Solutions
1. Van der Waals Fluid
(a) Show that the specific heat at constant volume (CV ) is a function of T only.
Remember the
CV = T ·
∂S
.
∂T V
To show that CV is a function of T only, we need to show that
∂CV
∂V
= 0.
T
So:
∂CV
∂V
T
=
∂
∂V
T·
∂S
∂T V
T
=T·
∂2S
∂V ∂T
=T·
∂
∂T
V
∂S
∂V
T
Using
F (T, V ) = −S dT − p dV =
⇔
∂F
∂T V
∂S
∂V T
dT +
=
∂p
∂T
∂F
∂V
V
Plugging this into the equation above, we can find:
∂CV
∂V
=T·
T
∂2p
(∂T )2 V
Using
p=
⇔
RT
V −b
∂CV
∂V
−
T
1
a
V2
=0
T
dV and
∂2F
∂V ∂T
=
∂2F
∂T ∂V
(b) Is the results still true if the fluid obeys the Dieterici equation of state?
No, for the Dieterici equation of state, we find that:
∂2p
(∂T )2 V
1
1
e− RT V
R2 V 2 T 3 (V −b)
=
6= 0.
(c) Find the entropy S(T, V ) for the van der Waals fluid.
Remember that we can write:
dS =
∂S
∂T V
dT +
∂S
∂V
dV =
T
1
C
T V
dT +
∂p
∂T
Using:
∂p
∂T
=
R
V −b
we can find for the entropy S:
⇔S=
R
1
C
T V
CV T1 dT
dS =
dT +
R
V −b
dV
+ R ln(V − b)
If CV is independent of T , this can be simplified to:
S = CV ln T + R ln(V − b) + constants
2
V
dV
2. Simple paramagnet
(a) Show that the thermodynamic potential
Using the partition function Z for the spin- 12 system:
Z = eβµHext + e−βµHext = 2 cosh (βµHext ), for one particle.
For N particles:
ZN = [2 cosh (βµHext )]N
Using G = −kB T ln ZN :
G = −KB T ln [2 cosh (βµHext )]N = −N kB T ln [2 cosh (βµHext )]
This means that a = N kB and b =
µ
.
kB
(b) Find the expression of the magnetization M and the magnetic enthalpy H = G + T S
in terms of Hext and T. Verify that H = −M Hext .
The magnetization M is given by
M = µ̄z · N .
To find µ̄z , we use:
µ̄z =
P
s
e
µz (s) · P (s) = µP↑ − µP↓ , where P↑ =
−
P↓ =
e
µHext
kB T
Z
−
=
µHext
kB T
Z
µHext
=
e kB T
µH
2 cosh k ext
T
B
µHext
e kB T
µH
2 cosh k ext
T
.
B
⇒ µ̄z =
µ
µH
2 cosh k ext
T
e
µHext
kB T
−e
−
µHext
kB T
ext
= µ tanh µH
.
kB T
B
So,
ext
M = N µ tanh tanh µH
.
kB T
3
and
The entropy S is given by:
S=−
∂G
∂T V
Using the results from part (a), we find that:
S=−
∂G
∂T V
ext
= N k ln 2 cosh muH
−
kB T
N µHext
T
ext
tanh muH
kB T
Using H = G + T S:
ext
ext
ext
ext
tanh muH
H = −N KB T ln 2 cosh muH
+ N KB T ln 2 cosh muH
− T N µH
kB T
kB T
T
kB T
⇔ H = −M · Hext
(c) Show that the expression of the entropy S obeys the Planck formulation of the third
principle.
We have shown in part (b) that the entropy S is given by:
S=−
∂G
∂T V
ext
= N k ln 2 cosh muH
−
kB T
N µHext
T
ext
tanh muH
kB T
The third principle states that the specific heat will go to 0 for T approaching 0K.
we can use:
∂S
CV = T ∂T
∂S
⇔ T ∂T
=
2
N µ2 Hext
kB T
cosh2
1
muHext
kB T
Since cosh x → ∞ for large x, and CV is proportional to
4
1
,
coshx
the third principle is fulfilled.
