)

dSdt

= (npvdtdScos )(sin dd)(Fv(v)dv)

4dSdt

= 1

4 npvFv(v)dv cos sin dd

= 1

4np

vFv(v)dv

=v

/20

cos sin d

20

d

= 1

2np v

1

2cos2

/20

= 1

4np v

= p2mkBT

Starting point: The particle flux is de-fined as the number of particles strikingper unit area dS, per unit time dt.

For a particle density ofnp m3, dN isgiven by:

dN = NCN(Fv(v)dv)

where NC = npvdtdScos

N = d

4 =

sin dd

4

NC is the number of particles within thecylinder (refer to diagram); Fv(v)dv isthe fraction of those with the correct ve-locity (betweenv andv + dv); N is the

fraction of those within the correct solidangle, d.

Integrating over all space, we find theexpression for the flux, , in terms ofnpandv.

From the ideal gas law, p = npkBT, andfrom worked example 1.6,

v =

8kBT

m

Therefore, we can express in terms ofpressure and temperature.


19/94

1.5. ESTABLISHING EQUILIBRIUM 19

a collision with another particle. If we define thescattering cross-section, [m2], as

= d2

where d is the effective particle diameter, then

the mean free path of an ideal gas with particlenumber density np, is given by Eq. 1.35 (Deriva-tion 1.7).

Mean free path: (Derivation 1.7)

= 1

2 np (1.35)

where the scattering cross-section is

= d2

1.5 Establishing equilibrium

The ideal gas law holds for systems in equilibrium.A system in equilibrium is one whose macroscopicproperties are not changing with timefor ex-ample, constant pressure, constant temperatureand constant volume. Equilibrium is an impor-

tant concept in thermal physics, since the subjectofthermodynamics, which deals with the thermalproperties of matter exclusively on macroscopicscales, only holds for systems in equilibrium.However, since kinetic theory treats systems onmicroscopic scales, we can use it to determinethe transport properties of systems that arent inequilibrium.

Thermodynamic equilibrium has three compo-nents:

1. Chemical equilibrium: constant particledensity. This is achieved by particle diffusionfrom regions of high particle concentration tolow concentration.

2. Mechanical equilibrium: constant pres-sure. This is achieved by viscous forces,which transport particle momentum from re-gions of high momentum to low momentum.

3. Thermal equilibrium: constant tempera-ture. This is achieved by thermal conduc-

tivity, which is the transport of heat from a

region of high temperature to a region of lowtemperature.

The microscopic behaviour governing these pro-cesses is very similar in each case. Below, we will

look in turn at how each of these equilibria areestablished.

1.5.1 Chemical equilibrium: particle

transport

Chemical equilibrium is established when the par-ticle concentration, np, is uniform throughoutthe system. Therefore, if there is a gradient inthe concentration, dnp/dz, particle transport, ordiffusion, occurs down the concentration gradi-ent. We will consider two situations: thesteady-state situation, where the concentration gradientis constant in time, and the non-steady situation,where the concentration gradient varies with re-spect to time.

Steady-state diffusion

Ficks First Law of diffusion states that forsteady-state particle diffusion, the particle flux,6

J [m2s1], is proportional to the gradient of theparticle number density, np. In one-dimension,for example along the z -axis, Ficks law is there-fore:

Jz = D npz

where D [m2s1] is the diffusion coefficient. An

expression for the diffusion coefficient is foundby considering the number of particles crossing agiven surface per unit time, as shown in Deriva-tion 1.8. We find that D depends on the averageparticle speed,v, and the mean free path, (Eq. 1.37 and Derivation 1.8).

6The particle flux is the number of particles crossing

unit area per unit time.


20/94


Table 1.7: Derivation of particle mean free path (Eq. 1.35)

vRt

Cylinder volume = vRt

= d2

d

particle density = np

d

dEffectivecollision area

= dx

dN

= vdt

np d

= vdt

np vR dt 1

2np

Starting point: The mean free path, ,is the mean distance travelled before acollision, per particle: = dx/dN.

If the time between collisions is dt, thenthe mean distance travelled by a particlewith velocity v is:

dx= v dt

In a time dt, the collision volume isd = vR dt, where vR is the relativevelocity between colliding particles (referto diagram). If the particles have veloci-tiesv1and v2, wherevR = v1v2, thenthe magnitude of the relative velocity is

|vR| =

vR vR

= (v1 v2) (v1 v2)=

v21+ v

22 2v1 v2

Therefore, the average relative velocity,for particles where|v1| |v2| v, andtravelling in opposite directions, is

vR =

v21

+

v22 2 v1 v2

0

= v21 + v22

2 v2

2v


21/94


vdt

dS

z0

z0-

z0+

Particle flux, JzShear stress, Pzxor Heat flux, jzz

Figure 1.8: (Refer to Derivations 1.81.10.) Assume that particles striking an area dS come froma sphere of radius , centred at dS. The number of particles striking dS in a time dt is dN =NCN(Fv(v)dv), where NC=npvdtdScos is number of particles within the cylinder; N= d/4=

sin dd is the fraction of particles within the correct solid angle; and Fv(v)dv is the fraction ofparticles with the correct velocity (betweenv and v + dv). If we consider a reference plane at z0, then,in a time dt, the particles crossing the plane from below come from the plane at z =z0 , whereasthose crossing the plane from above come from z+ = z0 + . Is this right?? and are derivations1.81.10 correct? I should probably draw a hemisphere/semi-circle instead of straight planes...

