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University of Warwick, Department of Physics
PX 119 Matter: some associated notes
Introduction to Gases Liquids and Solids
This course is principally concerned with matter made up from atoms (and
molecules). At time of writing most of the confirmed mass of the Universe is in stars,
and primordial gas clouds (of hydrogen) out of which stars condense. In a star the
matter is too hot for charged electrons and nucleons to be bound into atoms: the
resulting interacting fluid of charged particles is termed a ‘plasma’, and is outside the
scope of this course. However around our sun (and some 50 other stars to date) we
find planets of much cooler matter.
A quick tour of our Solar System confirms three principal phases of matter: gas,
liquid and solid. A principle issue of this course is how these three phases arise.
As you will all know (but the Greeks who first identified four phases of Earth-AirFire-and Water did not), most simple substances can exhibit all three phases of matter
provided one explores a wide enough range of conditions. This is conventionally
expressed in terms of a Phase Diagram:
triple point,
where three phases coexist
pressure
p
solid
liquid
gas
critical point,
around which liquid and gas merge
temperature
T
The idea of Thermal Equilibrium will rather dominate our answers as to where the
phases come from. Thermal equilibrium you probably first met in terms of measuring
and matching temperature – a large scale or ‘macroscopic’ point of view. In this
course we will to some extent sneak up on the idea in the case of gases to gain a
microscopic point of view in terms of Boltzmann’s law. This is so central I am
determined to make it equation number one:
 E( X ) 
p ( X )  exp  