(d) Is the expression of the isotheral susceptibility χT in terms of Hext and T consistent
with the postulates of thermodynamics?
The isothermal susceptibility is give by:
χT =
For
µHext
kB T
1, cosh
µHext
kB T
∂M
∂Hext T
=
N µ2
kB T
cosh2
1
µHext
kB T
→ 1, which means that in the limit of high temperatures:
χT =
In the limit of low temperature,
µHext
kB T
N µ2
kB T
Curie’s Law.
→ ∞, cosh
µHext
kB T
lim χT = 0
T →0
5
→ ∞, so that:
3. Degenerate Fermi Gas
(a) For the low temperature limit (i.e. T = 0K), find the Fermi energy EF and the total
energy U0 of all electrons in the conduction band (per unit volume). Express your answer in
eV.
The number of electrons per unit volume N is given by:
N=
R∞
g(ε)f (ε) dε.
0
At T = 0, all the states are completely filled up to the Fermi-energy (εf ):
N=
εRf
g(ε)f (ε) dε =
0
εRf
g(ε) dε =
0
εRf
0
⇔ εf =
1
2π 2
h̄2
2m
2m
h̄2
3/2
ε1/2 dε =
1
2π 2
2m
h̄2
3/2
2 3/2
ε
3 f
3/2
(3π 2 N )
Assuming h̄ = 1.05 × 10−34 Js, me = 9.1 × 10−31 kg, and N = 8.46 × 1028 m−3 :
εf = 1.1 × 10−18 J ≈ 7 eV
The total energy U0 is given by:
U0 =
εRf
g(ε)ε dε =
0
1
2π 2
2m
h̄2
3/2 εRf
ε3/2 /, dε
0
2
N εf
5
U0 =
=
2 1
5 2π 2
2m
h̄2
3/2
5/2
εf
(b) Calculate the energy interval that is occupied by the electrons that participate in the
current flow at room temperature (i.e. T = 300K).The occupancy at non-zero temperature
is given by:
n̄(ε) =
1
ε−εf
+1
e kB T
So, the lower limit is given by:
0.9 =
1
ε1 −εf
+1
e kB T
ε1 = εf − kB T ln 9
and the upper limit:
6
0.1 =
1
ε2 −εf
+1
e kB T
ε2 = εf + kB T ln 9
This will result in a total energy interval at room temperature:
∆ε = 0.11 eV
(c) Using your answers to part (b), find the fraction of “current carrying” electrons N1 over
the total number of electrons in the conduction band Ntotal (i.e.
N1
).
Ntotal
First, find the number of “current carrying” electrons N1 in the energy window εf ±
using ∆ε = 0.11 eV and εf = 7 eV :
∆ε
,
2
εf + ∆ε
2
R
N1 =
n̄(ε)g(ε) dε
εf − ∆ε
2
For a small energy window, we can assume that g(ε) is independent of energy ε (i.e. g(εf ±
∆ε
) = 32 εNf , and n̄ varies linearly with energy, i.e. n̄(εf ± ∆ε
) = 12 − 4kεB T (using only the
2
2
linear term in the series expansion of the Fermi-Dirac distribution).
Using these approximations, we can find for N1 :
εf + ∆ε
2
N1 =
R
n̄(ε)g(ε) dε ≈
εf − ∆ε
2
3N
2εf
∆ε
2
R 1
2
− ∆ε
2
So, the fraction of charges contributing to the current
N1
N
=
3N
∆ε
3N εf
N1
N
=
ε
4kB T
−
N1
N
dε =
3N 1
∆ε
2εf 2
is now given by:
3∆ε
4εf
= 0.012
This means that only 1.2% of all the electrons in the conduction band contribute to the
current in Cu at 300K.