Particle transport: (Derivation 1.8)Ficks First Law:

Jz =

D

np

z

(1.36)

Diffusion coefficient:

D = 1

3v (1.37)

Non-steady diffusion

The rate of change of particle density between zand z+ dz is given by the difference in particle

flux entering at z and leaving atz + dz:

npt

dz= J(z) J(z+ dz)

Therefore, using a Taylor expansion on J(z + dz)together with Eq. 1.36, we recover Ficks SecondLaw, which gives the rate of change of particledensity (Eq. 1.38):

npt

dz = J(z) J(z+ dz)

= J(z) J(z) J

z dz

= D2npz2

dz

The solution to Eq. 1.38 is found have the form:

np(z, t) = n04Dt ez2

/4Dt

which is a Gaussian in z (this can be shown bysubstitution). Therefore, since the variance ofa Gaussian of the form ex

2is 1/2, np has a

variance that grows linearly with time (Eq. 1.39).The standard deviation, , which is the squareroot of the variance, is called the diffusion length,LD [m], and it provides a measure of the parti-cle propagation distance in a timet (Eq. 1.40 andFig. 1.9).

Particle transport:Ficks Second Law:

npt

= D2npz2

(1.38)

Variance and diffusion length:z2

= 2 = 2Dt (1.39)

LD = =

2Dt (1.40)


22/94


Table 1.8: Derivation of diffusion coefficient (Eq. 1.37, refer to Fig. 1.8)

dJz = dN(z)

dSdt cos

= 14

np(z)vFv(v)dv cos2 sin dd

Jz = 1

4np(z0 )

vFv(v)dv

=v

/20

cos2 sin d

=1/3

20

d

=2

= 1

6np(z0 ) v

Jz = J+z

Jz

= 1

6np(z0 ) v 1

6np(z0+ ) v

16

v

n(z0) npz

n(z0) + np

z

= 13

v =D

npz

D = 1

3v

Starting point: The z-componentof particle flux, for particles inci-dent at an angle , is the numberof particles, dN, per unit areadS,per unit timedt. Since the particleconcentration is not constant,N isa function ofz. (Refer to Fig. 1.8for expression for dN)

The net upward flux, J+, anddownward flux,J, is found by in-tegrating over all angles in a hemi-sphere, and over all velocities. Theupward flux is a result of particlescoming from the planez =z0,whereas the downward flux is from

particles in the plane z+ =z0+(Refer to Fig. 1.8).

The total flux, Jz, is thereforethe difference between upward anddownward flux. We can approxi-mate using a Binomial expansionaboutz0.

By comparing the expression forJz to Ficks First Law in Eq. 1.36,we find an expression for the dif-fusion coefficient, D .


23/94


t1

t2> t1

z

n(z)

1=2Dt1

2=2Dt2

Figure 1.9: Particle diffusion in 1D (Ficks SecondLaw, Eq. 1.38). The standard deviation increaseswith time. The diffusion length equals the stan-dard deviation. Draw in mathematica.

1.5.2 Mechanical equilibrium: mo-

mentum transport

We can use a similar steady-state approach toFicks First Law to determine the how pressurereaches equilibrium. Pressure gradients are dueto gradients in viscous forces between particles.These are shearing forces per unit area, and arereferred to as shear stress, Px [Pa]. Shear stressacts parallel to the surface on which it acts, simi-lar to friction between solids. The perpendicularshear stress gradient, Pzx, is found to be propor-

tional to the z-gradient of particle velocity in thex-direction,ux:

Pzx = ux(z)z

(Refer to the figure in Derivation 1.9). The co-efficient of proportionality, [Pa s], is called theviscosity. Viscosity is often defined in terms ofthis relation, as the shear stress per unit velocitygradient. The value of gives a measure of the

resistance of a substance to fluid flow. Highly vis-cous substances, such as honey, have a slow flowrate and therefore have a large resistance to shearstress, whereas less viscous fluids, such as water,have lower resistance to shear stress. Using amicroscopic approach, an expression for viscositycan be calculated (Eq. 1.42 and Derivation 1.9).The quantity has a similar form to the diffusioncoefficient (compare Eq. 1.42 with Eq. 1.37).

Shear stress: (Derivation 1.9)

Pzx = uxz

(1.41)

Viscosity:

= 1

3npm v (1.42)

1.5.3 Thermal equilibrium: heat

transport

The final equilibrium well consider is thermalequilibrium, which acts to equalise temperaturedifferences. This is achieved by energy transfer

heat flowdown temperature gradients. Thereare three types of heat transfer: conduction, con-vection and radiation. (See Section 4.6.4 for ra-diation.)

Conduction

Conduction is the transfer of heat through mat-ter, from particle to particle. Higher energy par-ticles transfer some of their energy to lower ener-getic particles via collisions, which results in a net

heat flow from hotter to cooler regions. Equilib-rium is achieved when the temperature is uniformthroughout the substance. Conduction occurs insolids, liquids and, to a lesser extent, in gases.

