Physics 304
Thermodynamics Notes
An Introduction to Thermal Physics, D. V. Schroeder, Addison Wesley Longman, 2000
© J Kiefer 2009
First “Law” of Thermodynamics ................................................................................ 2
A. Temperature Sec 1.1 .............................................................................................. 2
1. Energy ................................................................................................................. 2
2. Thermal Equilibrium ........................................................................................... 2
3. Thermometers ..................................................................................................... 4
B. Work ....................................................................................................................... 4
1. First “Law” of Thermodynamics Sec 1.4 .......................................................... 4
2. Compressive Work Sec 1.5 ................................................................................ 5
3. Other Works ........................................................................................................ 8
C. Heat Capacity Sec 1.6 .......................................................................................... 10
1. Changing Temperature...................................................................................... 10
2. Heat Capacity and Degrees of Freedom ........................................................... 11
II. Second “Law” of Thermodynamics .......................................................................... 14
A. Combinatorics ....................................................................................................... 14
1. Two State Systems Sec 2.1 ............................................................................. 14
2. Einstein Solid Sec 2.2, 2.3 .............................................................................. 15
B. Entropy.................................................................................................................. 17
1. Large Systems Sec 2.4 .................................................................................... 17
2. Second “Law” Sec 2.6 .................................................................................... 18
C. Creating Entropy ................................................................................................... 19
1. Temperature Sec 3.1, 3.2 ................................................................................ 19
2. Pressure Sec 3.4 .............................................................................................. 21
3. Chemical Potential Sec 3.5 ............................................................................. 22
4. Expanding & Mixing Sec 2.6........................................................................... 23
III.
Processes ............................................................................................................... 25
A. Cyclic Processes.................................................................................................... 25
1. Heat Engines & Heat Pumps Sec 4.1, 4.2 ........................................................ 25
2. Otto, Diesel, & Rankine Sec 4.3 ...................................................................... 27
B. Non-cyclic Processes or Thermodynamic Potentials ............................................ 30
1. Thermodynamic Potentials Sec 5.1.................................................................. 30
2. Toward Equilibrium Sec 5.2 ............................................................................ 33
3. Phase transformations Sec 5.3 ......................................................................... 34
IV.
Statistical Mechanics ............................................................................................ 40
A. Partition Function.................................................................................................. 40
1. Boltzmann Sec 6.1 .......................................................................................... 40
2. Probability Sec 6.2 ........................................................................................... 43
3. A Couple of Applications ................................................................................. 44
B. Adding up the States Sec 1.2, 1.7, 2.5, 6.6, 6.7 ...................................................... 46
1. Two-State Paramagnet Sec 6.6 ....................................................................... 46
2. Ideal Gas Sec 6.7............................................................................................. 47
3. Thermodynamic Properties of the Ideal Monatomic Gas ................................. 49
4. Solids Sec 2.2, 3.3, 7.5 ..................................................................................... 51
5. Photons Sec 7.4 ................................................................................................ 53
6. Specific Heat (Heat Capacity) of Solids Sec. 7.5 ............................................ 55
I.
1
I.
First “Law” of Thermodynamics
A.
Temperature Sec 1.1
1.
Energy
a. Definition of energy
Energy is a fundamental physical concept. My favorite dictionary gives as its 4th
definition of energy: 4. Physics. Capacity for performing work. So, now we go to the W
section for the 13th definition of work: 13. Mech. The transference of energy by a process
involving the motion of the point of application of a force, as when there is movement
against a resisting force or when a body is given acceleration; it is measured by the
product of the force and the displacement of its point of application in the line of action.
[How about the definition of work number 10. The foam or froth caused by fermentation,
as in cider, in making vinegar, etc.]
That definition of work is adequate as a definition of mechanical work, but that definition
of the word energy is nearly useless. Of course, that’s what dictionaries do, define words
in terms of other words in endless circles—they are about usages more than meanings. A
fundamental concept cannot be defined in terms of other concepts; that is what
fundamental means.
b. Conservation of energy
We can list the forms that energy might take. In effect, we are saying that if such and
such happens in or to a system, then the energy of the system changes. There is potential
energy, which is related to the positions of parts of the system. There is kinetic energy,
which is related to the movements of parts of the system. There is rest energy, which is
related to the amount of matter in the system. There is electromagnetic energy, chemical
energy, and nuclear energy, and more.
We find that for an isolated system, the total amount of energy in the system does not
change. Within the system, energy may change from one form to another, but the total
energy of all forms is a conserved quantity. Now, if a system is not isolated from the rest
of the universe, energy may be transferred into or out of the system, so the total energy of
such a system may rise or fall.
2.
Thermal Equilibrium
a. Temperature
Consider two objects, each consisting of a very large number of atoms and/or molecules.
Here, very large means at least several multiples of Avogadro’s Number of particles. We
call such an object a macroscopic object. Consider that these two objects (they may be
two blocks of aluminum, for instance, though they need not be the same material—they
might be aluminum and wood, or anything) are isolated from the rest of the universe, but
are in contact with each other.
2
We observe that energy flows spontaneously from one block (A) to the other (B). We
say that block A has a higher temperature than block B. In fact, we say that the energy
flow occurs because the blocks have different temperatures. We further observe that after
the lapse of some time, called the relaxation time, the flow of energy from A to B ceases,
after which there is zero net transfer of energy between the blocks. At this point the two
blocks are in thermal equilibrium with each other, and we would say that they have the
same temperature.
b. Heat
The word heat refers to energy that is transferred, or energy that flows, spontaneously by
virtue of a difference in temperature. We often say heat flows into a system or out of a
system, as for instance heat flowed from block A to block B above. It is incorrect to say
that heat resides in a system, or that a system contains a certain amount of heat.
There are three mechanisms of energy transfer: conduction, convection, and radiation.
Two objects, or two systems, are said to be in contact if energy can flow from one to the
other. The most obvious example is two aluminum blocks sitting side by side, literally
touching. However, another example is the Sun and the Earth, exchanging energy by
radiation. The Sun has the higher temperature, so there is a net flow of energy from the
Sun to the Earth. The Sun and the Earth are in contact.
c. Zeroth “Law” of Thermodynamics
Two systems in thermal equilibrium with each other have the same temperature. Clearly,
if we consider three systems, A, B, & C, if A & B are in thermal equilibrium, and A & C
are in thermal equilibrium, then B & C are also in thermal equilibrium, and all three have
the same temperature.
3
3.
Thermometers
a. Temperature scales
What matter are temperature differences. We can feel that one object is hotter than
another, but we would like to have a quantitative measure of temperature. A number of
temperature scales have been devised, based on the temperature difference between two
easily recognized conditions, such as the freezing and boiling of water. Beyond that, the
definition of a degree of temperature is more or less arbitrary. The Fahrenheit scale has
180 degrees between the freezing and boiling points, while the Celsius scale has 100.
Naturally, we find 100 more convenient than 180. On the other hand, it turns out that the
freezing and boiling points of water are affected by other variables, particularly air
pressure. Perhaps some form of absolute scale would be more useful. Such a scale is the
Kelvin scale, called also the absolute temperature scale. The temperature at which the
pressure of a dilute gas at fixed volume would go to zero is called the absolute zero
temperature. Kelvin temperatures are measured up from that lowest limit. The unit of
absolute temperature is the kelvin (K), equal in size to a degree Celsius. It turns out that
0 K = -273.15 oC. [The text continues to label non-absolute temperatures with the degree
symbol: oC, etc., as does the introductory University Physics textbook. The latter also
claims that temperature intervals are labeled with the degree symbol following the letter,
as Co. That’s silly.]
b. Devices
Devices to measure temperature take advantage of a thermal property of matter—material
substances expand or contract with changes in temperature. The electrical conductivity
of numerous materials changes with temperature. In each case, the thermometer must
itself be brought into thermal equilibrium with the system, so that the system and the
thermometer are at the same temperature. We read a number from the thermometer scale,
and impute that value to the temperature of the system. There are bulb thermometers,
and bi-metallic strip thermometers, and gas thermometers, and thermometers that detect
the radiation emitted by a surface. All these must be calibrated, and all have limitations
on their accuracies and reliabilities and consistencies.
B.
Work
1.
First “Law” of Thermodynamics Sec 1.4
a. Work
Heat is defined as the spontaneous flow of energy into or out of a system caused by a
difference in temperature between the system and its surroundings, or between two
objects whose temperatures are different. Any other transfer of energy into or out of a
system is called work. Work takes many forms, moving a piston, or stirring, or running
an electrical current through a resistance. Work is the non-spontaneous transfer of
energy. Question: is lighting a Bunsen burner under a beaker of water work? The hot
gasses of the flame are in contact with the beaker, so that’s heat. But, the gases are made
hot by combustion, so that’s work.
4
b. Internal energy
There are two ways, then, that the total energy inside a system may change—heat and/or
work. We use the term internal energy for the total energy inside a system, and the
symbol U. Q and W will stand for heat and work, respectively. Energy conservation
gives us the First “Law” of Thermodynamics: U  Q  W .
Now, we have to be careful with the algebraic signs. In this case, Q is positive as the heat
entering the system, and W is positive as the work done on the system. So a positive Q
and a positive W both cause an increase of internal energy, U.
2.
Compressive Work Sec 1.5
a. PV diagrams
Consider a system enclosed in a cylinder oriented along the x-axis, with a moveable
piston at one end. The piston has a cross sectional area A in contact with the system. We
may as well imagine the system is a volume of gas, though it may be liquid or solid. A
force applied to the piston from right to left (-x direction) applies a pressure on the gas of
F
P  . If the piston is displaced a distance  x , then the work done by the force is
A
W  Fx. If the displacement is slow enough, the system can adjust so that the pressure
is uniform over the area of the piston. In that case, called quasistatic, the work becomes
W  PAx  PV .
Now it is quite possible, even likely, that the pressure will change as the volume changes.
So we imagine the compression (or expansion) occurring in infinitesimal steps, in which
case the work becomes an integral:
Vf
W    P(V )dV
Vi
Naturally, to carry out the integral, we need to have a specific functional form for P(V).
On a PV diagram, then, the work is the area under the P(V) curve. In addition, the P(V)
curve is traversed in a particular direction—compression or expansion, so the work will
5
be positive or negative accordingly. Notice over a closed path on a PV diagram the work
is not necessarily zero.
b. Internal energy of the Ideal Gas Sec 1.2
As an example of computing compressive work, consider an ideal gas. But first, we need
to address the issue of the internal energy of an ideal gas. Begin with the empirical
equation of state: PV  NkT  nRT . In an ideal gas, the particles do not interact with
each other. They interact with the walls of a container only to collide elastically with
them. The Boltzmann Constant is k  1.381  1023 J / K , while the Gas Constant is
R  8.315 J / mol  K ; N is the number of particles in the system, n is the number of
moles in the system.
On the microscopic level, the atoms of the gas collide from time to time with the walls of
the container. Let us consider a single atom, as shown in the figure above. It collides
elastically with the wall of the container and experiences a change in momentum
mvx  m2vx . That is, the wall exerts a force on the atom in the minus x direction. The
atom exerts an equal and opposite force on the wall in the positive x direction. The timem2v x
averaged force exerted by a single atom on the wall is Fx 
. Now, in order that the
t
atom collides with the wall only once during t , we set t  2 L , whence the force
vx
2
x
becomes Fx  m v . Next, the pressure, P, is the force averaged over the area of the wall:
L
6
P
mvx2 mvx2 , where V is the volume of the container. Normally, there are many atoms

