Problem 1 (2 marks): The first excited energy level of a hydrogen atom has energy of 10.2 eV, if
we take the ground-state energy to be zero. However, the first excited level is really four
independent states (microstates), all with the same energy. We can therefore assign it an entropy
of S = k ln 4 , since for this macrostate the multiplicity is Ω = 4 . For what temperatures is the
Helmholtz free energy of a hydrogen atom in the first excited level positive, and for what
temperatures is it negative?
Solution: The Helmholtz free energy of the first excited level of hydrogen atom is
F = U − T S = 10.2 eV - T k ln 4 . At low temperatures this is positive, so the atom would rather be
in the ground state, which has F = 0 . At high temperatures, however, F for the excited level
becomes negative, so this level becomes preferred over the ground state. The transition
temperature is
k Tt =
10.2
= 7.36 eV,
ln 4
Tt =
7.36 eV
= 8.5 ⋅10 4 K .
−5
8.62 ⋅10 eV/K
Problem 2 (5 marks): By subtracting µ N from U , H , F , or G , one can obtain four new
thermodynamic potentials. Of the four, the most useful one is the grand free energy (or grand
potential),
Φ ≡U −T S − µ N .
(a) Derive the thermodynamic identity for Φ , and the related formulas for the
partial derivatives of Φ with respect to T , V , and µ .
(b) Prove that Φ = − PV .
(c) Prove that for a system in thermal and diffusive equilibrium (with a reservoir
that can supply both energy and particles), Φ tends to decrease.
Hint: Study lecture 17.
Solution: (a) Using the product rule for derivatives, an infinitesimal change in Φ can be written
as
dΦ = dU − T dS − S dT − µ dN − N dµ .
Taking into account the first thermodynamic identity, dU = T dS − P dV + µ dN , we obtain
dΦ = − S dT − P dV − N dµ .
Therefore, the natural variables for the function Φ are temperature, volume and chemical
potential. Holding any two of these fixed yields
 ∂Φ 
S = −
 ,
 ∂T V , µ
 ∂Φ 
P = −
 ,
 ∂V T , µ
 ∂Φ 
N = −
 .
 ∂µ T ,V
(b)
Φ ≡ U − T S − µ N = G − PV − µ N = µ N − PV − µ N = − PV .
(c)
As in lecture 17, consider an infinitesimal change in the total entropy of the system plus its
environment (the “reservoir”):
dS total = dS + dS R
Applying the first thermodynamic identity, dU = T dS − P dV + µ dN , to dS R and assuming that
V is fixed, we obtain
dS R =
µ
1
dU R − R dN R .
TR
TR
The system is in thermal and diffusive equilibrium with a reservoir, hence: TR = T , µ R = µ .
The system and a reservoir is an isolated system as a total. Therefore, dU R = −dU , and
dN R = − dN . As a result,
dS R = −
1
µ
dU + dN .
T
T
Therefore,
dS total = dS −
1
µ
1
dΦ
,
dU + dN = − (dU − T dS − µ dN ) = −
T
T
T
T
where we took into account that dΦ = dU − T dS − µ dN provided that T and µ are constant.
Thus, the increase in the total entropy of the universe (system + reservoir in thermal and diffusive
equilibrium) is the same thing as a decrease in the grand free energy, Φ .
Problem 3 (3 marks): For a mole of aluminum, CV = aT + bT 3 at T < 50 K;
a = 1.35 ⋅10 −3 J/K 2 , b = 2.48 ⋅10 −5 J/K 4 . The linear term – due to mobile electrons, the cubic term
– due to the crystal lattice vibrations. Find a formula for the absolute entropy of a mole of
aluminum as a function of temperature. Evaluate the absolute entropy at T = 1 K, at T = 10 K.
Hint: Study lecture 20.
Solution: As the temperature of the aluminum increases from 0 to some final value T f , its
entropy increases by
Tf
S f − S0 =
∫
CV
dT =
T
Tf
0
∫
aT + bT 3
b
dT = aT f + T f3 .
3
T
0
If we choose the constant value of entropy at the absolute zero, S 0 , to be zero, we define the
absolute entropy of the aluminum (see lecture 20). Therefore,
S abs (1K ) = 0.00136 J/K,
S abs (10 K ) = 0.0218 J/K.
Problem 4 (4 marks): Consider a system of two Einstein solids, A and B each containing 10
oscillators, sharing a total of 20 units of energy. Assume that the solids are weakly coupled, and
that the total energy is fixed.
(a) How many different macrostates are available to this system?
(b) How many different microstates are available to this system?
(c) Assuming that this system is in thermal equilibrium, what is the probability of finding
all the energy in solid A?
(d) What is the probability of finding exactly half of the energy in solid A?
Solution: (a) As usual, when we are talking about weakly interacting systems, the word
macrostate refers to the state of the combined system, specified by the temporary constrained
values of U A and U B . Of the 20 units of energy, anywhere from 0 to 20 could be in solid A.
Each possibility from 0 to 20 defines a different macrostate, so there are 21 macrostates.
(b) The combined system has 20 oscillators and 20 units of energy, so the total number of
microstates is
 20 + 20 − 1
39!
 =
Ω (20, 20 ) = 
= 6.89 ⋅1010 .
20