kT 

This relates the probability p, to find a system of particles in some particular
configuration X, to the energy E for that configuration. Here k  1.2  10 -23 JK -1 is
Boltzmann’s constant, more properly – but cumbersomely - denoted as kB. What I
want you to remember now is that all this is of central importance!
How are we going to handle atoms? For the purposes of this course, atoms will be
treated as point masses with energies of interaction between them. Almost all the
mass is in the nucleus, essentially a point (10-15 m) compared to atomic size (10-10 m),
so this defines the natural centre. The detail of interactions comes mainly from the
electrons, but as long as stay bound to the atoms what total energy this gives can still
be expressed in terms of the positions of the atomic nuclei. We will also widely use
the approximation that the interaction energy is additive over pairs of interacting
atoms.
University of Warwick, Department of Physics
PX 119 Matter: some associated notes
For any given pair of atoms the interaction (potential) energy U will be a function of
the distance r between their nuclei: this is often referred to as the ‘pair potential’ and
typically looks as follows.
repulsive core
U(r )
attractive tail
r
mechanical equilibrium
So far this begins to sound rather like Chemistry, but here comes a crucial point very
much in the style of Physics. The above model of point masses with a (pair)
interaction potential can apply just as well to objects bigger than atoms – so we expect
analogous physical behaviour in the following.
The C60 ‘buckyball’ molecule fairly obviously approximates to an atom 10 times
bigger than usual. One of the puzzles we have to explain is why it does not exhibit
any liquid phase. The same issue arises with protein molecules, where it has been
suggested that passing via an unstable liquid state favours good crystallisation, the
key to protein structure determination by X-ray scattering.
Microemulsion droplets and colloidal particles (10-5 m) are further steps up in size
with the same physics applying.
As we go further up in size it gets harder to maintain thermal equilibrium on Earth
without contrivance: gravity segregates larger objects, and artificial mechanical
systems such as billiard tables suffer dissipation (friction). There is however one
spectacular example in globular star clusters. Here we have stars interacting at
distance through gravity, and behaving as a nearly ideal gas!
Solids it is harder to be sweepingly general about, because the arrangement of atoms
in a solid is (almost) fixed – and it matters where! In the case of crystals there is
symmetry to the arrangement of atoms, and we can relate microscopic detail (the
arrangement of atoms) to macroscopic detail in scattering patterns.
To form a solid is not the only way atoms and molecules can order their arrangement,
as you can see upon inspecting a bottle of bath oil or playing with a magnet. From
this perspective, the crystalline solid is just one of many ordered phases which are
studied in ‘condensed matter physics’.
Fascinating though all these equilibrium states of matter may be, dynamically they
are rather boring: they just stay the same. Equilibrium is where things evolve
University of Warwick, Department of Physics
PX 119 Matter: some associated notes
towards, and a very rich area of investigation is how they approach it. You will see
that things like thermal conductivity and diffusion can be viewed in this way, and so
this whole area is also known as Transport Theory. Viscosity is another example,
and will bring us to a final topic: the glass, which as a liquid which has ceased to be
able to flow is a state of matter essentially dynamical in origin.
There are some final introductory remarks which I should like to add about classical
vs quantum mechanics. To localise a particle of mass m within distance l entails an
energy of order
h2
8m 2
(where Planck's constant h  6  10 -34 Js ) and at this minimum energy the particle
will be constrained to occupy the lowest quantum state [you will cover this in the
Quantum Phenomena module]. Up to a few times this energy quantum mechanical
details will be important, but by several times this energy classical (Newtonian)
mechanics will be a good approximation. For an electron ( me  10 -30 kg ) localised to
an atom (   2  10 -10 m ) this energy is about 10-18 J. The same energy using the
mass of even a light atom (e.g. mHe  10 -26 kg ) is about 10-22 J, so we can say that if
the energy available per particle is in the range
10-22 J < E < 10-18 J
then electrons will be stuck in their lowest atomic quantum states, whereas the atoms
themselves will move classically on the scale of atomic distances.
Now you are probably facing the dilemma: what are these energy values (in Joules)
worth? How can we think of them? I want to give you two different answers using
two different ‘energy currencies’. The first is electrical. We can simply ask: what
Voltage difference would give one electron worth of charge ( e  1.6  10 -19 C ) this
much energy? The voltage difference is just E/e so we obtain (in round
numbers 10 -3 V  E / e  10V . In this case scientific practice is to multiply through by
the symbol e to obtain
10-3 eV < E < 10 eV,
where you can now just think of eV, or ‘electron volts’, as a unit of energy equal
to 1.6  10 -19 J . In other words, up to a few volts will not dislodge electrons from their
atoms, whereas anything more than millivolts will move atoms quite classically on the
atomic scale.
The second way to look at it is in terms of equivalent thermal energy. Boltzmann’s
law shows that probabilities (in thermal equilibrium) are not sensitive to energy
differences small compared to kT , or if you prefer, that this much energy is readily
available. So now we can ask: what are the corresponding temperatures at which our
energy limits become thermally available? The temperature being just E/k we then
obtain
0.1 K < E/k < 103 K.
In other words, at temperatures above fractions of a Kelvin atoms move classically
over atomic distances, whilst electrons remain quantum mechanically bound up to of
order 1000 Kelvin.
University of Warwick, Department of Physics
PX 119 Matter: some associated notes
1. Ideal Gases and Thermal Equilibrium
Equation of State
Just about the most basic (and practical) question you can ask about a substance is:
what do we need to do to contain it? The answer is that you need a container with
walls which (as well as being leak proof) must oppose the outwards pressure of the
material. This force exerted per unit area of the containment is the key way the
substance communicates mechanically with its surroundings, and characterises in
mechanical terms the state of the material. Hence the (substance-dependent) law for
how the pressure depends on experimental conditions is know as the Equation of State
and it takes the form
p  f ( n, T )
where n is the density of molecules, T is the temperature (we will always use absolute
temperature in this course), and f is a function to be measured and for which we
would like to have theories.
Notice that we do not expect the pressure to depend separately on the number of
molecules N and the volume occupied V because these both scale up with the size of
sample measured; only the combination n  N can enter the equation of state.
V
You have probably met the ‘ideal gas equation’ in some form like
pV  NkT or equivalently pV  N moles RT
where R is the Universal gas constant, which is just Avogadro’s number
6.02  10 23 times Boltzmann’s constant k. (I will consistently work with number of
molecules and k, but if you prefer to be a little old fashioned and work with number
of moles and R that is fine – this is really only a choice of units) It is obviously trivial
to divide the gas equation through by V to obtain the ideal gas equation of state
p  nkT .
The ideal gas equation subsumes two important experimental results and the first
definition of absolute temperature. At fixed temperature, pV = constant is Boyles’
Law (1660), and at fixed pressure V  T is Charles’ Law (1787) – with the
extrapolation to zero volume setting an absolute zero of temperature. With the
appreciation that the constant of proportionality (k in our notation) is universal if you
figure the density by number rather than by mass, the beauty of this law became a
standard for temperature determination.
Kinetic Theory
Having taken on board three hundred years of experimental results, we now come to
the theory. We will assume:
gas molecules behave as point masses, mass m
they are uniformly distributed through space with (number) density n
they have some random distribution of velocities
they undergo specular reflection at the walls (as light off a mirror), but
otherwise have negligible interactions.
University of Warwick, Department of Physics
PX 119 Matter: some associated notes
The key to the calculation is Newton’s third law
dP
,
F
dt
so to get the force per unit area on the wall we calculate its rate of gain of momentum
per unit area from the gas.
I will give you two approaches to the calculation: the first will seem more
straightforward to some, but I hope you will appreciate that the second – using
probability distributions - is more powerful.
Approach 1
Let us consider a rectangular box with two parallel walls of area A distance L
apart, and the contribution to the pressure on these walls from just ONE
molecule. If x is the direction normal to the walls, then the time between
collsions with the right hand wall is 2 L / u where u  v x is the magnitude
of the x component of particle velocity.
(Notice u does not change upon any reflection).
Every collision with the RH wall gains the wall an amount of momentum
equal to that lost by the particle, which is
mu - m(-u)  2mu
Hence the rate of gain of momentum by the wall is given by
2mu mu 2