A L
V
in the gas, say N of them. Each atom has a different velocity. Therefore,
N
P
i 1
 
m vx2
V
i

mN  vx2  , where  v 2  is the square of the x-component of velocity,
x
V
averaged over the N atoms. Finally, we invoke the ideal gas law, PV  NkT .
1
1
PV  Nm  vx2   NkT , whence m  vx2   kT .
2
2
3
The translational kinetic energy of one atom is  K trans   kT , since
2
1
 K trans   m vx2    v y2    vz2   .
2
For an ideal gas of spherical particles (having no internal structure) there is no potential
energy and the internal energy is just the kinetic energy.
1
3
U  NfkT  NkT
2
2
c. Isothermal & adiabatic processes
Imagine a system in thermal contact with its environment. The environment is much
larger than the system of interest, so that heat flow into or out of the system has no effect
on the temperature of the environment. We speak of the system being in contact with a
heat bath so that no matter what happens to the system, its temperature remains constant.
If such a system is compressed slowly enough, its temperature is unchanged during the
compression. The system is compressed isothermally. As an example, consider an ideal
gas:
dU  Q  W
3

 NkT   Q  W
2

T  0
Vf
Vf
dV
V
Vi
0  Q  W  Q   PdV  Q  NkT 
Vi
Q  NkT ln
Vf
Vi
On the other hand, the compression may be so fast that no heat is exchanged with the
environment (or, the system is isolated from the environment) so that Q = 0. Such a
process is adiabatic. Naturally, the temperature of the system will increase. Staying with
the example of an ideal gas,
7
dU   PdV
3
NkT
NkdT   PdV  
dV
2
V
3 dT
dV

2 T
V
Vf
3 Tf
ln
  ln
2 Ti
Vi
3
3
V f T f 2  Vi Ti 2
[Notice that the text uses f for final and for degrees of freedom.]
5
5
Substituting for T with the ideal gas law, we can write V f 2 Pf  Vi 2 Pi . That exponent of
V is called the adiabatic exponent,  . In general,  
f 2
.
2
An isotherm is a curve of constant temperature, T, on the PV diagram. An arrow
indicates the direction that the system is changing with time. For an ideal gas, an
1
isotherm is parabolic, since P  ; that’s a special case. A curve along which Q = 0 is
V
called an adiabat.
3.
Other Works
In our discussion of energy conservation, we spoke of work as being any energy flow into
or out of the system that was not heat. We spoke of compressive work (sometimes called
piston work) and “all other forms of work.” The all other forms of work included stirring
(called shaft work) and combustion and electrical currents and friction. It would also
include any work done by external forces beyond the compressive work, particularly
work done by the force of gravity. We also have been assuming that the center of mass
8
of the system is not moving, so there was no kinetic energy associated with translation of
the entire system. A general form of the First “Law” of Thermodynamics ought to
include all the energy of the system, not only its internal energy. Thus for instance, the
1
total energy of a system might be E  U  MVcm2  MgZ cm .
2
a. Steady flow process
We might consider a situation in which a fluid is flowing steadily without friction, but
with heat flow into the fluid and a change in elevation and changes in volume and
pressure and some stirring.
E  Q  W
U  E K  E P  Q  Wcompress  Wother


1
U 2  U1  M Vcm2 , 2  Vcm2 ,1  Mg Z 2  Z1   Q  P1V1  P2V2  Wshaft
2

 

1
1
U 2  MVcm2 , 2  MgZ 2  P2V2   U1  MVcm2 ,1  MgZ1  P1V1   Q  Wshaft
2
2

 

In engineering real devices, all the various sources of work have to be taken into account.
In any specific device, some works can be neglected and other works not.
b. The turbine
In a turbine, a fluid flows through a pipe or tube so quickly that Q = 0, and normally the
entry and exit heights are virtually the same. In the case of an electrical generator, the
moving fluid turns a fan, so that the shaft work is negative. The energy balance equation
for a volume element of the fluid having a mass, M, would look something like this:
9

 

1
1
U exit  MVcm2 , exit  PexitVexit   U entry  MVcm2 , entry  PentryVentry   Wshaft
2
2

 

1
Wshaft  U  M Vcm2 , exit  Vcm2 , entry  PexitVexit  PentryVentry
2
An equation like this tells us how to design our turbine to maximize the shaft work.

 

c. Bernoulli’s Equation
Suppose both Q and Wshaft are zero.

 

1
1
U exit  MVcm2 , exit  MgZ exit  PexitVexit   U entry  MVcm2 , entry  MgZ entry  PentryVentry   0
2
2

 

1
U  MVcm2  MgZ  PV  constant
2
If the fluid is incompressible, then the volume is constant, and we can divide through by
V to obtain Bernoulli’s Equation. The internal energy is also constant because Q = 0 and
no compressive work is done. [  is the mass density of the fluid.]
1
P  Vcm2  gZ  constant
2
C.
Heat Capacity Sec 1.6
1.
Changing Temperature
a. Definition
By definition, the heat capacity of an object is C 
Q
. The specific heat capacity is the
T
C
. This definition is not specific enough, however,
m
since Q  U  W . A heat capacity could be computed for any combination of
conditions—constant V, constant P, constant P & V, etc.
heat capacity per unit mass, c 
b. Constant pressure heat capacity
If pressure is constant, then
 U  W 
 U   PV  
 U 
 V 
CP  
  
  
  P
 .
T
 T  P 
P
 T  P
 T  P
10
The second term on the right is the energy expended to expand the system, rather than
increase the temperature.
c. Constant volume heat capacity
 U  W 
 U 
 U 
CV  
  
  
 , since V  0 .
 T V  T V
 T V
2.
Heat Capacity and Degrees of Freedom
a. Degrees of freedom
A degree of freedom is essentially a variable whose value may change. In the case of a
physical system, the positions of the particles that comprise the system are degrees of
freedom. For a single particle in 3-dimensional space, there are three degrees of freedom.
Three coordinates are required to specify its location. We are particularly interested in
variables that determine the energy of the system—the velocities determine the kinetic
energy, the positions determine the potential energy, etc. In other words, we expect to
associate some kinetic energy and some potential energy with each degree of freedom.
In effect, this text treats the kinetic and potential energies as degrees of freedom. An
isolated single particle, having no internal structure, but able to move in threedimensional space, has three degrees of freedom which may have energy associated with
them: the three components of its velocity. Since the particle is not interacting with any
other particle, we do not count its position coordinates as degrees of freedom. On the
other hand, a three-dimensional harmonic oscillator has potential energy as well as
kinetic energy, so it has 6 degrees of freedom. Molecules in a gas have more degrees of
freedom than simple spherical particles. A molecule can rotate as well as translate and its
constituent parts can vibrate. A water molecule is comprised of three atoms, arranged in
the shape of a triangle. The molecule can translate in three dimensions, and rotate around
three different axes. That’s 3+3 = 6 degrees of freedom for an isolated water molecule.
Within the molecules, the atoms can vibrate relative to the center of mass in three distinct
ways, or modes. That’s another 2x3 = 6 degrees of freedom. Now, finally if the water
molecule is interacting with other water molecules, then there is interaction between the
molecules, and the degrees of freedom are 2x3+2x3+2x3 = 18. Notice that if we regard
the molecule as three interacting atoms, not as a rigid shape, there are 3x2x3 = 18 degrees
of freedom.
A system of N particles, such as a solid made of N harmonic oscillators, has 6 degrees of
freedom per particle for a total of 6N degrees of freedom.
The idea is that each degree of freedom, as it were, contains some energy. The total
internal energy of a system is the sum of all the energies of all the degrees of freedom.
Conversely, the amount of energy that may be transferred into a system is affected by
how many degrees of freedom the system has.
b. Exciting a degree of freedom
11
Consider a harmonic oscillator. The energy required to raise the HO from its ground
state to the first excited state is h , where h is Planck’s Constant and  is the oscillator
frequency. If the system temperature is such that kT  h , then that oscillator will never
be excited. It’s as if that degree of freedom does not exist. We say that the degree of
freedom has been frozen out.
The Equipartition Theorem says that the average energy in each quadratic degree of
1
freedom is kT . By quadratic degree of freedom, we mean that the kinetic and potential
2
energy terms all depend on the position and velocity components squared. If that is the
case, the internal energy of a system of N harmonic oscillators in a solid is
Nf
U
kT ,
2
where f is the number of degrees of freedom per particle, which we would expect to be 6.
From this, we obtain the constant volume heat capacity of Dulong-Petit,
Nfk
 U 
.
CV  
 