 20! 19!
(c) For the macrostate with all the energy in solid A, the multiplicity of solid A is
 20 + 10 − 1
29!
 =
Ω A (10, 20 ) = 
= 10 7
20

 20! 9!
and the multiplicity of solid B is 1. Assuming that the system is in equilibrium, i.e. all possible
microstates are equally possible, we obtain
(
)
probability = 107 / 6.89 ⋅ 1010 = 1.45 ⋅ 10− 4 .
(d) For the macrostate with half the energy in each solid, the multiplicity of combined system is
2
Ω = Ω AΩ B
10 + 10 − 1 10 + 10 − 1  19! 
 = 8.534 ⋅10 9 .
 
 = 
= 
10
10


  10! 9! 
Therefore, in equilibrium
(
) (
)
probability = 8.53 ⋅10 9 / 6.89 ⋅1010 = 0.124 .
Problem 5 (6 marks): Consider a system of two Einstein solids with N A = 300 , N B = 200 ,
q total = 100 .
(a) Compute the entropy of this system assuming that all of the microstates are accessible
(the system’s entropy over long time scales).
(b) Compute the entropy of the most likely macrostate of this system (the system’s entropy
over short time scales).
Then, consider a system of two Einstein solids with N A = 10 23 , N B = 10 23 , q total = 2 ⋅10 23 .
(c) Compute the entropy of this system assuming that all of the microstates are accessible
(the system’s entropy over long time scales).
(d) Compute the entropy of the most likely macrostate of this system (the system’s
entropy over short time scales).
(e) Is the issue of time scales really relevant to the entropy of this system?
Solution: To answer part (a) and (b) we can use Figure 2.5 in the text that I reproduced in the
beginning of lecture 26.
(a) If we calculate entropy over all possible microstates (long time scale entropy), we obtain
(
)
S = k ln 9.3 ⋅10115 = k [ln(9.3) + 115 ⋅ ln(10 )] = 267 k .
(b) The most likely macrostate of the system is the macrostate with q A = 60 , q B = 40 . If we
calculate entropy over all microstates corresponding to this macrostate (short time scale entropy),
we obtain
(
)
S = k ln 6.9 ⋅10114 = k [ln (6.9 ) + 114 ⋅ ln(10)] = 264.4 k .
To answer part (c) and (d) we need to use the formula for the multiplicity of any large Einstein
solid:
q
q+N  q+N 
 
Ω( N , q ) = 

 q   N 
N
N
.
2 π q (q + N )
(c) To calculate entropy over all possible microstates we need to substitute 2N for both N and q:
 2N + 2N 
Ω( N , q ) = 

 2N 
2N
 2N + 2N 


 2N 
2N
2N
1
24N
2N 2N
=2 2
=
.
2 π ⋅ 2 N (2 N + 2 N )
8π N
8π N
Then,
 24N
S = k ln
 8π N


 = k 4 N ln 2 − ln 8 π N = k 2.77 ⋅10 23 − 28.1 .


(
) (
)
(d) The most likely macrostate of the system is the macrostate with q A = 10 23 , q B = 10 23 . If we
calculate entropy over all microstates corresponding to this macrostate (short time scale entropy),
we obtain
(
 24N 
 = k (4 N ln 2 − ln(4 π N )) = k 2.77 ⋅10 23 − 55.5
S = k ln

 4π N 
)
(e) As we can see, the issue of time scales is obviously irrelevant for such a large system.