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1
CONTENTS
Contents
1 Introduction
3
2 Atoms
2.1 The atomic hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 The kinetic theory of an ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.2 Problem: Marianas Trench . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.3 The kinetic theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.4 Counting collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2.5 Number density of air molecules . . . . . . . . . . . . . . . . . . . . . . . .
2.2.6 Problem: molecular beam epitaxy . . . . . . . . . . . . . . . . . . . . . . .
2.3 Temperature and the Boltzmann factor . . . . . . . . . . . . . . . . . . . . . . . .
2.3.1 Boltzmann factor and chemical reaction rates . . . . . . . . . . . . . . . . .
2.4 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.4.1 Problem: calorimetric determination of the latent heat of vapourisation of
water . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 The velocity distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.5.1 Problem: Estimate the spectral width of the Na D-line emission from a street
lamp. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6 How fast do molecules fly? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.1 Speed of ambient air molecules . . . . . . . . . . . . . . . . . . . . . . . . .
2.6.2 Cold atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7 The equation of state of an ideal gas . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.1 Ideal gas temperature: the constant volume gas thermometer . . . . . . . .
2.7.2 How big is one mole? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.7.3 Problem: candle flame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.8 Internal energy and the classical equipartition theorem . . . . . . . . . . . . . . . .
2.8.1 Specific heat capacity of a monatomic gas . . . . . . . . . . . . . . . . . . .
2.8.2 Specific heat capacity of a simple solid . . . . . . . . . . . . . . . . . . . . .
2.9 Diatomic gases: failure of the equipartition theorem . . . . . . . . . . . . . . . . .
2.10 Thermal equilibrium and electromagnetic radiation . . . . . . . . . . . . . . . . . .
2.10.1 Radiation balance between the sun and the earth . . . . . . . . . . . . . . .
2.11 Mean free path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.11.1 Mean free path of ambient air molecules . . . . . . . . . . . . . . . . . . . .
2.12 Transport processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.12.1 Example: thermal conductivity and mean free path of argon gas . . . . . .
2.13 Interatomic forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.13.1 Young’s modulus and bulk modulus . . . . . . . . . . . . . . . . . . . . . .
2.13.2 Thermal expansivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.14 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3 The energy principle
3.1 The 1st Law . . . . . . . . . . . . . . . . . . .
3.2 Reversible work: the indicator diagram . . . .
3.2.1 Problem: compression of an ideal gas.
3.2.2 Isothermal compression of an ideal gas
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2
CONTENTS
3.3
3.4
3.5
3.6
3.7
4 The
4.1
4.2
4.3
4.4
Principal specific heats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.3.1 Adiabatic compression of an ideal gas . . . . . . . . . . . . . . . . . . . . .
Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.4.1 Problem: Prove two relationships between the thermal properties of a hydrostatic system. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Heat engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.5.1 Problem: Efficiency of an ideal gas Rankine engine . . . . . . . . . . . . . .
Heat pumps and refrigerators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
entropy principle
The 2nd Law of Thermodynamics . . . . . . . . . . . . . .
Carnot’s theorem . . . . . . . . . . . . . . . . . . . . . . .
The Carnot cycle . . . . . . . . . . . . . . . . . . . . . . .
The Clausius inequality . . . . . . . . . . . . . . . . . . .
4.4.1 Example: Theoretical limiting efficiency of a steam
4.5 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.5.1 Example: Hot object cooled in water . . . . . . . .
4.5.2 Isothermal compression of an ideal gas . . . . . . .
4.5.3 Joule expansion of an ideal gas . . . . . . . . . . .
4.5.4 Entropy change during a change of phase . . . . .
4.6 Statistical interpretation of entropy . . . . . . . . . . . . .
4.7 1st and 2nd law for an infinitesimal change . . . . . . . .
4.7.1 Entropy of an Ideal Gas . . . . . . . . . . . . . . .
4.7.2 Two systems in thermal equilibrium . . . . . . . .
4.7.3 Problem: Thermodynamics of a rubber band. . . .
4.8 The availability of work . . . . . . . . . . . . . . . . . . .
4.9 Thermodynamic potentials . . . . . . . . . . . . . . . . .
4.9.1 Example: gas cylinder . . . . . . . . . . . . . . . .
4.9.2 Example: Release of heat in combustion . . . . . .
4.9.3 Example: graphite versus diamond . . . . . . . . .
4.10 Real gases: interatomic forces and phase transitions . . .
4.10.1 Example: Melting point of ice . . . . . . . . . . . .
4.10.2 Example: Boiling point of water . . . . . . . . . .
4.11 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1 INTRODUCTION
1
3
Introduction
These notes define the content and level of the First Year Energy and Matter Core Course,
PHYS1013. They must be used together with textbooks that give a fuller discussion of the material. About 3/4 of the course material is covered in the custom Pearson text: University Physics
Semester 2. For those topics which go beyond Pearson, I recommend Concepts in Thermal Physics,
by Stephen J. Blundell and Katherine M. Blundell, Oxford (2006).
Bring these notes with you to lectures. They contain blanks to fill in, for example around unlabelled diagrams used to set up key derivations, and left as space to working problems that we
shall solve in lectures.
Every week at the top of your Problem Sheet there will be a reading assignment, directing you
to the parts of Pearson and Blundell that are relevant for that week’s work. There may also be
references to other texts: these are optional, but you will find them helpful if you want to go
further. I particularly recommend The Feynmann Lectures on Physics (vol. 1), by Feymann,
Leighton and Sands: they are inspiring but also hard work. To keep up with the course and get the
most out of the lectures you must tackle the reading assignment in advance of the lectures.
Every week you must hand in solutions to both of the questions on your Problem Sheet. One
of these will be marked: you will not know in advance which one. Problem solving is a key skill
for physicists. Present your work as if you were a consultant preparing a report: your customer
needs your account to be as short as possible but very clear and easy to understand, with your
assumptions and your logic explained at each step. Assessment of the course is based 70% on the
written examination; 20% on the problem sheets and 10% on the mid-semester test.
You will find all the course materials on the course website at
http://www.phys.soton.ac.uk/teach/year1/notes/phys1013/home1013.html
Here is a list of fundamental constants and conversion factors for quick reference. Values are rounded to 3 significant figures: this is almost always sufficient for the problems we shall
tackle.
1 atmosphere (atm) = 1.01 × 105 Pa
1 psi = 1 lb/in2 =6.90×103 Pa
1 torr = 1 mm Hg at 0 deg C = 133 Pa
1 mbar = 100 Pa = 0.987 × 10−3 atm
Unified mass constant (u) =1.66 × 10−27 kg
Boltzmann’s constant: kB = 1.38 × 10−23 JK−1
Planck’s constant: h = 6.63 ×10−34 Js
Avogadro’s number: NA = 6.02 × 1023
Gas constant: R = NA kB = 8.31 JK−1 mol−1
Zero degrees Centigrade = 273 K
Stefan-Boltzmann constant: 5.67 ×10−8 Wm−2 K−4
Anne Tropper
February 2010
2 ATOMS
2
4
Atoms
By convention there is bitterness, by convention there is sweetness, colour, heat and cold: in
reality there are only atoms and empty space.
Democritus, 450 BCE.
The purpose of this section is
• to introduce the atomic hypothesis;
• to use the kinetic theory to make the connection between the motion of atoms and the
large-scale physical properties of a gas;
• to define the absolute temperature of a system in thermal equilibrium and introduce the
Boltzmann distribution;
• to derive the equation of state of an ideal gas;
• to study the classical equipartition theorem;
• to investigate the transport properties of ideal gases using the kinetic theory; and
• to introduce the relation between interatomic forces and the properties of matter.
2.1
The atomic hypothesis
“If, in some cataclysm, all of scientific knowledge were to be destroyed, and only one
sentence passed on to the next generations of creatures, what statement would contain
the most information in the fewest words? I believe it is the atomic hypothesis (or
the atomic fact, or whatever you wish to call it) that all things are made of atoms little particles that move around in perpetual motion, attracting each other when they
are a little distance apart, but repelling upon being squeezed into one other. In that one
sentence, you will see, there is an enormous amount of information about the world, if
just a little imagination and thinking are applied.”
Richard P. Feynman, The Feynman Lectures in Physics, volume 1, 1963
2.2
The kinetic theory of an ideal gas
The aim of this section is to prove a relationship between the pressure of a gas, P , and the motion
of its individual atoms:
1
(1)
P = nmhv 2 i,
3
where n is the average number density of atoms in the gas, m is the mass of one atom, and hv 2 i is
the average of the squared speed of the atoms. We need to find out whether the atomic hypothesis
can explain the observed properties of stuff on the large scale - the macroscopic properties. First
we must understand what we mean by pressure.
2 ATOMS
2.2.1
5
Pressure
Pascal’s Law states that the pressure applied to a confined fluid at any point is transmitted undiminished throughout the fluid in all directions, and acts upon every part of the confining vessel at
right angles to its interior surfaces, equally upon equal areas.
Suppose we have some gas is in a cylinder of cross-section area A, retained by a piston which
is held down by a weight of mass M . The cylinder is inside an evacuated chamber, so there is no
surrounding atmosphere to exert force on the piston. In equilibrium the weight pushes down with
force M g, and the gas pushes back with an equal and opposite force, uniformly distributed over
the whole area of the piston. The gas exerts a pressure P , where P = M g/A.
The SI unit of pressure is the N/m2 , called the pascal (Pa); but we find many other units in
Figure 1: On the left: Diagram showing forces on mass supported by gas pressure. On the right: Forces on
a layer of fluid in equilibrium under gravity.
practical use:
1
1
1
1
atmosphere (atm) = 1.013 × 105 Pa
psi = 1 lb/in2 =6.895×103 Pa
torr = 1 mm Hg at 0 deg C = 133.3 Pa
mbar = 100 Pa = 0.987 × 10−3 atm.
Pressure increases with depth. Consider the thin layer of fluid, thickness dh, shown in the ves-
6
2 ATOMS
sel on the right of the figure. In equilibrium, the net force exerted by the pressure of the fluid above
and below must support the weight of the layer:
ρAgdh = AP (h) − AP (h + dh).
If the layer is very thin, the difference in pressure above and below is very small, and we can write
P (h + dh) = P (h) + dP,
where dP is an infinitesimal quantity. Combining these expressions we obtain the equation of
hydrostatic equilibrium:
dP
= −ρg.
(2)
dh
Now we can understand Archimedes’ principle and buoyancy. Consider an object immersed in
fluid. From Pascal’s Law, the total Hforce that the fluid exerts on the object is given by an integral
over the surface of the object: F = P dS. This force is just enough to balance the weight of fluid
with the same volume V as the object: F = −ρV g. It follows that the buoyancy force pushing up
on an immersed object is equal and opposite to the weight of the fluid that the object displaces.
This is the principle of Archimedes.
2.2.2
Problem: Marianas Trench
Prove that in an incompressible fluid pressure increases linearly with depth.
Estimate a value for the pressure of the ocean at the deepest point of the Marianas Trench (western
Pacific Ocean), 11 km below the surface. You may assume that seawater is an incompressible fluid,
density 1000 kg m−3 , and take atmospheric pressure to be 100 kPa.
7
2 ATOMS
2.2.3
The kinetic theory
The kinetic theory model of an ideal gas is based on six assumptions:
1. A gas is composed of a large number of molecules.
2. The molecules are small compared to their separation.
3. The molecules are uniformly distributed and move randomly.
4. The molecules obey Newton’s laws of motion.
5. The molecules feel no force except during collisions with other molecules or the walls of the
container; they are “hard spheres”.
6. Molecules collide elastically, and the walls are smooth.
Assumptions 2 and 5 fail at high densities when the molecules are close together. Thus kinetic
theory cannot be expected to describe either liquids or solids.
Assumption 4 is wrong: quantum mechanics rules. In reality the behaviour of matter expresses
wave-particle duality: a particle of momentum p = mv is characterised by its de Broglie
wavelength:
h
λ= ,
(3)
p
where h is Planck’s constant; h = 6.62607544 × 10−34 J s. But quantum predictions tend to
those of classical physics if the de Broglie wavelength of the atoms is much less than their mean
separation. This is true if the if the atoms are fast enough, or if their number density is low enough.
(These are just the conditions needed for assumptions 2 and 5 to hold as well.) Quantum effects
can be very important and remarkable (e.g. the amazing properties of liquid helium), but they are
for the most part beyond the remit of this course. The important point to realise is that kinetic
theory is approximate and applies best to gases at low density and high temperature.
2.2.4
Counting collisions
Figure 2: Molecule bouncing off a wall.
Gas exerts pressure on the walls that contain it because gas atoms or molecules repeatedly
bounce off the wall, and transfer momentum to it with each bounce. From Newton’s 2nd Law
8
2 ATOMS
we know that the force on the wall must be equal to the momentum absorbed per unit time.
A single elastic molecular collision transfers momentum 2mv cos θ to the chamber wall (assumption 4), where m is the mass of the molecule, v is its speed, and θ is the angle of incidence with
which it approaches the wall.
In our model gas molecules travel randomly in all directions, with speeds varying from stationary
Figure 3: On the left: diagram defines the angles which specify the orientation of the velocity of a molecule.
On the right: diagram shows the cylinder of length vdt which contains all the molecules of a particular
orientation and speed that will strike the wall in the next interval of time dt.
to very fast. We begin our calculation by finding the contribution to the pressure, dP , made just
by the subset of the molecules which are approaching the wall in the specific direction defined by
the angles θ and φ, as explained in the left-hand sketch in the figure. Actually, we specify a very
narrow range of orientations, dθ and dφ, so that some well-defined fraction of our molecules will
have directions within this range. The fraction of our molecules with velocities oriented in this
range is the same as the fraction of the area of the sphere that lies within the little box. If the box
is small enough, it is very close to a rectangle, with height (along a circle of longitude) rdθ, and
width (circle of latitude) r sin θdφ, where r is the radius of the sphere.
Thus the fraction of (θ, φ) molecules in our gas is
rdθ r sin θdφ
sin θdθdφ
=
2
4πr
4π
(4)
9
2 ATOMS
Now consider the cylinder sketched on the right of Fig. 3. All the (θ, φ) molecules travelling
with speed v inside this cylinder will hit the horizontal wall within the next interval of time dt.
The volume of this cylinder is Avdt cos θ. Let us define the number density, n; the number of
molecules per unit volume of the gas. Using assumptions 1 and 3 of the kinetic theory we can treat
n as constant for a particular equilibrium state of the gas; fluctuations in n are very small indeed,
despite the incessant molecular motion. Then at any moment there will be nAvdt cos θ molecules
inside the cylinder, of which the fraction given by the expression in Eq. 4 are (θ, φ) molecules. Only
some of these molecules will have speed close to v: let the fraction of all molecules with speed in
the range v to v + dv be f (v)dv.
Putting all this together, we can say that the contribution to the pressure from just our special (θ, φ) molecules with speed v will be given by:
number density of all the molecules,
× fraction of those molecules within our chosen range of (θ, φ),
× fraction of those within dv of our chosen speed,
× volume of our little cylinder,
÷ time dt it takes all the molecules in the cylinder to hit the wall,
× momentum that each molecule transfers,
÷ cross-section area of the cylinder.
In algebra it looks like this:
dP = n ×
sin θdθdφ
Avdt cos θ 2mv cos θ
× f (v)dv ×
×
.
4π
dt
A
(5)
Some cancelling simplifies this expression:
dP =
1
nmv 2 f (v)dv sin θ cos2 θdθdφ.
2π
To find the total pressure, P , we have to add together contributions from molecules travelling at
all possible angles, with all possible speeds. We therefore integrate this equation with respect to
angle and speed. Angle φ can vary from zero to 2π. Angle θ can vary from zero (normal incidence)
to π/2 (glancing incidence). The speed v can vary from zero (stationary) to infinite - we have to
assume that any sensible speed distribution, f (v), will give zero probability of infinite speed.
1
nm
P =
2π
Z∞
2
v f (v)dv
0
π/2
Z
sin θ cos2 θdθ
Z2π
0
0
The integral over φ is easy:
Z2π
dφ = 2π.
0
So is the integral over θ:
π/2
Z
0
1
sin θ cos2 θdθ = − cos3 θ
3
·
¸π/2
0
1
= .
3
dφ.
10
2 ATOMS
The integral over the molecular speed distribution
gives us an average over the square of the
­ 2®
molecular speed; the mean square speed v :
Z
D
E
v 2 f (v)dv = v 2 .
(6)
Our final result is very simple:
D E
1
P = nm v 2 .
(7)
3
We have derived it without knowing anything about how fast the molecules fly through space
between collisions, or what the speed distribution looks like. This is the next gas property that we
need to think about.
2.2.5
Number density of air molecules
We shall find out in section 2.6.1 how to calculate the average speed of a gas molecule; for now
take it on trust that at ambient temperature the root mean square speed of a nitrogen molecule,
defined by
q
(8)
vrms = hv 2 i,
is roughly 500 ms−1 . For comparison, the Eurofighter Typhoon has a top speed at sea level of
Mach 1.2, or 350 m/s, though at altitude it can get up to Mach 2. The mass of N 2 is 28u, where
‘u’ is the symbol for the unified mass unit, formerly known as the atomic mass unit or amu:
1 unified mass unit (u) =1.661 × 10−27 kg.
Now we can use Eq. 7 to estimate the number density of molecules in air. For simplicity assume that air contains only N2 molecules, and that it behaves like an ideal gas. Take ambient
pressure to be 100 kPa.
n≈
3P
3 × 105 [Nm−2 ]
≈
≈ 2.5 × 1025 m−3
m hv 2 i
(28 × 1.7 × 10−27 )[kg] × 5002 [m2 s−2 ]
Now estimate the number density of air molecules at ambient temperature in an ultra-high vacuum
(UHV) system at a pressure of about 10−10 torr.
n≈
3P
3 × (10−10 × 133)[Nm−2 ]
≈
≈ 3.3 × 1012 m−3
m hv 2 i
(28 × 1.7 × 10−27 )[kg] × 5002 [m2 s−2 ]
so remember that gas molecules are very light, very fast; and that, even in the cleanest systems
that technology can devise, the UHV chambers in which computer chips and nanostructures are
made, there are enormous numbers of them.
2.2.6
Problem: molecular beam epitaxy
The technique of molecular beam epitaxy (MBE) laid the foundations of nanotechnology, making
it possible to lay down layers of different atoms in a single crystal, one layer of atoms at a time.
MBE takes place in a UHV chamber; the source of atoms is a Knudsen cell, in which a heated
solid exerts a vapour pressure. Atoms or molecules effuse out of the cell into the UHV chamber
through an aperture, and part of this plume is deposited on the substrate, a highly polished thin
11
2 ATOMS
disc. A mechanical shutter can open and close the Knudsen cell, controlling the thickness of each
layer. In this example we think about MBE of gallium arsenide (GaAs): the laser that operates a
DVD drive is made from alloys of this semiconductor material.
Consider an MBE machine with a Ga source which is a Knudsen cell containing solid Ga at 1310
K. The vapour pressure of Ga atoms in the cell is 1 Pa, the atoms have rms speed 680 ms −1 , and
the aperture of the cell has area 24 mm2 . 60% of the effused atoms are deposited uniformly on the
surface of a 4-inch diameter substrate, where they line up in a regular square array, with repeat
distance 0.25 nm. The mass of a Ga atom is 70u, and you may assume that
hvi =
r
8
vrms = 0.92 vrms .
