Download Statistical Physics Exam - FU Berlin

Statistical Physics Exam
17th April 2015
Name
Problem 1
Problem 2
Useful constants
gas constant R
Boltzmann constant kB
Avogadro number NA
speed of light c
Student Number
Problem 3
Problem 4
8.31J/ (K · mol)
1.38·10−23 J/K
6.02·1023 mol−1
300·106 m/s
1
Total
Percentage
Mark
(25P) Problem 1
1. (1P) Write down the ideal gas equation of state. P V = N kB T
2. (2P) Find the entropy of one coin toss, and find the entropy of one dice rolling.The entropy S
reads
n
X
1
S = kB
pi ln ,
p
i
i=1
where pi is the probability that the system is in the i-th microstate, and n is the total number
of the microstates.
For the coin toss one finds pi = 1/2 and n = 2, yielding S = kB ln 2 . For the dice rolling one finds
pi = 1/6 and n = 6, yielding S = kB ln 6 .
1. (4P) Which quantities are intensive: volume, temperature, particle number, pressure, chemical
potential, entropy, density and mass? temperature, pressure, chemical potential, and density
2. (1P) Write down the value of the thermal energy at 300 K in units of pN·nm. kB × T = 0.0138 pN · nm/K × 300K = 4
3. (3P) Which quantities are constant in a
(a) canonical ensemble ? N, V, T
(b) grand-canonical ensemble ?µ, V, T
(c) isobaric-isothermal ensemble?N, P, T
4. (1P) How are Gibbs free energy and chemical potential related?µ =
energy per particle
∂G
∂N p,T ;
µ: Gibbs free
5. (2P) How does the temperature of a solvent in a mixture change compared to the pure solvent
(a) for boiling? higher in mixture
(b) for melting ? lower in mixture
6. (2P) In a phase diagram, what is meant by a
(a) triple point? Point (pressure and temperaure) where solid, liquid and gas phase coexist.
(b) critical point? Point (pressure and temperaure) where liquid and gas phase cannot be
distinduished (refelxion point in van der Waals isothermes).
∂S
= ∂V
7. (3P) Derive the Maxwell relation − ∂P
∂T P . You can assume the number of particles
T
N to be fixed. We use
∂G
∂G
dG =
dPT +
dT
∂P
∂T P
= V dP − SdT
(1)
The mixed derivatives, which are equal for state functions are
∂S
∂V
=
∂P T
∂T P
2
(2)
8. (1P) How does the heat capacity of a classical ideal monoatomic gas depend on temperature?
Cv = 32 N kB no temperature dependence.
9. (3P) What corrections, compared to the ideal gas, does the van der Waals gas introduce ? Sketch
the van der Waals Potential (Energy vs. distance). Gas particles have finite volume such that
the total volume is reduced. This is a repelling interaction. In addition, there is an attractive
interaction that can be understood as increased pressure
N2
(3)
p + a 2 (V − bN ) = N kB T
V
10. (2P) Name an impotant assumption in the Einstein Theory for crystal phonons. All normal
modes (global motions of all atoms together) have the same frequency. This Einstein frequency
is different for different materials.
3
(25P) Problem 2
Consider a system of N spins subject to a magnetic field B. The spins are non-interacting, distinguishable, and have the non-degenerated energy eigenvalues m = −µBm + 0 per one spin, where µ
is the magnetic moment per one spin, and m = −1 or 1.
1. (5P) Calculate the partition function ZN . The single-spin partition function Z1 :
X
Z1 =
eβµBm−β0
m=−1,−1




= e−β0 eβµB + e−βµB 
|
{z
}
2 cosh(βµB)
=
2e
−β0
cosh(βµB).
(4)
The N -spin partition function ZN simply leads to
ZN
=
=
Z1 N
−β0
N
2e
cosh(βµB) .
(5)
N
or = e−N β0 eβµB + e−βµB .
(6)
2. (5P) Calculate the average energy E (2 points). For this, you may calculate E using ZN , or
you may calculate E from the average of the total energy eigenvalues. The average energy is a
monotonically increasing function of kB T : Find two asymptotic values of E in the limit of T → 0
and T → ∞ (2 points), and sketch E(kB T ) (1 point). The average energy E:
∂ ln(ZN )
E = −
∂β
N,V
=
N [0 − Bµ tanh(βµB)] .
(7)
N (−Bµ − 0 ) e−βBµ−β0 + (Bµ − 0 ) eβBµ−β0
.
e−βBµ−β0 + eβBµ−β0
or = −
(8)
or, using the average of the total energy eigenvalues,
E
=
=
=
N h−µBm + 0 i
P
βµBm−β0
m=−1,1 (−µBm + 0 )e
P
N
βµBm−β0
m=−1,1 e
N (−Bµ − 0 ) e−βBµ−β0 + (Bµ − 0 ) eβBµ−β0
−
.
e−βBµ−β0 + eβBµ−β0
or = N [0 − Bµ tanh(βµB)] .
For T → 0:
E → −N µB + N 0 .
4
(9)
For T → ∞:
E → N 0 .
(10)
The average energy is plotted as below.