2
 T V
This result is independent of temperature. The measured heat capacity for a solid is not
the same for all temperatures. In fact, as temperature decreases, the heat capacity
decreases toward zero. Working backward, it would appear that as temperature
decreases, the number of degrees of freedom available decreases. At low temperature,
energy cannot be put into the degrees of freedom that have been frozen out. The reason
is that the quantity kT is smaller than the spacing between discrete energy levels. This
applies not only to a solid. For instance, one of the intramolecular vibrational modes of
the water molecule is shown here in the sketch.
Its frequency is in the neighborhood of 1014 Hz. Assuming harmonic vibration, the
energy level spacing for that mode is about h  4.136  1015 eV  sec  1 1014 Hz  0.41eV .On
the other hand, at T = 300 K, the quantity kT  8.617  10 5 eV  300K  0.03eV ; we would
K
not expect the intramolecular modes to be excitable, as it were, at 300K.
As an example of computing the constant volume heat capacity of something other than
an ideal gas, consider liquid water and ice. An effective potential energy function is
assumed to represent the interaction between water molecules. The total kinetic and
potential energy is computed at a range of temperatures, with the volume kept fixed.
12
E
. In this
T
set of molecular dynamics simulations, the intermolecular energy, E, leaves out the
kinetic energies of the hydrogen and oxygen atoms with respect to the molecular center
of mass as well as the intramolecular potential energies—that is, the degrees of freedom
of the atoms within each molecule, such as the vibrational mode shown above, are frozen
out. The results shown on the graph are CV  1.03cal/g/K for the liquid phase and
0.66cal/g/K for the solid phase. The purpose of the study was to test the effective
potential function—would it show a melting transition at the correct temperature, and
would it give correct heat capacities.
The CV is estimated by numerically evaluating the slope of the graph: CV 
CV ,liquid  1 cal/g/K ; CV , solid  0.5 cal/g/K
13
II.
Second “Law” of Thermodynamics
A.
Combinatorics
1.
Two State Systems Sec 2.1
a. Micro- and macro-states
Consider a system of three coins, as described in the text. The macrostate of this system
is described by the number of heads facing up. There are four such macrostates, labeled
0, 1, 2, & 3. We might even call these energy levels 0, 1, 2, & 3.
Specifying the orientation of each individual coin defines a microstate. We can list the
microstates, using H for heads and T for tails: TTT, HTT, THT, TTH, HHT, HTH, THH,
HHH.
Now, we sort the microstates into the macrostate energy levels.
energy level
microstates
multiplicity, 
0
TTT
1
1
HTT, THT, TTH
3
2
HHT, HTH, THH
3
3
HHH
1
The multiplicity is the number of distinct ways that a specified macrostate can be
realized. The total multiplicity of the system is the total of all the possible microstates.
For these three coins, that’s all   8 .
b. Two-state paramagnet
Consider a large number of non-interacting magnetic dipole moments, let’s say N of
them. These dipoles may point in one of only two ways: up or down. If an external
uniform magnetic field is applied, say in the up direction, each dipole will experience a
torque tending to rotate it to the up direction also. That is to say, parallel alignment with
the external field is a lower energy state than is anti-parallel alignment.
The energy of the system is characterized by the number of dipoles aligned with the
external field, q. But, we don’t care which q dipoles of the N total are in the up state.
Having q dipoles up specifies the energy macrostate, which may be realized by the
selection of any q dipoles out of N to be up. The number of microstates for each
macrostate is just the number of combinations, the number of ways of choosing q objects
from a collection of N objects.
N
N!
q     
 q  q!N  q !
Now, what are the odds of observing this paramagnet to be in a particular energy
macrostate? Assuming every microstate is equally likely, then we have
14
( q )
.
(all )
Notice that the total multiplicity is (all )  2 N because each dipole has only two
possible states.
P(q) 
Here is a microstate for a system of N = 10 dipoles, with q = 6 (6 dipoles point up).
The probability function, P(q), for this system looks like this:
0.3
Probability, P
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
9
10
macrostate, q
The most probable macrostate has one-half the dipoles pointed up, q = 5.
2.
Einstein Solid Sec 2.2, 2.3
a. More than two states
A harmonic oscillator has energy levels that are uniformly spaced in steps of h , where
h is Planck’s Constant and  is the frequency of the oscillator. We imagine a solid made
of N such harmonic oscillators. The total energy of the solid is qh , where q is an
integer which in this case may well be greater than N. As shown in the text, the
multiplicity of the macrostate having energy qh is the number of ways q items can be
selected from q + N – 1 items.
 q  N  1 q  N  1!
 
q   
q

 q!N  1!
Then (all ) is the sum of all the q  .
b. Interacting systems
We are interested in the transfer of energy from one such Einstein solid to another. Now
we want the multiplicity of q energy units distributed over both systems.
15
Let’s say we have NA, NB, qA, and qB, such that q  qA  qB . The total multiplicity for the
two systems in contact is
(all ) 
q

q A 0
A
B
 q A  N A  1 qB  N B  1


qA
q
q A 0 
B


q
 q  N A  1 q  q A  N B  1


   A
qA
q  qA
q A 0 



q
 
Assuming that all microstates are equally probable, then the macrostate having the
greatest multiplicity is the most probable to be observed. As q, NA, and NB are made
larger, the multiplicity curve is taller, and more narrowly peaked at (if NA = NB) q A 
q
.
2
Say that initially, qB  qA . Then over time, as energy is exchanged more or less
randomly between oscillators in the two systems, there will be a net flow of energy from
system B to system A, from a macrostate of lower multiplicity to a macrostate of greater
multiplicity.
The text has one numerical example on page 57. In that example, N A  N B  3 and
q
q  6 . The maximum multiplicity occurs at q A   3 . Let’s look at a case in which
2
N A  N B , namely N A  6 and N B  8 and q  8 .
A
1
6
21
56
126
252
462
792
1287
qB
8
7
6
5
4
3
2
1
0
B
6435
3432
1716
792
330
120
36
8
1
all  
total
6435
20592
36036
44352
41580
30240
16632
6336
1287
P(qA)
0.031623
0.101194
0.17709
0.217957
0.204334
0.148607
0.081734
0.031137
0.006325
203490
16
0.25
0.2
0.15
P
qA
0
1
2
3
4
5
6
7
8
0.1
0.05
0
0
1
2
3
4
qA
5
6
7
8
The probability peaks at about q A  3 , rather than qA  4 , that is, at q A  q 
NA
 3.4 .
N A  NB
[I did the calculation of  using the COMBIN function in Excel.]
As we increase the numbers, the  s become very large very quickly, as illustrated by the
text example on pages 58 & 59.
B.
Entropy
1.
Large Systems Sec 2.4
a. Very large numbers
Macroscopic systems contain multiples of Avogadro’s number, 6.02  10 23 , perhaps
many, many multiples. The factorials of such large numbers are even larger—very large
numbers. We’ll use Stirling’s Approximation to evaluate the factorials:
N! N N e  N 2 N .
Ultimately, we will want the logarithm of N!: ln N! N ln N  N .
b. Multiplicity function
Consider an Einstein solid with a large number of oscillators, N, and energy units, q.
 q  N  1 q  N  1! q  N !
 
The multiplicity function is N , q   
.

q
q! N!

 q!N  1!
Take the logarithm, using Stirling’s formula.
 q  N ! 
  ln q  N ! ln q! ln N!
ln   ln 
 q! N! 
 q  N ln q  N   q  N   q ln q  q  N ln N  N


Now further assume that q >> N. In that case, ln q  N   ln q1  N   ln q  N .
 
The ln  becomes ln   N ln
q 
q
2
q
N
q
N
 N ln  N , whence
N
q
N
N
 eq 
q
     e N  
N
N
N
c. Interacting systems
The multiplicity function for a pair of interacting Einstein solids is the product of their
separate multiplicity functions. Let’s say N A  N B  N and q  q A  qB  N .
N
N
 eq   eq 
e
Then    A B   A   B    
 N   N 
N
2N
q AqB N .
If we were to graph this function
vs. qA, what would it look like? Firstly, we expect a peak at q A 
17
q
with a height of
2
2N
 e  q
 max     
 N  2
2N
. That’s a very large number. How about the width of the curve?
 N  2 x 
 q
2
In the text, the author shows that the curve is a Gaussian:    max e
, where
q
q
x  q A  . The origin has been shifted to the location of q A  . The point at which
2
2

q
  max occurs when x 
. Now, this is a large number, but compared to the
e
2 N
scale of the horizontal axis q A  0 to q , that peak is very narrow, since N is a large
1
number in itself. That is, the half width of the peak is
of the whole range of the
2 N
independent variable.
The upshot is that as N and q become large, the multiplicity function peak becomes
narrower and narrower. The most probable macrostate becomes more and more probable
relative to the other possible macrostates. Put another way, fluctuations from the most
probable macrostate are very small in large systems.
2.
Second “Law” Sec 2.6
a. Definition of entropy
We define the entropy of a system to be S  k ln  . The units of entropy are the units of
the Boltzmann Constant, J/K.
The total entropy of two interacting systems, such as the two Einstein solids above, is
S  k ln(  AB )  S A  S B .
The Second “Law” of Thermodynamics says: Systems tend to evolve in the direction of
increasing multiplicity. That is, entropy tends to increase. This is simply because the
macrostate of maximum multiplicity is the most probable to be observed by the time the
system has reached thermal equilibrium.
18
b. Irreversible
The concept of entropy was introduced originally to explain why certain processes only
went one way spontaneously. When heat enters a system at temperature, T, its entropy
increases by S  Q . When heat leaves a system, the system’s entropy decreases.
T
Consider two identical blocks of aluminum, initially one hotter than the other. When
brought into thermal contact, heat will flow from the warmer block(A) to the cooler(B)
until they have reached the same temperature. The total entropy of the two blocks will
C dT
C dT
Q
Q
have increased. Incrementally, dS  A  B   V A  V B . In effect, because TA
TA TB
TA
TB
is greater than TB, the entropy of block A decreases less than the entropy of block B
increases.
Processes that create new entropy cannot happen spontaneously in reverse. Heat flow
from a warmer object to a cooler object creates entropy, and is irreversible. Mixing two
different gases creates entropy, and is irreversible. Rapid expansion or compression
creates entropy. On the other hand quasistatic volume change can be reversible,
depending what other processes are taking place at the same time. Very slow heat flow
produces very little new entropy and may be regarded as practically reversible.
An irreversible process may be reversed by doing work on the system, but that also
increases the total amount of entropy in the universe.
C.
Creating Entropy
1.
Temperature Sec 3.1, 3.2
a. Thermal equilibrium
When two objects are in thermal equilibrium, their temperatures are the same. According
to the Second “Law”, their total entropy is at its maximum.
Consider two objects in contact, exchanging energy. The total energy of the two objects
is fixed. At equilibrium,
Stotal S A S B
S
S


 A  B 0
U A U A U A U A U B
S A
S
 B
U A U B
The quantity
S
has units of K-1, so perhaps we can define the temperature in terms of
U
1
 S 
the entropy as T  
 .
 U  N ,V
19
b. Heat capacities
We cannot measure entropy directly, but we can measure changes in entropy indirectly,
through the heat capacity. For instance, if no work is being done on the system,
dU Q CV dT
dS 
 