3π
(9)
Estimate the time that it takes for this source to lay down a single monolayer of Ga over the whole
area of the substrate.
Figure 4: Molecules of Ga metal effusing out of a Knudsen cell and being deposited onto a substrate. The
process takes place inside an ultra-high vacuum chamber.
12
2 ATOMS
2.3
Temperature and the Boltzmann factor
How fast do gas molecules fly? There is no single answer; some molecules are faster, some slower;
the range of speeds is wide. There will be some slow molecules even in a very hot gas, and vice
versa; but we do suspect that on average molecules fly faster in hotter gases. The purpose of this
section is to introduce a statistical definition of the absolute temperature of a gas.
Before we can define what we mean by temperature, we must ask whether our gas actually has
a temperature at all. Think about the instant of a gas explosion. At the centre of the explosion
there are some very fast molecules; shock waves radiate outwards. The molecular velocities are not
random; they are correlated with position relative to the centre and with time. But if we wait long
enough the shock waves propagate out and die away; the faster molecules and the slower molecules
collide incessantly with each other and with solid boundaries, and these collisions randomise the
molecular momenta and velocities. Eventually (if nothing else explodes or changes) the gas reaches
a state of thermal equilibrium, and we know when it has got there because nothing on the
macroscopic scale is changing any more. In thermal equilibrium the gas isn’t going anywhere, so
the molecules are just as likely to be travelling left as right; there cannot be a concentration of fast
molecules in one particular place, because then we could see energy from these molecules being
transferred outwards from this place by collisions, and the gas would be changing over time. In
thermal equilibrium the molecular velocities are random.
Systems in thermal equilibrium obey a principle that is known as the zeroth law of thermodynamics:
If system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium
with system C; then system A and system B will also be in thermal equilibrium with each other they can be brought into contact and the state of neither will change. A, B and C are all at the
same temperature.
Think about a single molecule inside a gas which is in thermal equilibrium. Our chosen molecule
continually suffers collisions; sometimes it gains energy from a collision and rebounds at high speed;
sometimes it is briefly brought almost to a standstill. Our molecule (the system) is continually exchanging energy with the gas (the reservoir) - we say that its microstate is changing, even though in
the macroscopic world all the measurable physical properties of the gas as a whole stay the same in
time. Although we cannot predict the velocity of our molecule at any moment, we can say precisely
that the relative probability of a state of velocity v is given by the Boltzmann distribution:
pr (v) = exp(−E(v)/kB T )
(10)
where E(v) is the kinetic energy of the molecule:
´
1
1 ³
E = m |v|2 = m vx2 + vy2 + vz2 .
2
2
(11)
In the 2nd year you will derive the Boltzmann distribution from statistical principles: for now we
accept it without proof as an axiom of thermal physics. It is not restricted to ideal gases - or to
kinetic energy; it holds for any system in the classical limit, where the de Broglie wavelength of
an individual atom or molecule is short compared to the spacing between atoms. Remember that
this distribution tells us only about the relative probability that a molecule has a particular energy.
Absolute probabilities have to add up to unity when they are summed over all possible states of a
13
2 ATOMS
1.0
0.8
relative probability
hot gas
0.6
cold gas
0.4
0.2
0.0
0
10
20
30
40
energy
Figure 5: The Boltzmann distribution, plotted for a ‘cold’ gas, at a temperature of 5 energy units, and a
‘hot’ gas, at a temperature of 35 energy units.
molecule - and there will generally be large numbers of different states with the same energy.
The temperature T in the Boltzmann distribution is the absolute temperature. This absolute temperature scale has a zero with a very special physical meaning: a classical system at zero
absolute temperature has every atom in a state of zero energy, completely motionless.
The SI unit of temperature is called the Kelvin (K), and its magnitude is fixed with respect
to a special state of water, the water triple point, where the solid, liquid and vapour phases
of water are all in mutual equilibrium. We define the temperature of the water triple point to
be 273.16 K; this choice gives us (almost exactly) 100 degrees difference between the boiling and
freezing points of water; same size of degree as the historic Centigrade scale. The freezing point
of water under a pressure of 1 atm is a smidgen colder than the triple point, at 273.15 K. The
significance of the water triple point will become clearer later on when we study phase transitions:
for now just understand that it represents a far more precisely reproducible temperature than the
ice point, which varies with applied pressure.
If physics had developed in a different historical sequence, we might conceivably have started out
by measuring temperature in units of energy; this would have made perfect sense, and we should
not have needed an additional physical constant in the Boltzmann distribution. As it is, we need
the Boltzmann constant kB , which multiplies temperature in Kelvins to convert it to energy in
Joules:
kB = 1.380658(12) × 10−23 JK−1 .
2.3.1
Boltzmann factor and chemical reaction rates
The Boltzmann factor has important consequences for the rates of chemical and other reactions.
Consider a mixture of hydrogen and oxygen. At room temperature it is stable, and can last for
years in this state. However a tiny spark can set off an explosive reaction resulting in the formation
of water and the liberation of a great deal of heat. The water molecule therefore represents a
more stable system than free H and O atoms, with lower potential energy: but this stable state
14
2 ATOMS
cannot necessarily be reached, because a certain (positive) amount of energy is required to start
the reaction. This is the activation energy, EA . For a system like hydrogen and oxygen, it is clear
that at ambient temperature EA À kB T , or the reaction would start spontaneously. In this case
the Boltzmann factor e−E/kB T can make rates very sensitive to temperature.
If the activation energy, EA = 100 kB T , and T = 300 K, by how much does the Boltzmann
factor change if T is raised by 20 K?
The initial value of the Boltzmann factor is e−100 . For T = 320 K, EA /kB T = 100(300/320) =
93.75, so the new value is e−93.75 . The ratio is
e−93.75
= e100−93.75 = e6.25 = 518.
e−100
Thus a 20-K increase in temperature causes a 500-fold increase in the Boltzmann factor. This is
why chemical reaction rates are so sensitive to temperature. Catalysts are substances that can
reduce activation energies and allow reactions to take place at much lower temperatures; they are
essential in biology.
2.4
Heat
The purpose of this section is to introduce the idea of heat.
We define heat as energy that is transferred from one system to another as a result of a temperature gradient. The quantity of heat transferred in a process that raises the temperature of
mass m of a substance with specific heat capacity c by an amount ∆T is
Q = mc∆T.
(12)
Within an isolated system, the flow of heat irons out temperature differences; over time the system
reaches thermal equilibrium at a uniform temperature.
Substances can often exist in different phases, and make phase transitions in response to the
transfer of heat. Phase transitions typically occur at sharply defined temperatures, used as fixed
points for practical temperature scales.
Some phase transitions involve latent heat. For example, heat must be supplied to ice at 0 degC
to convert it to water at 0 degC. In general, to melt mass m of a solid with latent heat of fusion
L, requires heat
Q = mL.
(13)
The latent heat of fusion of ice is 3.34 × 105 J/kg.
The experimental technique for measuring heat capacity and latent heat is called calorimetry;
it involves mixing known masses of hot and cold materials in a calorimeter, which is a thermally
isolated container, from which no heat can leak in or out.
2 ATOMS
2.4.1
15
Problem: calorimetric determination of the latent heat of vapourisation of
water
A calorimeter contains 0.6 kg of water at 12 deg C. Steam at 100 deg C is passed through the
calorimeter for several minutes; then the steam is shut off, and the contents of the calorimeter are
allowed to reach thermal equilibrium. The calorimeter is then found to contain 0.69 kg of water
at a temperature of 68 deg C. Use these measurements to calculate a value for the latent heat of
vapourisation of water. You may assume that the specific heat capacity of water is 4190 J/(kg deg
C), and does not vary over the temperature range of the experiment.
16
2 ATOMS
2.5
The velocity distribution
The purpose of this section is to show, based on the Boltzmann distribution, that the velocity distribution of gas molecules has a Gaussian shape. (Please note: not the speed distribution - you will
study this next year.) The velocity distribution is important because we see it experimentally in
the spectrum of light emitted by gases. Thus we have direct experimental evidence for the validity
of the Boltzmann distribution.
Consider the translational kinetic energy of gas molecules. (The molecules may also have other
forms of energy; they may be spinning or vibrating, but this does not affect our conclusions about
their velocity distribution.) Each molecule, mass m, has (time-varying) velocity
v = (vx , vy , vz )
(14)
and energy
1
1
1
E = mvx2 + mvy2 + mvz2 .
(15)
2
2
2
We can use the Boltzmann distribution to derive the velocity distribution for the gas. The relative
probability that, at some instant, one chosen molecule has velocity v is given by:
Ã
pr (v) = exp −
Ã
mvx2
= exp −
2kB T
!
1
2
2 m(vx
mvy2
exp −
2kB T
Ã
!
+ vy2 + vz2 )
kB T
Ã
mvz2
exp −
2kB T
!
!
.
We see that, because the kinetic energy of the molecule is a sum of contributions from motion
in the x, y and z directions, the relative probability for some particular velocity factorises into
the product of three independent probability distributions. For molecules in thermal equilibrium,
there is no correlation between the different velocity components; the motion is random. Of course,
molecules that have just emerged into a UHV chamber in a beam do not move randomly - and are
therefore not in thermal equilibrium. But a very short period spent bouncing off the walls of the
chamber, each other, and the huge number of air molecules that were there already is enough to
randomise their motion; and their energy distribution reverts to the Boltzmann factor.
Lets think about the x-component of velocity in isolation. At any instant the molecule has momentarily a well-defined value of vx , such that
−∞ < vx < +∞.
We can plot the probability
distribution
for vx : it is the bell-shaped Gaussian distribution, with
¢
¡
the functional form exp −X 2 , sketched in Fig.3. Notice first that the most probable value for
vx is zero. Do not let this confuse you: the most probable value for the speed of the molecule,
³
´1
v = vx2 + vy2 + vz2 2 is NOT zero, but the reason for this apparent contradiction will become clear
to you when you study the Maxwell-Boltzmann speed distribution in the 2nd year. For the
time being we stick with the velocity distribution, and avoid the statistical subtleties as best we can.
17
2 ATOMS
probability distribution
0.4
0.2
0.0
-6
-4
-2
0
2
4
6
x-component of velocity
Figure 6: The probability distribution for one (any) component of the velocity of a molecule in a gas in
thermal
equilibrium. This distribution is normalised ; the area under the curve is 1. The units of speed are
q
kB T
m .
Now, it can be shown that the total area under a Gaussian bell curve has a finite value:
+∞
Z
−∞
exp −X 2 dX =
√
π.
(16)
This means that we can multiply pr (vx ) by some constant, K, to give us the absolute probability
distribution for vx . The area under the absolute probability distribution has got to be 1: it is
the integral of the absolute probability over all possible values of vx , which is the same as the
probability that there is some - any - value of vx that characterises the molecule at a given instant.
And that is a certainty. Let us do the analysis and find out the value of K.
Ã
mvx2
p (vx ) = Kpr (vx ) = K exp −
2kB T
If
Z
then
K=
"Z
!
p (vx ) dvx = 1,
Ã
mvx2
exp −
2kB T
!
dvx
#−1
.
The technique we need here is that of change of variable: define
X = vx
r
m
2kB T
dX = dvx
Hence
Z
Ã
mvx2
exp −
2kB T
!
dvx =
r
s
m
.
2kB T
2πkB T
.
m
18
2 ATOMS
So here is our final result: an expression for the absolute distribution of probability that a molecule
of mass m in a gas in thermal equilibrium at temperature T will have the a velocity with xcomponent vx :
!
Ã
r
m
mvx2
p (vx ) =
.
(17)
exp −
2πkB T
2kB T
It is this normalised probability distribution that is plotted in Fig. 3. We see that the distribution
√
falls to 1/ e, or 60% of its peak value symmetrically either side of zero at the points
vx = ±
s
kB T
.
m
This means that for a hotter gas, with larger T , the distribution is broader, which means that the
molecules on average are faster. On the other hand, heavier molecules at a given temperature have
a narrower distribution than lighter ones, so these on average fly slower.
2.5.1
Problem: Estimate the spectral width of the Na D-line emission from a street
lamp.
A street lamp contains Na atoms that, when excited by an electric discharge, emit yellow resonance
radiation at 509 THz. Calculate the Doppler linewidth of this light, assuming that the temperature of the Na gas in the lamp is 350 K. You may assume that, if the x-direction is the line of sight
from the observer to the lamp, then the frequency ω measured by the observer of light emitted by
a particular atom will be related to the resonance frequency ω0 of the atom at rest by
ω = ω0
µ
vx
,
1+
c
where vx is the x-component of the atomic velocity.
¶
(18)
19
2 ATOMS
2.6
How fast do molecules fly?
Now that we understand that there is a distribution of molecular speeds in a gas, we can make
life a bit simpler by defining a typical speed for a molecule of given mass, in a gas at thermal
equilibrium. Our ‘typical’ speed is a statistical average. Go back to our velocity distribution,
Eq. 17. The average value of vx is zero, because at thermal equilibrium vx is positive as often as
it is negative. So this is not a very useful quantity. If, however, we average the square of v x , then
every molecule makes a positive contribution to the sum, and the number we get starts to be a
useful measure. What we mean literally by the average of the square of the x-component of the
velocity of our gas molecules is
D
³
E
2
2
2
+ ... + vxN
vx2 = N −1 vx1
+ vx2
´
where we have N molecules altogether, the nth molecule has vxn , and we use the angle brackets
to denote a time average. Given the likely magnitude of N , this threatens to be a very long sum
indeed. Our preferred approach, fortunately, only takes a minute or two; we use the probability
distribution of Eq. 17 and write
D
vx2
E
=
Z
vx2 p(vx )dvx
=
r
m
2πkB T
Z
vx2 exp
Ã
mvx2
−
2kB T
!
dvx .
The integral looks nasty, but isn’t, because
Z
´
³
x2 exp −Ax2 dx = −
d
dA
Z
´
³
exp −Ax2 dx = −
d
dA
µr
π
A
¶
=
1
2A
r
π
.
A
(Note: this is the sort of cowboy mathematics that gets physicists a terrible bad name - swapping
the order of differentiation and integration between infinite limits without any heed to the convergence of anything. That is why you are taught mathematics only by professional mathematicians
and proper theoretical physicists. Here we get away with it because the Gaussian function is so
beautifully well-behaved.)
Now put all this together:
D
E
vx2 =
r
m
2kB T
×
×
2kB T π
2m
s
2πkB T
kB T
=
.
m
m
(19)
We are not there yet: we want a measure of the speed of a molecule, v, where
v 2 = vx2 + vy2 + vz2 .
Now, averaging is a linear process, so the mean square speed of the molecule is just the sum of the
three mean square velocity components:
D
E
D
E
D
® D
E
E
D
E
v 2 = vx2 + vy2 + vz2 .
The molecular velocities are random, so vx2 , vy2 , and vz2 must all have the same magnitude,
which we calculated in Eq. 19. So here we have our final result, and it is very important:
­
D
E
v2 =
3kB T
.
m
­
®
(20)
20
2 ATOMS
The root-mean-square (rms) speed of the molecules is just the square root of this quantity.
The most interesting result of all is that the average kinetic energy of the molecule depends
only on the absolute temperature of the ideal gas:
1 D E 3
hK.E.i = m v 2 = kB T.
(21)
2
2
This result follows from the classical equipartion theorem, which we shall explore shortly. Notice that the right-hand side of Eq. 21 is the product of two factors: 12 kB T , which is a measure of
mean thermal energy at temperature T , and the factor 3, which is there because our molecule has
3 degrees of freedom; the three components of its velocity in a space of three dimensions.
This is our most important conclusion. The macroscopic property of ‘hotness’, or temperature,
corresponds at the nanoscale of individual atoms and molecules, to a measure of the average internal energy per molecule.
2.6.1
Speed of ambient air molecules
What is the rms speed of a nitrogen molecule in air at 290 K? The mass of N 2 is 28u.
From the equipartition theorem, we know that a nitrogen molecule, which has 3 degrees of freedom
for spatial translation, will have an average kinetic energy of
1 D E 3
hK.E.i = m v 2 = kB T.
2
2
in a gas in thermal equilibrium at temperature T . Hence for air at 290 K
vrms =
2.6.2
s
3kB T
=
m
s
3 × 1.38 × 10−23 × 290
ms−1 = 508 ms−1 .
28 × 1.66 × 10−27
Cold atoms
In 1997 the Nobel Prize for Physics was awarded to Steven Chu, Claude Cohen-Tannoudji and Bill
Phillips ‘for the development of methods to cool and trap atoms with laser light’. The ‘methods’
they used were based on Doppler line broadening; they tuned the frequency of their laser to interact
selectively with fast atoms flying towards the beam, so that continually absorbing photons head-on
slowed the atoms down, as if they were travelling in ‘optical molasses’. This field of research is now
very topical and important; it has brought us new insights into the nature of quantum coherence,
and it makes it possible to ‘freeze out’ the Doppler line-broadening mechanism, so that very precise
spectral measurements can be made.
The coldest place in Hampshire is Tim Freegarde’s lab in the basement of the Physics building,
where there is a magneto-optical trap, populated by rubidium (Rb) atoms at a temperature of 10
mK. Find the rms speed of these atoms, given that the mass of Rb is 85u.
From the equipartition theorem
vrms =
s
3kB T
=
m
s
3 × 1.38 × 10−23 × 10 × 10−3
ms−1 = 1.7 ms−1 .
85 × 1.66 × 10−27
So these molecules travel at a brisk walking speed.
21
2 ATOMS
2.7
The equation of state of an ideal gas
The purpose of this section is to prove that an ideal gas containing N molecules or atoms obeys
pV = N kB T,
(22)
and to introduce the idea of an equation of state.
Suppose we write our kinetic theory expression for the pressure of an ideal gas in terms of absolute temperature rather than mean squared speed, using Eq. 20:
1
3kB T
P = nm ×
= nkB T.
3
m
(23)
The result is a useful relationship between the pressure, temperature and molecular number density
of a gas. Notice that m, the molecular mass has cancelled out of this equation: we have a universal
relationship that is true for any gas, no matter what the weight of the molecules, provided only
that it is ideal, in the sense defined in Section 1.1.
This relationship was first discovered experimentally by the great chemists of the late 18th and
early 19th centuries, who first stated the gas laws. They had no way of detecting atoms directly,
let alone counting them; but they observed that, provided a given sample of gas was not too dense,
its volume at fixed temperature was inversely proportional to its pressure (Boyle’s Law), and its
volume at fixed pressure was directly proportional to its temperature (Charles’ Law). Suppose
the sample of gas in question has volume V , and contains a number N = nV of molecules. If we
multiply both sides of Eq. 23 by V , we obtain an equation that predicts the gas laws explicitly:
pV = N kB T
(24)
Eq. 24 expresses a result known as Avogadro’s hypothesis, the brilliant conceptual breakthrough
published in 1811 by Amedeo Avogadro, that allowed scientists to unify the gas laws into a single
equation of state of an ideal gas, valid for all gases in the ideal limit.
Avogadro’s hypothesis states that equal volumes of gas at the same temperature and pressure contain the same number of molecules.
This hypothesis made sense of Dalton’ observation that, when gases reacted with fixed initial and
final pressure and temperature, the volumes of the reacting gases and their products were always in
simple integer ratios. Thus, long before the advent of mass spectrometers and electron microscopes,
the ‘atomic hypothesis’ was understood to be an atomic fact. From then on it became natural to
measure quantities in chemistry, not in units of mass, but in units of number - Avogadro’s number.