L
2 L/u
p( 1 )A
where the (time averaged) pressure due to this molecule is
2
p( 1 )  m u
AL
It follows that the total pressure , summing over all the molecules, is given by
2
2
p  m N/V  v x   n m  v x 
The frustrating feature of this argument is we had to think about each
individual molecule and details of the box geometry which are intuitively not
relevant. We would like an approach that dwells on the relevant variables.
Approach 2
Consider a wall facing backwards along the x-direction as shown. Upon
collision a molecule changes its velocity from (v x , v y , v z ) to (v x , v y , v z ) ,
where vx must be positive to hit this wall. Since the value vx is the crucial
property, let us identify the fraction of all molecules with velocity (xcomponent) in the range vx to vx +d vx as
f (v x ) d v x
(the formal way to say this is that f (v x ) is the probability distribution of
molecules over vx ).
The rate at which such molecules strike unit area of wall is just
(velocity towards wall)(total concentration) (fraction in range of vx) =
University of Warwick, Department of Physics
PX 119 Matter: some associated notes
v x n f (v x ) dv x
and upon each collision the momentum gained by the wall is, from the
momentum lost by the molecule,  2mvx in the x-direction.
Therefore the rate of gain of momentum per unit area of wall is
2mvx v x n f (v x ) d v x
and to get the total pressure we have to add (integrate) the contributions from
all velocity ranges, leading to

p  2nm v x f (v x ) d v x .
2
0
Notice: we can take the constants m and n outside the integral, and the range
of integration includes only positive velocities - because only these are
towards the wall in question.
Assuming that positive and negative velocities are equally likely, we can
absorb the factor of two by including negative velocities in the integral (or if
you prefer, average the results for the pressure on a right hand and a left hand
wall) to obtain

p  nm  vx f (vx ) d vx  nm  vx 
2
2

where the angular brackets <> denote an average over all the molecules.
The above formula is fundamental, in that it is x-velocities which determine the
pressure on an x-facing wall, but the standard result is to express this in terms of
molecular speeds. Averaging over the results for walls facing in x, y, and z gives
p  13 nm  c 2 
where c is molecular speed and
2
2
2
c 2  vx  v y  vz .
(Reminder! Velocity is a vector and speed is its magnitude.)
The second approach above argument may not seem so easy to many of you: we have
used mechanics, probabilty and the calculation of a rate. For a related account, in
slightly different notation, see Tabor’s book, section 3.2 (pages 53-5 in the third
edition); the library has plenty of copies. Also Podesta, pages 54++.
Now let us compare theory with experiment for the pressure:
2
p  nkT  nm  v x 
They agree if
2
2
2
 12 mvx    12 mv y   12 mvz   12 kT
that is if the mean kinetic associated with motion in each direction is 12 kT . We will
see later that this fits with Boltzmann’s law (lecture 1) and is an example of the
Equipartition Theorem. The total kinetic energy of each atom in a monatomic ideal
gas is then
 12 mc 2   32 kT
University of Warwick, Department of Physics
PX 119 Matter: some associated notes
where monatomic means that each free particle is a single atom (He, Ne, A …)
rather than a multi-atom molecule which will have further kinetic energy associated
with its rotational and vibrational motions (see later).
This kinetic energy is the only internal energy stored in the gas, so
Emonatomic  32 kT per atom and the corresponding heat capacity (at constant volume) is
found from CV  d E d T to give
CV ,monatomic  32 k per atom.
This is the ‘atomic’ heat capacity, that is per atom. The term specific heat capacity is
often used, and this refers to the heat capacity per unit mass, for which our result
becomes 32 k / m .
The specification of constant volume is important, because otherwise work is done in
moving the walls. This extra effect shows up in the other standard heat capacity, the
heat capacity at constant pressure. For the ideal gas there is a very simple result for
the difference, as we calculate below.
At constant pressure the energy we supply as heat is not all stored in the gas, because
as the gas expands it does work against the surroundings and that comes out of the
energy budget:
dQ  C p dT  dE  pdV
dV
dV
, so we need to calculate p
for the gas expanding at
dT
dT
constant pressure. This is particularly simple from the ideal gas equation pV  NkT :
differentiate both sides (with respect to T, with p staying constant) to obtain
dV
p
 Nk . Hence on a per molecule basis
dT
C p  CV  k .
We will return to use this result after discussing the equipartition theorem later.
Hence C p  CV  p