T
T
T
Tf
S 
CV
T
dT
Ti
Of course, we need to know CV as a function of T. This is obtained by measuring Q or U
vs. T. In general, the heat capacity decreases with decreasing temperature. At higher
temperatures, the heat capacity approaches the constant
f
k
(Dulong-Petit). For
2
instance, the CV vs. T for a monatomic substance would look like this:
The Third “Law” of Thermodynamics says that CV  0 as T  0 , or alternatively, that
S = 0 when T = 0K. In reality, there remains residual entropy in a system at T = 0K—
near absolute zero, the relaxation time for the system to settle into its very lowest energy
state is very, very long.
Now notice, if indeed CV  0 as T  0 , then absolute zero can not be attained in a
finite number of steps, since S  0 as T  0 . It’s like the famous example of
approaching a wall in a series of steps, each one half the previous step.
For example, let us say that we wish to cool an ideal gas to absolute zero. We’d have to
“get rid” of the entropy in the gas in a series of steps.
i) isothermal compression—heat and entropy is transferred to a reservoir
ii) adiabatic expansion—temperature decreases, entropy is constant, Q = 0
repeat
Now if we were to graph these S(T) points we have generated we would see two curves.
But the curves are not parallel; they appear to converge at T = 0. As a result, the S gets
smaller for each successive two-stage step, the closer we get to T = 0.
20
In practice, a real gas would condense at some point. The text describes three real-life
high-tech coolers. In any case, there will be a series of ever smaller steps downward
between converging curves on the S(T) graph, toward absolute zero.
2.
Pressure Sec 3.4
a. Mechanical equilibrium
Consider two systems whose volumes can change as they interact. An example might be
two gases separated by a moveable membrane. The total energy and volume of the two
systems are fixed, but the systems may exchange energy and volume. Therefore, the
entropy is a function of the volumes as well as the internal energies. However, we will
be keeping the numbers of particles in each system fixed.
At the equilibrium point,
S total
S total
 0 and
 0.
U A
V A
S total S A S B S A S B




V A
V A V A V A VB
S A S B

V A VB
As we did with temperature, we can identify the pressure with the derivative of entropy
 S 
with volume, thusly: P  T 
 .
 V U , N
21
b. Thermodynamic identity
Now if we envision a system whose internal energy and volume are changing, we would
write the change in entropy (a function of both U and of V) as follows:
 S 
 S 
dS  
 dU  
 dV
 U V
 V U
1
P
 dU  dV
T
T
dU  TdS  PdV
c. Creating entropy with mechanical work
Remember that compressive work ( PdV ) is just one form of work. If the compression is
slow, and no other form of work is done on the system, then the volume change is
quasistatic, and W  PdV . In such a case, we are allowed to combine the First “Law”
with the thermodynamic identity to obtain
Q  TdS .
But, if the work done on the system is greater than  PdV , then Q  TdS . In other
words, the amount of entropy created in the system is more than that accounted for by the
heat flow into the system. This might happen, for instance, with a compression that
occurs faster than the pressure can equalize throughout the volume of the system. It will
happen if other forms of work are being done, such as mixing, or stirring. In a similar
vein, if a gas is allowed to expand freely into a vacuum, no work is done by the gas, and
no heat flows into or out of the gas. Yet the gas is occupying a larger volume, so its
entropy is increased.
3.
Chemical Potential Sec 3.5
Now consider a case in which the systems can exchange particles as well as energy and
volume.
a. Diffusive equilibrium
S total S A S B S A S B




N A
N A N A N A N B
S A S B

N A N B
22
 S 
Define the chemical potential as   T 
 . Evidently, the minus sign is attached
 N U ,V
so that particles will tend to diffuse from higher toward lower chemical potential.
b. Generalized thermodynamic identity
For infinitesimal changes in the system,
 S 
 S
 dU  
dS  
 U V , N
 V

 S 
 dV  
 dN
U , N
 N U ,V
1
P

dU  dV  dN
T
T
T
dU  TdS  PdV  dN
This equation contains within it all three of the partial-derivative formulas for T, P and
for  . For instance, assume that entropy and volume are fixed. Then the

 U 
thermodynamic identity says dU  dN , whence we can write   
 . To apply
 N  S ,V
the partial-derivative formulae to a particular case, we need specific expressions for the
interdependence of the variables, i.e., U as a function of N.
4.
Expanding & Mixing Sec 2.6
a. Free expansion
Imagine a container of volume 2V, isolated from its surroundings, and with a partition
that divides the container in half. An ideal gas is confined to one side of the container.
The gas is in equilibrium, with temperature T and Pressure P. Now, imagine removing
the partition. Over time, the gas molecules will diffuse to fill the larger volume.
However, in expanding the gas does no work, hence the phrase free expansion. Because
the container is isolated, no heat flows into or out of the gas, nor does the number of
molecules, N, change..
U  Q  W  0  0  0
2V
 Nk ln 2 .
However, the entropy increases. S  Nk ln
V
[The expression for the entropy of an ideal gas is derived in Sec. IV B 3. It is

N
5
S  Nk  ln
  , where vQ is a constant.]
 NvQ 2 


23
b. Entropy of mixing
In a similar vein, we might imagine a container divided into two chambers, each with a
different ideal gas in it. When the partition is removed, both gases diffuse to fill the
larger volume. Since the gases are ideal, one gas doesn’t really “notice” the presence of
the other as they mix. The entropy of both gases increases, so the entropy change of the
2V
 2 Nk ln 2 , assuming of course that we started with the
whole system is S  2 Nk ln
V
same numbers of molecules of both gases, etc.
24
III. Processes
A.
Cyclic Processes
1.
Heat Engines & Heat Pumps Sec 4.1, 4.2
A heat engine is a device that absorbs heat from a reservoir and converts part of it to
work. The engine carries a working substance through a PVT cycle, returning to the state
at which it starts. It expels “waste” heat into a cold reservoir, or into its environment. It
must do this in order that the entropy of the engine itself does not increase with every
cycle.
a. Efficiency
The efficiency of the heat engine is defined as the ratio of work done by the engine to the
heat absorbed by the engine.
W
e
Qh
We’d like to express e in terms of the temperatures of the hot and cold reservoirs. The
Q Q
First “Law” says that Qh  Qc  W . The Second “Law” says that c  h . Putting
Tc Th
T
these together, we obtain e  1  c . Firstly, notice that e cannot be greater than one.
Th
T
Secondly, e cannot be one unless Tc = 0 K, which cannot be achieved. Thirdly, 1  c is
Th
the greatest e can be—in practice, e is less than the theoretical limit, since always
Qc Qh

.
Tc Th
b. Carnot cycle
Can a cycle be devised for which e  1 
Tc
? That’s the Carnot cycle, which uses a gas
Th
as the working substance.
25
i) the gas absorbs heat from the hot reservoir. To minimize dS, we need Tgas  Th ; the gas
is allowed to expand isothermally in order to maintain the Tgas  Th .
ii) the gas expands adiabatically, doing work, and cools from Th to Tc.
iii) the gas is compressed isothermally, during which step heat is transferred to the cold
reservoir.
iv) the gas is compressed adiabatically, and warms from Tc to Th.
Now, for the total change in entropy to be very small, the temperature differences
between the gas and the reservoirs must be very small. But that means that the heat
transfers are very slooow. Therefore, the Carnot cycle is not very useful in producing
useful work. [Empirically, the rate at which heat flows is proportional to the temperature
Q
 T .]
difference —
t
c. Heat pump
The purpose of a heat pump is to transport energy from a cold reservoir to a hot one by
doing work on the working substance. The work is necessary because the temperature of
the working fluid must be raised above that of the hot reservoir in order for heat to flow
in the desired direction. Likewise, at the other side of the cycle the working fluid must be
made colder than the cold reservoir.
Rather than efficiency, the corresponding parameter for a heat pump is the coefficient of
performance,
Q
COP  c
W
26
The First “Law” says Qh  Qc  W . The Second “Law” says
together, we obtain COP 
1
Th
1
Tc