Avogadro’s number, NA is defined to be the number of atoms or molecules in 22.4 × 10−3
m−3 of an ideal gas at ‘standard’ temperature and pressure, i.e. 1 atm and 273 K, or, equivalently,
the number of atoms (molecules) in 1 atomic weight (formula weight) in grammes of a substance.
In the 19th century it was a huge challenge to find out exactly how many atoms that was - the
first estimate was made by James Clerk Maxwell in 1865, using the kinetic theory to interpret
the relationship between diffusion and viscosity in gases. With a mass spectrometer we can weigh
atoms and molecules individually. The best known value for Avogadro’s number is
2 ATOMS
22
NA = 6.022 136 7(36) ×1023 .
The amount of a substance that contains Avogadro’s number of molecules or atoms is defined
to be 1 mole (mol). We can write the equation of state of 1 mol of an ideal gas in a particularly
simple and easy-to-remember form:
P V = RT ,
(25)
where the universal gas constant R = kB NA = 8.31441(26) JK−1 mol−1 .
We have focussed on the atomic hypothesis and the counting of atoms here; let us finish this
section by thinking about the meaning of an equation of state. Our ideal gas has a particularly
simple equation of state - ideal for working elementary thermodynamics problems - however, every
thermodynamic system in principle is described by its own equation of state, an equation that
expresses the relationship between the state variables of the system in thermal equilibrium. Each
system has its characteristic state variables. for example, a block of magnetic solid would have as
state variables pressure volume and temperature, but also applied magnetic field and magnetisation. The state variables are the macroscopic quantities needed to define completely an equilibrium
state of the system. The equation of state tells us that the state variables cannot all be changed
independently. In the case of a gas, we can fix the volume and we can fix the temperature, but
then the pressure is fixed for us - we do not have a third ‘knob’ that we can turn. (Or we can fix
the pressure and the temperature - but then the volume can only have one value.)
A helpful way to visualise the significance of the equation of state is to think of the state variables as labels on orthogonal axes in space. For example, for the ideal gas we have three state
variables, P , V , and T , and we can plot these on the x, y, and z axes respectively to make a
P − V − T diagram. Any point in the space of the P − V − T diagram might in principle represent
a state of the system; but most points will not represent states of thermal equilibrium allowed by
the equation of state. The equation of state is the equation of a surface in the P − V − T diagram.
As the ideal gas is heated, cooled, compressed - whatever - between different equilibrium states, it
moves from point to point on this surface.
2.7.1
Ideal gas temperature: the constant volume gas thermometer
A very important application of the ideal gas equation of state is that it gives us an experimental
method for determining the absolute temperature of a system. A constant volume gas thermometer is a device with a bulb containing gas that can be immersed in the system for which
the temperature is to be determined. When the bulb and the system have reached equilibrium, the
total volume of the gas can be set to a standard value by raising or lowering the head of mercury
(Hg). The pressure on the gas, P can then be determined in torr (mm Hg). A reference reading, with the same amount of gas in the thermometer, is then made with the bulb immersed in a
triple-point cell, to determine the reference pressure, P273.16 . If the gas in the thermometer were
ideal, one could immediately use the ideal gas equation of state to determine the temperature of
the system under test:
P
T =
× 273.16 K.
(26)
P273.16
Since the gas is real, it is necessary to take slightly more trouble.
23
2 ATOMS
P273.16
813.00
626.00
405.00
P
902.37
694.67
449.12
In an experiment to determine the melting point of gallium metal, the pressure of a constantvolume gas thermometer is recorded at standard reference volume with the bulb immersed first in
Ga at its melting point, and then in a triple point cell. The measurement is repeated twice, each
time with less gas inside the thermometer. The results are shown in the table at the top of the page.
find a value for the absolute temperature of the melting point of gallium. The method here is to use
Eq. 26 to determine a ‘temperature’ from each pair of pressures. Since the gas is non-ideal, all these
approximations to the absolute temperature of the Ga melting point will be slightly different. The
gas, however, gets closer and closer to ideal behaviour as its pressure gets lower, and the distance
between molecules gets larger. We can therefore find a value for the absolute temperature of the
Ga by extrapolating the data to zero pressure. The relevant graph is shown in the figure, which we
shall discuss in lectures.
303.4
(pressure ratio)x273.16 [K]
303.2
303.0
302.8
302.6
302.4
0
100
200
300
400
500
600
700
800
triple point pressure [torr]
Figure 7: Effective temperature of Ga melting point as a function of gas pressure
2.7.2
How big is one mole?
How many moles of N2 gas are contained in a gas cylinder that is 1 m long, has an internal diameter
of 20 cm, and is pressurised to 20 atm at a temperature of 290 K?
Suppose the cylinder contains nm moles of gas. Assuming that the N2 approximates well to an
ideal gas, we can write
PV
nm =
.
RT
The volume of the cylinder is given by
V = 1 × π × 0.12 m3 = 0.031 m3 .
24
2 ATOMS
Thus
nm =
20 × 105 × 0.031
= 26 mol.
8.31 × 290
How many moles of water are there in 300 ml - one glassful? The density of water is 10 3 kg m−3 ,
and the mass numbers of the H atom and the O atom are 1 and 16 respectively.
1 mol of any substance weighs the formula weight in grammes. The formula weight of water,
H2 O, is 18; thus 1 mol of water weighs 18 × 10−3 kg. Since 1 ml = 10−6 m3 , 300 ml of water weigh
0.3 kg and contain 0.3/(18 × 10−3 ) mol, or 16.7 mol.
2.7.3
Problem: candle flame
A candle burns in air under a cover, as shown in the figure, with access to outside air sealed off
by a water surface. It is observed that the candle flame grows weaker and goes out; at the same
time the water level under the cover rises to a steady final value. Discuss the mechanism for this
effect, and estimate the percentage by which the volume of gas around the candle is reduced after
it has ceased to burn. As background information, consider that candle wax typically consists of
straight chain hydrocarbons, with average length of 20 - 30 atoms, and that the composition of air
is about 78 % nitrogen, 21 % oxygen, and 1 % of other gases including carbon dioxide.
25
2 ATOMS
2.8
Internal energy and the classical equipartition theorem
The purpose of this section is to introduce the idea of internal energy, and to state the classical
equipartition theorem.
If the atomic hypothesis is correct, and matter is made up from huge numbers of nanoscale particles
in perpetual motion, then it must be true that material systems store internal energy associated
with this motion. We have seen that the kinetic theory model represents an ideal gas as a system in
which there is a significant amount of hidden energy stored as kinetic energy of individual molecules
or atoms.
The classical equipartition theorem states that, in a system in thermal equilibrium at temperature T , the average energy of an atom or molecule with f degrees of freedom is
f
kB T.
2
(27)
A degree of freedom is any dynamical variable that contributes a squared term to the expression
for the total energy of the molecule.
We have already seen how this works for the special case of the translational energy of an atom or
molecule. For a gas molecule, vx must correspond to a degree of freedom as defined above because
the total energy of the molecule includes a contribution 21 mvx2 . When we used the Boltzmann distribution to calculate the average value of 21 mvx2 , the answer came out to be 12 kB T - it did not depend
on the mass m of the molecule, but only on the temperature of the gas in thermal equilibrium.
Any term in the total energy of the molecule that depended quadratically on a dynamical variable
would yield the same average value.
The classical equipartition theorem is interesting because it fails: for some systems it predicts
the internal energy correctly, and for others it is wrong, but in a way that is revealing about the
nature of the world. We can test experimentally whether or not the classical equipartition theorem
predicts the internal energy of a particular system correctly by measuring specific heat capacities.
In section 2.4 we defined specific heat capacity by the relation c = Q/(m∆T ). Let us assume experimental conditions where all the heat transferred into the system goes into the internal energy;
then
1 dU
c=
.
(28)
m dT
We shall consider two examples where the equipartition theorem makes sound predictions, and then
look at the way in which it fails.
2.8.1
Specific heat capacity of a monatomic gas
Let us apply the classical equipartition theorem to deduce a value for the specific heat capacity of
the monatomic gas helium (He), firstly in units of JK−1 mol−1 , and then in units of JK−1 kg−1 .
Since He gas is composed of atoms, it has no energy associated with molecular rotation or vibration, and the total energy of an atom is just its translational kinetic energy. He atoms
in a gas move in 3 dimensions, thus the kinetic energy of a He atom is the sum of three terms
26
2 ATOMS
corresponding to vx , vy and vz , which are the degrees of freedom of the atom. The equipartition
theorem states that in a gas of atoms at thermal equilibrium at temperature T , with 3 degrees
of freedom per atom, the average energy per atom will be 32 kB T . 1 mole of He gas contains Avogadro’s number, NA , of atoms. Hence we can write an expression for the internal energy of he gas
at temperature T :
3
3
U = NA × kB T mol−1 = RT mol−1
(29)
2
2
where R is the gas constant, R = NA kB . Notice that the internal energy of an ideal gas
depends only on temperature, and not on the volume or the pressure of the gas. Now we can
deduce the heat capacity of the gas from Eq. 28:
c=
dU
3
= R JK−1 mol−1 = 12.5 JK−1 mol−1 .
dT
2
(30)
The equipartition theorem therefore predicts that the specific heat capacities per mole of all
monatomic gases will be the same: helium, argon, neon... The weight of a mole in each case
is different, because the mole measures a particular number of atoms or molecules, not a weight.
The atomic mass number of He is 4; thus the mass of 1 mole of He is 4g, or 4 × 10 −3 kg. Hence for
helium
12.5
JK−1 kg−1 = 3.1 JK−1 kg−1 .
c=
4 × 10−3
2.8.2
Specific heat capacity of a simple solid
Consider a simple model of an elemental solid in which each atom behaves like a 3-dimensional harmonic oscillator, tightly bound to its site in the crystal structure of the solid by harmonic restoring
forces. Use the equipartition theorem to predict a value of the molar specific heat capacity.
The total energy of a 1-dimensional classical harmonic oscillator may be expressed as the sum
of a potential energy term and a kinetic energy term:
1
1
E = mv 2 + κx2
2
2
where m is the mass of the oscillating particle, v is its speed, x is its displacement from its equilibrium position, and κ is the restoring force constant. Thus v represents a degree of freedom for
this particle, and so does x; both appear as squared terms in the expression for the energy. Since
a 1-dimensional oscillator has 2 degrees of freedom, a 3-dimensional oscillator must have 6 degrees
of freedom. We use the equipartition theorem to write an expression for the internal energy of 1
mol of this solid in thermal equilibrium at temperature T:
1
U = 6 × NA × kB T Jmol−1 = 3RT Jmol−1
2
where R is the gas constant. Hence:
c=
dU
= 3R JK−1 mol−1 .
dT
2 ATOMS
2.9
27
Diatomic gases: failure of the equipartition theorem
Monatomic gases and simple solids are described well by the classical equipartition theorem, but
diatomic gases reveal the failure of the classical approximation. The specific heat capacity of many
diatomic gases at ambient temperature - for example nitrogen and oxygen - corresponds to an
internal energy of 25 kB T per molecule. What are these five degrees of freedom? Three of them
must correspond to translational motion of the diatomic molecule. The other two correspond to
molecular rotation. Suppose we choose a z-axis along the line joining the centres of the two atoms
in the molecule. Then the molecule can make two independent kinds of spinning motion; about the
x-axis and the y-axis. Each of these has an associated kinetic energy, proportional to the square of
the vector component of the angular momentum of the molecule, which therefore represents a
degree of freedom.
What happened to the vibrational degrees of freedom? A diatomic molecule is a harmonic oscillator, with the chemical bond joining the two atoms acting as a spring. The optical spectra of
diatomic molecules show very clearly that interaction with electromagnetic radiation can set this
bond oscillating.
Figure 8: Molar heat capacities at constant volume of gaseous hydrogen and chlorine plotted against
temperature
In some hot diatomic gases, the vibrational degrees of freedom do reveal themselves thermally.
Fig. 8 shows data for chlorine (Cl2 ) gas: at 1500 K it has a specific heat capacity of 7R/2 mol−1 ,
however when it is cooled below 1000 K, the specific heat capacity starts to fall - and shows signs
of leveling out at 5R/2 mol−1 , except that it liquefies at about 250 K, before it has quite got there.
What we are detecting is a failure of the classical assumptions that we used to derive the equipartition theorem. In a classical world, the energy of a system is a continuous variable: it changes
smoothly in arbitrarily small increments. We can increase the energy of an oscillator, or a rotor,
by a very tiny amount just as easily as by a very large amount. But the world is not classical. The
energy of a system is quantised. We can only add or subtract energy in multiples of some quantum, with magnitude fixed by parameters like the size of the system, and the type of energy involved.
28
2 ATOMS
Systems tend to behave in a classical way when the number of energy quanta that take part
in any process are very large. In an intense light wave, for example, the number of energy quanta
is huge, and the behaviour of the light may be almost indistinguishable from that of a classical
wave. But when the light is so faint that only a few quanta - photons - are present, then strange
non-classical effects appear.
A diatomic molecule is a quantum harmonic oscillator: we can only add or subtract energy
from its motion in multiples of the quantum hν, where h is Planck’s constant (again), and ν is the
resonance frequency of the oscillator. At temperatures such that
kB T < hν
the ‘typical’ thermal energy of the jiggling environment is too small to add a single quantum to
the energy of vibration of the oscillator. We observe that these degrees of freedom (potential and
kinetic) are frozen out: they do not contribute to the observed internal energy or heat capacity
of the molecule. If we can raise the equilibrium temperature of the gas, then we may get it hot
enough that we start to excite the oscillator, so that we see the heat capacity begin to increase. Or
we may just make the diatomic molecules dissociate into atoms...
What about the 3rd rotation axis? - why can the diatomic atom not store kinetic energy by
spinning around its long axis? There are two ways of looking at this. Books often say that if
the molecule is ‘smooth’, then there is no way that a collision can excite such a rotation mode.
Alternatively, one can notice that the quantum of rotational energy is inversely proportional to the
moment of inertia of the rotor about its axis. You will not study moments of inertia until the 2nd
year of your course - this is just general knowledge for now. But the moment of inertia of a long
thin rotor about its long axis is small - so the quantum of the associated rotational energy is large.
The motion will be frozen out.
Fig. 8 also shows some data for diatomic hydrogen gas: here even the two rotational degrees
of freedom appear to freeze out around 200 K. (The moment of inertia of the hydrogen molecule
is small even about an axis perpendicular to the H-H bond.) We don’t ever see the three degrees
of translational freedom freeze out, however, - that could only happen if the atoms were cooled
so far that their wavelength was comparable with the diameter of the containing vessel. Atomatom interactions would have taken control of any real system long before that point - the ideal
approximation would not hold.
2.10
Thermal equilibrium and electromagnetic radiation
Hot matter emits light. The microscopic constituents of matter are in a continual state of motion,
and accelerated charges radiate electromagnetic waves. It is impossible to have a system in thermal
equilibrium consisting only of matter - there must be radiation as well.
The characteristics of radiation in thermal equilibrium play a central role in physics. Quantum
physics grew out of the ultraviolet catastrophe; the complete inability of classical physics to
explain the shape of the spectrum of thermal radiation. The subject is, however, inherently confusing: why are white-hot incandescent objects described as ‘black’ bodies; why do cavities emit
blackbody radiation; why are blue stars very hot, and blue light-emitting diodes (LEDs) rather
2 ATOMS
29
Figure 9: Thought experiment: a cavity made in two halves with an adiabatic partition
cold? The purpose of this section is to introduce the properties of thermal radiation.
Imagine a block of some solid, isolated from the world and kept in thermal equilibrium at temperature T , with a hole - a cavity inside it. The cavity contains a vacuum, and also electromagnetic
waves, which are continually emitted from the walls, reflected, scattered and reabsorbed. Our question is, what is the spectrum of the radiation in the cavity - what fraction of the energy in the cavity
is at red frequencies, or ultraviolet, or microwave? In particular, does the spectrum depend on the
optical properties of the solid we used - shiny yellow gold, or white quartz, or carbon black? The
zeroth law of thermodynamics (see section 1.2) says that it does not. Suppose, for example
that we made our cavity in two halves, as in Fig. 9 above, of two different materials, with a partition
between the two halves that is opaque to radiation - an adiabatic wall. The two sides of the cavity
are independently kept in thermal equilibrium at temperature T , so nothing can change when we
pull the partition away. We must therefore be able to define the spectral energy density of
radiation in thermal equilibrium; the energy, W (ν, T ), that is stored in the cavity radiation per
unit volume between frequencies A and R. W (ν, T ) is the Planck distribution, for which you
will derive a mathematical expression next year: it plays a role in physics just like a fundamental
constant, only it is not a constant but a function of two variables.
If we drill a tiny hole in the cavity so that it acts as a source of radiation - such a tiny hole that it
does not disturb the spectrum of the radiation in the cavity - then we can calculate that the spectral exitance of this source, the power that it radiates per unit area per unit frequency interval
E(ν, T ), is given by
1
E(ν, T ) = cW (ν, T )
(31)
4
where c is the speed of light. (Notice that there is an analogy between the black body radiation
emitted by a tiny hole in a hot cavity, and the molecules that effuse from a hole that is so small
that the velocity distribution of the molecules is unchanged.) Eq. 31 defines the spectrum of black
body radiation.
It is important to understand that the cavity emits black body radiation through its tiny aperture,
even though the walls of the cavity are not black. To understand this, think about absorption and
emission of light by surfaces. The wall of the cavity is characterised by its absorptance, A, which is
the fraction of incident power absorbed by the wall, and its reflectance, R, which is the fraction of
incident power reflected by the wall. A and R both vary with the frequency of incident radiation
in a way that is characteristic of the optical properties of the specific material that we have used
30
2 ATOMS
spectral exitance [W /m^2 Hz]
6000 K
100.0n
visible spectrum
50.0n
4000 K
2000 K
0.0
0.0
200.0T
400.0T
600.0T
800.0T
frequency [Hz]
Figure 10: Spectral exitance of a black body versus frequency
for the wall, and since these optical properties change with temperature, A and R are functions of
T as well. Since the walls are thick, and transmit no radiation, from the principle of conservation
of energy we know that
A(ν, T ) + R(ν, T ) = 1.
(32)
Let E(ν, T ) be the spectral exitance of some portion of the cavity wall. In thermal equilbrium
nothing is changing; it cannot heat up or cool down, so the energy that it radiates per unit area
in a given frequency range must be equal to the energy that it absorbs per unit area in the same
frequency range. This reasoning leads to Kirchoff ’s radiation law:
1
E(ν, T ) = A(ν, T ) cW (ν, T ).
4
(33)
A surface looks black if it absorbs all the radiation falling on it - if A(ν) ≈ 1 for all ν. But from
Kirchoff’s law we see that the spectral exitance of a black surface must be the same as that of a
black body, defined by Eq. 31. If a beam of light enters a cavity through a hole that is very small
compared to the dimensions of the cavity, the light will be reflected and scattered many times
before any of it passes back out through the hole. Even if the light is only weakly absorbed by a
single reflection inside the cavity, a large enough number of reflections will absorb virtually all the
power.
