Solutions to PHY294S Problem Set 6
(Dated: April 18, 2012)
* Sections 1 and 4 have been marked for a total of 20.
SECTION 1 [10 MARKS]
Schroeder 6.49
The rotational energy of a diatomic molecule at room temperature is kT, corresponding to
two degrees of freedom f = 2. The total energy of a mole of N2 is:
U = 32 NkT + NkT = 52 NkT = 52 nRT = 52 (1 mol)(8.31 J/mol · K)(298 K) = 6190 J [1 mark]
The enthalphy is just U + PV = U + nRT.
Since nRT = (1 mol)(8.31 J/mol · K)(298 K) = 2480 J, the enthalpy is H = 8670 J.
[1 mark]
We need to express the partition function to compute the other quantities (F,G,etc.). The
partition function is rotational:
Zint = Zrot =
kT
2✏
=
(8.617⇥10 5 eV/K)(298 K)
2(0.00025) eV
= 51 [1 mark]
The quantum volume is equal to:
h
vQ = ( p2⇡mkT
)3 = ( p
6.63⇥10 34 J· s
2⇡(28)(1.66⇥10 27 kg)(1.38⇥10
23
J/K)(298 K)
)3 = 6.98 ⇥ 10
The average volume per particle is:
V
N
=
kT
P
=
(1.38⇥10 23 J/K)(298 K)
1.01⇥105 N/m2
= 4.07 ⇥ 10
26
m3 . [1 mark]
We compute the logarithm term in the Helmholtz free energy:
26
int
ln( VZ
) = ln( (4.07⇥10
NvQ
6.98⇥10
m3 )(51)
)
m3
33
= 19.5. [1 mark]
The Helmholtz free energy is therefore:
33
m3 . [1 mark]
2
int
nRT[ln( VZ
) + 1] =
NvQ
F=
(2480 J)[19.5 + 1] =
50.8 kJ. [1 mark]
The Gibbs free energy is:
G = F + PV =
50.8 kJ + 2480 J =
48.3 kJ. [1 mark]
The entropy can be obtained from the definition F=U-TS:
S=
U F
T
=
( 6190 J) (50,800 J)
298 K
= 191 J/K. [1 mark]
To calculate the chemical potential, use Eqn. 6.93, derived by di↵erentiating F with
respect to N (at fixed T and V):
µ=
int
kTln( VZ
)=
NvQ
(1.38 ⇥ 10
23
J/K)(298 K)(19.5) =
8.03 ⇥ 10
20
J=
0.501 eV.
[1 mark]
SECTION 2
Schroeder 7.6
It’s easiest to start from the right-hand side of the desires relation and work backwards:
X @
kT X N(s) [E(s) µN(s)]/kT
[E(s) µN(s)]/kT
kT @Z
kT
=
e
=
e
Z @µ
Z
@µ
Z s kT
s
X
X
1
N(s) e[E(s) µN(s)]/kT =
N(s)P(s) = N.
Z
s
s
Similarly, makine use of the second line above,
X
1 X
(kT)2 @ 2 Z
[E(s) µN(s)]/kT
kT @
=
N(s)e
=
[N(s)]2 e[E(s)
2
Z @µ
Z @µ
Z s
s
µN(s)]/kT
=
X
s
[N(s)]2 P(s) = N2 .
Notice that these two results can be combined to obtain a formula for N 2 in terms of N :
N2 =
(kT)2 @ NZ
Z @µ kT
=
kT @N
Z
Z @µ
+ N @Z
= kT @N
+ (N)2 .
@µ
@µ
The standard deviation of a quantity can be calculated as the square root of the average of
the squares minus the square of the averages:
q
q
2
2
N
(N ) = kT @N
N =
@µ
3
For an ideal gas, µ =
kT ln(V Zint /N vQ ) where vQ and Zint are functions of temperature
only and the N could just as well be N if the number of particles in the gas fluctuates.