Qh Qc

. Putting these
Th Tc
Tc
. A Carnot cycle running in reverse will give
Th  Tc
the maximum COP.
2.
Otto, Diesel, & Rankine Sec 4.3
Real heat engines need to produce work at a more rapid rate than a Carnot engine.
Consequently, their efficiencies are lower than that of a Carnot engine. Of course, real
engines do not achieve even their theoretical efficiencies due to friction and conductive
heat loss through the cylinder walls and the like.
a. Otto cycle
The Otto cycle is the basis for the ordinary 4-stroke gasoline engine.
i) air-fuel mixture is compressed adiabatically from V1 to V2; pressure rises from P1 to P2.
ii) air-fuel mixture is ignited, the pressure rises isochorically from P2 to P3.
iii) combustion products expand adiabatically from V2 to V1; pressure falls from P3 to P4.
iv) pressure falls isochorically from P4 to P1.
27
The temperatures also change from step to step. The efficiency is given by
V 
e  1   2 
 V1 
 1
1
T1
T
1 4
T2
T3
V1
is the compression ratio. The greater the compression ratio, the greater
V2
is the efficiency of the engine. However, so is T3 greater. If T3 is too great, the air-fuel
mixture will ignite prematurely, before the piston reaches the top of its stroke. This
reduces power, and damages the piston and cylinder. Up to a point, chemical additives to
the fuel can alleviate the premature detonation.
The quotient
Notice that there is no hot reservoir per se; rather the heat source is the chemical energy
released by the combustion of the fuel.
b. Diesel cycle
The Diesel cycle differs from the Otto cycle in that the air is first compressed
adiabatically in the cylinder, then the fuel is injected into the hot air and ignited
spontaneously, without need of a spark. The fuel injection takes place as the piston has
begun to move downward, so that constant pressure is maintained during the fuel
injection. Since the fuel is not in the cylinder during the compression, much higher
compression ratios can be used, leading to greater efficiencies.
28
c. Rankine cycle
In some ways the steam engine is a more nearly exact example of a heat engine than is
the Otto engine. No chemical reaction or combustion takes place within the working
fluid, and at least in principle the working fluid is not replaced at the beginning of each
cycle.
i) water is pumped to a high pressure into a boiler.
ii) the water is heated at constant pressure and changes to steam (water vapour).
iii) the steam expands adiabatically, driving a piston or a turbine, and cools and begins to
condense.
iv) the partially cooled steam/water mixture is cooled further by contact with the cold
reservoir.
The efficiency of the steam engine is e  1 
Qc
. At constant pressure, Q  H ,
QH
H 4  H1
. Now, H 2  H1 since the water is not compressed as it is
H3  H2
pumped, and only a little energy is added to the water (e.g., it’s not accelerated). So we
look up the enthalpies on the tables of enthalpy & entropy vs. temperature & pressure—
page 136.
whence, e  1 
d. Throttling and refrigerators Sec 4.4
For a refrigerator to work, the temperature of the working fluid must be made less than
that of the cold reservoir. This is done through what is called a throttling process.
The working fluid passes through a narrow opening from a region of high pressure into a
region of low pressure. In doing so, it expands adiabatically (Q = 0) and cools. As the
fluid expands, the negative potential energy of interaction among the atoms/molecules
increases and the kinetic energy decreases.
29
From the First “Law”
U  U f  U i  Q  W  0  Wleft  Wright
U f  U i  PiVi  Pf V f
U f  Pf V f  U i  PiVi
In a dense gas or a liquid, U potential  U kinetic  PV  constant . Therefore, as the gas
expands, U potential  0 so that U kinetic  0  the gas/liquid cools.
Subsequently, the chilled fluid absorbs heat from the cold reservoir and vaporizes.
Therefore, the working fluid must be a substance with a low boiling point. The
compressor does the work of compressing the gas to raise its temperature, as well as
maintains the pressure difference required for the throttle valve to work.
B.
Non-cyclic Processes or Thermodynamic Potentials
1.
Thermodynamic Potentials Sec 5.1
A number of thermodynamic quantities have been defined—useful under differing
conditions of fixed pressure, volume, temperature, particle number, etc. These are the
enthalpy, the Helmholtz free energy, and the Gibbs free energy. Together with the
internal energy, these are referred to as thermodynamic potentials.
a. Enthalpy Sec 1.6
The total energy required to create a system of particles at sea level air pressure would
include the expansive work done in displacing the air.
We define the enthalpy to be H  U  PV . The enthalpy is useful when a change takes
place in a system while pressure is constant.
30
H  U  PV  Q  W  PV  Q   PV   Wother  PV
H  Q  Wother
Now, if no other work is done, then H  Q exactly. In practice, tables of measured
enthalpies for various processes, usually chemical reactions or phase transitions are
compiled. The text mentions the enthalpy of formation for liquid water. Evidently, when
oxygen and hydrogen gases are combined to form a mole of liquid water, the change in
enthalpy is -286 kJ. In other words, burning hydrogen at constant pressure releases this
much energy.
Qc
H  H1
. Now, H 2  H1 since the water
 1 4
Qh
H3  H2
is not compressed as it is pumped, and only a little energy is added to the water (it’s not
accelerated). So, we look up the enthalpies on tables of H & S vs. T & P—page 136.
PV diagram…………………………………………
Efficiency of a steam engine: e  1 
COP of refrigerators:
Qc
H  H3
H1  H 4
H1  H 4
COP 


 1
Qh  Qc H 2  H 3  H 1  H 4  H 2  H 3  H 1  H 4 H 2  H 1
PV diagram…………………………………………………
Problem 4-29
Pi  12 bar
Pf  1 bar
From Table 4.3, H  116 kJ per kg. At 12 bar, the boiling point is 46.3oC.
a) Pf = 1 bar and Hliquid = 16 kJ
T = -26.4oC and Hgas = 231 kJ.
b) Starting with all liquid at Pi,
116 kJ  x  16 kJ  1  x   231 kJ
x  0.53  53% liquid at Pf and T f  26.4 o C.
31
Q43  0
b. Helmholtz
Let’s say the system is in contact with a heat bath, so that the temperature is constant.
The pressure may not be constant. To create the system, some of its total energy can be
taken from the environment in the form of heat. So the total work required to create the
system is not all of U, but less than U. Define the Helmholtz Free Energy of the system
as F  U  TS .
Any change in a system at constant temperature will entail a change in F.
F  U  TS  Q  W  TS ,
Where W is all the work done on the system.
c. Gibbs
Now, if the system is at constant pressure as well as constant temperature, then the extra
work needed to create the system is the Gibbs Free Energy, G  U  TS  PV .
If pressure is constant, we use the Gibbs free energy:
G  U  TS  PV  Q  W  TS  PV
G  H  TS
Again, W is the total work done on the system.
.
In a paragraph above, we burned some hydrogen. The 286 kJ released could be used to
run an Otto cycle for instance. The theoretical efficiency of an Otto engine is about 56%.
So, at most 160 kJ are used to drive the car. It’s possible to run that reaction in a more
controlled way and extract electrical work, in a hydrogen fuel cell.
G  H  TS
 286 kJ  298 K  163 J/K 
 237 kJ
That TS  49 kJ has to be expelled to the environment, and the efficiency of the fuel
cell alone is 83%. The fuel cell generates current which can run an electrical motor or
charge a battery. Of course, in both instances, there are numerous losses of energy along
the way to driving the car.
d. Identities
If we envision infinitesimal changes in thermodynamic variables, we can derive
thermodynamic identities for the thermodynamic potentials. We have already, the
thermodynamic identity for internal energy
dU  TdS  PdV  dN .
Now, consider the enthalpy, H.
H  U  PV
dH  dU  PdV  VdP  TdS  PdV  dN  PdV  VdP
dH  TdS  VdP  dN
32
 H 
For instance, if dP = 0 and dN = 0, then we could write T  
 , which is
 S  P , N
 U 
equivalent to the T  
 that we obtained earlier.
 S V , N
We can do the same for F and for G.
F  U  TS
dF  dU  TdS  SdT  TdS  PdV  dN  TdS  SdT
dF   SdT  PdV  dN
 F 
From this equation we can derive relationships like S  
 .
 T V , N
G  U  TS  PV
dG  dU  TdS  SdT  PdV  VdP  TdS  PdV  dN  TdS  SdT  PdV  VdP
dG   SdT  VdP  dN
2.
Toward Equilibrium Sec 5.2
a. System and its environment
An isolated system tends to evolve toward an equilibrium state of maximum entropy.
That is, any spontaneous rearrangements within the system increase the entropy of the
system. Now, consider a system which is in thermal contact with its environment. The
system will tend to evolve, by exchanging energy with the environment. The entropy of
the system may increase or decrease, but the total entropy of the universe increases in the
process.
Let’s say that the system evolves toward equilibrium isothermally. The environment is
such a large reservoir of energy that it can exchange energy with the system without
changing temperature. It’s a heat bath. The total change in entropy involved with an
exchange of energy would be
dStotal  dS  dS R
Assuming the V and N for the environment are fixed, an recalling that dUR = - dU and T =
TR , then
33
dS R 
1
P

1
dU R  R dVR  R dN R  dU R
TR
TR
TR
TR
1
1
1
1
dU R  dS  dU   dU  TdS    dF
TR
T
T
T
The increase in total entropy under conditions of constant T, V, and N is equivalent to a
decrease in the Helmholtz free energy of the system.
dStotal  dS 
In a similar vein, if the system volume is not fixed, but the pressure is constant, then we
have
1
P
1
1
dS total  dS  dU  dV   dU  TdS  PdV    dG
T
T
T
T
The increase in total entropy under conditions of constant T, P, and N is equivalent to a
decrease in the Gibbs free energy of the system.
system condition
isolated—constant U, T, V, & N
constant T and V and N
constant T and P and N
system tendency
entropy increases
Helmholtz free energy decreases
Gibbs free energy decreases
b. Extensive & Intensive
The several properties of a system can be divided into two classes—those that depend on
the amount of mass in the system, and those that do not. We imagine a system with
volume V in equilibrium. The system is characterized by its mass, number of particles,
pressure, temperature, volume, chemical potential, density, entropy, enthalpy, internal
energy, Helmholtz and Gibbs free energies. Now imagine slicing the system in half,
forming two identical systems with volumes V/2. Some properties of the two systems are
unchanged—temperature, pressure, density, and chemical potential. These are the
intensive properties. The rest are extensive—they are halved when the original system
was cut in half.
The usefulness of this concept is in checking the validity of thermodynamic relationships.
All the terms in a thermodynamic equation must be the same type, because an extensive
quantity cannot be added to an intensive quantity. The product of an intensive quantity
and an extensive quantity is extensive. On the other hand, dividing an extensive quantity
by another yields an intensive quantity, as in mass divided by volume gives the density.
3.
Phase transformations Sec 5.3
We are familiar with water, or carbon dioxide or alcohol changing from liquid to vapour,
from solid to liquid, etc. We are aware that some metals are liquid at room temperature
while most are solid and melt if the temperature is much higher. These are familiar phase
changes.
34
More generally, a phase transformation is a discontinuous change in the properties of a
substance, not limited to changing physical structure from solid to liquid to gas, that takes
place when PVT conditions are changed only slightly.
a. Phase diagram
Which phase of a substance is the stable phase depends on temperature and pressure. A
phase diagram is a plot showing the conditions under which each phase is the stable
phase.
For something like water or carbon dioxide, the phase diagram is divided into three
regions—the solid, liquid, and gas (vapour) regions. If we trace the P,T values at which
changes in phase take place, we trace the phase boundaries on the plot. At those
particular values of P,T the two phases can coexist in equilibrium. At the triple point, all
three phases coexist in equilibrium. The pressure on a gas-liquid or gas-solid boundary is
called the vapour pressure of the liquid or solid.
Notice that the phase boundary between gas and liquid has an end point, called the
critical point. This signifies that at pressures and/or temperatures beyond the critical
point there is no physical distinction between liquid and gas. The density of the gas is so
great and the thermal motion of the liquid is so great that gas and liquid are the same.
Other sorts of phase transformations are possible, as for instance at very high pressures
there are different solid phases of ice. Similarly for carbon, there is more than one solid
phase—diamond and graphite. Diamond is the stable phase at very high pressure while
graphite is the more stable phase at sea level air pressure. The glittery diamonds people
pay so much to possess are ever so slowly changing into pencil lead.
Still other phase transformations are related not to pressure, but to magnetic field strength
as in the case of ferromagnets and superconductors.
Here’s a phase diagram for water, showing the several solid phases. They differ in
crystal structure and density, as well as other properties such as electrical conductivity.
35
D. Eisenberg & W. Kauzmann, The Structure and Properties of Water, Oxford Univ. Press, 1969.
The phase diagram for water shown in the text figure 5.11 is a teensy strip along the T
axis near P = 0 on this figure. [One bar is about one atmosphere of air pressure, so a kbar
is 1000 atm.]