A similar argument explains why gas in a discharge tube may emit radiation with a line spectrum, while the same gas in a star emits blackbody radiation. The difference is that the gas in the
discharge tube is optically thin, whereas the gas in the star is optically thick. Fig. 11 shows schematically the line spectrum from an optically thin column of gas in a discharge tube (bottom curve, or
blue curve if you have a colour print), and then the spectrum that we would see if the column of
31
2 ATOMS
Figure 11: Line radiation spectrum from optically thin and optically thick sources, with the black body
spectrum (dotted) for comparison
gas in the discharge tube was made much thicker (middle, or red curve). By Kirchoff’s law, Eq. 33,
the frequencies at which the gas emits most strongly are also the frequencies at which it absorbs
most strongly. As the gas gets thicker, radiation from the wings of the line is favoured more than
radiation from the centre. Near the line centre, radiation will be absorbed and re-emitted many
times before it finally escapes from the gas. In the limit of large thickness, the spectral emittance
of the gas will tend to the blackbody value.
Once you have learned to derive the Planck distribution next year, you will find it easy to prove
two famous results of statistical thermodynamics. The first is the Stefan-Boltzmann law, which
says that the total power radiated per unit area by a black body emitter - the spectral exitance
integrated over all frequencies - is proportional to the 4th power of its absolute temperature:
P/A = σT 4 Wm−2
(34)
where the constant of proportionality is called the Stefan-Boltmann constant,
σ = 5.6697 × 10−8 Wm−2 K−4 .
(35)
The second is Wien’s law for the wavelength, λmax , at which the spectral exitance of a black body
at temperature T peaks:
λmax T = 2.8978 × 10−3 mK.
(36)
At what wavelength is your spectral exitance maximum?
2.10.1
Radiation balance between the sun and the earth
The temperature of the Earth’s surface is maintained by radiation from the Sun. In the approximation that the Sun and the Earth radiate as black bodies, we can show show that the ratio of the
32
2 ATOMS
Earth’s temperature to that of the Sun is given by
TEarth
=
TSun
s
RSun
2D
where RSun is the radius of the Sun, and the Earth-Sun separation is D.
In the black body approximation, we can use Stefan’s Law to write an expression for the luminosity, L, of the Sun - the total power that it radiates:
2
4
L = 4πRSun
× σTSun
.
At a distance D from the Sun, its radiated power is distributed uniformly over a sphere with surface
2
area 4πD 2 , and the Earth is only able to capture this power over its projected area πR Earth
. Thus
the total solar power incident on the Earth is
L×
Ã
2
πREarth
4πD 2
!
.
From Stefan’s Law again, the power radiated away by the Earth is
2
4
4πREarth
× σTEarth
The Earth comes to thermal equilibrium at a temperature such that the power it receives is equal
to the power that it radiates away. Equating these quantities gives
2
4πRSun
×
4
σTSun
×
Ã
2
πREarth
4πD 2
!
2
4
= 4πREarth
× σTEarth
,
and we can rearrange this equation to give the desired result.
Putting in the numbers RSun = 7 × 108 m, D = 1.5 × 1011 m and TSun = 5800 K yields TEarth =
280 K, which is not bad for such a simplified model.
2.11
Mean free path
We shall now turn our attention back to gases, and think in more detail about the motion of a single
molecule. In a gas at ambient temperature and pressure, the number density of molecules is huge,
and the typical speed is high. If our molecules are not point masses - if they have a finite diameter then it must follow that they undergo frequent collisions. If our gas is ideal, then between collisions
a molecule flies freely, and does not interact with the rest of the gas.
How long does a molecule survive between collisions? This is another question that has a statistical answer - sometimes the flight is longer, sometimes shorter. We can use simple statistical
arguments to derive an expression for the probability, P (t), that a molecule will fly for a time t
without colliding.
We shall model the interaction between molecules using a hard sphere potential. In other
words, each molecule is represented as a sphere of radius R and diameter D = 2R. When two
such spheres touch they rebound rigidly off each other; if they miss by even a tiny distance, they
33
2 ATOMS
Figure 12: Geometry of molecular collisions
do not interact. The geometry is illustrated in Fig. 12, which shows a snapshot of a gas at some
instant in time. For the moment, suppose that just the shaded molecule is moving with speed v.
In the short interval of time dt the moving molecule will collide with any molecule positioned with
its centre inside a short section, length vdt, of the the cylinder of radius D drawn with solid line
edges. Notice carefully that we use a cylinder of radius D, not R: this takes account of the fact
that both molecules in a collision have a finite size. The probability that a collision occurs in time
dt is thus just the number of molecules in length vdt of the big cylinder: it is equal to
nπD 2 vdt.
We can make dt as small as we like, so this probability can be very much less than 1.
We have defined the probability, P (t), that a molecule will fly for a time t without colliding:
it follows that the probability P (t + dt) that the molecule will fly for the slightly longer time t + dt
without a collision must be given by P (t) multiplied by the probability that the flight survives for
a further increment of time dt:
³
´
P (t + dt) = P (t) 1 − nπD 2 vdt .
In the limit of small dt, this relationship can be written as a differential equation that controls how
P (t) changes with time:
1 dP
= −nπD 2 v
P dt
which has the solution
P (t) = exp(−nπD 2 vt).
34
2 ATOMS
The different probability of surviving without a collision for time t,and then colliding in the next
time interval dt is therefore given by
exp(−nπD 2 vt) nπD 2 vdt.
This expression gives us the probability that the flight lasts for time t - the probability distribution
for the scattering time. If we add together the probabilities for all the different possible values of t
we expect to get unity: so we check that our expression is normalised correctly Z∞
0
h
exp(−nπD 2 vt) nπD 2 vdt = −exp(−nπD 2 vt)
i∞
0
=1
- which it is.
Now we have derived everything we need to find an expression for the mean scattering time, τ ,
of our molecules - that is the average time for which they fly freely between collisions.
τ=
Z∞
0
t exp(−nπD 2 vt) nπD 2 vdt =
1
nπD 2 v
Z∞
x exp(−x)dx =
0
1
.
nπD 2 v
You should work through the calculus carefully here: we change variables to get all the physical
parameters, diameter, speed and so on, out of the integral, which can then be evaluated using
integration by parts.
We have an expression for τ , but it is inversely proportional to a speed that we introduced with
the rather peculiar assumption that only one molecule in our gas was moving. To make our model
more realistic, we should let all the molecules move, and take v as the average relative speed
of any two molecules. This is actually rather easy to calculate: suppose molecules 1 and 2 have
velocities v 1 , v 2 respectively. Their relative velocity is then v r = v 2 - v 1 . The dot product of this
relative velocity vector with itself is equal to the square of the relative speed of the two molecules;
it contains terms in the squares of the individual speeds, and a cross term in the form of a dot product. Suppose we now average the relative-speed-squared; the dot product part vanishes because
the orientation of the two velocities is random, and we are left with
hvr2 i = hv12 i + hv22 i = 2hv 2 i.
In other words, the root-mean-square relative speed of two molecules is larger by a factor of
than their rms speed.
√
2
Finally, we can put this all together, and write expressions for the mean scattering time and
mean free path of our molecules:
1
τ=√
,
(37)
2πD 2 nvrms
and
1
,
(38)
λ ≈ vrms τ = √
2πD 2 n
where we have derived the approximate expression for λ by assuming that v rms is the average speed
of the molecules - actually this is not quite right, but the error is less than 10%.
35
2 ATOMS
2.11.1
Mean free path of ambient air molecules
Lets calculate the mean free path of air molecules at standard temperature and pressure
(STP), i.e. 273 K and 1 atm, assuming that a N2 molecule may be modelled as a hard sphere of
diameter 0.37 nm.
First calculate the number density using P = nkB T :
n=
1.01 × 105 [Pa]
= 2.7 × 1025 m−3 .
−1
−23
1.38 × 10 [JK ] × 273[K]
Now use the equation for the mean free path:
1
1
√
λ= √
=
= 61 nm
2πD 2 n
2π(0.37 × 10−9 )2 [m2 ] × 2.7 × 1025 [m−3 ]
Our answer is a length of order 10−7 m, or 100 nm. This is the smallest distance over which
these gas molecules ‘communicate’, to exchange energy and establish thermal equilibrium. It is
meaningless, for example, to talk about sound waves with a wavelength that is shorter than the
mean free path. Remember also that when gas escapes through a hole into a vacuum, the character
of the escape process depends on the size of the hole. If the diameter of the hole is less than the
mean free path in the gas, the escape process is called effusion, it does not perturb the equilibrium
state of the gas, and it has distinctive properties, see problem 2.2.6. If the diameter of the hole is
bigger than the mean free path, then the pressure and other properties of the gas in the container
are affected.
2.12
Transport processes
Transport processes occur when a system is not in thermal equilibrium. If one end of a system is
hot and the other cold, then molecular motions will transport energy from the cold side to the hot
side - this is thermal conduction. If domestic gas escapes from a leak, there is locally a high
concentration of methane molecules around the leak, and diffusion of methane away to regions
of lower concentration will take place - this is transport of mass. Or if a solid object moves with
a certain speed through a fluid, molecules close to the solid acquire momentum, which is then
transferred out to more distant molecules - this is viscosity, which is transport of momentum.
Each of these processes is described empirically by a transport equation, which has the form of a
linear relationship between a gradient and a flux - flux is the amount of the transported quantity
that crosses unit area of some reference plane in unit time. The constant of proportionality is a
coefficient that characterises the transport process in a particular substance or system.
It is possible to make reasonable predictions of the transport properties of gases using the kinetic theory model of an ideal gas. The calculation highlights the physical significance of the mean
free path, as defined in the previous section. We therefore conclude our study of kinetic theory by
predicting the thermal conductivity of an ideal gas.
The heat transport equation is called Fourier’s Law: it takes the form
dT
dQ
= −κA ,
dt
dx
(39)
36
2 ATOMS
where dQ/dt is the energy flow rate across area A down a temperature gradient of dT /dx - the
minus sign reminds us that heat is conducted away from hot bodies. Eq. 39 defines the thermal
conductivity κ, which has SI units of Wm−1 K−1 . Our objective is to derive an expression for the
thermal conductivity of an ideal gas. We use the following simple assumptions:
• 1/6 of the molecules travel in each of the directions ±x, ±y, ±z at speed hvi.
• Molecules have the average properties of the position of their last collision, an average distance
λ away from the plane through which the property of interest is being transported.
Consider an ideal gas in which there is a linear temperature gradient along the x-axis. We imagine a
reference plane perpendicular to the x-axis. The average energy per molecule, E(x) is x-dependent.
If the average molecular speed is hvi, then we expect that there are roughly
1
nhvi
6
molecules crossing unit area of the reference plane from left to right in unit time, and the same
number crossing from right to left. The gas, however, is hotter on the right than on the left, so
the molecules travelling from right to left carry on average more energy than those moving in the
opposite direction. We estimate that if the reference plane is at position x, the molecules travelling
to the right will have on average energy E(x − λ), and the molecules travelling to the left will have
on average energy E(x + λ), where we are making use of the assumption that molecules ‘remember’
the state of the gas for one mean free path on average.
The net energy flow from left to right across area A is therefore given by
dQ
1
1
dE
= nhvi (E(x − λ) − E(x + λ)) × A = − nAhvi2λ
.
dT
6
6
dx
This is not quite in the form of Fourier’s Law, which involves a temperature gradient rather than
a molecular energy gradient. However we know that the average energy of a molecule in an ideal
gas depends only on T , hence
dE
dE dT
=
.
dx
dT dx
The product ndE/dT is the rate at which the internal energy of unit volume of our gas increases
with increasing temperature: this is the specific heat capacity per unit volume, denoted by c. We
put these relations together to yield the equation
dQ
1
dT
= − chviλA .
dt
3
dx
(40)
Thus we have used the kinetic theory model to derive Fourier’s law for an ideal gas, and shown
that the thermal conductivity is related to the properties of the gas by
1
κ = λhvic.
3
(41)
In the argument above, you might legitimately have worried that the temperature gradient causes
n and hvi to vary with position x. However, with the temperature gradient in a steady state, the
rates at which molecules cross our reference plane in either direction must be the same, and these
37
2 ATOMS
are proportional to the product nhvi, as we saw in Example 1.1.2. Hence we needed only to consider
the position dependence of E, the energy per molecule.
The result that we have derived, Eq. 41, has a counter-intuitive property; it predicts that the
thermal conductivity of a gas will be independent of the number density - and hence independent of
the pressure of the gas. To see how this comes about, remember that c is proportional to n, and λ
is inversely proportional to n - all the n-dependence in the expression for κ cancels. If the thermal
conductivity of a gas does not drop with decreasing pressure, what is the point of the vacuum space
in a thermos flask, for example? The answer is that, as we drop the pressure in a gas, the mean
free path of the molecules increases, and at some point the molecules can fly freely from one side of
their container to the other without colliding en route. At this point the mean free path in Eq. 41 is
limited by the size of the container, and it does not increase any further as the pressure decreases.
In this regime the specific heat capacity term c dominates, falling linearly with n and pressure.
2.12.1
Example: thermal conductivity and mean free path of argon gas
Question: The thermal conductivity of argon gas at STP (see Example 1.9.1) is 1.6 × 10 −2
Wm−1 K−1 . Use kinetic theory to estimate a value for the mean free path of an Ar atom at STP,
and the effective radius for collisions of Ar atoms. The atomic weight of Ar is 40u. You may neglect
the difference between the mean and rms speeds of the atoms.
Answer: Using the kinetic theory model (Eq. 41), we estimate that
λ=
3κ
,
hvic
where κ is the thermal conductivity of the gas, hvi is the mean speed of the atoms and c is the
heat capacity per unit volume. From the equipartition theorem we know that the average kinetic
energy per atom of gas atoms moving in 3 dimensions with 3 degrees of freedom is given by
hKEi =
whence
hvi ≈ vrms =
3
kB T,
2
sµ
3kB T
.
m
¶
From the equipartition theorem again, the heat capacity of the gas per atom is
d
dT
µ
3
kB T
2
¶
3
= kB .
2
We convert this to a heat capacity per unit volume of the gas by multiplying by the number of
atoms in unit volume, n. The value of n can be related to pressure and temperature using the
equation of state of an ideal gas:
n=
P
105
=
= 2.7 × 1025 m−3 .
kB T
1.38 × 10−23 × 273
It follows that
c=
3P
.
2T
38
2 ATOMS
Combining these results, we find an expression for the mean free path:
sµ
2T
2κT
×
=
3P
P
m
3kB T
¶
2 × 1.6 × 10−2 × 273
κ=
×
105
s
λ = 3κ ×
sµ
m
.
3kB T
¶
Finally we put in numerical values:
40 × 1.7 × 10−27
= 220 nm.
3 × 1.38 × 10−23 × 273
Now we must work back from this value for the mean free path to estimate the effective collision
radius (Eq. 38):
D=
³√
2πnλ
´− 1
2
=
³√
2π × 2.7 × 1025 × 220 × 10−9
´− 1
2
m = 0.19 nm.
Is this value plausible? Solid argon has a close-packed cubic structure; this is the structure that
you get if you pack a box with spheres, so that each layer has 6-fold symmetry, and every 3rd layer
is vertically in line. The geometrical property of packing solid spheres in this way is that they fill
74% of the available volume in space. If the Ar atom is something like a hard sphere, with size
corresponding to the effective collision radius, then we expect that the density of solid argon will
be
m
,
ρ=η× 4
3
3 πD
where η is the packing fraction. Substituting in η = 0.74, D = 0.19 nm, and m = 40u we get ρ =
1.8 ×103 kg m−3 . The experimental value is 1.6 ×103 kgm−3 , thus our model is not too far wrong.
2.13
Interatomic forces
Atoms and molecules experience negligibly small mutual interaction forces when they are far apart;
this is why the ideal gas equation of state can describe real gases so well. At closer range they
experience weak attractive forces, but at very close range the mutual force becomes repulsive and
very strong. There is a balance point between attraction and repulsion at which the potential
energy is minimised; this corresponds to the equilibrium spacing of the atoms in the solid phase.
Solids are often characterised by a high degree of symmetry; the atoms are arranged in a periodic
crystalline array that minimises the potential energy of the system. In liquids, the spacing between
atoms is not very different from solids; but the symmetry has been lost, and there are lots of gaps
in the packing that allows for fluid behaviour. The strong short range repulsive forces between
atoms, that are quantum mechanical in origin, make solids and liquids highly incompressible. In
this last part of the section on atoms, we review some basic material properties that reflect the
nature of the underlying interatomic forces.
2.13.1
Young’s modulus and bulk modulus
We have studied pressure in fluids: the equivalent quantity in a solid structure is called stress; it is
a vector with dimensions of force per unit area. Solids are elastic; they deform in response to stress,
and the fractional change in dimensions that results is called strain. We only consider very simple
geometries here. If both stress and strain are small enough, then they are proportional to each
39
2 ATOMS
Figure 13: A bar subjected to 1-dimensional tensile stress.
other, with a constant of proportionality that is called Young’s modulus, Y . This relationship is
know as Hooke’s Law:
tensile stress
F/A
Y =
=
(42)
tensile strain
∆l/l0
Typical values of Young’s modulus are of order 1010 Pa - steel has Y of about 200 GPa
When a solid object is submerged in a fluid, it experiences a bulk or hydrostatic stress; all
its dimensions are strained. The resulting volume strain is defined to be the fractional change in
volume, ∆V /V0 . We define the bulk modulus, B, by
B=−
∆P
,
∆V /V0
(43)
where ∆P is the amount by which the pressure has changed. The bulk modulus of steel is about
160 GPa.
We can also define the bulk modulus for liquids; they are more compressible than solids, so their
values of B are less - about 2 GPa for water. The inverse quantity, 1/B, is called the compressibility; it is a measure of squashiness.
Gases are much more compressible than liquids - they have large distances between atoms, so
the short range repulsive forces are not in play. We can use the equation of state of an ideal gas to
find the value of its bulk modulus at constant temperature:
B = −V
d
RT
dP
= −V
(RT /V ) =
= P.
dV
dV
V
An ideal gas thus does not follow Hooke’s Law; its bulk modulus is equal to its pressure.
(44)
40
2 ATOMS
2.13.2
Thermal expansivity
The restoring forces which bind a pair of atoms to their equilibrium separation get weaker when
the atoms vibrate more energetically. Thus simple solids expand when they are heated. (Note,
however, that polymers like latex shrink when heated.) Thermal expansion is an example of a
thermometric property, used, for example in the mercury-in-glass thermometer and in bimetallic strips.
Consider a solid bar of length L, that expands to length L + ∆L when the temperature rises
by ∆T . We define the coefficient of linear expansion, α, of the material of the bar by
α=
∆L
1
×
;
L
∆T
the fractional increase in length per unit temperature rise. The coefficient of volume expansion
is called β; it is the fractional increase in volume of the material per degree temperature rise. For
a solid which expands isotropically we have the relation
β=
((L + ∆L)3 − L3 )
1
3L2 ∆L
1
×
≈
×
= 3α.
L3
∆T
L3
∆T
Typical values of β for solid materials are or order 10−5 /deg C. Liquids expand more; β for liquid
water at 20 deg C is 5 × 10−4 /deg C. Close to the freezing point water behaves anomalously, and
expands as it cools; thus ice floats.
2 ATOMS
2.14
Summary
41
3 THE ENERGY PRINCIPLE
3
42
The energy principle
It is well known that heat may be used as a cause of motion, and that the motive power which
may be obtained from it is very great. The steam-engine, now in such general use, is a manifest
proof of this fact.