The one-dimensional heat flux, jz [Jm2s1],

is defined as the heat flow per unit area per unittime. This obeys a similar differential equationto those for particle flux and shear stress, wherethe heat flux is proportional to the temperaturegradient:

jz = T

z

The coefficient of proportionality, [Wm1K1],is called the thermal conductivity, and it hasa similar form to the diffusion coefficient fromEq. 1.37 and viscosity from Eq. 1.42 (Eq. 1.45 andDerivation 1.10).

Multiplying jz by the cross-sectional area, A,we obtain the heat conduction formula, whichgives the rate of heat flow (Eq. 1.44). Draw adiagram to show whatjz andA are? Can maybe

include this in the derivation box?


24/94


Table 1.9: Derivation of viscosity coefficient (Eq. 1.42, refer to Fig. 1.8)

x-direction drift velocity, ux(z)

A molecule carries momentum px(z)=mux(z)

z-direction shear stress, Pzx

dPzx = dpx(z)

dSdt cos

= dN

dSdtmux(z)cos

= 1

4npmux(z)vFv(v)dv cos

2 sin dd

Pzx = 1

4npmux(z0

)

vFv(v)dv

=v

/2

0

cos2 sin d

=1/3

2

0

d

=2

= 1

6npmux(z0 ) v

Pzx = P+zx Pzx

= 1

6npmux(z0 ) v 1

6npmux(z0+ ) v

16

npm v

ux(z0) uxz

ux(z0) + ux

z

= 13

npm v =

uxz

= 1

3npm v

Starting point: The z-component of shear stress,dPzx, for particles incidentat an angle , is given bythe particle impulse in the xdirection, Fx = dpx/dt, perunit area dS.

The particle momentum in thex-direction is proportional tothe drift velocity per particle,ux(z):

dpx(z) =dNmux(z)

(Refer to Fig. 1.8 for expressionfor dN)

The net upward shear stress,P+zx, and downward shear

stress, Pzx, is found by in-tegrating over all angles ina hemisphere, and over allvelocities.

The total shear stress, Pzx,is therefore the difference be-tween upward and downwardcomponents. We can approxi-mate using a Binomial expan-sion about z0.

By comparing the expressionfor Pzx to Eq. 1.41, we find anexpression for the viscosity co-efficient, .


25/94


Thermal conduction: (Derivation 1.10)

jz = Tz

(1.43)

dQ

dt = AdT

dz (1.44)

Thermal conductivity:

= 1

3npC

NVv (1.45)

Thermal conductivity is a measure of a sub-stances ability to conduct heat. Materials withhigh thermal conductivity, for example silver andother metals, are good conductors of heat, mean-

ing that they rapidly transport heat and warmup throughout when in contact with a hot object,such as a flame. Conversely, materials with lowthermal conductivity, such as wood or rubber, arethermal insulators, which means that they arepoor conductors of heat.

Newtons law of cooling

For a constant temperature-gradient over a dis-

tance L, the heat conduction equation (Eq. 1.44)isdQ

dt = A (TL T0)

L

where T0 and TL are the temperatures at z =0 and z = L respectively. From the definitionof specific heat capacity (Eq. 1.22), the rate ofchange of heat can be written as

dQ

dt =mc

dT

dt

Therefore, equating these two expressions fordQ/dt, we find that the rate of change of temper-ature of an object is proportional to the tempera-ture difference between it, T(t), and its surround-ings, Ts. This statement is known as Newtonslaw of cooling (Eq. 1.46). The general solution toEq. 1.46 is an exponential decay, with time con-stant = 1/K. The exact solution depends onboundary conditions. Worked example?

Newtons law of cooling:

dT

dt = A

mcL(T(t) Ts)

K(T(t)

Ts) (1.46)

where K is a positive constant.

Convection

Convection occurs only in liquids and gases, andit involves the bulk transport of parcels of fluidor gas. Natural convection occurs when hot,less dense parcels rise, while cooler, more denseparcels sink to take their place, which are then inturn heated and rise. This repeated circulationis called a convection current. Forced convectionoccurs when a fan, pump or draft, for example,assists convection.

The differential equations governing convectionare similar to those for conduction (Eqs.1.44 and1.46). The difference is that the constant of pro-portionality is not the thermal conductivity, ,but is called the heat transfer coefficient, h. Thisis not solely a material property, but depends alsoon external properties of the flow, such as geom-etry, temperature and flow velocity. For this rea-son, h is usually an experimentally-determinedquantity.


26/94


Table 1.10: Derivation of thermal conductivity (Eq. 1.45, refer to Fig. 1.8)

djz = dQ(z)

dSdt cos

= dN

dSdtq(z)cos

=

1

4 npq(z)vFv(v)dv cos2

sin dd

jz = 1

4npq(z0 )

vFv(v)dv

=v

/20

cos2 sin d

=1/3

20

d

=2

= 1

6npq(z0 ) v

jz = j+z jz

= 1

6 npq(z0 ) v 1

6 npq(z0+ ) v 1

6np v

q(z0) q

z

q(z0) +

q

z

= 13

np v qz

= 13

np v qT

T

z

= 13

np CNVv

=

T

z

= 13 np CNVv

Starting point: The z-component of heat flux,djz, for particles incident at anangle , is given by the heatgradient, dQ(z), per unit areadS, per unit time dt.

