Statistical Mechanics - Physics at Oregon State University

44 CHAPTER 3. VARIABLE NUMBER OF PARTICLES
and the free energy is a minimum if the chemical potentials are equal, µ1 = µ2.
This, of course, is similar to the conditions we found before for temperature and
pressure. Although we could have defined the chemical potential as an arbitrary
function of the partial derivative, the definition above is the only choice which
is consistent with the experimentally derived chemical potential. The definition
in terms of a derivative only holds for large values of N, in general one should
write
µ(T, V, N) = F (T, V, N) − F (T, V, N − 1) (3.3)
which tells us that the chemical potential equals the free energy needed to subtract
one particle to the system. This is another reminder that of a difference
between thermodynamics and statistical mechanics. In thermodynamics, the
number of moles N of a material is a continuous variable. In statistical mechanics,
the number of particles N is a discrete variable, which can be approximated
as continuous in the thermodynamic limit (showing up again!). Finally, if different
types of particles are present, on specifies a chemical potential for each
type according to
∂F
µi(T, V, N1, N2, ...) =
∂Ni
Chemical potential, what does it mean?
T,V,Nj=i
(3.4)
What is this chemical potential? The following example will help you to
understand this important quantity. If the systems 1 and 2 are not in equilibrium,
the Helmholtz free energy is reduced when particles flow from the system
with a high value of the chemical potential to the system with the lower value,
since ∆F = (µ2 − µ1)∆N. We can also add some other potential energy to
the systems. For example, the systems could be at different heights are have
a different electric potential. This potential energy is Φi. If we bring ∆N
particles from system 1 to system 2, we have to supply an amount of energy
∆E = (Φ2 − Φ1)∆N to make this transfer. Since this is a direct part of the
energy, the total change in the Helmholtz free energy is given by
∆F = (ˆµ2 + Φ2 − ˆµ1 − Φ1)∆N (3.5)
where ˆµ is the chemical potential evaluated without the additional potential
energy. Therefore, if we choose Φ such that
ˆµ2 + Φ2 = ˆµ1 + Φ1
(3.6)
the two systems will be in equilibrium. Of course, the presence of an external
potential Φ leads to a term ΦN in the internal energy U and hence in F. As a
result