Entropy of the Einstein Solid
We consider the high-temperature case of q≫N.
lnΩ = ln ( q + N ) !
q!N!
"
# $ %
&
' = ln q + N ( ) !− ln q!− ln N!
≈ 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 N
N
N e q N
eq
!! "
#
$$ %
&
!! =
"
#
$$ %
&
≈ Ω
S = k ln ⌦
S total = k ln ⌦ total = k ln ⌦ A ⌦ B = S A + S B U = ~!q
S = N k + N k ln U N k ln( ~!N)
http://www.compadre.org/stp/items/detail.cfm?ID=8685 Consider now two Einstein solids with N
A=300, N
B=200, and q
total=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.
@S
total@q
A= 0 @S
total@U
A= 0
q A max = N A
N A + N B q = 60
@S A
@U A = @S B
@U B
Equilibrium condition:
S A
S B
S total
S total
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 q
Athat 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)
−1has 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.
1
T ⌘ @S
@U 1
T ⌘
✓ @S
@U
◆
N,V
We consider the high-temperature Einstein solid (q≫N).
S = N k + N k ln U N k ln( ~!N) 1
T =
✓ @S
@U
◆
N,V
= N k
U U = N kT
consistent with the equipartition theorem Now we take the monoatomic ideal gas.
If no work is performed (e.g., V=const)
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?
S = N k
"
ln V N
✓ 4⇡mU 3N h
2◆
3/2!
+ 5 2
#
T =
✓ 3
2 N k U
◆ 1
U = 3
2 N kT
dS = dU
T = dQ T
dS = C V dT
T S =
Z Tf
T i
C V
T dT S(T ) S(0) =
Z T 0
C V
T dT
Clausius 1865
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.
Left: Absolute zero can be reached in a finite number of steps if S(0,X
1)≠S(0, X
2).
Right: An infinite number of steps is needed since S(0,X
1)= S(0,X
2).
The temperature of a substance can be reduced in an isentropic process by
changing the parameter X from X
2to X
1.
S(T ) S(0) =
Z T 0
C V
T dT
the heat capacity of all substances must go to zero at absolute zero (otherwise the integral would diverge; S is always non-negative)
96 Chapter 3 Interactions and Implications
A B
500 K 300 K
1500 J
∆SA =−3 J/K ∆SB = +5 J/K
Figure 3.4. When 1500 J of heat leaves a 500 K object, its entropy decreases by 3 J/K. When this same heat enters a 300 K object, its entropy increases by 5 J/K.
Object A loses entropy, because heat is flowing out of it. Similarly, the entropy of B changes by
∆S
B= +1500 J
300 K = +5 J/K. (3.24)
Object B gains entropy, because heat is flowing into it. (Notice that the traditional entropy unit of J/K is quite convenient when we compute entropy changes in this way.)
Just as I often visualize energy as a “fluid” that can change forms and move around but never be created or destroyed, I sometimes imagine entropy, as well, to be a fluid. I imagine that, whenever energy enters or leaves a system in the form of heat, it is required (by law) to carry some entropy with it, in the amount Q/T . The weird thing about entropy, though, is that it is only half-conserved: It cannot be destroyed, but it can be created, and in fact, new entropy is created whenever heat flows between objects at different temperatures. As in the numerical example above, the entropy that is “carried by” the heat is more when it arrives at the cooler object than it was when it left the hotter object (see Figure 3.5). Only in the limit where there is no temperature difference between the two objects will no new entropy be created. In this limit, however, there is no tendency of heat to flow in the first place. It’s important to remember that fundamentally, the net increase
Q
S
Q
S
Q
S
Q
S
Q
S
Q
S
Q
S
Q
S
Q
S
Q
S
Figure 3.5. Each unit of heat energy (Q) that leaves a hot object is required to carry some entropy (Q/T ) with it. When it enters a cooler object, the amount of entropy has increased.