Write the heat gradient interms of the heat gradient perparticle, q(z):

dQ(z) =dN q(z)

(Refer to Fig. 1.8 for expressionfor dN)

The net upward heat flux, j+z ,and downward heat flux, jz ,are found by integrating overall angles in a hemisphere, andover all particle velocities.

The total heat flux, jz, is there-fore the difference between up-ward and downward compo-nents. We can approximate us-ing a Binomial expansion aboutz0.

We can express the heat gradi-ent as a temperature gradient:

q

z =

q

T

T

z =CNV

T

z

where CNV is the constant vol-ume heat capacity per particle.

By comparing the expressionfor jz to Eq. 1.43, we find anexpression for the thermal con-ductivity, .


27/94

Chapter 2

Classical Thermodynamics

Classical thermodynamics is concerned withthe interchange of matter and energy between asystem and its surroundings, and how the sys-tem responds to such interchanges. The birth ofthe subject coincided with the start of the indus-

trial revolution around the start of the nineteenthcentury, motivated by the interest to develop me-chanical power and engines from heat, such as thesteam engine. At that time, Dalton was only be-ginning to propose the atomic structure of mat-ter. Therefore, whereas kinetic theory uses themicroscopic properties of individual particles toexplain the macroscopic properties of a system,classical thermodynamics treats systems exclu-sively on a macroscopic scale. Although thermo-dynamics is more general, and handles systems

other than ideal gases, there is good agreementbetween the two descriptions.

2.1 Thermodynamic systems

Classical thermodynamics is based on four laws,known collectively as the Laws of Thermodynam-ics. Before presenting these laws in Section 2.2,we will first provide an overview of the terminol-ogy of thermodynamics.

A thermodynamic system is defined by threecomponents: the system, the surroundings, andthe boundary between them (Fig. 2.1). The sys-tem is the part whose properties we are inter-ested in studying, for example an ideal gas ina cylinder with a piston; the surroundings in-volve everything else. The system can interactwith its surroundings by exchanging matter or en-ergy across the boundary. Thermodynamic sys-tems fall into one of three categories dependingon what can be exchanged between the system

and its surroundings: (i) in an open system, both

System

Surroundings

Boundary

Interchange of energy andmatter across boundary

Figure 2.1: Schematic of a thermodynamic sys-tem. In an open system, both matter and en-ergy can be exchanged between the system and

the surroundings; in a closed system only energyexchange is possible; and in an isolated system,neither matter nor energy is exchanged.

matter and energy can be exchanged with the sur-roundings, (ii) in aclosed system, only energy canbe exchanged, therefore the number of particleswithin the system remains constant, and (iii) inan isolated system, the system is totally isolatedfrom its surroundings, meaning that neither mat-

ter nor energy can be exchanged. In this chapter,well deal primarily with closed systems. In Sec-tion 2.3.4, we will look briefly at the thermody-namics of open systems, since this is importantfor understanding phase transitions.

Heat, work and internal energy

Recall from kinetic theory that the internal en-ergy, U [J], of a system is the sum of kineticand potential energies of the particles that make

it up. For example, the kinetic energy includes

27


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28 CHAPTER 2. CLASSICAL THERMODYNAMICS

translational, rotational and vibrational motion,whereas the potential energy includes chemicalenergy stored in atomic bonds, nuclear energystored in nuclear bonds, and latent energy associ-ated with the state of matter (solid, liquid or gas).Therefore, the internal energy is the sum of allenergies within the system, but does not includethe energy the system possesses as a function ofits overall motion or position. For example, thekinetic energy of a ball rolling down a hill is notconsidered as internal energy.

The combination of kinetic energy and latentenergy is often called thermal energy. The tem-perature, T [K], of a system provides a measureof the average kinetic energy of its particles; thegreater the average molecular kinetic energy, the

higher the systems temperature.The internal energy of a system changes if thereis a net exchange of energy with its surround-ings. The two ways in which energy is exchangedacross a boundary are via heat flow, Q [J], orwork done, W [J]. These are both processes ofenergy transfer from one body to another. Heatflow refers specifically to the transfer of thermalenergy between two bodies at different tempera-tures, whereas work is the general term to de-scribe energy transfer ofany nature, other than

heat. Since classical thermodynamics was devel-oped before the microscopic behaviour of parti-cles within a system was understood, the internalenergy of a system is usually understood in termsof its change with heat and work inputs and out-puts, as described by the First Law of Thermo-dynamics (see Section 2.2.2).

Thermodynamic equilibrium

An important concept in thermodynamics is that

of equilibrium. A system is in thermodynamicequilibriumwith its surroundings when its macro-scopic properties, such as temperature, pressure,and volume, are not changing with time. In clas-sical thermodynamics, we deal exclusively withsystems in equilibrium. To describe systems thatarent in equilibrium, we require a microscopicapproach, such as kinetic theory. If a system isin thermodynamic equilibrium, then it is simul-taneously in (i)mechanicalequilibrium (constantpressure) (ii)chemicalequilibrium (constant par-

ticle concentration), and (iii)thermalequilibrium

(constant temperature). Thermal equilibrium isachieved when a system and its surrounding arein thermal contact, which means that heat canflow between them. A boundary that allowstwo systems to be in thermal contact is called adiathermal boundary, whereas a boundary thatthermally isolates two systems, such that heatdoes not flow between them, is called anadiather-mal boundary.