Therefor @µ/@N = kT /N or @N /@µ = N /kT , and the standard deviation is simply
q
p
kT (N /kT ) = N .
N =
The appearance of the square root is ubiquitous in formulas for fluctuations, so we could
have guessed this result (up to a numerical factor) by dimensional analysis. For a gas of
about 1023 molecules, the number will typically fluctuate by less than 1012 or about one
part in a hundred billion of the total.
Schroeder 7.9
The mass of an N2 molecules is about 28 atomic mass units, so it’s quantum volume at
room temperature is:
vQ =
3/2
h2
2⇡mkT
=
(6.63⇥10 34 J.s)2
2⇡(28)(1.66⇥10 27 kg)(1.38⇥10
(Since the atoms are 10
10
23 J/k)(300K)
= 6.9 ⇥ 10
33
m3 = (1.9 ⇥ 10
V
N
m)3
metes wide, this is substantially smaller than physical volume
of the nitrogen molecule.) Now we can use Boltzman statistics whenever Zi
V
11
N , that is
N vQ . But at standard temperature and pressure,
=
kT
P
=
(1.38⇥10 23 J/k)(300K)
105 Pa
= 4.1 ⇥ 10
26
m3
This is greater than the quantum volum by a factor of about 6 million, so Boltzmann
statistics should be very accurate. On the other hand, Boltzmann statistics would break
down when vQ semieqV /N . HoldingV /N fixed, this would require that T 3/2 be lower by a
factor of 6 million or that T be lower by a factor of about 30000. That’s 1/100K. In other
words, quantum statistics is irrelevant to an ordinary gas at this density provided that the
temperature is higher than 0.01K
4
SECTION 3
Schroeder 7.15
For a system of particles obeying the Boltzmann distribution, the total number of particles
should be:
P
P
N = all s n̄Boltzmann = s e(
✏s µ)/kT
= eµ/kT
P
s
e(✏s /kT .
The sum in the last expression is just the single-particle function, Z1 . Therefore,
N
Z1
kTln ZN1 .
= eµ/kT or µ = kTln ZN1 =
Schroeder 7.19
The density of copper is 8.93 g/cm3 , and the atomic mass is 63.5 g/mol. Let’s consider a
chunk of copper that contains one mole of copper atoms. Then the mass of the chunk is
63.5 g and its volume is:
V=
mass
density
=
63.5 g
8.93 g/cm3
= 7.11 ⇥ 10
6
m3
Assuming that each atom contributes one conduction electron, the number of conduction
electrons per unit volume is:
N
V
=
6.02⇥1023
7.11⇥10 6 m3
= 8.47 ⇥ 1028 m 3 .
Therefore, the Fermi energy is:
✏F =
h2 3 N 23
( )
8m ⇡ V
= 1.13 ⇥ 10
18
J = 7.05 eV.
The Fermi temperature is:
TF =
✏F
k
= 82, 000 K.
This is 720 times greater than room temperature. Therefore, room temperature is
sufficiently low for this electron gas to be considerate degenerate.
The degeneracy pressure for this system is (eqn. 7.44):
5
2N
✏
5V F
P=
= 3.8 ⇥ 1010 N/m2 = 3.8 ⇥ 105 atm.
The contribution of the degeneracy pressure to the bulk modulus is (eqn. 7.45):
2N
✏
3V F
B=
= 6.33 ⇥ 1010 N/m2 = 6.33 ⇥ 105 atm.
SECTION 4 [10 MARKS]
Schroeder 7.28
a) [3 marks]
The allowed energy levels in 2D are:
✏=
h2
(nx 2
8mL2
+ ny 2 ). [1 mark]
At T=0, the fermions occupy the lowest unfilled levels, filling a quarter circle of radius
nmax . The Fermi energy is the highest filled level: ✏F = h2 nmax 2 /8mA, and the total
number of fermions is: N = 2 · ⇡nmax 2 /4, for fermions with spin 1/2 and two allowed states
for each spatial wave function. Solving for nmax 2 , will give the following Fermi energy:
✏F =
h2 2N
( )
8mA ⇡
=
h2 N
.
