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In the study of heat, and of thermodynamics in general, there was no need to know
the microscopic nature of the substance which was being
heated or cooled and so on. Instead, we reasoned
using general principles such as temperature, work, and energy conservation
and explained a number of important phenomenon. We then introduced
the idea of entropy, which is needed for explaining the physically observed
irreversibility of the world around us. The Second Law of
Thermodynamics was then shown to imply a number of new results, including
the fact that there can never be a perpetual motion machine.
The idea of energy is well understood from mechanics. To explain thermodynamics
the new and profound idea of entropy had to be invented. It is the idea of entropy that is
central and unique to the study of heat and thermodynamics, and the branch of
statistical mechanics has resulted from the attempt to understand entropy
from a microscopic point of view.
Recall from (9.41) we have a microscopic definition of
entropy given by
The microscopic definition unavoidably led to assumptions as to what
matter is made out of, namely what is the microscopic composition
of matter. For example, in applying the equation for entropy to the case of an
ideal gas, we had to
take into account the microscopic nature of the gas, in
particular, that it is made out of an enormous collection of
microscopic objects that we identified with atoms.
The entire field of statistical mechanics was
founded by Boltzmann in the late nineteenth century. As a
historical aside, it is worth recording that it was
in order to understand the concept of entropy
from a microscopic point of view, that Boltzmann had postulated the
existence of atoms well before their discovery in the twentieth
In sum, the challenge posed by thermodynamics was the following:
how can we reconcile ideas such as temperature, entropy and so on
with the ideas of (Newtonian) mechanics? In particular, if any sample
of matter that we observe in daily life is made out of an
inordinately large number of atoms, approximately
, how can we
apply the laws of mechanics to this large collection of particles?
Clearly, it is hopeless to try and describe how every single
particle is moving, as this would involve specifying, at each
number of positions and velocities. So
what is the way out of this impasse?
We know for a fact that all matter is composed out of small
particles called atoms. Statistical mechanics is that branch
of physics which explains the thermodynamic properties of nature
starting from a microscopic point of view. In particular, we will
attempt to apply classical mechanics to a large collection of particles,
and in this way derive all the results of thermodynamics.
We know that classical (Newtonian) mechanics cannot explain why
atoms even exist, let alone explain its properties for which
quantum mechanics is necessary. So can we, at all, classically analyze a
large collection of atoms before understanding quantum mechanics?
The answer, surprisingly enough, is yes. The reason is the
following. In studying statistical mechanics, we will be concerned
with the object's bulk (macroscopic) properties such as temperature,
energy and so on. These properties result mainly from the
interaction of atoms (and molecules) with each other.
Recall atoms are electrically neutral, and are composed of a positively
charged nucleus and negatively charged electrons which are distributed
outside the nucleus. Let there be two atoms at positions and .
The distance between them is then given by
; the so
potential results from the quantum mechanical interaction of the charges
and angular momentum that is carried by the atoms. The potential due
to a typical atom or molecule is given by
where is a constant which depends on charge, and is the Lennard-Jones (LJ) radius,
and is shown in Figure 10.1. Note that there is a minimum value in the inter-atomic
potential at a distance of from the atom.
As long as the atoms are moving slowly, and are farther
away from each other than distance , they can be treated
as hard spheres of radius that behave as classical particles.
For typical atoms and molecules the LJ-radius is around 3 to 5A
(A= Angstrom =). For example, for the argon atom,
the LJ radius is 3.5A, and is 5A for a large molecule such as propane.
However, in some cases the LJ radius is not
suitable for determining the effective classical size of an atom.
For example the molecule has an LJ radius
of 0.7A, and is too small a distance to be taken as the classical
radius of the molecule.
As long as the object being analyzed is at temperatures
and densities that are not very high or very low, the
atoms are not squeezed together closer than the distance of the LJ radius, and we
can treat the atoms as classical billiard balls.
However at very low temperatures and high densities, this is not
true and the classical analysis needs to replaced by quantum
mechanics. At very high temperatures, the inner structure of
the atoms, composed as it is out of a nucleus and electrons, needs to be
taken into account, and requires an analysis which goes beyond
We can now return to the problem at hand, namely that we have a
number of classical
particles, thought of as hard spheres of radius , having mass , and at
temperature . We would like to derive all the thermodynamic properties
of the object in question starting from Newtonian mechanics.