Thermodynamic states

The thermodynamic stateof a system is a broadterm that encompasses all of the macroscopicproperties that define it. These properties arecalled state variables. For example, volume, tem-perature and internal energy are state variables,

whereas heat and work are not. State variablesare often related to one another with an equa-tion of state. For example, the ideal gas law,

pV = nRT, is the equation of state that relatespressure, volume and temperature. A state vari-able that depends on the physical size of the sys-tem, such as volume, is called an extensive vari-able, whereas one that is independent of the sys-tems size, such as temperature and pressure, isanintensivevariable. Said another way, an inten-sive variable has the same value whether we are

looking at the whole system, or only at a part ofthe system, but this is not true for an extensivevariable.

Thermodynamic processes

When a thermodynamic system changes state, wesay that it undergoes a thermodynamic process.Processes are classified as either reversible or ir-reversible. In irreversible processes, energy is per-manently lost from the system, due to dissipative

forces such as friction. Irreversible systems ex-hibit hysteresis, namely the change of state de-pends on the path taken. Since the dissipatedenergy cannot be recovered by changing the sys-tem back to its original state, then the processis irreversible. Conversely, reversible processesdo not exhibit hysteresis, and the original stateof the system can be recovered by reversing theoperations on the system. In practice, perfectlyreversible processes are idealisations. However,they can be achieved by carrying out the process

in infinitesimal steps, with each step performed


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2.2. THE LAWS OF THERMODYNAMICS 29

over a time period much slower than the responsetime between the system and surroundings. Thisensures that they are always in thermal equilib-rium with each other, and therefore the interme-diate states of the system correspond to definitevaluesthat is, ones that can be known throughmeasurementof its macroscopic properties. Aprocess carried out in this way, through a seriesof equilibrium states, is called a quasi-staticpro-cess.

Since the intermediate states are, in theory,known, reversible and quasi-static processes canbe plotted on p-V diagrams. From this, we canintegrate the curve to determine total changes.In contrast, only the initial and final states of ir-reversible processes may be plotted since they do

not proceed via equilibrium states, and thereforethe integration path between them is unknown(Fig. 2.2).

In this Chapter, we will deal primarily with re-versible processes since they are easier to treatmathematically, and also because they representthe idealised experimental outcomes since energyis not irrecoverably lost from the system. Themost common reversible gaseous processes aresummarised in Table 2.1. Should this be a newsubsubsection, with more description?

2.2 The Laws of Thermodynam-

ics

Classical thermodynamics is concerned with thechanges in a systems macroscopic properties, andits energy inputs and outputs, when it undergoesa thermodynamic process (a change of state). Inpractical terms, it deals with quantifying the ex-traction of useful work from a system during achange in state. The fundamental results of clas-

sical thermodynamics are encapsulated in fourlaws, known as the Laws of Thermodynamics.These are summarised in Table 2.2. Below, wewill look at each in turn.

2.2.1 The Zeroth Law

When a hot and a cold system are put in thermalcontact with each other, heat flows from the hotto the cold system. This decreases the tempera-ture of the hot system, whilst increasing that of

the cold one. When the two systems are at the

same temperature, then no more heat flows andthe systems are said to be in thermal equilibrium.This is the basis of the Zeroth Law of Thermo-dynamics (Table 2.2). The Zeroth Law impliesthat systems in thermal equilibrium are at thesame temperature as each other, and that theirtemperature remains constant in time. Histori-cally, the Zeroth Law was introduced after theother three laws. However, it is considered morefundamental than the others and was thereforelabelled the Zeroth Law rather than the FourthLaw. It provides the basis for the definition oftemperature in terms of thermal equilibrium.

2.2.2 The First Law

The First Law of Thermodynamics(Eq. 2.5) is a

statement of the conservation of energy. The lawis often stated in terms of an incremental changeof internal energy, dU (Eq. 2.6), rather than thenet change, U (Eq. 2.5). The incremental formis more useful because we can find expressions forthe elementsdU,dQ and dW in terms of changesin the systems state variables, such as volumeand temperature changes, as shown below. How-ever, the expressions derived below only hold forreversible processes; they do not hold for irre-versible processes because we do not know how

much energy is lost to the environment.

Work done, dW

Gases and liquids exert pressurea force per unitareaon the walls of their container. From amolecular viewpoint, this pressure results fromthe change in momentum of molecules collidingagainst the container walls. When the systemsvolume decreases, work must therefore be doneagainst these molecular forces. (Recall that for

non-dissipative processes, the work done over adistance,L, is related to force byW = L Fdl.)By considering the work done,W, on a system bycompressing it with a piston, we find an expres-sion for W in terms of the pressure and changein volume of the system (Eqs. 2.82.9 and Deriva-tion 2.3). Work is doneon the system if its vol-ume decreases (dV is negative), and work is doneby the system if its volume increases (dV is pos-itive). Equation 2.9 also shows that (i) no workis done on or by the system if its volume remains

constant, and (ii) the area under a p-V curve of a


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p

V

p

V

p

VReversible Quasistatic Irreversible

Figure 2.2: p-V diagrams of reversible, quasi-static and irreversible processes, occurring through twoequilibrium states, A and B . Show A andB on diagram.