Here’s a phase diagram for a ferromagnet, subjected to an external magnetic field, B .
The phase boundary is just the straight segment along the T axis. The segment ends at
the critical point, at T  Tc .
b. van der Waals model
There are phases because the particles interact with each other, in contrast to an ideal gas.
The interactions are complicated (quantum mechanics), so we create simplified, effective
models of the interparticle interactions in order to figure out what properties of the
interactions lead to the observed phase diagrams. For instance, the van der Waals model:
The model of a non-ideal gas is constructed as follows. Firstly, the atoms have nonzero
volume—they are not really point particles. So, the volume of the system cannot go to
zero, no matter how great the pressure or low the temperature. The smallest V can
possibly be, let’s say, is Nb. It’s like shifting the origin from V = 0 to V – Nb = 0.
Secondly, the atoms exert forces on each other. At short range, but not too short, the
forces are attractive—the atoms tend to pull one another closer. This has the tendency to
reduce the pressure that the system exerts outward on its environment (or container). We
36
N
introduce a “correction” to the pressure that is proportional to the density   and to the
V 
N 2
number of atoms in the system, N. That is, P  P  2 .
V
With the new V and P, the gas law becomes the van der Waals equation of state:

N 2 
NkT
N 2
 P  2 V  Nb  NkT or P 


V 
V  Nb V 2

Now, the b and  are adjustable parameters, whose values are different for different
substances. They have to be fitted to empirical data.
There are countless other equations of state. For instance, there is the virial expansion,
which is an infinite series,
 B(T ) C (T )

PV  NkT 1 
 2   .
V
V


There is the Beattie-Bridgeman equation of state
 

PV  nkT   2  3 .
V V
V
All represent “corrections” to the ideal gas equation of state.
c. Gibbs free energy—Clausius-Clapeyron—PV diagram
Which phase is stable at a given temperature and pressure is that phase with the lowest
Gibbs free energy. On a phase boundary, where two phases coexist, there must be a
common Gibbs free energy. That is, on the boundary between liquid and gas,
Gliquid  Ggas . Imagine changing the pressure and temperature by small amounts in such a
way as to remain on the phase boundary. Then the Gibbs free energy changes in such a
way that
37
dGliquid  dGgas
 SliquiddT  VliquiddP  liquiddN   S gasdT  VgasdP   gasdN
dP S gas  Sliquid

dT Vgas  Vliquid
We’ve assumed that dN = 0. This result is the slope of the phase boundary curve on the
PT diagram.
Commonly, we express the change in entropy in terms of the latent heat of the
transformation, thusly
dP
L

dT TV
This is the Clausius-Clapeyron relation, applicable to any PT phase boundary.
Finally, we compute the Gibbs free energy for the van der Waals model at a variety of
temperatures and pressures to determine which phase is stable in each case.
dG   SdT  VdP  dN
Let dN = 0, and fix the temperature, varying only P.
dG  VdP
 G 
 P 

 V

 V  N ,T
 V  N ,T
 
 V 
 V


Integrate
 NkT
N 2  

 2  
 V  Nb V   N ,T
NkTV
2N 2

V  Nb2 V 2
NkT V  Nb NkT Nb 2N 2


V  Nb2 V  Nb2 V 2
NkT Nb  2N
G   NkT ln V  Nb 
2
 c(T )
V  Nb
V
We have now expressions for both P and G as functions of V, at a fixed temperature, T.
Firstly, we plot G vs. P at some fixed V. This yields a graph with a loop in it. The loop
represents unstable states, since the Gibbs free energy is not a minimum. Integrating dG
around the closed loop should give zero.
 G 
 dG    P T dP  VdP
We plot out on a PV diagram the same points of free energy, and obtain an isotherm
something like that shown on the diagram below right.
38
The pressure at which the phase transition occurs, at temperature T, is that value of P
where the two shaded areas cancel. So, tracking along an isotherm from right to left, the
gas is compressed and pressure rises until that horizontal section is reached. At that
point, further compression does not increase the pressure because the gas is condensing to
liquid. When the phase transition is complete, further compression causes a steep
increase in pressure, with little decrease of volume, as the liquid is much less
compressible than the gas. During the transition, both gas and liquid phases coexist.
If the temperature is high enough, there is no phase transition as V decreases. On a PT
diagram, we see a phase boundary between the liquid and gas phases, up to the critical
point, where the boundary terminates. Above the critical point, there is no distinction
between gas and liquid. That would correspond to the isotherms having no flat segment
on the PV diagram.
The van der Waals model is not very accurate in reality, but it does illustrate how the
observed phase behavior arises from the interactions among the atoms or molecules of
the substance.
39
IV.
Statistical Mechanics
A.
Partition Function
1.
Boltzmann Sec 6.1
a. Multiplicity
Consider a system, in contact with an energy reservoir, consisting of N weakly interacting
identical particles. The energy of each particle is quantized, the energy levels labeled by
Ei. Previously, we associated a multiplicity,  i , with each different energy level. But,
we could just as well list each microstate separately. That is, each particle energy level,
Ei, has multiplicity of 1, but some energy values occur  i times in the list. At
equilibrium, the total internal energy of the system of N particles is constant (apart from
small fluctuations)
U   N i Ei .
i
Where Ni is the number of particles in state Ei. If the particles are indistinguishable, the
multiplicity of the whole system is
1

i Ni !
1
  Ni
Ni
i
i
On the other hand, if the particles are distinguishable, as in a solid, the multiplicity of the
whole system is
1
  N!
i Ni !
ln    Ni ln
1
 N ln N
Ni
i
The energy levels available to the particles, Ei, are not necessarily easy to determine. If
we assume an ideal gas, then the energy is just the kinetic energy of a particle confined in
a finite volume. Interacting particles will have also potential energy. The single particle
energy states are influenced by the presence of N – 1 other particles. We normally make
some simplifying assumptions. We might assume that in a dilute gas, the particle states
are the same as those of an isolated particle, unaltered by interaction between particles.
We might assume that the fact that level E8 is occupied by 28 particles, rather than by
465, has no affect on the energy levels. In fact, the particle energy levels of a system of
N particles are ever so slightly different that those of an otherwise identical system
having N – 1 particles.
ln    N i ln
b. Equilibrium
The equilibrium state of the system is the state that maximizes  , subject to the
constraints N = constant and U = constant. We want to solve for the Ni that maximize  .
40
For indistinguishable particles,
 1 
1
dN i   N i2   2 dN i  0
 N 
Ni
i
i
i 

dN   dN i  0
d ln    ln
i
dU   Ei dN i  0
i
We’ll apply the method of undetermined multipliers to determine the equilibrium
distribution of particles among the states, that is, the Ni.
1
i ln N dNi  0
i


a   dN i  0
 i



b   Ei dN i  0
 i

[At this point, notice that because dN = 0, the d ln  for distinguishable particles is
exactly the same as for indistinguishable particles. So, we only have to do this once.]
Add the three equations.




dNi  a   dNi   b   Ei dNi   0
i
i
 i

 i

Each term must be zero separately.
1
ln
 a  bEi  0
Ni
Solve for Ni.
 ln Ni  a  bEi
1
 ln N
Ni  exp( a  bEi )  ea ebEi
Now we determine the multipliers by applying the constraints. The first one is easy.
 Ni   eaebEi  N
i
i
N
N
ea 

bEi
Z
e
i
The second one, b, is a bit more complicated. If the system were to exchange a small
amount of energy, dU, with the reservoir, the entropy would change as small alterations
occur in the {Ni}.
41
d ln    ln
i
1
dN i
Ni
   a  bEi dN i
i
 adN  bdU
 bdU
b
d ln 
1 d k ln  
1  S 
1

  
  
dU
k dU
k  U V , N
kT
The equilibrium numbers of particles in each energy state finally are
E
N  i
N i   e kT
Z
E
1  kT
This is the Boltzmann probability distribution. P ( E )  e . The greater the energy of
Z
a particle state, the less likely a particle is to be in that state. [For atoms, we are making
the ground state zero, E  0 .]
As T increases, though, the likelihood of a higher-energy state being occupied increases
as well. The exponential decay of P is slower at higher temperatures.
c. Partition function
The sum over states (Zustandsumme) is called the partition function. In principle, it’s a
sum over all the particle states of a system, and therefore contains the statistical
information about the system. All of the thermodynamic properties of the system are
derivable from the partition function.
Z  e

Ei
kT
i
Remember, the partition function is written as a sum over all microstates, rather than over
energy macrostates, so that the multiplicity factor does not appear explicitly.
All of the macroscopic thermodynamic quantities are obtainable from the microscopic
statistics of the single particle energy states. For instance, internal energy:
42
 E   i Z
 Z 

    i 2 e kT 
Z
i  kT 
 T V
Z

 Ei N i
kT 2 i
ZU

kT 2
  ln Z 
U  kT 2 

 T V
E
Entropy:
S  k ln   k  N i ln N i  kN
i
 N E 
  k  N i  ln  i   kN
i
 Z kT 
Z U
 Nk ln   Nk
N T
Pressure:
TdS  dU  PdV
P


 
dU
dS
U  TS       NkT ln Z  NkT  
T
 
dV
dV
N
 V
T
 T
 V 
  ln Z 
P  NkT 

 V T
2.
Probability Sec 6.2
a. Ensembles (awnsombles)
microcanonical
In an isolated system, every microstate has equal probability of being occupied. The total
energy is fixed. The collection of all the possible microstates of the system is called the
microcanonical ensemble of states.
canonical
If the probability distribution is the Boltzmann distribution, the collection of energy states
is called the canonical ensemble. We’ve seen that such a distribution applies to a system
at constant temperature and fixed number of particles, in contact with an energy
reservoir. The internal energy is not fixed, but we expect only small fluctuations from an
equilibrium value.
grand canonical
43
If the number of particles is allowed to change, then we have to sum also over all possible
numbers of particles, as well as all possible energy states, giving the grand canonical
ensemble.
b. Average values
The probability that a particle will be observed to occupy a particular energy state is
given by the Boltzmann distribution.
e
P( E ) 

E
kT
Z
The average value of the particle energy would be computed in the usual way,