To the agency of heat may be ascribed those vast disturbances which we see occuring everywhere
on the earth; the movements of the atmosphere, the rising of mists, the fall of rain and other
meteors, the streams of water which channel the surface of the earth, of which man has succeeded
in utilising only a small part. To heat are due also volcanic eruptions and earthquakes. From
this great source we draw the moving force necessary for our use.
Sadi Carnot, Réflexions sur la Puissance Motrice du Feu, 1824
The purpose of this section is
• to state the 1st Law of thermodynamics;
• to define the term function of state;
• to introduce the concept of reversible work;
• to analyse work done compressing ideal gases;
• to introduce the methods of thermodynamics; and
• to define the concept of a heat engine.
3.1
The 1st Law
The textbf1st Law of thermodynamics states that in any process in which a system changes from
an initial to a final state of thermal equilibrium, the change in the internal energy of the system is
given by
∆U = Q + W
(45)
where Q is the total heat transferred to the system during the process, and W is the total work
done on the system.
The 1st Law asserts that energy is conserved: it asserts further that thermodynamic systems
store energy internally in hidden molecular degrees of freedom, and that internal energy is a
function of state.
We define a function of state as any physical quantity that has a well-defined value for each
equilibrium state of the system. In particular, the value of a function of state does not depend
on the history of the system; the path by which the system arrived at its equilibrium state. It
follows that when a system undergoes a process from an initial to a final state, the change in some
quantity which is a function of state depends only on these end-points, and not on the details of
the process that brings about the change. Notice that the 1st Law is a statement about changes
that begin and end in equilibrium: the process between these states can take the system very far
from equilibrium, with shock waves, extreme thermal gradients, and any sort of mayhem.
From our modelling of gases, we have acquired an intuitive understanding of the internal energy
of a system, as the sum of the individual energies of all the atoms and molecules of which it is
43
3 THE ENERGY PRINCIPLE
composed. It is therefore not hard to accept that internal energy has the properties of a thermodynamic function of state, as defined above. It is vital in addition to understand that heat is not
a function of state: nor is the work done on the system.
We shall more often need to write the 1st law as an equation describing an infinitesimal change
in the state of a system, such as, for example, the change of a gas from a state (T, V ) to a state
(T +dT, V +dV ). We use a special notation for infinitesimal amounts of heat and work to emphasise
that these quantities are not functions of state:
dU = d¯Q + d¯W,
(46)
where the bar through the d tells us that these quantities are not exact differentials. This is an
important point to which we shall return.
Heat is not a function of state: this confused early scientists horribly. The great Sadi Carnot,
who used pure reasoning to figure the truth about heat engines, while the pioneering engineers of
the day were totally in the dark, was himself confused about the difference between heat and internal energy. You can guess this just from reading the extract from his essay, quoted at the beginning
of this section. Such a confusion inevitably arose when people modelled heat as a fluid (‘caloric’) and suspected that the atomic hypothesis was a ‘convenient fiction’. The scientist who cut through
this muddle was James Joule, and he did it not with reasoning alone, but with exquisitely precise
experimental measurements. In these experiments, he eliminated heat from the process undergone
by his system, and used only work to change its state.
Figure 14: One of Joule’s many paddle wheel experiments
We define work as any process that may be accomplished by lowering a weight. The falling weight
might drive a rotating paddle wheel, as in the Joule experiment shown in Fig 14, or it might drive
a dynamo generating an electric current. If the weight has mass M , and falls through a height h,
then we know that an amount of work W = M gh is done. The point about the Joule paddle wheel
experiment is that the paddle wheel churned the contents of the tank, and did a precisely defined
amount of work on the system. Using skillfully made mercury-in-glass thermomenters, which could
3 THE ENERGY PRINCIPLE
44
resolve tiny temperature changes, Joule was able to show, in effect, that internal energy was a
function of state. If his system started in a given state, specified by the mass of water in the
chamber and its temperature, then the final temperature - the final state of the system - depended
only on how much work had been done, in other words only on how far the weight had fallen; not
how fast or how slowly. There was therefore no such thing as ‘caloric fluid’, only energy, with heat
and work as interchangeable forms.
The Joule paddle wheel experiments also laid the basis for measuring heat. Whereas work can
in principle be measured at the point where it is delivered, for example M gh from a falling weight,
or I 2 Rt from electrical power dissipated by a current I flowing for time t in a resistor R (also proved
by Joule), there is no direct way of monitoring the transfer of heat energy down a temperature
gradient. All we can measure is the amount by which the temperature of a system changes as heat
is transferred. Thus the original unit of heat was the calorie, defined as the amount of heat needed
to raise the temperature of 1 g of water by 1 K. Joule measured the mechanical equivalent of heat
experimentally - in modern terms, the conversion factor from calories into SI units of energy - later
called joules in his honour.
3.2
Reversible work: the indicator diagram
Figure 15: Compressing gas in a cylinder: calculation of reversible work
Consider the system shown in Fig. 15; we have some gas in a cylinder with a piston, and we
propose to compress it. This requires work ; the gas has a certain pressure, and exerts a force on
the piston, which we have to work against to reduce the volume that the gas occupies. One option
might be to fire a bullet into the back of the piston; the kinetic energy of the bullet could then be
used to do work on the gas. How much work? - well; this would be a tough calculation, because
some undefined part of the bullet’s energy would be used up making things hot, ricocheting around,
45
3 THE ENERGY PRINCIPLE
etc. If we tried to set up a model to calculate directly how much work had been done on the gas,
we should have to represent things like shock waves - not easy at all.
A much simpler option would be to push the piston slowly, so that no shock waves were generated, and the gas stayed at all times very close to thermal equilibrium, with a well-defined value
of pressure. The piston could, in fact, move pretty fast, and still stay in this regime: remember
that gas molecules are likely to be flying at hundreds of metres per second, and making collisions
every hundred nanometres or so. It does not take the gas long to readjust to a change in volume.
If a system undergoes a process in which it never significantly departs from an equilibrium state,
then this process is said to be reversible. Reversibility is one of the most important concepts in
thermal physics: this idea will be staying with us until the end of the course.
We can calculate the reversible work done on a gas precisely. Suppose that the piston shown
in Fig. 15 moves to the left by an infinitesimal distance dx. The force on the piston, which has
cross-sectional area A, is P A; thus the work which must be done on the gas to move the piston by
this amount has magnitude P Adx. The product Adx is just the amount by which the volume of
the gas has changed. This gives us our key result:
d¯WR = −P dV.
(47)
Because the change in volume is vanishingly small, the pressure of the gas changes only negligibly
in this infinitesimal compression. The minus sign is very important: we do positive work +W
ON the gas when we compress it and its volume gets smaller. We put a subscript R on the work
term to remind ourselves that this equation is only true for reversible changes. The indicator or
Figure 16: Process takes a system from state 1 to state 2 by alternative paths A and B, shown on a P-V
or indicator diagram
P − V diagram gives us a graphical technique for representing reversible processes of gas systems.
Suppose our system consists of a specific quantity of a specific gas. The thermodynamic state of
46
3 THE ENERGY PRINCIPLE
this system can be specified by two variables; its pressure and its volume. (Remember that once
we have specified P and V we do not need to specify T - this is fixed by the equation of state.)
If we draw a graph of pressure versus volume, then every thermodynamic equilibrium state of the
system will correspond to a unique point in the P − V plane. A reversible process thus draws out
a path on the plane.
Consider the indicator diagram shown in Fig. 16, on which two possible paths, A and B, are
shown, connecting an initial state (P1 , V1 ) with a final state (P2 , V2 ). The system is travelling from
right to left on the indicator diagram; the gas is being compressed into a smaller volume. On path
A we push on the piston confining the gas, its volume decreases and its pressure - and probably its
temperature - goes up. The total work that we do to compress the gas via path A is equal to the
area under the curve - the region that is hatched in with vertical lines. To see this, we use Eq. 47,
and integrate it along path A:
Z
Z
WA =
d¯W = −
P dV .
Path B represents a totally different process, although with the same start and finish points: this
time we first heat up the gas at constant volume so that its pressure goes high: then we start to
compress it keeping its pressure constant. This will only be possible if we allow heat to escape out
of the gas. Finally we get the volume of the system down to the target value, V 2 , but the pressure
is too high, so we have to do some more cooling at constant volume to reach the final state. The
work that we have done in this process is also the area under the path, shown shaded gray in the
figure.
The amount by which the internal energy of the gas has changed in going from state 1 to state 2 is
∆U = U2 − U1 ,
the same for path A as for path B. This is what we mean by a function of state; U has a well-defined
value for every state - every point on the indicator diagram, and it does not depend on the path by
which the system arrived at that state. The work that we did, however, was much greater along
path B than along path A. From the 1st Law it is therefore clear that the heat supplied to the
system must be much smaller for path B than for path A - it must be a negative amount to get
rid of energy from the extra work that was done.
Fig. 16 thus illustrates why neither work nor heat is a function of state.
3.2.1
Problem: compression of an ideal gas.
Consider path B in Fig 16, where the system consists of 1 mol of an ideal monatomic gas, compressed from 1 atm to 2 atm, with initial and final volumes of 0.025 m3 and 0.019 m3 respectively.
The highest pressure reached on path B is 3.5 atm. Calculate (a) the change in internal energy of
the system; (b) the amount of work done on the gas, and (c) the amount of heat supplied to the
system in this process.
(a) From the equipartition theorem we know that gas atoms have an internal energy of 3RT /2
mol−1 , corresponding to 3 degrees of translational kinetic freedom. To calculate the change in
internal energy U2 − U1 we need to know the initial and final temperatures, and we calculate these
47
3 THE ENERGY PRINCIPLE
using the equation of state of an ideal gas.
3R
3
3 × (2 × 0.019 − 0.025) × 105
(T2 − T1 ) = (P2 V2 − P1 V1 ) =
= 1.95 kJ.
2
2
2
∆U = U2 − U1 =
(b) No work is done while the gas is heated or cooled at constant volume, because dV is zero. Work
is done on the gas at a constant pressure, P3 = 3.5 atm, hence
W =−
ZV2
V1
P dV = −P3 (V2 − V1 ) = 3.5 × 105 × (0.025 − 0.019) J = 2.10 kJ
(c) Finally we use the 1st Law to deduce Q, the heat supplied to the system:
Q = ∆U − W = (1.95 − 2.10)kJ = −150 J.
The minus sign means that this amount of heat is transferred OUT of the system.
3.2.2
Isothermal compression of an ideal gas
Let us find an expression for the work done on 1 mol of an ideal gas when it is compressed
isothermally and reversibly from initial volume V1 to final volume V2 at temperature T . A
process is said to be isothermal when the temperature of the system remains constant: the path
that an isothermal process follows on the indicator diagram is called an isotherm. Fig. 17 shows
pressure
isotherm: P = RT/V
P
dV
V1
V2
volume
Figure 17: Isotherm on indicator diagram
a sketch of an ideal gas isotherm. The shape of the curve is a hyperbola, with P ∝ V −1 , from the
equation of state. Again we use Eq. 47 to calculate the work done on the gas:
W =
Z
d¯WR = −
ZV2
V1
P dV = −RT
ZV2
V1
dV
V1
.
= −RT [ln V ]VV21 = RT ln
V
V2
·
¸
48
3 THE ENERGY PRINCIPLE
This is a useful result that we shall need when we study heat engines. Notice that the signs work for
expansion too: if the final volume is bigger than the starting volume, then we have the logarithm
of a fraction that is < 1, and W is negative.
When we compress an ideal gas isothermally, where does the energy that we supply to the system
by doing work on it go to? The internal energy of an ideal gas depends only on temperature; on an
isotherm U is constant. Thus from the 1st Law we know that in isothermal compression Q must
be equal and opposite to W : all the energy we put in by doing work must flow straight back out
as heat.
3.3
Principal specific heats
The purpose of this section is to make our concept of specific heat capacity more precise.
pressure
It should be apparent from the discussion in the last section that Eq. ?? does not fully define
the heat capacity. What is d¯Q? - its value depends on the precise path that the system follows
to raise its temperature by dT . The difficulty we face is illustrated in Fig. 18. Suppose we wish
isotherm T + dT
isotherm T
volume
Figure 18: Ideal gas isotherms separated by dT on indicator diagram
to determine the heat capacity of a gas system. For simplicity we assume we have 1 mol of gas,
and can save ourselves the trouble of including factors of nm . Our system is initially in the state
represented by the dot on the indicator diagram, with the T isotherm running through it. How
much heat need we supply to raise the temperature of the system by dT ? The final state for this
process is not defined; it could be any state on the T + dT isotherm. Even when we have made
some choice of final state, we still have to choose the path by which we take the system there. We
could, in fact, find paths that required arbitrarily large amounts of heat.
We solve this problem by defining two standard paths from our starting point: one is vertical,
49
3 THE ENERGY PRINCIPLE
and follows a path of constant volume, and the other is horizontal and follows a path of constant pressure. The corresponding values of the specific heat capacity are called the principal
specific heats; the specific heat at constant volume cv , and the specific heat at constant
pressure cp .
First apply the 1st law to the vertical path from isotherm T to isotherm T + dT . Since the
volume of the system does not change, the work done on the system is d¯WR = −P dV = 0. By
definition, the heat supplied to the system is d¯Qv = cv dT , and this must be equal to the change in
internal energy of the system:
dU = cv dT.
(48)
Thus cv has fundamental significance: it measures the rate of change of the internal energy of the
system with respect to temperature, as we stated in section 1.6, though without carefully defining
what we meant by a heat capacity. But cv also has a fundamental problem; it is almost impossible
to measure. When you heat the gas, you heat the box it is in, and solids expand with increasing
temperature. For a gas there are cunning ways of compensating for this effect, but if our system is
a solid there is a more acute difficulty. We can only measure cv for a solid if we apply pressure to
prevent the thermal expansion, and solids, unlike gases, are very incompressible - we shall need to
apply unfeasibly large forces.
This is where cp comes in; it is practical to measure. The atmosphere acts as a pressure reservoir; small changes in the volume of the system under study have no effect on ambient pressure,
and it is relatively easy to keep the pressure constant during a measurement.
For an ideal gas it is easy to prove a simple, famous, relationship between c p and cv . We apply the 1st Law to the horizontal path joining the two isotherms in Fig. 18;
d¯Qp = dU − d¯WR ,
which we may rewrite like this;
cp dT = dU + P dV.
For an ideal gas, the internal energy depends only on temperature. All the states on the same
isotherm have the same internal energy. It follows that dU = cv dT for any path that starts on the
lower isotherm and finishes on the upper one. If we substitute in this result, we get
(cp − cv )dT = P dV.
(49)
We have not yet used the equation of state of an ideal gas. For an infinitesimal change, we need to
put the equation of state into differential form:
d(P V ) = d(RT )
(50)
P dV + V dP = RdT.
(51)
On the constant volume path dV is zero. On the constant pressure path dP is zero, and therefore
P dV = RdT . If we substitute this result into Eq. 49, and cancel the factor of dT that now appears
on both sides, we achieve our final result:
cp − c v = R
(52)
50
3 THE ENERGY PRINCIPLE
where R is the gas constant, and the principal specific heat capacities are molar. In other words,
it takes more heat to raise the temperature of an ideal gas at constant pressure than at constant
volume. Although the internal energy changes by the same amount irrespective of pressure and
volume, when the gas expands it does work, and we have to supply that energy as heat. Gases
expand a lot when they are heated: the difference is large. Solids expand relatively little, so the
difference can generally be neglected in practice.
3.3.1
Adiabatic compression of an ideal gas
When a bicycle tyre is pumped up by hand, the barrel of the pump gets noticeably warm. Compressing the gas heats it: this is evidently not an isothermal process such as we analysed in Example
2.2.2. To compress gas isothermally we must be extremely slow, and allow time for the transport
of heat out of the system. In this example we analyse a different sort of idealised compression
process; one that is closer to everyday examples such as the bicycle pump. This is adiabatic
compression, in which the system is thermally isolated so that no heat can flow in or out. The path
of an adiabatic process on an indicator diagram is called an adiabat.
Let us derive a relationship between the pressure and volume of an ideal gas in an adiabatic
process, and hence find an expression for the work that must be done on 1 mol of an ideal gas to
compress it from an initial state (P1 , V1 ) to a final state (P2 , V2 ).
Apply the 1st law to an infinitesimal step along an adiabatic path, for which by definition d¯Q = 0.
We have
dU = d¯WR
or, equivalently
cv dT = −P dV.
Use Eq. 50 to eliminate dT from this equation, so that we have a relation involving only P and V :
cv
(P dV + V dP ) = −P dV.
R
Rearrange to get all the V s on one side and the P s on the other:
µ
cv
cv dP
dV
+1
=−
.
R
V
R P
¶
The constant multiplying the left-hand side has the form
cp
cv + R
= ,
R
R
where we have used Eq. 52. Now we define a new constant, γ, called (cumbersomely) the ratio of
the principal specific heats:
cp
(53)
γ= .
cv
We can integrate the differential relationship between V and P :
γ ln V = − ln P + constant.
51
3 THE ENERGY PRINCIPLE
Take the exponential function of both sides of the equation:
Vγ =
constant
.
P
This is the equation of an ideal gas adiabat on the indicator diagram:
P V γ = constant.
(54)
With the help of Eq. 54 we can calculate how much work is done in an adiabatic compression that
starts from an initial state (P1 , V1 ) and finishes in a final state (P2 , V2 ). We know that at any
intermediate point on the path
P V γ = P1 V1γ = P2 V2γ .
Thus
W =−
ZV2
P dV =
V1
−P1 V1γ
ZV2
V1
"
dV
V 1−γ
= −P1 V1γ
γ
V
1−γ
#V2
.
V1
All that is left is to tidy up the algebra:
P1 V1
W =
γ−1
"µ
V1
V2
¶γ−1
#
−1 .
(55)
We shall use this result when we come to study the Carnot cycle.
3.4
Thermodynamics
The purpose of this section is to introduce the methods of thermodynamics.
The most important application of thermodynamics is to derive rigorous relationships connecting the derivatives of thermodynamic functions of state with experimentally measurable system
properties. In your 2nd year statistical mechanics course you will learn how to construct expressions for the thermodynamic functions of state of a system, starting from a quantum mechanical
model of the interactions of the constitutent atoms or molecules. Since we can’t measure the state
functions directly, it is important to figure out how experimental quantities, such as principal specific heat capacities, bulk modulus, and coefficient of thermal expansion are related to them.
The equilibrium states of our system are described by an equation of state; an equation connecting values of P , V and T , which can be written quite generally as
f (P, V, T ) = 0.
(56)
This is the equation of a surface in a 3-dimensional space whose x−, y− and z− axes represent P ,
V and T . Now: so long as the squashing, or expanding, or whatever is done reversibly, and the
system stays moreorless in an equilibrium state, then it is following a path on this surface. This
implies a rigorous mathematical relationship between the possible partial derivatives of P , V and
T with respect to each other, because they all represent the gradient of some cut through the surface.
In Fig. 19 a tiny portion of the surface that represents some equation of state in P − V − T
space is shown - a triangular portion, bounded by cuts made by surfaces of constant P , V and T
52
3 THE ENERGY PRINCIPLE
Figure 19: Surface in P − V − T space
that meet in point A. If we take a small enough piece of surface it looks flat, and the gradient of
the line on which the surface intersects the plane of constant V , for example, can be expressed as
the partial derivative (∂P/∂T )V . We can write expressions for all three of the boundary lines:
µ
∂P
∂T
¶
V
=−
AD
;
AC
µ
∂T
∂V
¶
P
=−
AC
;
AB
µ
∂V
∂P
¶
T
=−
AB
.
