Statistical Mechanics - Physics at Oregon State University

18 CHAPTER 1. FOUNDATION OF STATISTICAL MECHANICS.
and hence the entropy of the total system is
S = SA + SB
(1.60)
which shows that for two systems in thermal equilibrium the entropy of the total
system is equal to the sum of the entropies of subsystems. Hence the entropy
is an extensive quantity. Since U is also extensive, it follows from the definition
of T that T is intensive.
At this point we have two definitions of entropy. In thermodynamics we
defined entropy by making the differential for heat flow exact. In statistical
mechanics we define entropy related to the number of quantum states. In order
for statistical mechanics to be a useful theory, these definitions have to be the
same. This one can show as follows. The flow of energy between the two
systems we considered was not due to work, hence must be the exchange of
heat. Thermodynamics tells us that two systems are in equilibrium if they
have the same temperature, when only heat can flow between them. Therefore,
the quantity T defined in statistical mechanics must be a function of the real
temperature only, since equilibrium is independent of volume and number of
particles. Hence if we define
1
=
T SM
SM ∂S
where the superscript SM denotes defined in statistical mechanics, we find
∂U
N,V
(1.61)
T SM = f(T ) (1.62)
Also, by construction the entropy defined in statistical mechanics is a state
function. For each value of U,V, and N we have a unique prescription to calculate
the entropy. Therefore, the differential of this entropy is exact and we have
¯dQ = T dS = f(T )dS SM
(1.63)
As we saw in thermodynamics, the only freedom we have in making an exact
differential out of the heat flow is a constant factor. Hence
T SM = αT (1.64)
S SM = 1
S (1.65)
α
The entropy is therefore essentially the same in both thermodynamics and
statistical mechanics. The factor α is set equal to one, by choosing the correct
value of the prefactor kB in the statistical mechanical definition. A change in
this value would correspond to a change in temperature scale. In order to show
this, we have to consider a simple model system. A gas of independent particles
is a good test case, and the calculations show that the equation of state derived
from a microscopic model is the same as the experimental version, if we take
NAvogadrokB = R.