Ei
kT
Ei e
Ni 
i
 E   Ei

.
N
Z
i
The internal energy of the system of N particles is U = N<E>. We are averaging over a
collection of particles whose energies are distributed according to the Boltzmann
distribution.
3.
A Couple of Applications
a. Equipartition Theorem Sec 6.3
A particle’s kinetic energy is proportional to the square of its velocity components. In
Cartesian coordinates, K  vx2  vy2  vz2 . In a similar vein, the rotational kinetic energy of
a rigid body is also proportional to the square of the angular velocity, thus K rotation   2 .
In the case of a harmonic oscillator, the potential energy is also proportional to a square,
namely the displacement components, U potential  x2  y 2  z 2 . Very often, we
approximate the real force acting on a particle with the linear restoring force of the
harmonic oscillator. Let us consider a generalized quadratic degree of freedom,
E q   cq 2 . Each value that q takes on represents a distinct particle state. The energy is
quantized, so the q-values are discrete, with spacing  q .
The partition function for this “system” is a sum over those q-states
Z  e

E q 
kT
 e
q

cq 2
kT
.
q
In the classical limit  q is small, and the sum goes over to an integral

Z  e
0

cq 2
kT

dq  kT  e  cy dy  kT  
2
0
3
 1  1  2
1 Z
1
1
The average energy is  E  

      kT .
Z  1 
2
kT    2  kT 
 
 kT 
44
So, each quadratic degree of freedom, at equilibrium, will have the same amount of
energy. But, this equipartition of energy theorem is valid only in the classical limit and
high temperature limits. That is, when the energy level spacing is small compared to kT.
We saw earlier that degrees of freedom can be “frozen out” as temperature declines.
b. Maxwell Speed Distribution Sec 6.4
A distribution function is rather like a histogram. When we counted the multiplicity for
the energy states of an Einstein solid, we obtained a histogram, or bar chart that was low
on either side and peaked in the middle. If the bins of the histogram are very, very
narrow, we have a continuous function—the distribution function. A distribution
function can be created for any variable. Let’s consider the distribution of speeds of the
atoms in an ideal gas. We want the relative probability that an atom is moving with a
particular speed, v. This is equivalent to asking what fraction of the atoms are moving
with that speed. The probability that an atom’s speed lies in an interval (v1,v2) is the
integral of the distribution function.
v2
P (v1 , v2 )   D (v)dv
v1
We need to figure out what D(v) is for an ideal gas in equilibrium.
The distribution function has two parts. The first is the probability of an atom having
speed, v. That’s given by the Boltzmann probability distribution. The second factor is a
multiplicity factor—how many velocity vectors have the same magnitude, v.
Dv  probabilit y of v number of velocity vectors with v
The number of velocity vectors that have the same magnitude is obtained by computing
the surface area of a sphere of radius v in velocity space. That is Av  4v 2 .
mv 2
2 kT
Dv   C  4v  e
The C is a proportionality constant, which we evaluate by normalizing the distribution
function.

2


 Dv dv   C  4v
0
0
 m 
C  

 2kT 
3
2
45
2
e

mv 2
2 kT
dv  1
3
mv 2

 m  2
Dv   
 4v 2e kT
 2kT 
B. Adding up the States Sec 1.2, 1.7, 2.5, 6.6, 6.7
1.
Two-State Paramagnet Sec 6.6
The specifics of computing the partition function for a system depend on the nature of the
system—the specifics of its energy levels. For instance, each dipole moment in an ideal
two-state paramagnet in an external magnetic field has two discrete states.
a. Single dipole
There are two states in a system consisting of a single magnetic dipole, Ei   B .
Therefore, the partition function is
E
B
B
2
 i

B
.
Z1   e kT  e kT  e kT  2 cosh
kT
i 1
b. Two or more dipoles
If the system consists of two non-interacting, distinguishable dipoles, then there are four
states: , , ,  .
4
Z2   e

Ei
kT
e

B  B
kt
e

B  B
kT
e
i 1

 B  B
kT
e

 B  B
kT
 Z1(1) Z1( 2)  Z12
Now if the dipoles are indistinguishable, there are fewer distinct states, namely three for
N = 2, so Z 2  Z12 . This is because the states ,  are the same state if the dipoles are
indistinguishable.
Extending to N dipoles, Z N  Z1N for distinguishable dipoles; Z N 
indistinguishable dipoles, if N is large.
46
Z1N
for
N!
2.
Ideal Gas Sec 6.7
a. One molecule
Z1   e
i

Ei
kT

e

translational
states, i
Ei
kT

e

Ej
kT
rotational
states, j

e

En
kT
 Z transZ rot Z int
internal
states, n
b. Two or more molecules
If the molecules are not interacting, then as before, the partition function for N molecules
ZN
is just Z N  Z1N or Z N  i depending on whether the molecules are distinguishable or
N!
not. In the ideal gas, the molecules are not distinguishable one from another. If the
molecules were in a solid, then they would be distinguishable because their positions
would be distinctly different.
[Note that in the text Section 6.7, the rotational partition function is lumped in with the
internal partition function.]
c. Internal partition function
The internal partition function sums over the internal vibrations of the constituent atoms.
We would usually approximate the energy levels by harmonic oscillator energy levels.

Z int   e
i

nh i
kT
n 1
The index i labels the vibrational modes, while n labels the uniformly spaced energy
levels for each mode. For instance, the water molecule has three intermolecular modes.
A molecule having more atoms has more modes. A diatomic molecule has just one mode
of vibration.
d. Rotational partition function
A molecule is constrained to a particular shape (internal vibrational motions apart), which
we regard as rotating like a rigid body. The angular momentum, and therefore the
2
rotational kinetic energy, is quantized, thusly E j  j  j  1 , where I is the moment of
2I
inertia of the molecule about the rotational axis. Classically, if  is the angular velocity
and L is the magnitude of the angular momentum, then the kinetic energy of rotation is
I 2  L2 . Quantum mechanically, the angular momentum is
1
KErot  I 2 
2
2I
2I
2
quantized, so that L  j  j  1 2 with j equaling an integer.
Now, this applies to each distinct axis of rotation. In three dimensions, we start with
three axes, but the symmetry of the molecule may reduce that number. The water
molecule has three axes, but a carbon monoxide molecule has only one. Basically, we
look for axes about which the molecule has a different moment of inertia, I. But it goes
beyond that. If the symmetry of the molecule is such that we couldn’t tell, so to speak,
47
whether the molecule was turning, then that axis does not count. That’s why there are no
states for an axis that runs through the carbon and oxygen atoms of carbon monoxide.
Therefore, a rotational partition function will look something like this for three axes
3
Z rot   e
i 1

j  j 1 i
kT
.
j
e. Translational partition function
In an ideal gas, the molecules are not interacting with each other. So the energy
1 2 p2
associated with the molecular center of mass is just the kinetic energy, mv 
. The
2
2m
molecule is confined to a finite volume, V, so that kinetic energy is quantized also.
First consider a molecule confined to a finite “box” of length Lx on the x-axis. The wave
2L
function is limited to standing wave patterns of wavelengths n  x , where
nx
nx = 1, 2, 3, 4, . . . This means that the x-component of the momentum is limited to the
h hnx

discrete values px , n 
. The allowed values of kinetic energy follow as
n 2 Lx
h 2 nx2
.
8mL2x
Naturally, the same argument holds for motion along the y- and z-axes.
En 
Ztr   e
nx
ny
 n2
n 2y
n2
h 2  x 2 
 z
 8 mLx 8 mL2y 8 mL2z


kT




nz
Unless the temperature is very low, or the volume V is very small, then the spacing
between energy levels is small and we can go over to integrals.
2 2
h 2 n 2y
   h 2 nx2
 1
 
  h nz
2

 
2
8 mLy
8 mLx
8 mL2z
Z tr    e
dn x  e
dn y  e
dn z  e kT
0

0
0



2mkT
2mkT
2mkT
 Lx
 Ly
 Lz
2
2
h
h
h2
 2mkT 
 V 

2
 h

3
2
3
 h2  2
  vQ is the quantum volume of a single molecule. It’s a box
The quantity 

2

mkT


whose side is proportional to the de Broglie wavelength of the molecule. In terms of that,
V
the Ztr 
.
vQ
48
[Actually, in the classical form of the partition function, we are integrating over the
possible (continuous) values of particle momentum and position.
Z tr , x 
Lx 
e

p x2
2 mkT
dpx dx  Lx 2mkT
0 0
The classical partition functions differ from the classical limit of the quantum mechanical
partition functions by factors of h. Because h is constant, this makes no difference in the
derivatives of the logarithm of Z.]
Putting the parts all together, for a collection of N indistinguishable molecules
1
Ztr Z rot Zint N
Z
N!
3.
Thermodynamic Properties of the Ideal Monatomic Gas
a. Helmholz free energy
Sec 6.5
Consider the derivative of the partition function with respect to temperature.

1 Z
ln Z 

T
Z T

1
e

Ei
kT
 e
T
Ei
kT
1
Ei  kTi
U
e
 2

2
Z
kT
kT
E

On the other hand, recall the definition of the Helmholtz free energy.
 F 
F  U  TS  U  T 

 T V
 F 
F  T 
  U
 T V
  F 
   
 T 
T 2     U
T





V
  F 
   
U
 T 
 T   T 2





V
Evidently, we can identify the Helmholtz free energy in terms of the partition function
thusly,
F  kT ln Z .
N
1 V 
For the monatomic ideal gas, Z    , whence (using the Stirling approximation)
N!  vQ 
49




F  kT N ln V  N ln vQ  N ln N  N  NkT ln N  1  ln vQ  ln V .
b. Energy & heat capacity
  ln vQ 
  ln Z 
1 vQ
  NkT 2
U  kT 2 
  NkT 2 

vQ T
 T V
 T V
 2mkT 
 NkT 

2
 h

3
 2mkT 
 NkT 

2
 h

3
2
  2mkT 


T  h 2 
2
3  2mkT 


2  h 2 
2
2
CV 
5

d. Entropy and chemical potential
2
2
 2mk  3
 2   NkT
 h  2
 3
U
 3

 NkT   Nk
T T  2
 2
c. Pressure
Of course, we’re going to get the ideal gas law.
  NkT ln N  1  ln vQ  ln V
 F 
P  
  
V
 V T , N 

3

  ln V
  NkT  V

T , N
 

  NkT ln N  1  ln vQ  ln V
 F 
S  
  
T
 T V , N 
 ln vQ
S  Nk ln N  1  ln vQ  ln V  NkT
T



NkT
 
V
T , N

V , N


 h2 

S  Nk ln N  1  ln vQ  ln V  NkT

2

mkT






3
2
 h2 

S  Nk ln N  1  ln vQ  ln V  NkT 

2

mkT


3
S  Nk ln N  1  ln vQ  ln V  Nk
2
  V  5
 
S  Nk  ln 
  NvQ  2 

 

50
  2mkT 


T  h 2 
2
2
3
 3  2mk
   2
 2 h
2


  NkT ln N  1  ln vQ  ln V
 F 
  
N
 N T ,V 
 ln N
 kT ln N  1  ln vQ  ln V  NkT
N
 V 

 kT ln 
 NvQ 


  

4.