Table 2.1: Common reversible processes for ideal gases (where pV =nRT):

Isothermal: constant temperature

T = 0

pV = constant (2.1)

Isochoric: constant volume

V = 0

p

T = constant (2.2)

Isobaric: constant pressure

p = 0

V

T = constant (2.3)

Adiabatic: no heat exchange

Q = 0

pV = constant (2.4)

p

V

Isobaric

Isothermal

AdiabaticIsochoric

REDO. Needs more explanation. eg adiabaticstuff


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Table 2.2: The Laws of Thermodynamics:

Zeroth Law: Thermal equilibriumIf two thermodynamic systems are at the same time in thermal equilibrium with a third system,then they are also in thermal equilibrium with each other.

First Law: Conservation of internal energy

The change in internal energy of a thermodynamic system is given by the sum of work and heatexchanged with its surroundings:

U = Q + W (2.5)

dU = dQ + dW (2.6)

Conventions:: net change; : incremental change+Q: heat into the system;Q: heat out of the system+W: work done on the system;W: work done bythe system

Second Law: Law of increasing entropyNo process is possible where the total entropy of the universe decreases. The total change inentropy always greater than or equal to zero:

Reversible process: S = 0

Irreversible process: S > 0

S 0 always (2.7)

Third Law: Absolute zero temperature

The entropy of a system approaches zero as its temperature approaches absolute zero.


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reversible process represents the total work doneduring the change.

Work done for a reversible process:(Derivation 2.3)

dW = p dV (2.8)

W = V2V1

p dV (2.9)

Heat exchange, dQ

The heat capacity of a system is defined as

C=Q

T

whereis the variable that is kept constant whileheat is added. For example, for an isobaric pro-cess, p, whereas for an isochoric process, V. Rearranging, we can find an expressionfor an element of heat flow into or out of a sys-tem (Eq. 2.10). An increase in temperature corre-sponds to heat flowingintothe system, whereas adecrease in temperature corresponds to heat flow-

ing out of the system. For reversible processes,the total heat input or output is equal to the inte-gral of heat capacity over temperature, betweenthe initial and final temperature limits (Eq. 2.11).

Heat exchange for a reversible process:

dQ = C dT (2.10)

Q =

T2T1

C dT (2.11)

Internal energy change, dU

The internal energy change is found by combiningthe expressions for dW and dQ using the FirstLaw of Thermodynamics, as shown in Deriva-tion 2.4 (Eqs. 2.122.13). Even though these ex-pressions are derived for an isochoric process,Eq. 2.12 is true for all ideal gas processes, even

when the volume is not kept constant. It shows

that the internal energy of an ideal gas dependsonly on its temperature. (For real gases, the in-ternal energy depends on its volume as well, seeSection??).

Internal energy for a reversible process:(Derivation 2.4)

dU = CVdT (2.12)

U =

T2T1

CVdT (2.13)

Using the First Law

Worked example 2.3 uses the above expressions tocalculate W, Q and U for reversible processeswith ideal gases. The First Law can also be usedto derive the adiabatic relationship, pV = con-stant, as shown in Derivation 2.5.

2.2.3 The Second Law

The Second Law of Thermodynamics introducesanother function of state, called the entropy, S[JK1]. It states that, for any process within a

closed system, such as the universe, entropy can-not decrease. Entropy is often interpreted as thedegree of disorder of a system. For example, itcan be thought of as a measure of the numberof different ways that quanta of energy in a sys-tem can be arranged between the particles thatmake it up, without changing the total energy ofthe system. A systems entropy changes if heatis transferred into or out from it. The changein entropy when a reversible quantity of heat,dQrev , flows into a system at temperature T, is

defined by Eq. 2.14. This definition follows fromtheClausius inequality, which is described below.

Entropy change for a reversible process:

dS = dQrev

T (2.14)

Whereas the First Law puts restrictions onwhat processes areenergeticallypossible, the Sec-

ond Law tells us the direction in which they can


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Table 2.3: Derivation of work done for a reversible process (Eqs. 2.82.9)

dx

F=pA

dW = F dx= pA dx=

p dV

W = V2V1

p(V) dV

Starting point: Consider the work done onthe gas in a cylinder by moving a pistonthrough a length dx.

The opposing force results from the sys-tems pressure, where F =pA

The element of volume is dV =A dx For reversible processes, the integration

path is defined, therefore we can determinethe total work done by integrating from V1to V2.

Table 2.4: Derivation of internal energy change for a reversible process (Eqs. 2.122.13)

dU = dQ + dW

= dQ

= CVdT

U =

T2T1

CV dT

Starting point: From the First Law of Ther-modynamics, the change in internal energyis the sum of heat in and work done on asystem.

Consider an isochoric process, where dV =0 (V = constant). From Eqs. 2.8 and 2.10,

dW = 0and dQ = CV dT

For reversible processes, the integrationpath is defined, therefore we can determinethe total internal energy change by integrat-ing over temperature, between the initialand final values


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Figure 2.3: Worked example: applying the First Law of Thermodynamics to various re-versible processesAlso include pV diagrams to show the processes, work done, and heatin or out.