AD
It follows that at every point on the surface these three partial derivatives are related:
µ
∂P
∂T
¶ µ
V
∂T
∂V
¶ µ
P
∂V
∂P
¶
T
= −1.
(57)
There are other ways of writing this expression, for example
µ
∂P
∂T
¶ µ
V
∂T
∂V
¶
+
P
µ
∂P
∂V
¶
= 0.
T
Eq. 57 is one of two important results that we need to derive thermodynamic relationships.
The second important result is the expansion of an exact differential in partial derivatives. We
know, for example, that the internal energy of a system is a function of state. Thus we can write
U = U (T, V ),
where we choose the two independent variables T and V to define the state of the system. (Remember that only two of the three state variables can be independent because they are connected
by an equation of state.) If we change the state of the system by an infinitesimal amount, we can
write an exact expression for the resulting change in U :
dU =
µ
∂U
∂T
¶
V
dT +
µ
∂U
∂V
¶
dV.
T
(58)
3 THE ENERGY PRINCIPLE
53
Remember that work and heat are not functions of state, that d¯W and d¯Q are therefore not exact
differentials, and that there is no equivalent of Eq. 58 for W or Q.
We could equally well choose U = U (T, P ) or U = U (P, V ) as our starting point, and derive
alternative expansions for dU .
Armed with Eq. 57 and Eq. 58 we can derive all sorts of results.
3.4.1
Problem: Prove two relationships between the thermal properties of a hydrostatic system.
A thermodynamic system whose state can be completely specified by P , V and T is sometimes
called a hydrostatic system.
(a) By considering the internal energy of a hydrostatic system to be a function of T and V , prove
that
µ
¶
∂U
Cp − C v
=
− P.
∂V T
Vβ
(b) By considering the internal energy of a hydrostatic system to be a function of P and V , prove
that
¶
µ
Cv
∂U
.
=
∂P V
βBT
54
3 THE ENERGY PRINCIPLE
3.5
Heat engines
A heat engine is a machine that converts heat into work in a cyclical process. The most familiar
example is probably the petrol engine. The heat is supplied by internal combustion of hydrocarbon
fuel mixed with air in a cylinder. The heat released develops a high pressure, driving a piston
which performs work on the surroundings via a crankshaft. The piston subsequently sweeps all the
residual hot gases out of the cylinder so that the cycle can start again with fresh air and fuel from
the same initial state.
Figure 20: Physicist’s view of a heat engine
The essential thermodynamics of this process is represented by the simple model shown in Fig. 20.
A heat engine operates between a hot reservoir and a cold reservoir. It operates in a cycle thus:
1. it takes an amount of heat Q1 from the hot reservoir;
2. it performs an amount of work W ;
3. it dumps an amount of heat Q2 to the cold reservoir;
4. it returns to exactly the same state as at the start of the cycle.
The last point is crucial because it means that the internal energy of the engine is the same at
the end of the cycle as at the beginning. Thus over one cycle ∆U = 0, and from the 1st Law of
thermodynamics we can write, quite generally, for any heat engine
W = Q 1 − Q2 .
(59)
When we study the 2nd Law, we shall focus on the thermodynamic significance of the cold reservoir. For now, just note that all heat engines generate hot exhaust gases, and that this waste heat
dumping process is an essential feature that we cannot leave out of the model.
55
3 THE ENERGY PRINCIPLE
We define the efficiency, η, of a heat engine as the ratio of W , the work that we want, to Q 1 , the
heat that it costs us. From Eq. 59
η=
W
Q1 − Q 2
Q1
=
=1−
.
Q1
Q1
Q2
(60)
The indicator diagram is a useful tool for analysing heat engines. To illustrate this, the indicator
Figure 21: Indicator diagram of heat engine cycle - the Rankine cycle
diagram in Fig. 21 shows the working cycle of a steam engine. (This analysis is credited to William
Rankine, in 1859). The working substance - the system - is a small mass of water, which
undergoes the following processes:
1. Liquid water is compressed adiabatically up to the pressure of the boiler; 1 → 2. Notice how
incompressible a liquid is; its volume changes very little in this process.
2. The water is heated to its boiling point, vapourised into saturated steam, and superheated
to a temperature higher than the boiling point, all at constant pressure; 2 → 3. This is the
power stroke.
3. The superheated steam expands adiabatically, becoming wet steam at a lower temperature
and pressure; 3 → 4.
4. The steam is cooled at constant pressure and condensed back into water at the starting
temperature; 4 → 1.
The net work done by the working substance in one cycle is the area under the path 2 → 3 → 4
minus the area under the path 4 → 1 → 2; it is the area inside the loop.
56
3 THE ENERGY PRINCIPLE
Heat Q1 is taken in on stroke 2 → 3; heat Q2 is dumped on stroke 4 → 1. The other two
steps are adiabatic, so no heat is transferred to or from the working substance.
3.5.1
Problem: Efficiency of an ideal gas Rankine engine
Suppose we construct a heat engine where the working substance is an ideal gas following the
Rankine cycle: adiabatic compression; isobaric expansion; adiabatic expansion; isobaric compression. We label the path 1 → 2 → 3 → 4 as in Fig. 21, where point 1 has coordinates (P 1 V1 ) and so
on. Show that the efficiency of this engine is
η =1−
P1 (V4 − V1 )
.
P2 (V2 − V2 )
This heat engine follows the Rankine cycle, but it is not a steam engine, so all the complications
of vapourizing and condensing the steam are eliminated. Let Cp be the heat capacity at constant
volume of the working substance. We may write
Q1 = Cp (T3 − T2 ) = cp
cp
(P3 V3 − P2 V2 )
= P2 (V3 − V2 ),
R
R
and
(P4 V4 − P1 V1 )
cp
= P1 (V4 − V1 ),
R
R
where we have used the equation of state of the ideal gas. If we put these expressions for Q 1 and
Q2 into Eq. 60 we obtain the desired result.
Q2 = Cp (T4 − T1 ) = cp
3.6
Heat pumps and refrigerators
Figure 22: Physicist’s view of a heat pump
If we reverse the directions of the arrows in Fig. 20, then we have a thermodynamic representation of a heat pump. A heat pump is a machine on which we do work in order to extract heat
57
3 THE ENERGY PRINCIPLE
from a cold reservoir and dump it into a hot reservoir. Again, the 1st Law of thermodynamics
requires that
W = Q 1 − Q2 .
Heat pumps can be used to extract heat from ground water and transfer it into the heating system
of a building. In this case the thing that we want is the heat, Q1 , and what it costs us is the work
W . We therefore define the efficiency of a heat pump by
ηpump =
Q1
Q1
=
.
W
Q1 − Q 2
(61)
Fig. 22 could equally well represent a refrigerator, where the thing that we want is the heat Q 2
that we extract from the cold reservoir. So here we define the efficiency like this:
ηfridge =
Q2
Q2
=
.
W
Q1 − Q 2
(62)
Heat engines, heat pumps and refrigerators are closely related: this is where we begin our study of
the 2nd Law.
3 THE ENERGY PRINCIPLE
3.7
Summary
58
4 THE ENTROPY PRINCIPLE
4
59
The entropy principle
Available energy is the main object at stake in the struggle for existence and the evolution of
the world.
Ludwig Boltzmann, quoted in D’A W Thompson On Growth and Form (Cambridge 1917)
The purpose of this section is
• to show how the 2nd law of thermodynamics restricts the evolution of systems and defines
the arrow of time;
• to prove Carnot’s theorem;
• to introduce the Clausius inequality;
• to define entropy as a thermodynamic function of state;
• to understand how entropy is linked to the atomic hypothesis; and
• to introduce the concepts of irreversibility and the availability of work.
4.1
The 2nd Law of Thermodynamics
Consider the following problems in thermal physics:
• An ice cube forms spontaneously in my cup of hot coffee. Use the 1st Law of thermodynamics
to calculate the increase in temperature of the remaining liquid coffee in terms of the mass
and temperature of the ice cube, the initial mass of coffee, and the specific heat capacity and
latent heat of fusion of water.
• A copper block, with specific heat capacity c, is initially at rest on a hard surface. It then
spontaneously leaps into the air, reaching a height h. Use the 1st Law of thermodynamics to
calculate the amount by which the temperature of the block falls in this process.
Our experience of the world tells us that neither of these processes would ever happen, but they are
not forbidden by the 1st Law. There is another Law of thermodynamics which tells us something
about the direction in which spontaneous thermal changes go - about the arrow of time.
The Clausius statement of the 2nd Law asserts that it is impossible to construct a device
that, operating in a cycle, produces no effect other than the transfer of heat from a colder to a
hotter body.
The Kelvin-Planck statement of the 2nd Law asserts that it is impossible to construct a
device that, operating in a cycle, produces no effect other than the extraction of heat from a single
body at uniform temperature and the performance of an equivalent amount of work.
Clausius and Kelvin-Planck just give us two alternative ways of stating the same physical law.
A device which violates one of these statements will also violate the other. To see this consider
Fig. 23. The diagram on the left shows a heat engine which disobeys the Kelvin-Planck statement;
it takes heat Q from a hot reservoir and does work W = Q without discarding any waste heat to the
cold reservoir. Suppose we now use all this work to drive any old heat pump, shown schematically
on the right of Fig. 23. The total effect of these two engines working together is to transfer heat
from the cold reservoir to the hot reservoir without the need for any work to be done, violating the
Clausius statement.
4 THE ENTROPY PRINCIPLE
60
Figure 23: An impossible heat engine driving a possible heat pump
4.2
Carnot’s theorem
Carnot’s theorem states that, for a given pair of hot and cold reservoirs, no heat engine running
between them can be more efficient than a reversible heat engine.
The concept of reversibility was introduced in Section 2.2; it plays a vital role in understanding
the 2nd Law. To recap: in a reversible process the system never significantly departs from an equilibrium state. A gas undergoing a reversible process follows a defined path on the P − V diagram
- remember that only an equilibrium state of the gas can be represented by a point on the P − V
diagram. It follows that if we do work W on the system to take it reversibly from state A to state
B along a particular path, then we can get the same amount of work W out again by taking the
system backwards from B to A along the same path.
Any heat engine can in principle be run backwards as a heat pump. But if the heat engine is
reversible, then the values of W , Q1 and Q2 will be the same for the device running backwards as a
heat pump as for the device running forwards as an engine. We can see that reversibility is an ideal
property. For any real engine, friction will see to it that we need to do more work on it to run it
backwards than we get out of it running forwards. Using this definition of a reversible heat engine,
we can show that if a heat engine obeys the 2nd law of thermodynamics, it also obeys Carnot’s
theorem. This can be proved by considering the two engines, A and B shown in Fig. 24. Engine B
is reversible and is run as a heat pump using the power generated by engine A, which may or may
not be reversible. Carnot’s theorem can be expressed as ηB ≥ ηA . Since B is reversible, then we
can say
W
ηB = 0 .
(63)
Q1
We could not say this if B was not reversible because then the magnitudes of Q01 and Q02 would not
remain the same through the switch from engine to heat pump. The efficiency of engine A, which
61
4 THE ENTROPY PRINCIPLE
Figure 24: The setup used to prove Carnot’s theorem. Engine A, which may or may not be reversible, is
used to run the reversible engine B as a heat pump
is running as an engine, is given as usual by
ηA =
W
.
Q1
(64)
All the work produced by A is used to run B, thus the two engines together form a complete system
which produces no work, but just transfers heat from one reservoir to another. By the Clausius
statement of the 2nd law, the heat must flow from the hot to the cold reservoir. Therefore we must
have that
Q1 − Q01 ≥ 0.
(65)
Therefore Q1 ≥ Q01 , which implies that
W
W
≤ 0,
Q1
Q1
(66)
ηA ≤ η B .
(67)
which, from the relations above, shows that
This is Carnot’s theorem.
Had A been reversible, we could have reversed the argument to show that η B ≤ ηA . For both
to be correct, we must have ηA = ηB .
We have shown that reversible engines are the most efficient of all engines, and that all reversible
engines operating between the same reservoirs have the same efficiency.
4 THE ENTROPY PRINCIPLE
4.3
62
The Carnot cycle
The discussion of heat engines has been a little abstract. We need to convince ourselves that we
can at least contemplate a real example of a reversible heat engine. Carnot provided such an
example. This is an engine that operates in a cycle of four reversible steps, which together are
known as the Carnot Cycle. The Carnot cycle is based upon the expansion and compression of an
ideal gas. Remember that we are free to choose as simple a system as possible, because Carnot’s
theorem shows that all reversible heat engines operating between the same reservoirs have the same
efficiency. Starting with the gas in thermal contact with the hot reservoir, at a point labelled A,
Figure 25: The left-hand part of the figure shows the configurations of pistons and reservoirs at the change
points in the Carnot cycle. The Carnot cycle is shown on the indicator diagram on the right. Between points
B and C, and also between points D and A the piston is not in contact with either reservoir.
shown in Fig. 25, the steps of the Carnot cycle are as follows:
• A→B
While in contact with the hot reservoir, the gas is expanded isothermally, doing work W 1
while absorbing heat Q1 . As it is ideal, U = U (T ), so ∆U = 0. Therefore W1 = Q1 (NB: W1
here defined as work done by the gas, not on it ).
• B→C
The gas is removed from contact with the hot reservoir and expanded adiabatically until
it is in thermal equilibrium with the cold reservoir (its temperature drops during adiabatic
expansion). During the expansion it performs work W2 , while Q = 0. Therefore ∆U2 = −W2 .
• C→D
The gas is placed into thermal contact with the cold reservoir and isothermally compressed.
Work W3 is done on the gas during the compression, and it gives up heat Q2 to the cold
reservoir. As in step 1, ∆U = 0, so Q2 = W3 .
63
4 THE ENTROPY PRINCIPLE
• D→A
Finally the gas is removed from thermal contact with the cold reservoir and adiabatically
compressed until it is once more in thermal equilibrium with the hot reservoir, and it has the
volume that it started with. Work W4 is done on the gas during this stage, while Q = 0, so
∆U4 = W4 .
The end result is that the gas has completed a cycle, ending in the same state as it started. Q 1
has been extracted from the hot reservoir, while Q2 has been dumped to the cold reservoir, and
the net work extracted is:
W = W 1 + W2 − W3 − W4 = Q1 − Q2
(68)
Fig. 25 shows the configurations of the piston and reservoirs at each end of the steps of the Carnot
cycle. The right-hand side shows the indicator diagram of the cycle. Since areas in indicator
diagrams represent work, the shaded area in Fig. 25 is the work extracted from one cycle of the
Carnot cycle. Note that when drawing indicator diagrams, the two lines representing the adiabatic changes, P V γ = constant, rise more steeply than those representing the isothermal changes,
P V = constant, and that the isotherms cross the adiabatics, but isotherms do not cross each other,
nor do adiabatics. This should be obvious for isotherms (the same state cannot have two different
temperatures); it will only become “obvious” for adiabatic changes later.
The efficiency of this Carnot cycle engine is given (Section 2.5) by
η =1−
Q2
W3
=1−
.
Q1
W1
Using the expression for the work done on a gas during an isothermal compression derived in Section
2.2.2, we can write
³ ´
RT2 ln VV43
³ ´.
η =1−
RT1 ln VV21
We also know that on an ideal gas adiabat P V γ = constant, or equivalently T V γ−1 = constant.
For the adiabat BC
T1 V2γ−1 = T2 V3γ−1
and for the adiabat DA
T1 V1γ−1 = T2 V4γ−1
where T1 and T2 are the temperatures of the hot reservoir and the cold reservoir respectively. From
these equations we see that
V2
V3
=
.
V1
V4
It follows that the efficiency of the Carnot cycle engine is given by the simple expression
η =1−
T2
.
T1
(69)
We have proved that the maximum theoretical efficiency for any heat engine is given by Eq. 69,
set only by the temperatures of the hot and cold reservoirs, and achieved in the ideal limit where
the engine operates reversibly.
64
4 THE ENTROPY PRINCIPLE
4.4
The Clausius inequality
The 2nd Law of thermodynamics is expressed mathematically by the Clausius inequality, which
asserts that for any system undergoing a cyclical process
I
d¯Q
≤0
T
(70)
where d¯Q represents the heat transferred into the system from a reservoir at temperature T in one
infinitesimal step in the cycle. The circle on the integration sign means that we integrate over a
closed path that begins and ends with the system in the same state.
To understand why Eq. 70 represents the 2nd Law, suppose that we set up the following machinery
to drive our system around its cycle:
• In the nth infinitesimal step we supply the system with heat d¯Qn from a reservoir at temperature Tn . Notice that the temperature of the system at this point is not necessarily T n .
• We set up a Carnot cycle engine which uses the Tn reservoir as its cold reservoir. In one
cycle it dumps heat d¯Qn back into this reservoir to bring it back to the state in which it
started. We have one such engine for each tiny step around the cycle - a huge number of
Carnot engines and reservoirs at different temperatures. If, for some of the steps in our cycle,
d¯Qn is negative, then we run the corresponding Carnot engine backwards as a heat pump.
• All of our Carnot engines use the same hot reservoir at temperature T 0 .
• In the nth step of the cycle, the Carnot engine dumps heat d¯Qn into the reservoir at Tn and
draws heat
T0
d¯Q0n =
d¯Qn
Tn
down from the hot reservoir (Carnot’s theorem).
• Now consider the big composite engine which has our system at the centre, surrounded by a
large number of Carnot engines, all coupled to the same hot reservoir. The composite engine
contains everything except the hot reservoir. When we complete one cycle of our system, the
whole of the composite engine returns to its intial state. The total heat that we have drawn
down from the hot reservoir to complete one cycle is
X
d¯Q0n = T0
X d¯Qn
Tn
.
• The 1st Law of thermodynamics says that the total heat we have taken out of the hot reservoir
in one cycle must be equal to the total work that our composite engine has done. The system
itself may do work; so will the array of Carnot cycle engines. But in some steps we will do
work on the system, and sometimes the Carnot engines will be heat pumps, so we do work
on them too.
• The 2nd law of thermodynamics says that the total heat we take out of the hot reservoir
P
cannot be positive. If
d¯Q0n is positive, then we violate the Kelvin-Planck statement: our
composite engine has operated in a cycle, and produced no effect other than to convert some
quantity of heat into work.
65
4 THE ENTROPY PRINCIPLE
• If the steps in our cycle are tiny enough, then we can write the sums as integrals:
I
d¯Q0 = T0
I
d¯Q
≤0
T
from which the Clausius inequality follows because T0 is a constant.
Suppose the cycle through which we drive our system is reversible. Then we can run it backwards, and in each step d¯Qn turns into −d¯Qn . For the reversible cycle run backwards
I
d¯Q
≥ 0.
T
But the Clausius equality must also remain true. The only way these two conditions can be true
simultaneously is if
I
d¯QR
=0
(71)
T
where the subscript R denotes reversible.
Clausius’ inequality becomes an equality for reversible changes.
4.4.1
Example: Theoretical limiting efficiency of a steam turbine
Consider a power station in which the generator is driven by a steam turbine powered by superheated steam at 568 deg Centigrade. Waste heat is dumped to the atmosphere via cooling towers.