For the sake of concreteness, let us consider an ideal gas at
temperature , confined in a container of volume ,
and let us further suppose that the gas is in equilibrium.
By the gas being ideal, we mean that all the interactions of the
particles which compose the gas can be ignored. The energy of the
gas hence consists entirely of kinetic energy; let the three-dimensional
velocity of the -th particle be denoted by
. Since there are
particles, the total kinetic energy of the gas is simply the sum of
the kinetic energies of the individual particles (atoms). Hence
the energy of the gas is given by the following
Recall that by equilibrium we mean that the gas has attained a state of
maximum entropy, or equivalently, that there are no more changes
of temperature and other state variable taking place. By the statement
that the gas is at temperature , we mean that the gas in
question is in contact with a heat bath which is at a temperature
. The very fact that we have introduced the physical idea of
temperature already implies that the gas is not an isolated
system, but rather is part of a larger system which includes the
heat bath and the object at a given temperature.
How do we describe a gas, shown in Figure 10.2,
composed out of
particles, occupying a volume
and at temperature ?
There are simply too many particles to keep
track of. To provide a mechanical description of the gas,
we need to know the exact position and velocity of each and every
particle, and which in general, is called a microstate of the system. A description
of the microstate of any large object, containing about Avogardo's number of
atoms, is in practice too difficult.
And even more importantly, there is no need
since the questions asked in thermodynamics do not refer
to any single atom composing the gas, but rather, refer to the
properties of the gas taken as a whole, called the
bulk properties of the gas.
We now make a major conceptual leap. We postulate that having
a gas at a temperature means
that the gas is not in a definite (mechanical) microstate state. Instead,
the various (mechanical) microstates states of the gas are now taken to occur
with a certain probability. Hence, the description of the gas by its microstates,
that is, by the detailed knowledge of the position and
momentum of each and every atom of the gas,
is replaced by an ensemble of microstates. An ensemble is a collection
of all the possible microstates of the gas. The ensemble is called a
microcanonical, canonical or grand canonical depending on the way that
probabilities are assigned for the occurrence of the various microstate . We will return to
this question is some detail in Section .
Since we know nothing of the microstates of the gas, the most consistent
manner of assigning a probability of occurrence for the various allowed
microstates is to
assume that all the microstates of the gas are equally
likely; this is how a microcanonical ensemble is defined.
Our ignorance of the microstates is consequently given a complete expression in the
microcanonical ensemble which is defined as follows.
Given the parameters such as energy, volume and so on that specify
the macroscopic properties of the gas, in the microcanonical ensemble
all the microstates of the gas are equally likely.
One should note that the idea of ensemble reflects our
ignorance as to what is the microstate of the
gas. The gas is inherently not in a probabilistic state, but rather
it is our inability
to determine its state which has led us to the ideas of probability, and to the idea of
classical uncertainty. In quantum mechanics we will encounter
uncertainty which is not a function of our ignorance, but rather,
is an intrinsic property of nature.
In the language of probability theory, the positions and
velocities are all considered to be continuous random
variables. In other words, the velocity of the particle has no
definite value, but rather, its probability of occurrence is
determined by the ensemble that describes it. We will denote by
brackets the average value of a random variable. Hence, the
average value of the kinetic energy of the th particle is
. In Section 4 we will examine
more closely how to calculate the average value of various
physical quantities including kinetic energy.
As the first application of the idea of ensembles, we discuss the
kinetic theory of gases. This theory is applicable to a dilute
gas that, to a good degree of accuracy, can be considered to be an ideal gas.
Recall every atom of the gas is considered to be a free classical
particle with mass and having a random velocity . We will
use the term atom and particle interchangeably.
From the atomic point of view, how does pressure arise? Even before going into
the detailed mechanism, we expect that pressure should be a macroscopic
manifestation of microscopic motion.
frictionless piston, which has an area , contains a gas in some volume and
with total number of atoms given by . The piston is in equilibrium with the
gas at some temperature . Outside the piston is a perfect vacuum.
Gas in contact with a heat bath
The atoms of the gas are constantly bombarding the piston and this will cause a force
to be exerted on the piston, and to keep
the piston in place (stationary) we need to counter this force.
Consider for an atom traveling straight towards the piston with velocity
; we will show
later the case of the atom moving with arbitrary velocity gives the same result.