4⇡mA
[1 mark]
To compute the total energy, we add all the energies of the filled states and convert the
sum to an integral, integrated over a quarter-circle in polar coordinates:
Rn
R ⇡/2
R n ax 2 n2
2n
4
max
U = 2 0 max dn 0 d n✏(n) = ⇡ 0 m n h8mA
dn = ⇡h32mA
.
U=
⇡h2 2N 2
( )
32mA ⇡
=
h2 N 2
8⇡mA
= 12 ✏F . [1 mark]
b) [3 marks]
To find the density of states, we have to change variables to ✏ in the integral for the total
energy of the system (7.54). For the variable change from n to ✏, we use ✏ = h2 n2 /8mA.
This implies that d✏ = (h2 n/4mA)dn, or ndn = (4mA/h2 )d✏. [1 mark].
The integral for the internal energy at T = 0 will become:
R✏
U = 0 F ⇡( 4mA
)✏ d✏. [1 mark]
h2
6
This sum is the sum of all energies multiplied by the number of states: g(✏)d✏. Therefore,
the density of states is equal to:
g(✏) =
4⇡mA
.
h2
[1 mark]
d) [4 marks]
At nonzero temperature, the total number of particles is equal to:
R +1
R +1
1
N = 0 g(✏)n̄FD (✏) d✏ = g 0 e(✏ µ)/kT
d✏.
+1
By changing the variables to x = (✏
R +1
N = gkT µ/kT ex1+1 dx. [1 mark]
µ)/kT), the integral will become:
The integrand is a ex composite function. If it is multiplied by the derivative of ex (also
equal to ex ), we can integrate it by using another substitution. To get the integrand into
this form, we multiply the numerator and denominator by e x :
R +1
e x
N = gkT µ/kT 1+e
x dx. [1 mark]
Now we substitute for y = e x and dy = e x dx :
R +1 1
N = gkT eµ/kT 1+y
dy = gkT ln(1 + y)|0eµ/kT =
gkT ln( 1+e1µ/kT = gkT ln(1 + eµ/kT ).
[1 mark]
When kT ⌧ ✏F , the exponential eµ/kT is very large, and the 1 is negligible in comparison.
Therefore, the right side of the equation is approximately kT · epsilonF /kT = ✏F . When
kT
✏F , the exponential is only slightly larger than 1, and the argument of the logarithm
is less than 1. Therefore the chemical potential is negative, as expected. [1 mark]
SECTION 5
Schroeder 7.40
Notice that equation 7.83 can be written
R1
2
U = 0 ✏ 8⇡V✏
nPl (✏)d✏
(hc)3
7
This is in the form of an integral of ✏ times g(✏) times the appropriate distribution
function, provided that we identify:
g(✏) =
8⇡V✏2
(hc)3
as the density of state-the number of single-particle states per unit energy. This formula is
quadratic in ✏, so its graph ia a parabolaopening upwards.
Schroeder 6.20
• Taking the ground state energy to be zero, the partition function for a single
oscillator is:
Z = e0 + e
✏
2✏
+e
• Average energy is:
E=
1 @Z
Z@
!E=
✏e
1 e
=
(1
✏
✏
=
✏
+ ··· = 1 + e
e
✏
) @@ (1
e
✏
)
+ (e
1
=
✏ 2
✏ 3
) + ··· =
) + (e
(1
e
✏
)( 1)(1
e
1
1 e
✏
✏
) 2 (✏e
✏
)
✏
e
✏
1
• The total energy of N identical, independent oscillators is just N times the average
energy of one oscillator
U = NE =
N✏
e
✏
1
This is the same result as in problem 3.25(c).