Statistical Physics: September 26, 2012
Solution for the Homework 3
Problem 6.42: In Problem 6.20 you computed the partition function for a quantum harmonic oscillator: Zh.o. = 1/(1 − e−β ), where = hf is the spacing between energy levels.
(a) Find an expression for the Helmholtz free energy of a system of N harmonic oscillators.
N . So,
Solution: Let the oscillators are distinguishable. Then Ztot = Zh.o.
1
N
F = −kT ln Ztot = −kT ln Zh.o. = −N kT ln
.
1 − e−β
(1)
(b) Find an expression for the entropy of this system as a function of temperature. (Don’t worry,
the result is fairly complicated.)
Solution: Using the eq. 6.64,
∂
1
∂F
=
N kT ln
S = −
∂T V,N
∂T
1 − e−β
∂β
1
(1 − e−β )(−)(−e−β )
+ N kT
= N k ln
−β
∂T
1−e
1
= k ln Ztot − N T
(1 − e−β )e−β
kT 2
h
i
F
1
U
= k ln Ztot −
N (1 − e−β )e−β = − + .
T
T
T
(2)
(3)
(4)
(5)
This result is same with the equation 6.62.
Problem 6.43: Some advanced textbooks define entropy by the formula
X
S = −k
P (s) ln P (s),
(6)
s
where the sum runs over all microstates accessible to the system and P (s) is the probability of
the system being in microstate s.
(a) For an isolated system, P (s) = 1/Ω for all accessible states s. Show that in this case the
preceding formula reduces to our familiar definition of entropy.
Solution: Since the probability distribution is uniform,
S = −k
X
P (s) ln P (s) = −k
s
1
X1
X1
1
ln
= k ln Ω
= k ln Ω.
Ω Ω
Ω
s
s
(7)
Statistical Physics: September 26, 2012
(b) For a system in thermal equilibrium with a reservoir at temperature T, P (s) = e−E(s)/kT /Z.
Show that in this case as well, the preceding formula agrees with what we already know about
entropy.
Solution: Substitute P (s) to given equation:
S = −k
X
P (s) ln P (s) = −k
X e−E(s)β
Z
s
= kβ
Z
s
s
= −k
X e−E(s)β
!
[−E(s)β + ln Z]
#
"
X E(s)e−E(s)β
s
ln
e−E(s)β
Z
= kβU + k ln Z =
(9)
"
#
k ln Z X −E(s)β
−
e
Z
s
Z
(8)
U
F
− .
T
T
(10)
(11)
This result agree with the equations 6.58 and 6.59.
Problem 6.44: Consider a large system of N indistinguishable, noninteracting molecules (perhaps
in an ideal gas or a dilute solution). Find an expression for the Helmholtz free energy of this
system, in terms of Z1 , the partition function for a single molecule. (Use Stirling’s approximation
to eliminate the N!.) Then use your result to find the chemical potential, again in terms of Z1 .
Solution: Stirling’s approximation tell us that ln N ! ≈ N ln N − N . Using this,
F
= −kT ln
Z1N
= −kT (N ln Z1 − ln N !) ≈ −N kT (ln Z1 − ln N + 1) .
N!
(12)
Now, differentiate the Helmholtz free energy to N, then we can calculate the chemical potential
that is:
1
∂F
≈ −kT (ln Z1 − ln N + 1) + N kT
= −kT (ln Z1 − ln N ) . (13)
µ =
∂N T,V
N
Problem 6.48: For a diatomic gas near room temperature, the internal partition function is simply
the rotational partition function computed in Section 6.2, multiplied by the degeneracy Ze of
the electronic ground state.
(a) Show that the entropy in this case is
S = N k ln
V Ze Zrot
N vQ
7
+
.
2
(14)
Calculate the entropy of a mole of oxygen (Ze = 3) at room temperature and atmospheric
pressure, and compare to the measured value in the table at the back of this book.
Solution: The total partition function is
Z =
2
1
N!
N Zint
vQ
N
,
(15)
Statistical Physics: September 26, 2012
where Zint = Ze Zrot . Using this condition, keep following the equations from 6.85 to 6.92). Then
the result is
V
5
∂
S = N k ln
+
+
[N kT ln Zint ] .
(16)
N vQ
2
∂T
But, in this case, the temperature is quite high and so we can treat this problem as Zrot ≈ kT /.
Then the second term of above equation is
Z
∂
ZZe ∂Zrot
(17)
[N kT ln (Ze Zrot )] ≈ N k ln (Ze Zrot ) + N kT
Z
∂T
ZZe Zrot ∂T
k
(18)
= N k ln (Ze Zrot ) + N k T
k T = N k ln (Ze Zrot ) + N k.
(19)
So, the final result is
V
5
V Ze Zrot
7
S = N k ln
+
+ N k ln (Ze Zrot ) + N k = N k ln
+
. (20)
N vQ
2
N vQ
2
Now, put the values 1 mole of oxygen with 298K for temperature and 1 bar for pressure and
= 0.00018eV on this equation. Then, S ≈ 210J/K. This is similar to the experimental value,
205.14J/K.
(b) Calculate the chemical potential of oxygen in earth’s atmosphere near sea level, at room
temperature. Express the answer in electron-volts.
Solution: From the equation 6.93,
µ = −kT ln
V Ze Zrot
N vQ
.
(21)
Put the same values with the previous problem, then µ = −0.56eV .
Problem 6.50: Show explicitly from the results of this section that G = N µ for an ideal gas.
Solution: Use the equations 6.87, 6.88, 6.91, 6,92 and 6.93. Then we can easily show that the
Gibbs free energy is:
G = U − TS + PV
3 V
∂
5
= Uint + N
kT
−
N
kT
ln
+
+T
[−N kT ln Zint ] + N
kT
2
N
v
2
∂T
Q
V
∂β
∂
2
= Uint − N kT ln
− N kT ln Zint − N kT
ln Zint
N vQ
∂T
∂β
V Zint
N
kT2
−
E int
= Uint − N kT ln
N vQ
kT2
+ Nµ −
= Uint
N
E int
= N µ.
3
(22)
(23)
(24)
(25)
(26)