The particle hits the piston and bounces back. We
assume that the collision is elastic in that the energy of the atom is the same before
and after bouncing off the piston. The assumption of the collision being elastic means
that the atom does not lose any of its energy to the piston. This is
since if it lost energy to the piston, the piston would heat up; and once the piston reached
equilibrium with the gas, the assumption of collisions being elastic would be correct.
As shown in Figure 10.4, for a wall placed along the -axis, an elastic
collision leads to a velocity
. Hence, for
simplicity we consider only the special case of
Let the velocity of the atom after the collision be
Since the atom possesses
only kinetic energy, we have from energy conservation
Gas Inside the Piston and Vacuum Outside
Diagram v to -v
In other words, in colliding off the piston, the particle's
velocity changed from to , and hence the momentum imparted
to the piston is
Diagram v to -v
In time , how many atoms will bounce off the piston? All the
atoms with velocity can reach the piston in time if they are at a
distance less than from the piston. Consequently, all the atoms in a volume of
size will bounce off the piston. Hence, since the density (particles per
unit volume) is
, in time the momentum imparted
to the piston is
Since the force on the piston is nothing but the rate at which momentum changes on
the piston due to collisions of the gas atoms, we have
Recall pressure is defined to be force per unit area, and
hence the pressure on the piston due to the gas is
From the ensemble point of view, the velocity of
the atom is a random variable, and what the piston really
experiences is the average value over all possible
velocities that the atoms has as it bounces off the piston.
Hence, denoting as usual average values by , we have
The reason we have dropped the factor of in going from
(10.12) to (10.13) is that we need to perform the average
over only those particles which are heading towards the
piston and not away from it. Since we are taking the average value
of , we are over-counting by a factor of since we are
also including the particles moving away from the piston.
Recall we had considered a very special set of velocities, namely,
those heading straight for the piston, and hence with
. In general, the velocity of an arbitrary atom has the form
. Since all directions for the gas are
equivalent, we have
We finally have, from eq.(10.13), the following
The total energy, of the gas is solely composed of kinetic energy. Hence
From (10.13) and (10.15) and (10.17) we have
We have so far been able to define both the pressure and
energy of the gas from the atomic point of view. We now need
to define temperature.
Consider two gases in a cylinder separated by a frictionless piston. When the
piston reaches equilibrium there is no further change in the
system. We hence conclude that the temperature of both of the
gases must be the same, since the very definition of temperature
is that there will be heat flows, and consequently, no
equilibrium, unless and until the temperatures of the two gases
We now examine the conditions under which there will be
equilibrium. Let us label the gas on the left of the cylinder as
and that on the right as . For equilibrium, the pressure
exerted by both the gases on the piston must be equal. Hence, from
(10.16) we have
Piston Separating out Two Gases
Are the densities of the two gases the same on two sides of the
piston, that is
answer is yes, although to prove this is quite difficult. The
intuitive proof that the two densities are equal is that if there
was a difference in the densities, there would be a net ``osmotic'' pressure
on the piston forcing it to move, and consequently the
system would not be in equilibrium. Hence, in equilibrium, we have
and from (10.19) we obtain
We see that the equation above is simply a statement that the
average kinetic of the atoms in two gases which are in equilibrium
is the same. Hence temperature is defined to be
proportional to the average kinetic energy of a single atom of the gas.
Fixing the constant of proportionality to be the Boltzmann
constant we finally arrive at the following definition of temperature .
Temperature is a measure of how fast, on the average, that the
atoms of a gas are moving. At room temperature
The faster the atoms move, the
hotter the temperature. The sensation of burning that we have on
putting our hands into, say a fire, is because fast moving atoms
from the fire impart high amounts of kinetic energy to our hands,
causing atoms in our hand to move very fast and result in the
sensation of burning.
Hence, from eqns.(10.2) and (10.22)we have
Combining our results, from (10.16) and (10.22) we
finally obtain the ideal gas law
The result above has the remarkable implication that no matter
what the gas is composed of, for example, be it nitrogen, helium and so on,
equal volumes of the various gases at the same pressure and
temperature have the same number N of atoms. Note this
result follows directly from Newton's Laws as is seen by the
derivation given. This remarkable property of the ideal gas led
Avoagardo to postulate that one molar volume of any gas will have
the same number of atoms, given by Avagardo's number
Next: Quantum Theory
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Previous: Second Law of Thermodynamics