T ,V

Solids Sec 2.2, 3.3, 7.5
In a solid, the atoms/molecules are each confined to specific locations. Therefore, we
regard them as distinguishable particles.
a. States of the system
Each particle has an equilibrium position about which it can vibrate like a harmonic
oscillator. So, we regard the solid as a collection of three-dimensional harmonic
oscillators. For a single oscillator, the partition function is a sum over harmonic
oscillator states
Z  e

Ei
kT
i 0
This is an infinite series, but it converges.
Z  e
 1
 i   h
 2

kT
i 0
Z
e

e

h
2 kT
h
2 h
3 h





1  e kT  e kT  e kT  




h
2 kT
1 e
ln Z  

h
kT
h



1 h
 ln 1  e kT 


2 kT


b. Heat capacity
The total energy for a system of N three-dimensional oscillators is



h 
1
U  3NkT 2
ln Z  N  h  h

T
2

kT
e 1

The heat capacity is
h
h
2
2


 h 
U
  h 
1  h 
e kT
e kT
CV 
 3N
 3Nk 

2
2
  3N k  T 
h
T
T  hkT
kT   h







 e 1
 e kT  1
 e kT  1








51
This result assumes that all N oscillators have the same frequency. The expression
approaches the Delong-Petit limit for high temperatures, but decreases exponentially at
low temperatures. Experiment shows that CV decreases like T3 at low temperature. The
crucial incorrect assumption in the Einstein model of a solid is that the oscillators do not
interact. In reality, they do not oscillate independently.
c. Metals
In a metallic solid, the valence electrons are liberated from the atoms, and are free to
move about among the atoms. As a result, the conduction electrons behave similarly to
atoms in a gas. Their energy states are those of particles confined in the volume of the
crystal, rather than the electronic states of the atoms from which the electrons escaped.
However, this electron gas is not an ideal gas. For one thing, the electron gas is dense,
and the electrons obviously interact with each other. They also interact with the ions of
the crystal lattice. Finally, electrons obey the exclusion principle, meaning that no two
electrons can be in exactly the same energy microstate.
You see that the CV goes above the Delong-Petit level as T increases.
52
5.
Photons Sec 7.4
a. Photon gas
Consider electromagnetic radiation inside a box. We may regard the electromagnetic
field as a superposition of standing waves that fit between the walls of the box. The
system, then, consists of these standing waves, rather than literal atoms oscillating back
and forth. The energy of a standing wave of a particular frequency, f, is quantized
ihf


1
.
Ei  ihf . Therefore, the sum over states for a single frequency is Z   e kT 
hf

i 0
kT
1 e
The average energy associated with that frequency is
1 Z
hf
.
 E  f   
 hf
Z kT 
kT
e 1
This expression is known as the Planck Distribution. Since each quantum of EM energy
is hf, the average number of photons of frequency f is
1
.
nPl  hf
kT
e 1
In effect, we are treating the system as a gas composed of photons.
b. Total energy
We have the average energy of photons of frequency f in the box. The total energy
contained in the box is obtained by summing over the allowed frequencies. The
frequencies are restricted by the finite volume of the box. For instance, along the x-axis,
nc
the frequencies of standing waves that will fit in the length Lx are f n x  x . The
2 Lx
n hc
corresponding energies are En x  hf n x  x . Of course, the same applies to the y- and
2 Lx
hc
nx2  n y2  nz2 . (Let’s say the box is a cube of side
z-axes. In three dimensions, En 
2L
L.) The total energy of the photons in the box is
hc
1
.
U 2 
nx2  ny2  nz2 En
nx ,n y ,nz 2L
e kT  1
We are adding up the points in a spherical volume of radius n  nx2  n y2  nz2 . Since the
number of photons in the box is very, very large at any but very low temperatures, the
sum can go over to an integral.


hcn
U  n
L
0
1
2
e
hcn
2 LkT
1

2
2
dn  sin d  d 
0
0
Now let us change variables from n to En.
53
 hc 
2 L
n3

0
e
hcn
2 LkT
dn
1
3
En 
 2L 
hcn
2L
dE and n3  
, whence dn 
En  .
2L
hc
 hc 
 hc 

8L3
E3
U
dn

dE
3
E
2 L 0 2hcn

hc  0 kT
LkT
e
1
e 1

3
U U
8
E
 
dE
3  E
3
L V hc  0 kT
e 1
4
5
U 8 kT 

3
V
15hc 
We can now compute also the heat capacity and entropy of the photon gas.
 U 
8 5 k 4 L3 3
CV  
T
  4
3
15hc 
 T V
n3
3
C (T )
32 5  kT 
S (T )   V
dT  
V  k
T
45  hc 
0
T
c. Black body spectrum
What we have here is the energy per unit volume per unit energy, u ( E ) 
8
hc 3
E3
E
kT
,
u(E)
e 1
also called the spectrum of the photons. It’s named the Planck spectrum, after the fellow
who first worked it out, Max Planck.
0
2
4
6
8
10
12
E/kT
U
 T 4 , and that the spectrum peaks at E  2.82kT . These “Laws” had
V
been obtained empirically, and called Stefan-Boltzmann’s “Law” and Wein’s
Displacement “Law.”
Notice that
d. Black body radiation
Of course, the experimentalists were measuring the spectra of radiation from various
material bodies at various temperatures. Perhaps we should verify that the radiation
emitted by a material object is the same as the spectrum of photon energies in the oven.
So, consider an oven at temperature T, and imagine a small hole in one side. What is the
spectrum of photons that escape through that hole? Well, the spectrum of the escaping
54
photons must be the same as the photon gas in the oven, since all photons travel at the
same speed, c. By a similar token, the energy emitted through the hole is proportional to
T4.
Finally, we might consider a perfectly absorbing material object exchanging energy by
radiation with the hole in the oven. In equilibrium (at the same T as the oven), the
material object (the black body) must radiate the same power and spectrum as the hole,
else they would be violating the Second “Law” of thermodynamics.
6.
Specific Heat (Heat Capacity) of Solids Sec. 7.5
a. Einstein solid
The total energy for a system of N three-dimensional oscillators is


h 
1
2 
U  3NkT
ln Z  N  h  h

T
2

kT
e

1


The heat capacity is
h
h
2
2


kT
 h 
U
  h 
1  h 
e
e kT
CV 
 3N

3
N

3
Nk




h

2
2
 kT 

h
T
T  kT
k  T   h






 e 1
 e kT  1
 e kT  1








[The text uses   hf for h .] This result assumes that all N oscillators have the same
frequency. The expression approaches the Delong-Petit limit for high temperatures, but
decreases exponentially at low temperatures. Experiment shows that CV decreases like T3
at low temperature. The crucial incorrect assumption in the Einstein model of a solid is
that the oscillators do not interact. In reality, they do not oscillate independently.
You see that the CV goes above the Delong-Petit level as T increases.
55
b.
Debye theory of specific heat
The oscillators do not vibrate independently. Rather, there are collective modes of
vibration in the crystal lattice. we’ll treat the situation as elastic waves propagating in the
solid. We consider that the energy residing in the elastic wave of frequency  is
quantized. Quanta of elastic vibrations are called phonons. Secondly, there is an upper
limit of the frequency that can exist in the crystal—the cut-off frequency.
So, consider sound waves propagating in the crystal, with the dispersion relation   vs q .
The total vibrational energy of the crystal will be
U     g  d .

The average energy of a mode is still    
e
g   

kT
1
and in a continuous medium
3V  2
3V

. Therefore, U 
 2 
d . Now, what is the range of
2
3
2 3 
2 vs
2 vs
e kT  1
frequency? Not 0,   , but 0, D  , where  D is the Debye frequency, or cut-off
frequency. The cut-off frequency arises because the shortest possible wavelength is
determined by the inter-atomic spacing. Put another way, the maximum possible number
of vibrational modes in the crystal is equal to the number of atoms (let’s say a mole) in
the crystal, times 3. I.e.,
D
 g  d  3N
A
0
D

0
3V  2
d  3 N A
2 2 vs3
3V  D3
 3N A
2 2 3vs3
6 N A 2vs3
V
With this as the upper integration limit, the total energy becomes
D3 
3V

3V
U
 2 
d 
2 3 
2 vs
2 2vs3
e kT  1
In turn, the specific heat is
D

D
 3
e

0
kT
d .
1
3

 T  D   
e kT


d


9
R
d .
2
0   kT 2
     kT 




D  0 

kT
 1
 1
e
e




 D
In the last expression, the Debye temperature,  D 
, is introduced.
k
3V  2
 U 
Cv  


2 3
2
 T V 2 vs kT
 4e
kT
4
N 
      
   D     vs3  6 2 A 
V 
 k  k 
3 k 3 V
vs3  D2 3
6  N A
3
3
D
56
3
Let’s look at the high and low temperature limits.
For T   D ,

d  1   D  , so that Cv  3R .
3 T 
3
3
4 4
12 4  T 
  T 3 .
, so that Cv 
R
For T   D , D  , and   d 
15
5

 D
Evidently, the Debye version matches the experimental temperature dependence of the
specific heat at low temperatures, as the classical and Einstein theories do not.
We can relate the Debye temperature to the Young’s modulus of the crystal, through the
wave speed.
D   2 N A 
D 
 vs  6

k
k 
V 
Now, the density is  
c.
1
3
 Y  2 NA 

  6

k  
V 
N AM
, so qualitatively,  D 
V
1
3
Y
. (M is the atomic mass.)
M
Reduced temperature
T
is used when plotting the heat
D
capacity. The Debye temperature contains the information about the particular crystal
substance. When plotted versus reduced temperature, the CV curve is identical for all
substances.
Often, a quantity called the reduced temperature, T * 
57