Isothermal(dT= 0):

U = CVT = 0

W = V2V1

nRT1V

dV

= nRT1ln

V1V2

Q = W

= nRT1ln

V1V2

Isochoric(dV= 0):

U = CV(T2 T1)

W = pV = 0

Q = U

= CV(T2 T1)

Isobaric (dp= 0):

U = CV(T2 T1)

W = p1(V2 V1)

Q = U W= CV(T2 T1)

+p1(V2 V1) Cp(T2 T1)

Adiabatic: (pV =A)

U = CV(T2 T1)

W = AV2V1

dV

V

= A

( 1)

V2V2

V1V1

= (p2V2 p1V1)

(

1)

CV(T2 T1)

Q = 0

Table 2.5: Derivation of the adiabatic p-V relationship (Eq. 2.4) Include more on theadiabatic condition in earlier section, and reference equation

dU = dW

CVdT = pdV= nRT

V dV

= (Cp CV)TV

dV

CV

T2T1

dT

T = CV( 1)

V2V1

dV

V

ln(T2/T1) = ( 1)ln(V2/V1)

T2T1

=

V2V1

(1)T V(1) = constant

pV = constant

Starting point: For an adiabatic process,the heat exchange is zero, dQ = 0, there-fore, from the First Law, dU =dW.

Use Eqs.2.8 and 2.12 for expressions fordWanddU.

The difference and ratio of heat capacitiesare:

nR = Cp CV =

CpCV

(see Section 1.4.2)

Use the ideal gas law to eliminate T fromT V(1).


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occur. For example, we know from experiencethat a quantity of heat, Q, can spontaneouslyflow from a hot to a cold body. Although theFirst Law allows the reverse process to happen,namely heat Q to flow from a cold body to ahot body, the Second Law states that it cannothappen spontaneously since, from Eq. 2.14, thedecrease in entropy of the cool object would begreater than the increase in entropy of the hot ob-

ject, since Tcold < Thot. This would result in a netdecrease of entropy of the combined system (anduniverse), which violates the Second Law. Saidanother way, the natural direction for change istowards disorder.1

Heat engines and heat pumps

The Second Law of Thermodynamics describesthe operation and efficiencies of heat engines andheat pumps. Heatenginesare machines that con-tinuously consume heat to generate useful work.For example, a steam engine that drives a trainis a heat engine. Heat pumpsare machines thatcontinuously transfer heat from a cold object toa hot object, requiring work to be done on thepump since this is against the direction of spon-taneous heat flow. Refrigerators and air condi-

tioners are examples of heat pumps. Both heatengines and heat pumps involve cyclic processes,and therefore energy conversion is continuous.

Kelvin and Clausius found that a perfect heatengine, which converts all the consumed heatinto useful work, and a perfect heat pump, whichtransfers heat from a cool to hot object withoutusing work, were impossible to construct. Thereason is that both devices violate the SecondLawthe total entropy of the universe decreasesin both cases. The Second Law is therefore often

stated in terms of the operation of heat enginesand heat pumps, and makes no mention of en-tropy. These are known as the Kelvin and Clau-sius formulations of the Second Law of Thermo-dynamics:

Kelvins formulation: The complete con-version of heat into work is impossible; it isimpossible to construct a perfect heat engine.

1This issue raises some interesting philosophical ques-tions regarding the ultimate fate of the universe; for a good

discussion, refer to (refer to somewhere).

Hot reservoir

Cold reservoir

TC

TH

QH

QC

WHeat

Engine

Hot reservoir

Cold reservoir

TC

TH

QH

QC

WHeat

Pump

Figure 2.4: Schematic representations of heat en-gines (left) and heat pumps (right) operating be-tween two reservoirs at constant temperatures,TH and TC.

Clausius formulation: Heat cannot spon-taneously flow from a cold to a hot object;it is impossible to construct a perfect heatpump.

These statements are equivalent to one another,and to the entropy formulation of the SecondLaw.

The Second Law is satisfied for a heat engine ifsome of the consumed heat is lost to the environ-ment. For a heat pump, work must be provided totransport the heat against its natural direction offlow. These operations are shown schematicallyin Fig. 2.4. We consider a heat engine or pumpoperating between two large reservoirs, of con-stant temperature TH and TC. The heat flow toor from each reservoir, QH and QC, is small incomparison with the size of the reservoirs, there-fore their temperatures remain constant. Fromthe conservation of energy, we have

QH=QC+ W

both for heat engines and heat pumps. Figure 2.4illustrates that a heat pump can be thought of asa heat engine operating in reverse.

Efficiency and coefficient of performance

The efficiency of these machines is a measure ofthe useful energy output to the energy input. For

a heat engine, the useful energy output is W for


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an input energy, QH. Therefore the efficiency isgiven by Eq. 2.15, where W = QH QC. Forheat pumps, the useful quantity isQHfor a heat-ing device, and QC for a cooling device, for aninput energy of W (Eqs. 2.16 and 2.17 respec-tively). Since these ratios can be greater thanone, the term coefficient of performance, or CoP,is used instead of efficiency.

