Entropy of the Einstein Solid
We consider the high-temperature case of q≫N.
æ (q + N ) !ö
lnW = ln ç
÷ = ln ( q + N ) !- ln q!- ln N!
è q!N! ø
» ( q + N ) ln ( q + N ) - ( q + N ) - q ln q + q - N ln N + N
Now further assume that q ≫ N. In that case,
N
æ eq ö
æqö
W » çç ÷÷ = e N çç ÷÷
èNø
èNø
N
Consider now two Einstein solids with NA=300, NB=200, and qtotal=100
Aside from fluctuations that are normally much too small to measure, any isolated
macroscopic system will inevitably evolve toward whatever (accessible) macrostate
has the largest entropy.
Equilibrium condition:
http://www.compadre.org/stp/items/detail.cfm?ID=8685
Now let us ask how entropy is related to temperature. The most fundamental
way to define temperature is in terms of energy flow and thermal equilibrium:
Two objects in thermal contact are said to be at the same temperature if they
are in thermal equilibrium, that is, if there is no spontaneous net flow of
energy between them. If energy does flow spontaneously from one to the
other, then we say the one that loses energy has the higher temperature
while the one that gains energy has the lower temperature.
So the quantity that is the same for both solids when they are in equilibrium is
the slope dS/dq. Temperature must be some function of this quantity.
To identify the precise relation between temperature and the slope of the
entropy vs. energy, take qA that is larger than its equilibrium value. Here the
entropy graph for solid B is steeper than that for solid A, meaning that if a bit
of energy were to pass from A to B, solid B would gain more entropy than
solid A loses. Since the total entropy would increase, the second law tells us
that this process will happen spontaneously. In general, the steeper an
object’s entropy vs. energy graph, the more it “wants” to gain energy (in order
to obey the second law), while the shallower an object’s entropy vs. energy
graph, the less it “minds” losing a bit of energy. We therefore conclude that
temperature is inversely related to the slope dS/dU. In fact, the reciprocal
(dS/dU)−1 has precisely the units of temperature, so we might guess simply
In fact, this can be used as a definition of temperature:
The factor of Boltzmann’s constant k in the definition of entropy eliminates the
need for any further constants in this formula. To confirm that this relation
gives temperature in ordinary Kelvin units one must check a particular
example, as we will do in the following.
We consider the high-temperature Einstein solid (q≫N).
consistent with the
equipartition theorem
Now we take the monoatomic ideal gas.
If no work is performed (e.g., V=const)
Clausius 1865
This equation works even if T=const, e.g., during the phase transition. When T
is changing, we can write:
What about S at T=0?
Third law of thermodynamics (Nernst1906-1912)
The third law of thermodynamics is sometimes stated as follows:
The entropy of a perfect crystal at absolute zero is exactly equal to zero.
At zero kelvin the system must be in a state with the minimum possible energy,
and this statement of the third law holds true if the perfect crystal has only one
minimum energy state. Entropy is related to the number of possible microstates,
and with only one microstate available at zero kelvin, the entropy is exactly zero.
A more general form of the third law applies to systems such as glasses that may
have more than one minimum energy state (mathematically, the absolute entropy
of any system at zero temperature is the natural log of the number of ground
states times Boltzmann's constant):
The entropy of a system approaches a constant value as the temperature
approaches zero.
The constant value (not necessarily zero) is called the residual entropy of the
system. An example of a system which does not have a unique ground state is
one containing half-integer spins, for which time-reversal symmetry gives two
degenerate ground states. For such systems, the entropy at zero temperature is
at least ln(2)k. Physically, the law implies that it is impossible for any procedure to
bring a system to the absolute zero of temperature in a finite number of steps.
The temperature of a substance can be reduced in an isentropic process by
changing the parameter X from X2 to X1.
Left: Absolute zero can be
reached in a finite number of
steps if S(0,X1)≠S(0, X2).
Right: An infinite number of
steps is needed since
S(0,X1)= S(0,X2).
the heat capacity of all substances must go to zero at absolute zero
(otherwise the integral would diverge; S is always non-negative)