Find a value for the theoretical upper limit to the efficiency of the power station as a heat engine
- say on a day when the ambient air temperature is 27 deg C.
The theoretical upper limit to the efficiency of the power station is the Carnot efficiency, which
would be realised if the steam turbine was perfectly reversible; in a state of thermodynamic quasiequilibrium at all times, with no temperature gradients. Under these conditions
Q2
T2
300
=
=
Q1
T1
841
and the efficiency of the power station would become
η =1−
T2
= 0.64.
T1
The efficiency of a real power station is only about half this value. Reversible processes are intrinsically slow, especially isothermal expansion and compression. We would have to wait for the heat
to flow in and out of the gas along vanishingly small temperature gradients. It would be impossible
to generate power at a practical rate.
4.5
Entropy
The Clausius equality for a reversible cyclical process, Eq. 71, implies the existence of a new function
of state. To see this, consider the process shown in Fig. 26: the system goes reversibly from state
A to state B along path 1, and then back to state A along reversible path B. Eq. 71 becomes
I
d¯QR
=
T
ÃZ
B
A
d¯QR
T
!
+
1
ÃZ
A
B
d¯QR
T
!
= 0,
2
(72)
66
4 THE ENTROPY PRINCIPLE
Figure 26: Indicator diagram shows a system taken through a cycle from A to B via path 1, and then back
to A via path 2.
where the subscripts on the brackets around the integrals refer to the path traversed. Therefore
switching the limits on the second integral we have
ÃZ
B
A
d¯QR
T
!
=
1
ÃZ
B
A
d¯QR
T
!
.
(73)
2
The value of this integral is independent of the path taken; it is the same for all paths and can
therefore depend only on the initial and final states, A and B. It follows that there must exist a
function of state, S, such that
Z B
d¯QR
SB − SA =
.
(74)
T
A
We call this function of state entropy and denote it by S: from Eq. 74 we see that the amount by
which the entropy of a system changes in an infinitesimal reversible process is given by
d¯QR
,
(75)
T
where the R is a reminder that this is only true for reversible transfer of heat, and T is the temperature at which heat is supplied to the system.
dS =
It follows from Eq. 75 that in a change that is reversible and also adiabatic (d¯Q = 0), then
dS = 0. The entropy of a system therefore remains constant during reversible adiabatic changes.
Now imagine that path 1 in Fig. 26 is irreversible (so it can’t really be drawn on the diagram
because the system does not pass through a continuous set of equilibrium states!) while path 2
remains reversible. Then from Clausius’s inequality, Eq. 70, I can write
I
d¯Q
=
T
ÃZ
B
A
d¯Q
T
!
+
1
ÃZ
A
B
d¯QR
T
!
2
≤ 0.
(76)
67
4 THE ENTROPY PRINCIPLE
Using Eq. 74, the second term can be written
ÃZ
A
B
d¯QR
T
!
2
= SA − SB
and therefore in general
SB − SA ≥
Z
B
A
d¯Q
,
T
(77)
(78)
or for an infinitesimal heat transfer
d¯Q ≤ T dS,
(79)
where T is the temperature at which heat is supplied to the system. This is Clausius’ inequality
in a different form, and is a very important relation. For reversible heat transfer, it becomes an
equality. For an isolated system Q = 0, so Clausius’ inequality in the form of Eq. 79 shows that
dS ≥ 0.
(80)
In words, the entropy of an isolated system can never decrease. This is a very remarkable result,
as it imposes a direction in which processes can occur. Applied to the Universe as a whole, the
entropy will carry on increasing until it can no longer do so, by which point the Universe will be
in a state of equilibrium and it will no longer be possible to extract work from any process. This
is the so-called heat death of the Universe that was much discussed in the 19th century.
The increase of entropy gives us another way to decide whether a process is reversible or not:
reversible processes do not change the total entropy of the Universe
4.5.1
Example: Hot object cooled in water
Suppose you heat a lump of iron to temperature TFe , and then place it into a large quantity of
water of temperature TW . The final temperature will be ≈ TW . Discuss whether this process is
reversible or irreversible, and calculate the associated change in the entropy of the universe.
This process must be irreversible because when the hot iron goes into the water the combined
system is not in thermal equilibrium; there is a temperature gradient between the iron and the
water. We need to calculate the change in entropy of the water and the iron. The entropy change
for the water is relatively easy. Its temperature is constant, but heat
Q = mFe cFe (TFe − TW )
flows into it from the iron, giving a change in entropy of
∆SW = mFe cFe
TFe − TW
.
TW
The iron is trickier because as it loses heat, it cools down so the temperature T that we use in the
relation dS = d¯QR /T changes all the time. We therefore need to integrate. Remembering that
d¯Q = mFe cFe dT , then
Z TW
TFe
mFe cFe dT
∆SFe =
.
= −mFe cFe ln
T
T
TFe
W
68
4 THE ENTROPY PRINCIPLE
The total entropy change of the Universe is then
δS = ∆SW + ∆SFe = mFe cFe
µ
TFe − TW
TFe
− ln
TW
TW
¶
.
One can show that the term in brackets is always positive (or zero if TFe = TW ), so the entropy of
the Universe does indeed increase in this irreversible process.
4.5.2
Isothermal compression of an ideal gas
Calculate the change in the entropy of nm mol of an ideal gas when it is compressed reversibly and
isothermally from volume V1 to volume V2 . Show that the entropy of the universe does not change
in this process.
During isothermal compression of the ideal gas, its temperature is constant and therefore its internal
energy does not change. The work done on the gas must therefore be equal to the heat transferred
out of the gas to the reservoir at temperature T that maintains the isothermal condition. In symbols
d¯W = −P dV = −d¯Q = −T dS
and so
∆Sgas =
Z
V2
V1
P dV
.
T
From the ideal gas law, P V = nm RT , and so
∆Sgas = nm R
Z
V2
V1
V1
dV
= −nm R ln
V
V2
showing that ∆S is negative, and the entropy of the gas decreases as it is compressed. The reservoir
receives heat
Z 2
V1
d¯Q = nm RT ln
Q=
V2
1
during the compression, hence
Q
= −∆Sgas .
T
The change in the entropy of the Universe is given by
∆Sreservoir =
∆SUniverse = ∆Sgas + ∆Sreservoir = 0,
as we expect for a reversible process.
4.5.3
Joule expansion of an ideal gas
Question: Consider the same system as in the previous example, consisting of nm mol of an
ideal gas at temperature T . This time we have the gas initially occupying the smaller volume,
V2 . No piston moves; the expansion takes place simply by withdrawing a partition, or opening a
tap, and allowing the gas to fill a previously evacuated space, so that the total volume is now V 1 .
We surround the entire system by an adiabatic wall, so that no heat is transferred to or from the
environment. Calculate the change in entropy of the universe in this process.
69
4 THE ENTROPY PRINCIPLE
Answer: No heat is transferred into the environment, therefore
∆Senvironment = 0.
The system is the same as in the previous example, and the initial and final states are also the
same, although this time we have an expansion, not a compression. Since entropy is a function of
state, the change in entropy of the system must be the same, only opposite in sign. Hence
∆Sgas = nm R ln
V1
.
V2
(81)
This is therefore the change of entropy of the Universe, and since V1 > V2 , then ∆S > 0. The Joule
expansion is thus an irreversible process, causing the entropy of the Universe to increase. There
can be no such thing as a Joule compression - fortunately - it would violate the 2nd Law.
4.5.4
Entropy change during a change of phase
The change of a liquid to a gas or solid to liquid is called a change of phase. As we know this
involves “latent heat”, which is heat absorbed or emitted by the system at constant temperature,
while the atoms and molecules of the system are rearranging themselves into the new phase. The
latent heat reveals that the different phases of a system have different entropy.
Calculate the change in the entropy of a mass m of liquid at its boiling point Tb as it changes
into vapour at the same temperature. Let L be the specific latent heat of vaporisation.
The latent heat of vapourisation is all transferred into the liquid at the same temperature, T b ,
so the calculation is easy. To boil mass m of liquid requires heat input of
Q = mL,
at temperature Tb . Therefore the entropy of the system increases by
∆S =
mL
Tb
when it vapourises.
4.6
Statistical interpretation of entropy
Thermodynamic arguments establish the existence of a mysterious function of state called entropy,
but do not give us any intuitive insight into what kind of physical property it represents. We need
to think about physics on the nanoscale to understand the significance of entropy.
Consider a gas undergoing a Joule expansion. Its entropy increases: the process is irreversible.
In other words, the 2nd Law asserts that if we have N gas molecules randomly distributed in a box
of volume V , they will never spontaneously all move to one side, and occupy a volume V /2.
But surely this is nonsense isn’t it? If you had only 2 molecules, there would be a 25% chance at
any one time that both were in, say, the left-hand half of the box — nothing irreversible about
that. How about N molecules? The chance of finding all of these in the left-hand side of the box
70
4 THE ENTROPY PRINCIPLE
is 1/2N . For large N , this is tiny! There are 2N more ways to arrange N molecules in the whole
box than in only half of it. This number is typically enormous! To see just how big, note that
2N = 10(log 2)N = 100.3N .
For gas at STP, for which N ≈ 2.5 × 1025 m−3 , we therefore find that occupation of the whole of a
cubic metre is approximately
25
1010
(82)
times larger than occupation of half a cubic metre. Compare this with the age of the Universe,
which is of order 1017 seconds. If you looked at the box once every second for the age of the
Universe, the chance that you would find all the gas on one side is of order
25
1017−10 .
(83)
The “17” is tiny compared to 1025 . Thus a “Joule compression” will not happen in the age of the
Universe, or even many, many, many times the age of the Universe. For practical purposes, it will
indeed never happen.
We can investigate this some more. Let us count the number of ways in which we can distribute
N molecules into two halves of a box1 At the microscopic level, each molecule can be in either
the left half or the right half (but not both), and there are 2N possibilities in total: these are
distinct microstates of the system. At a macroscopic level, however, all that matters is how many
molecules are in each half: we can label these macrostates by the fraction of the molecules in, say,
the right half of the box (since the total number is fixed). The number of ways in which we can have
n molecules out of N in the right half is simply the number of combinations of n distinguishable
objects drawn from N , which is
N!
.
(N − n)! n!
If there are N = 10 molecules in total, for example, there are 11 possible macrostates, with
microstates distributed as follows:
macrostate
right left
0
10
1
9
2
8
3
7
4
6
5
5
6
4
7
3
8
2
9
1
10
0
number of microstates
1
10
45
120
210
252
210
120
45
10
1
With just 10 molecules we can begin to see that there are more microstates corresponding to
1
You can think of these molecules as classically identical, meaning that they all have the same properties, but can
be individually distinguished.
71
4 THE ENTROPY PRINCIPLE
Figure 27: Multiplicity distributions for states of N = 10, 100, 1000 and 10000 molecules distributed in
two halves of a box.
macrostates with equal or nearly equal numbers of molecules in the two halves. As N increases
it becomes less and less probable to have anything other than (almost) equal numbers in the two
halves. I have illustrated this in Fig. 27. I considered using, in turn, N = 10, 100, 1000 and
10 000 molecules. In each case I did 100 000 trials of distributing the N molecules at random and
with equal probability into the two halves of a box. For each set of trials I counted how often each
macrostate (expressed as the fraction of molecules ending up in the right half) occurred. The results
are presented as four graphs of the relative frequencies with which each macrostate occurred. What
you can see is that, as N increases, there is a sharper and sharper peak centred at 50% occupancy.
This should make it easy to believe that with N ≈ 1023 molecules you essentially always find a
macrostate with equal occupancy in the two halves of the box.
All this suggests that entropy may be related to the number of ways of arranging molecules, which
is known as the multiplicity, W . The multiplicity of a macrostate is the number of microstates
which correspond to it. For N molecules expanding from volume V2 to V1 (so V1 and V2 label the
macrostates), the multiplicity changes by
W1
=
W2
µ
V1
V2
¶N
.
Very large numbers are easier to handle by taking their logarithm. The natural logarithm of the
multiplicity increase is given by
W1
V1
ln
= N ln .
(84)
W2
V2
72
4 THE ENTROPY PRINCIPLE
This is very closely related to the relation of Eq. 77. Indeed if we suppose that S = k ln W , we
have
∆S = S1 − S2 = kB (ln W1 − ln W2 )
W1
V1
V1
V1
= kB ln
= kB N ln
= nm NA kB ln
= nm R ln ,
W2
V2
V2
V2
(85)
using Eq. 84, and that there are NA molecules in a mole. This is exactly the same as Eq. 77. The
relation
S = kB ln W
(86)
is one of the most famous in physics. It was proposed by Boltzmann and is carved on his gravestone.
It gives a very intuitive idea of entropy as a measure of the number of microscopic arrangements
or microstates of a system, always subject to constraints on energy, volume, etc. It is in this sense
that entropy is a measure of the disorder of a system.
Boltzmann’s relation is part of the foundation of statistical mechanics and makes it clear why
reversible mechanical laws nevertheless lead to macroscopic irreversibility. All this is fairly clear
for volume changes, but also applies to temperature increases. This however is much more difficult
to deal with, and is left to the second year thermal physics course.
Boltzmann’s interpretation makes the reason for the ever-increasing entropy of the Universe simple:
the Universe is always changing to a more probable configuration, in the sense of more available
microstates. The new configurations are so much more probable that the change is irreversible.
The chance of the entropy of the Universe decreasing by as little as 0.000001 J K −1 is so tiny as to
make the chance of being hit by a meteorite enormous by comparison.
4.7
1st and 2nd law for an infinitesimal change
We can combine the 1st and 2nd Laws into a single equation to describe infinitesimal changes. The
First Law takes the form
dU = d¯Q + d¯W.
(86)
For reversible work on a hydrostatic system we can write
d¯WR = −P dV.
From the Clausius inequality we have deduced that in a reversible infinitesimal process
d¯QR = T dS.
If we put this all together, we can write
dU = T dS − P dV.
(87)
This is sometimes referred to as the central equation of thermodynamics. It has a wonderful property: although we derived it specifically for reversible changes, it applies generally, to both irreversible and reversible processes. This is because all the variables in Eq. 87 are functions of state.
It is essential that the system starts and ends in equilibrium, so that it is meaningful to talk about
73
4 THE ENTROPY PRINCIPLE
pressure, temperature and entropy, but the result is independent of the path that the infinitesimal
change follows. In general, therefore, we have
d¯Q ≤ T dS
and
d¯W ≥ −P dV,
but the equality in Eq. 87 always remains true.
The central equation implies that the internal energy, U , of a general hydrostatic system changes
whenever S or V changes. We call S and V the natural variables of U : U = U (S, V ). It follows
(Section 2.4) that we can expand the differential dU of U like this:
dU =
µ
∂U
∂S
¶
dS +
V
µ
∂U
∂V
¶
dV.
(88)
S
Comparing this with Eq. 87, we can make the identifications
T =
µ
∂U
∂S
¶
V
and
∂U
.
∂V S
Using the methods discussed in Section 2.4, we can write further
P =−
P
∂U
=−
T
∂V
S
µ
∂S
∂U
¶
V
=
µ
∂S
∂V
¶
.
U
Lets use these results to solve some examples.
4.7.1
Entropy of an Ideal Gas
Find an algebraic expression for the entropy of an ideal gas in terms of its temperature, T , and its
volume V .
For one mol of ideal gas we have P = RT /V and dU = cV dT , since U for an ideal gas is a
function of temperature only. Therefore we can write Eq. 87 as
dS = cV
dT
dV
+R
.
T
V
This can be integrated to give
S = S0 + cV ln T + R ln V,
where S0 is a constant of integration. Note that this constant is not determined in classical thermodynamics, which makes statements only about changes in S, not about its absolute value. We have
shown that the entropy of an ideal gas increases with both T and V . In other words, increasing
temperature must increase the number of available states, which it does because states over a wider
range of energy become available to the atoms. Remembering that for a monatomic gas
3
3
c V = R = NA kB ,
2
2
74
4 THE ENTROPY PRINCIPLE
Boltzmann’s relation S = kB ln W implies that
W ∝ T 3NA /2 ,
or ∝ T 3/2 for one atom. This is a relationship that you will see explained by statistical thermodynamics in your 2nd year.
4.7.2
Two systems in thermal equilibrium
Figure 28: Isolated system divided into two halves by a flexible diathermal membrane.
Consider the system shown schematically in Fig. 28. The system is surrounded by a rigid
adiabatic wall that isolates it from its surroundings. Internally it is divided into two subsystems,
1 and 2, by a membrane. The membrane is diathermal and so allows the two sides to exchange
energy. It is also flexible, so that one of the systems can increase or decrease its volume at the
expense of the other. Find the conditions that must apply to this system when it reaches a state
of thermal equilibrium.
No work is done on the system and it receives no heat, so by the 1st Law its internal energy
is constant, and
dU = dU1 + dU2 = 0.
Also, the wall surrounding the system is rigid, therefore the volume of the system is constant, and
dV = dV1 + dV2 = 0.
From the 2nd Law, the entropy of an isolated system cannot decrease, so
dS = dS1 + dS2 ≥ 0.
The equality will hold once the system comes to a state of thermodynamic equilibrium. From the
combined equation for the 1st and 2nd Laws we can write
dS =
1
P
dU + dV.
T
T
75
4 THE ENTROPY PRINCIPLE
It follows that in equilibrium
1
P1
1
P2
dU1 +
dV1 +
dU2 +
dV2 =
T1
T1
T2
T2
µ
1
1
−
T1 T2
¶
dU1 +
µ
P1 P2
−
T1
T2
¶
dV1 = 0.
Since dU1 and dV1 are independent variables, this implies that T1 = T2 and P1 = P2 . If the subsystems are not initially in mutual equiilibrium, they will exchange energy across the diathermal
membrane, which will also deform to adjust the ratio in which the total volume is divided between
them, until in thermal equilibrium the temperature and pressure are the same on both sides of the
membrane.
4.7.3
Problem: Thermodynamics of a rubber band.
Use thermodynamic arguments to show that if a rubber band contracts when it is heated, it must
become warmer when it is stretched.
76
4 THE ENTROPY PRINCIPLE
4.8
The availability of work
Figure 29: A system in contact with an environment at temperature T0 and pressure P0 .
The 2nd Law of thermodynamics determines how much of the internal energy stored by any
system is available to do work in a given environment. To understand how the enviroment comes
into it, think about a plastic bottle containing air at ambient pressure. If you puncture the bottle
in a normal atmosphere, nothing happens - there is no work available from this system. But if you
puncture the bottle in the vacuum of outer space, then the escaping gas accelerates the bottle like
a rocket motor. Internal energy from the gas in the bottle is converted into work.
Suppose we want to find out from the 2nd Law how much work is available from a particular
system in an environment characterised by temperature T0 and pressure P0 . Let the internal energy, entropy and volume of the system be U , S, and V respectively. We define a function called
the availability, A of the system:
A = U − T0 S + P0 V.
(89)
Availability is not a function of state; it is just a book-keeping trick for keeping track of the 2nd
Law, with system variables and environment variables mixed up together. It does, nevertheless,
have a well-defined value for each equilibrium state of the system in a specific environment.
A small change of the system causes the availability to change by an amount
dA = dU − T0 dS + P0 dV
(90)
where we treat T0 and P0 as constants.
Now suppose that the environment does work d¯W on the system and transfers heat d¯Q to it.