7
4. The Ideal Gas
(a) Determine the kind of compression that took place here. Explain!
This is an adiabatic compression with ∆Q = 0.
(b) Find the temperature T2 and the volume V2 in terms of V1 , p2 and T1 after the thermodynamic equilibrium has been reached.
For adiabatic processes
∆U = ∆Q − pdV
Combining these two equations, we will get:
CV (T2 − T1 ) = −p2 (V2 − V1 )
or ∆U = CV ∆T . The equation of state for an ideal gas is given
We also know that CV = ∂U
∂T
by pV = N kB T and CP − CV = N kB .
This will lead to:
(T2 − T1 ) = − CpV2 (V2 − V1 ) ⇔ T2 = − CpV2 (V2 − V1 ) + T1
Now use p2 V2 = N kB T2 ⇔ V2 =
N kB T2
:
p2
T2 = − CpV2
N kB T2
p2
This can now be rearranged to:
V T1
T2 = − p2 V1 +C
Cp
8
− V1 + T1
(c) Find the final values of the temperature Tf and volume Vf in terms of p1 , V2 , and T2
after the thermodynamic equilibrium is reached again.
Rearranging the solution to part (a) will lead to:
1 V2
Tf = − CV T2C+p
p
and Vf =
N kB Tf
p1
Using CV ∆T + p∆V = 0 for the two parts of the experiments, we will find:
CV (T2 − T1 ) + p2 (V2 − V1 ) = 0
CV (Tf − T2 ) + p1 (V1 − V2 ) = 0
⇔ CV (Tf − T1 ) + p2 (V2 − V1 ) + p1 (Vf − V2 ) = 0
using pV = N kB T
∆(pV ) = p∆V + V ∆p = N kB ∆T
Using CV ∆T + p∆V = 0:
p∆V + V ∆p = N kB
p∆V
CV
Using CV + N kB T = CP , we can find:
CP p∆V + CV V ∆p = 0
Using this expression for the two experiments:
CP p2 (V2 − V1 ) + CV V2 (p2 − p1 ) = 0
CP p1 (Vf − V2 ) + CV Vf (pf − p2 ) = 0
⇔ p2 (V2 − V1 ) + p1 (Vf − V2 ) = CCVP [V2 (p1 − p2 ) + Vf (p2 − pf )]
Using pf = p1 and Vf = V1 , we can plug this into the equation above and find:
CV (Tf − T1 ) = CCVP [V2 (p1 − p2 ) − V1 (p1 − p2 )]
Tf − T1 = C1P (V2 − V1 ) (p1 − p2 )
Eliminating (V2 − V1 ), we can finally find:
Tf − T1 =
CV
(CP )2
9
(p1 − p2 )2
V2
P2
This result shows that the temperature difference Tf − T1 will always be positive, as required
by the thermodynamic principles. Also, it appears that the temperature difference is only a
2nd -order correction term, which will become very small for ∆P → 0.
10
5. Einstein Solid
(a) For the most probable macrostate, determine the number of units of heat energy in each
of the two systems. Explain your approach.
The most probable state will have the highest multiplicity Ω:
q+N −1
q
Ω=
!
=
(q+N −1)!
q!(N −1)!
and
Ωtotal = ΩA · ΩB
For NA = 4 and NB = 6, we can assemble the following table:
qA
8
7
6
ΩA
165
120
84
qB
12
13
14
ΩB
Ωtotal
6188 1.021 × 106
8568 1.028 × 106
11628 0.98 × 106
So, qA = 7 and QB = 13 will be the most probable state.
(b) Determine the probability of finding the system in the most probable macrostate.
To determine the probability of a macrostate, we need to calculate:
P =
ΩA ΩB
Ωtotal
using
Ωtotal =
q + Ntotal − 1
q
!
⇔P =
20 + 10 − 1
20
=
ΩA ΩB
Ωtotal
11
= 0.1
!
= 10.02 × 107