Efficiency and coefficients of performance:Heat engine efficiency:

= W

QH= 1

QCQH (2.15)

Heat pump coefficients of performance (CoP):

Heating: CoPH = QH

W (2.16)

Cooling: CoPC = QC

W (2.17)

Carnots theorem

Reversible engines do not dissipate heat energyvia friction. Carnots theorem states that re-versible heat engines are the most efficient, as

demonstrated in Proof 2.6. This is intuitively ob-vious, since if heat is lost via friction in a heatengine, then less energy is available to do work.A reversible heat engine follows what is called aCarnot cycle. The details of its operation areshown in worked example 2.5. A corollary ofCarnots theorem is that all reversible enginesoperating between the same two temperatureshave the same efficiency, which depends only onthe temperatures of the reservoirs (Eq. 2.18 andworked example 2.5). Equation 2.18 shows that

the efficiency of the engine is maximised by oper-ating at the lowest and highest possible tempera-tures of the cold and hot reservoirs, respectively.

Reversible heat engines (Carnot cycle):(refer to worked example 2.5)QHTH

=QCTC

R = 1 TC

TH(2.18)

In practice, no engine is perfectly frictionless,but it can be realised to a good approximationusing quasi-static processes. By considering theoperation of reversible engines, we can see wherethe definition of entropy as dS = dQ/T comesfrom, as well as the equivalence between the heatengine and the entropy formulations of the Sec-ond Law, as demonstrated below.

Clausius inequality and entropy

Another corollary of Carnots theorem is theClausius inequality, which shows that the ratioof heat entering the system to the temperatureat the point of heat entry, when integrated overa complete cycle, is less than or equal to zero(Eq. 2.19 and Derivation 2.7). If we define the

quantity dQ/Tas an element of entropy change,dS, then using this in the Clausius equality, wesee that entropy change can never be negative,implying that the entropy of a closed system can-not decrease (Eq. 2.20 and Derivation 2.8). Thisillustrates that the heat engine formulations ofthe Second Law are equivalent to the entropy for-mulation.

Clausius inequality and entropy:

(Derivations 2.7 and 2.8) dQ

T 0 (2.19)

S 0 (2.20)

2.2.4 The Third Law

The Third Law of Thermodynamics (Table 2.2)

provides a reference point for an absoluteentropyscale, rather than considering onlychangesin en-tropy. The reference point corresponds to abso-lute zero temperature, 0 K. An equivalent formu-lation of the Third Law is that the specific heatcapacity of all materials equals zero at absolutezero:

CV = dQ

dT (2.21)

= TdS

dT


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Table 2.6: Proof that reversible engines are the most efficient

Hot reservoir

Cold reservoir

TC

TH

Q`H

Q`C

WIrreversible

engine

QH

QC

Reversibleengine

Compositeengine

Starting point: Consider the composite en-gine in the adjacent diagram. The variablesW, Q

H,C and Q

H,Care positive in the di-

rections indicated. From Clausius formula-tion of the Second Law, there is a net flowof heat away from the hot reservoir:

QH QH 0QH QH

W

QH W

QH

Using the definition of efficiency from

Eq. 2.15:

where is the efficiency of the irreversibleengine and is that of the reversible engine.Therefore, reversible engines are the mostefficient:

R I


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Figure 2.5: Worked example: Reversible Carnot engine

p

V

1

2

3

4Qin

Qout

Work

V1TH

TC

V2

V3

V4

The Carnot engine is a cyclic, reversible en-gine, which means that it returns to its orig-inal state after one complete cycle. The de-tails of each process are described in Steps 14 opposite, which correspond to the strokes14 in the diagram above (refer to workedexample 2.3 for details). Other properties ofthe cycle, such as the work done per cycle andthe efficiency, are calculated below. The totalwork done per cycle is given by the enclosedarea. Is this layout clear?

Equating the adiabatic condition inSteps 2 and 4:

V3V2

1=

V4V1

1

V2V1

= V3

V4

Total work donebythe system (enclosedarea):

W = W1+ W2+ W3+ W4

= nR ln

V2V1

(TH TC)

Efficiency: depends only on the temper-atures of the reservoirs

= 1

QoutQin

= 1 TCTH

Step 1. Isothermal expansion atTH: (heatinto the system and work done by

the system)

U = 0

Qin = W1

=

V2V1

pdV

= nRTHln

V2V1

Step 2. Adiabatic expansion from TH toTC:

Q = 0

W2 = U

= CV(TH TC)

and TH

TC=

V3V2

1

Step 3. Isothermal compression at TC:(heat out of the system and workdone on the system)

U = 0

Qout = W3

=

V4V3

pdV

= nRTCln

V4V3

= nRTClnV3V4

Step 4. Adiabatic compression from TC toTH:

Q = 0

W4 = U

= CV(TH TC)

and THTC= V4V1

1


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Table 2.7: Derivation of the Clausius inequality (Eq. 2.19)

R I1 QCQHR 1 QCQHIQCQH

I

QCQH

R

TCTH

QCTC

QHTH

For positive QH and negative QC:

QHTH

+QCTC

0

ncycle

Q

T 0

Therefore, the Clausius inequality is: dQ

T 0

Starting point: From Carnots theorem, re-versible engines are more efficient than ir-reversible engines (Proof 2.6).

Use Eq. 2.15 for the efficiency of heat en-gines, and Eq. 2.18 for the efficiency of areversible engine.

If we take the heat entering the system,QH,to be positive, thenQC, leaving the system,is negative.

Generalising to a cycle with n processesduring which heat enters or leaves the sys-tem, we can write the inequality as a sumover the cycle.

In the limit where an element of heatdQen-ters the system at a temperature T, the sumbecomes an integralClausius inequality.


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Table 2.8: Derivation of the entropy for

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