Before we talk about gases, we need to understand a few more results from quantum mechanics. We will sometimes use the word "orbital" here to mean a state possible for one particle.

Every fundamental particle is of one of two types. A fermion is a particle with half-integer spin. For example, an electron has a spin of 1/2. A boson is a particle with integer spin. For example, a photon is a boson because it has spin 1.

The difference in the two types of particles can be characterized by what is known as the "Pauli Exclusion Principle", which insists that an orbital can only be occupied by 0 or 1 fermions. Bosons, on the other hand, can be fit without limit into a single orbital. This fact alone leads to radically different behavior under certain conditions, such as low temperature.

There is a bit of convention that must be established now. Instead of
writing
< *N*() >
for the average number of particles in a
particular orbital of energy
, we write
*f* ()
.
*f*
is known as the distribution function, and its value is of
course dependent on the type of system that we are talking about.

Notice that the difference between fermions and bosons has to do with
orbitals having an occupancy
*N*
larger than 1. For this reason, we
presume that fermions and bosons behave similarly for sparsely populated
orbitals; that is, for
*f* 1
. We call this condition the classical
regime, because it doesn't depend on the quantum distinctions between
particles.

It has been known for quite some time that the classical distribution function is given by:

Here we have used the symbol
*λ*
to mean
*e*
^{
μ/τ
}
.

We will start using the term ideal gas to mean a gas of particles
that do not interact with each other and are in the classical regime.
Another way of expressing that a system is in the classical regime comes
from the quantum concentration. We use
*n*
to mean
*N*/*V*
here.
Then if a gas is less dense than the quantum concentration,
*n*
_{Q} =
, we say it is in the classical regime.

Summing the particles over all orbitals of a system and setting this equal to
*N*
, the total number of particles, yields
*λ* =
. Expanding
*λ*
and solving for the chemical
potential gives us:

We spent a lot of time devising ways to relate the variables we need to
the energies. We can utilize that now. Recall that
*μ* =
. We can integrate
to solve for
*F*
, and we obtain:

We seek to get the pressure from the free energy. This is no problem
though, since we can recall or rederive that
*p* = -
.
Looking at the expression for
*F*
above, we see
that we can expand it to be the sum of many terms, most of which have no
*V*
dependence. The derivative becomes simple, and returns something
familiar:

This is the ideal gas law. If it doesn't look familiar, recall that the chemistry version uses number of moles instead of number of particles, and replaces the temperature as we've defined it with the temperature in Kelvin. You might wish to work out the conversion to assure yourself.

We use the relation
*σ* = -
to find the entropy from the free energy. Without much work, we come up
with:

Remember that the free energy can be defined in terms of the energy as
follows:
*F* = *U* - *τσ*
. We rearrange to solve for
*U*
, and plug in
our values for
*F*
and
*σ*
to find the simple result:

A measure of how much heat a gas can hold is the heat capacity.
There are two slightly different measures of the heat capacity. One,
the heat capacity at constant volume, is defined as
*C*
_{V}âÉá
. The other, the heat capacity at constant pressure, is defined as
*C*
_{p}âÉá
.

The only difference between the two definitions is in what is held constant in the derivative. The results for an ideal gas can be obtained by direct substitution and differentiation for the heat capacity at constant volume, and by the thermodynamic identity for the heat capacity at constant pressure. The results are:

Remember that these are in fundamental
units, and we need to multiply by the Boltzmann constant
*k*
_{B}
to change
to conventional units.

We define the ratio of the two heat capacities,
*C*
_{p}/*C*
_{V}
, to be
*γ*
. For an ideal gas,
*γ* = 5/3
.

There is a good shortcut to find the energy of any classical system,
known as equipartition. The theory states that every particle has
energy equal to
*τ*
for each degree of freedom of the
particle, which can be gleaned from the number of quadratic terms in the
expression for the energy.

Let us make the theory clearer by applying it to the ideal gas. Each
particle in the ideal gas has classical energy equal to
*mv*
^{2}
. Here, the velocity is a vector having 3 components. In
Cartesian, there are
*v*
_{x}
,
*v*
_{y}
, and
*v*
_{z}
. Therefore each particle
has energy
*τ*
. Summing up for all
*N*
particles in the
system gives the same answer we got before,
*U* =
*Nτ*
.