By the 1st Law, the internal energy of the system changes by dU = d¯W + d¯Q and the internal
energy of the environment changes by dU0 = −dU . The process also causes the volume of the system to change by dV , with a compensating change dV0 = −dV in the volume of the environment.
77
4 THE ENTROPY PRINCIPLE
Notice that we are assuming that the environment is huge compared to the system, so tweaking V 0
and U0 has negligible effect on P0 and T0 . We can rewrite Eq. 90 thus:
dA = −dU0 − T0 dS − P0 dV0 = −T0 (dS + dS0 ) ,
(91)
where we have used the central equation of thermodynamics, Eq.83, applied to the environment.
The 2nd Law tells us that the change in the entropy of the universe, which is the quantity in
brackets on the right-hand side of Eq. 95, can never be negative. This implies that the change in
A must always be negative - or zero for a reversible change. But the change will only be reversible
if the system is in thermal equilibrium with the environment, and at this point the system stops
changing. Thus we have a new way of writing the 2nd Law:
dA ≤ 0.
(92)
This means that the availability of any system in a given environment can only decrease, and will
reach a constant minimum value, Amin when the system comes to thermodynamic equilibrium with
its environment.
Next we see how this helps us to figure out how much useful work we can extract from any
given combination of system and environment. Let us define ‘useful work’: it is work done by the
system on the environment, corrected for the (not useful) work that the system does against P 0
when it expands a little:
d¯Wusef ul = −d¯W − P0 dV.
(93)
Applying the 1st and 2nd Laws, we see that
d¯Wusef ul = dU − d¯Q − P0 dV ≤ dU − T0 dS − P0 dV,
and with reference to Eq. 90 this can be expressed as
d¯Wusef ul ≤ −dA.
(94)
The 2nd law therefore defines a precise upper limit to the amount of useful work that we can extract
from any system in a particular enviroment: it is equal to A − Amin .
4.9
Thermodynamic potentials
For a system with specific constraints (constant V and T , for example), the general condition for
equilibrium, namely that the availability be a minimum, can be reduced to a simpler condition
involving a thermodynamic potential. There are four thermodynamic potentials; they have
dimensions of energy, and unlike availability they are functions of state. They do not contain
any new physics, but they help us to keep track of the constraints that the 2nd Law imposes on
equilibrium states. We list them here:
• Internal energy: U ;
• Enthalpy: H = U + P V ;
• Helmholtz free energy; F = U − T S;
• Gibbs free energy; G = U − T S + P V .
78
4 THE ENTROPY PRINCIPLE
We have shown that for any general system at equilibrium
dA = dU − T0 dS + P0 dV = (T − T0 )dS − (P − P0 )dV = 0,
(95)
where we have used the central equation of thermodynamics to expand dU . What does Eq. 95 tell
us about equilibrium in specific simple situations?
1. Thermally isolated system with fixed volume
If the system is thermally isolated behind an adiabatic wall, then in general T 6= T0 , and at
equilibrium we must have dS = 0 to ensure that dA = 0. Since the volume of the system is
constant, dV = 0, and
dA = dU = 0.
The availability of this system is just its internal energy, U : in equilibrium S is maximum
and U is minimum.
2. Thermally isolated system at constant pressure
Again T 6= T0 , so we require dS = 0 for A to be minimum. But now dV 6= 0, so in equilibrium
P = P0 and dP = 0. Thus we have
dA = dU + P0 dV = d(U + P V ) = dH.
The availability of this system is its enthalpy, H: in equilibrium S is maximum and H is
minimum.
Notice that the derivative of enthalpy with respect to temperature at constant pressure is c p :
cp =
d¯Qp
=
dT
µ
dU + P dV
dT
¶
=
p
µ
∂H
∂T
¶
.
p
Compare this relationship with its companion:
cv =
µ
∂U
∂T
¶
.
v
3. System not thermally isolated; volume fixed
Since the system is not thermally isolated, d¯Q 6= 0 and dS 6= 0. This implies that in thermal
equilibrium T = T0 to make dA = 0. There is no constraint on the equilibrium pressure of
our system because dV = 0. Thus
dA = dU − T0 dS = d(U − T S) = dF.
The availability of this system is its Helmholtz free energy, F : in equilibrium F is minimum
and the system acquires the temperature of its environment.
4. System not thermally isolated, and at constant pressure
With these constraints T = T0 , P = P0 and dT = dP = 0. Then
dA = dU − T0 dS + P0 dV = dU − T dS + P dV = d(U − T S + P V ) = dG.
The availability of this system is its Gibbs free energy, G: in equilibrium G is minimum and
the system acquires the temperature and pressure of its environment.
4 THE ENTROPY PRINCIPLE
Figure 30: Mnemonic for the thermodynamic potentials.
Figure 31: Summary of thermodynamic potentials.
79
80
4 THE ENTROPY PRINCIPLE
Thermodynamic potentials can be mastered quickly with a little memorizing and some logical deduction. Everything you need to memorise is in Fig. 30. If you can draw this figure from memory,
then you can work out Fig. 31 just by completing the patterns. Every potential starts with U . The
four boxes give you all possible combinations of with/without +P V and −T S.
The most important parts of Fig. 31 are the differential expressions given for the four potentials. You must memorise the combined 1st and 2nd Law equation for dU . It is then possible to
work out a differential expression for any other potential, for example
dF = d(U − T S) = dU − T dS − SdT = −SdT + P dV.
The row and column labels can be added last, based on the differential expressions for the potentials.
For example, the bottom row has dT in it and no dS: the potentials in this row are useful for
constant temperature systems where dT = 0.
4.9.1
Example: gas cylinder
Consider a gas cylinder of volume Vc = 0.5 m3 containing ideal gas at a pressure of Pc = 50 atm in
thermal equilibrium with its environment, which has temperature T0 = 300 K and pressure P0 =
1 atm. By considering the availability of this system, calculate an upper limit to the useful work
that can be done by the gas when the cylinder is opened.
The availability of the gas in the cylinder is
A = Uc − T0 Sc + P0 Vc .
When the cylinder is opened the gas comes to equilibrium at ambient temperature and pressure
with availability
Amin = Uf − T0 Sf + P0 Vf .
The maximum useful work done by the gas is
Wusef ul ≤ A − Amin = T0 (Sf − Sc ) − P0 (Vf − Vc ),
where we use the fact that the gas is ideal, so for a process that begins and ends at the same
temperature Uc − Uf = 0. From the result of Example 3.7.1 we can write
Sf − Sc = nm R ln(Vf /Vc ),
where nm = Pc Vc /RT0 is the number of moles of gas in the cylinder. Using the ideal gas equation
of state we obtain
d¯Wusef ul ≤ Pc Vc
4.9.2
µ
Pc
ln
P0
µ
¶
P0
− 1−
Pc
µ
¶¶
= 2.5 MJ × 2.93 = 7.3 MJ.
Example: Release of heat in combustion
Combustion is an example of an exothermic chemical reaction, in which atoms rearrange themselves into new compounds with reduced mutual potential energy. If the reaction takes place in a
thermally isolated vessel at constant pressure, then the equilibrium state will have minimum H.
By the 1st Law, the heat given off in the reaction, −Q, is
−Q = −(∆U + P0 ∆V ) = −∆H
81
4 THE ENTROPY PRINCIPLE
where we use the fact that ∆P = 0 at constant pressure. The heat we get out comes from the
internal energy reduction minus the bit of work done to push the environment back. Thermodynamic data books list “enthalpies of formation”, ∆Hf of common compounds, where the change
in enthalpy is specified relative to the stable form of the starting elements. Thus the enthalpies of
formation of hydrogen in the form of H2 molecules, and oxygen as O2 , are defined to be ∆Hf = 0.)
The enthalpy of formation of water (H2 O) is ∆Hf = −242 kJ mol−1 . Thus the formation of 1 kg of
water, starting from hydrogen and oxygen molecules, releases 242/0.018 = 13.4 MJ of heat (molar
mass of water =18 g).
How much heat is produced per mole of carbon monoxide (CO) when it is burnt to produce
carbon dioxide (CO2 )? The enthalpies of formation of CO and CO2 are −110.5 kJ mol−1 and
−393.5 kJ mol−1 respectively.
One mole of CO plus half a mole of O2 produces one mole of CO2 . The enthalpy of formation
of O2 is zero by definition, and therefore the heat released is given by
−∆H = −(∆Hf (CO2 ) − ∆Hf (CO)) = 283 kJ mol−1 .
4.9.3
Example: graphite versus diamond
Elemental carbon can exist in two solid forms in the earth’s crust; soft graphite and hard crystalline
diamond. The density of graphite is 2.25 × 103 kgm−3 while that of diamond is 3.51 × 103 kgm−3 .
Measured at 298 K, the enthalpy of formation of graphite is 0, while that of diamond is ∆H f =
1.897 kJmol−1 . The difference in entropies of diamond and graphite is
Sd − Sg = −3.36 JK−1 mol−1 .
Question:
(a) What is ∆G going from graphite to diamond at 298 K?
(b) Is diamond or graphite the stable form of carbon at 298 K and 1 atm?
(c) Assuming that graphite and diamond are incompressible, at what pressure are they in equilibrium at 298 K?
(a) Starting from G = H − T S, then
∆G = Gd − Gg = Hd − Hg − T (Sd − Sg )
= 1897 Jmol−1 − 298 K × (−3.36 JK−1 mol−1 ) = 2898 Jmol−1 .
(b) Since ∆G > 0, graphite is the stable form. Diamond could in principle convert spontaneously
to graphite at 298 K, since for that process ∆G < 0, but we never see it happen because
there is a large activation energy. Under ambient conditions, diamond is a metastable form
of carbon.
(c) We use the relation dG = −S dT + V dP from which
d(Gd − Gg ) = (Vd − Vg ) dP.
One mol of carbon has mass 0.012 kg, so the molar specific volumes are V g = 0.012 kg mol−1 /2250 kg m−3 =
5.333 × 10−6 m3 mol−1 and Vd = 3.419 × 10−6 m3 mol−1 . Therefore d(Gd − Gg ) = −1.91 ×
10−6 m3 mol−1 dP . The difference in G decreases with pressure, and will reach zero when
dP = 2898 J mol−1 /1.91 × 10−6 m3 mol−1 = 1.5 × 109 N m−2 , or 15,000 atm. Above this
pressure diamond is the stable form of carbon at 298 K.
4 THE ENTROPY PRINCIPLE
4.10
82
Real gases: interatomic forces and phase transitions
Real atoms interact: they exert forces on each other. At short range they repel each other strongly:
at larger distances the force become weakly attractive. The van der Waals equation is a modification of the ideal gas equation of state to model the effects of interatomic forces. It is thus an
approximate equation, which uses two adjustable parameters, the van der Waals coefficients a and
b. For 1 mol of a van der Waals gas in thermal equilibrium at temperature T , pressure P , and
volume V , we write
¶
µ
a
(96)
P + 2 (V − b) = RT.
V
The term in a causes the pressure of a van der Waals gas to be less than RT /V by a small amount
which varies as the square of the number density. This term represents the effect of the long-range
attractive forces. The term in b represents the effect of the short-range repulsive forces; each atom
has a finite effective size, reducing the total volume available to the gas.
These simple changes to the equation of state make the isotherms of the gas on the indicator
Figure 32: Isotherms of a van der Waals gas. The plot uses a = 0.138 N m4 and b = 3.18 × 10−5 m3 ,
appropriate for 1 mol of O2 gas. The critical isotherm is drawn dashed, corresponding to a temperature of
155 K. The other isotherms are drawn at temperature intervals of 8 K.
diagram look rather different. For given fixed P , volume V is determined by a cubic equation,
which may have more than one real root. The isotherms of a van der Waals gas are shown in
Fig. 32. For large V these look like ideal gas isotherms: the gas is approaching the ideal limit. For
small V the curves rise very steeply, showing the effect of short-range repulsion at high number
density.
In between these extremes there is a region where some of the calculated isotherms slope upwards over some range of V . Our van der Waals model is producing unphysical behavious here; a
real gas in a state where dP/dV > 0 would be unstable. Imagine the gas volume divided into two
halves; a perturbation that slightly compressed one half and expanded the other would run away,
because the pressure of the compressed gas would drop further, and that of the expanded gas would
4 THE ENTROPY PRINCIPLE
83
rise. Eventually the dense half would stabilise to the left of the minimum in the isotherm, and the
rareified half would stabilise to the right of the maximum. The gas would come into equilibrium
separated into two fluids - two different phases, one dense and one light. The van der Waals model
is very crude, but even so, it shows us how forces between atoms in a non-ideal gas give rise to
phase transitions.
The dashed line in Fig. 32 is the critical isotherm corresponding to the critical temperature,
Tc . The warmer isotherms are stable everywhere, with negative gradients. The colder isotherms
all have unstable regions of positive gradient. On the critical isotherm there is a point of inflection, where the gradient is zero. This behaviour is observed in real gases. There is some critical
temperature above which the gas will never liquefy, no matter how much pressure is applied. The
Figure 33: Isotherms of a pure substance such as H2 O or CO2 .
indicator diagram in Fig. 33 shows the typical shape of the isotherms measured for a pure substance. Suppose we start with the substance in state A, at a temperature colder than the critical
temperature. The substance is a gas; if we compress it isothermally its pressure rises until point
B is reached; here condensed vapour starts to appear. Between points B and C we have a mixture of liquid and saturated vapour, and we can change the volume of the system without any
corresponding change in its pressure, which is constant at the saturated vapour pressure. As
we move right along the isotherm, more and more saturated vapour condenses, until at point C
only liquid remains. The condensed liquid is highly incompressible, as the near-vertical slope of
the liquid isotherms indicates. As the temperature rises, the range of V over which the isotherm
is horizontal shrinks, until on the critical isotherm only a point of inflexion remains. This is the
critical point; a unique state, in which the substance has not yet separated into two phases, but
84
4 THE ENTROPY PRINCIPLE
at which
µ
∂P
∂V
¶
= 0,
T
so that huge density fluctuations occur, giving the gas a milky white appearance known as critical
opalescence.
Suppose our system is in a state somewhere on the line CB of the indicator diagram in Fig. 33;
a mixture of liquid and saturated vapour in thermal equilibrium with an environment at constant
temperature and pressure. Under these conditions the Gibbs free energy of the system has a
minimum value;
dG = −SdT + V dP = 0.
Now consider a process in which the volume of the system increases by dV , while dm l mol of liquid
changes to the vapour state. P and T remain constant; the system takes a tiny step to the right
along the horizontal isotherm. We use lower case letters, g, s and v to denote the molar Gibbs free
energy, entropy and volume of each phase of the system, with subscripts l and v to label liquid and
vapour respectively. In this process the Gibbs free energy of the system changes by
dG = −gl dml + gv dmv = 0.
Clearly the specific Gibbs free energies must be the same for the liquid and vapour phases. Now
suppose that we make small changes to T and P . To keep the system in equilibrium, g must change
by the same amount for each phase, so
−sl dT + vl dP = −sv dT + vv dP.
It follows that the rate at which the saturated vapour pressure varies with temperature is given by
sv − s l
dP
=
.
dT
vv − v l
(97)
This is the Clausius-Clapeyron equation. It shows that if the density of the substance is
different in the liquid and vapour phases, then the entropy will be different too. To vapourise 1
mol of the liquid substance reversibly, it will be necessary to provide heat
Qlatent = T (sv − sl ) = L,
(98)
where L is the latent heat of vapourisation. We can rewrite the Clausius-Clapeyron equation in
terms of latent heat:
dP
L
=
.
(99)
dT
T (vv − vl )
It is useful to represent phase equilibria on pressure-temperature, or P − T diagrams, because
we know that two phases in equilibrium will have the same P and T . Fig. 34 shows the typical
appearance of such a diagram. The P −T plane is divided up into regions where the system is solid,
liquid or gas. Each line shows the values of P and T for which two phases are in equilibrium. The
lines meet in a point where all three phases are in equilibrium. This is the triple point, where at
a uniquely defined temperature and pressure gs = gl = gv .
The liquid/vapour line stops at the critical point, (Pc , Tc ). For T > Tc and P > Pc the substance
4 THE ENTROPY PRINCIPLE
85
Figure 34: Phase diagram of a typical substance. The dotted line shows the ice/water equilibrium, which
is atypical.
becomes a supercritical fluid. This is a phase of matter with both liquid and gas properties; its
molecules interact strongly, and it can act as a solvent like a liquid; but its viscosity is very low,
and it fills its container completely like a gas with no meniscus. There is no equivalent point on
the solid/liquid line; solids and liquids appear to be intrinsically different, and there is no way to
make a smooth transition between them.
The slopes of the lines come from the Clausius-Clapeyron equation. The water/ice slope is unusual in having a negative gradient because in phase equilibrium the molar volume of ice is bigger
than that of water.
4.10.1
Example: Melting point of ice
Given that the latent heat of fusion of ice is L = 335 × 103 J−1 , while the specific volumes are
vl = 10−3 m3 kg−1 and vs = 1.09 × 10−3 3−1 for water and ice respectively, calculate the rate at
which the melting point of ice changes with pressure. (The volume per unit mass is simply the
inverse of the density: ice is less dense than water and hence has a larger specific volume).
We put the given data into the Clausius-Clapeyron equation:
dP
335 × 103 Jkg−1
= −13.6 × 106 N m−2 K−1 .
=
dT
273 K × (1.00 − 1.09) × 10−3 m3 kg−1
(100)
This equations shows that if the external pressure increases by 134 atm, the melting point of ice
drops by 1 K. It is the fact that ice is less dense than water that determines the sign of the change.
Most solids are denser than their equivalent liquids and their melting point increases with pressure.
The decrease of melting point with pressure for ice is thought to be important for the movement of glaciers: any obstruction to the flow of ice will cause an increase of pressure and hence
melting. It is also commonly said to be of importance to skating because of the melting caused
4 THE ENTROPY PRINCIPLE
86
by the pressure of the skate. However, taking a contact area of 2 × 10 −5 m2 for an 80 kg skater
gives a pressure of about 400 atm, which would decrease the melting point by about 3 C. This is
not enough to explain melting for figure skaters, who like the ice to be at −5.5 C, and ice-hockey
players who like it to be at −9 C. In fact, skating is possible at temperatures down to −30 C.
Estimates of frictional heating also cannot explain the melting. It is now thought that the surface
structure of ice includes a liquid-like layer, making it intrinsically slippery.
4.10.2
Example: Boiling point of water
Calculate the rate at which the boiling point of water increases with increasing pressure, given that
the latent heat of evaporation of water is L = 2.257 × 106 J kg−1 , while the specific volumes of
steam and water are vv = 1.673 m3 kg−1 and vl = 1.043 × 10−3 m3 kg−1 .
From the Clausius-Clapeyron equation, we have
dP
2.257 × 106
=
Nm−2 K−1 = 0.036 atm K−1 .
dT
373(1.673 − 1.043 × 10−3 )
(101)
This determines the rate at which the vapour pressure of water increases with temperature. Since
the boiling point of a liquid is defined by the point at which its vapour pressure equals the surrounding atmospheric pressure, we can deduce how the boiling point changes with pressure. For
instance, on top of Everest, the pressure is ≈ 0.65 atm less than at sea level, so the boiling point
of water is reduced by 0.65/0.036 = 18 C. Such a reduction significantly lengthens cooking times the chemical processes of cooking involve activation energies and temperature-sensitive Boltzmann
factors.
4 THE ENTROPY PRINCIPLE
4.11
Summary
87