Figure 1: The average energy E as a function of kB T
1. (6P) Sketch the heat capacity CB (kB T ). For this, you may calculate CB using E, or you may
calculate CB using the energy fluctuations, or you may use the sketch of E(kB T ). The heat
capacity CB :
∂E
CB =
∂T B
Bµ 0
0
0
Bµ
Bµ
Bµ
0
Bµ
0
0
kT − kT
N (−Bµ − 0 ) e− kT − kT + (Bµ − 0 ) e kT − kT
e− kT − kT kT
+
+
e
2
kT 2
kT 2 −
Bµ =
0
Bµ
0
e− kT − kT + e kT − kT 2
0
0
Bµ
Bµ
Bµ
Bµ
0
0
N (−Bµ − 0 ) e− kT − kT kT
+ (Bµ − 0 ) e kT − kT kT
2 + kT 2
2 − kT 2
−
0
0
Bµ
Bµ
e− kT − kT + e kT − kT
Bµ
kT 2
2Bµ
=
=
4B 2 µ2 N e kT
2
2Bµ
kT 2 e kT + 1
B 2 µ2 N sech2 Bµ
kT
kT 2
.
(11)
or, using the energy fluctuations,
CB
=
2
2
h∆E 2 i
2 h(−µBm + 0 ) i − h−µBm + 0 i
=
N
kB T 2
kB T 2
2Bµ
=
=
4B 2 µ2 N e kT
2Bµ
2
kT 2 e kT + 1
B 2 µ2 N sech2 Bµ
kT
kT 2
5
.
(12)
0.5
CB • kB
0.4
0.3
0.2
0.1
0.0
0
1
2
3
kBT•Bm
4
5
Figure 2: The heat capacity per the Boltzmann constant CB /kB as a function of kB T /Bµ
1. (4P) Write down the Helmholtz free energy A in terms of ZN and kB T (1 point). Calculate A
(3 points). The Helmholtz free energy A:
A
= −
=
ln(ZN )
β
−N kB T ln 2e−β0 cosh(βµB) .
(13)
2. (5P) Sketch the magnetization M(kB T ). For this, you may calculate the average of the total
magnetic moment h−N (m − 0 )/Bi, or you may calculate M using the Helmholtz free energy
A. The magnetization M:
∂A
M = −
∂B N,T
=
N µ tanh(βBµ).
(14)
or, using the average energy E,
M =
=
−(E − N 0 )/B
N µ tanh(βBµ).
6
(15)
1.0
M •N m
0.8
0.6
0.4
0.2
0.0
0
2
4
6
kBT•Bm
8
10
Figure 3: The magnetization per the total magnetic moment M/N µ as a function of kB T /Bµ
(25P) Problem 3
Consider the grand-canonical partition function for non-relativistic, ideal Fermi gases and Bose gases,
respectively, in logarithmic form
ln Ξ = ± (2s + 1) Σp ln (1 ± exp [−β (εp − µ)])
(16)
2
p
where εp = 2m
is the energy corresponding to momentum state p, m is the mass of a particle, µ is the
chemical potential, and β = kB1T with T temperature and kB Boltzmann’s constant
1. (4P) In the classical limit, the logarithm of the partition function can be expanded.
(a) What does “classical limit” mean? Give a description in words and as a relation. The
classical limit is low densities (large volume and/or low number of particles) and high
temperatures, exp [−β (εp − µ)] 1
(b) Write down the first two terms of the expansion. Expand the logarithm as ln (x ± 1) ≈
2
3
±x − x2 ± x3 · · · for small x 1
ln Ξ = ± (2s + 1)
X
exp [−β (εp − µ)] ∓
p
1
exp [−2β (εp − µ)] + · · ·
2
(17)
2. (6P) Show that the zeroth order term, assuming continuous space, for particles with s = 0, is
ln Ξ =
V
exp (βµ)
λ3
(18)
12
h2
with λ = 2πmk
where h is Planck’s number and V is the volume. [You need the integral
BT
2 √π
´∞
2
dxx exp −x = 4 .]
−∞
In continuous space
ˆ
X
V
≈ 3
d3 p
(19)
h
p
Such that in zeroth order
V
ln Ξ ≈ 3
h
ˆ
d3 p exp [−β (εp − µ)]
7
(20)
Plugging in βεp =
p2
2mkB T
and integrating over the sphere
V
ln Ξ ≈ 3 4π
h
We now substitute x2 =
p2
2mkB T
ln Ξ =
ˆ
dp · p exp −
2
p2
exp [βµ]
2mkB T
(21)
and obtain a standard integral
3
V
4π (2mkB T ) 2
3
h
ˆ
∞
dx · x2 exp −x2 exp [βµ]
−∞
|
{z
}
(22)
1√
π
4
such that
ln Ξ
=
=
3
V
(2πmkB T ) 2 exp [βµ]
h3
V
exp [βµ]
λ3
(23)
(24)
3. (2P) Use eq. 18 to express the chemical potential as a function of number of particles. For the
grand-canonical partition function we have
P V = kB T ln Ξ
and for an ideal gas
PV
kB T
= N , hence ln Ξ = N in the classical limit. Therefore
V
exp [βµ]
λ3
V
exp [βµ] =
N λ3
V
βµ = ln
N λ3
V
µ = kB T ln
N λ3
N
=
4. Use the first-order corrected logarithmic partition function
V
1 V exp (2βµ)
ln Ξ = ± (2s + 1) 3 exp (βµ) ∓ 5
λ
λ3
22
(25)
(a) (4P) Calculate a first-order corrected number of particles . We need
N=
1 ∂ ln Ξ
β ∂µ
(26)
and get
N
1 ∂
V
1V 1
=
± (2s + 1) 3 exp [βµ] ∓
exp [2βµ]
β ∂µ
λ
2 λ3 2 32
V
1 V
= ± (2s + 1) 3 exp [βµ] ∓ 3 3 exp [2βµ]
λ
22 λ
8
(27)
V
(b) (4P) Obtain a first-order corrected expression for the pressure P in the form NPkB
T for
1
PV
Fermi particles with s = 2 and for Bose gas particles with s = 0. We again use kB T = ln Ξ
and the N obtained above. to obtain the correction we determine the difference in the
first-order corrections ln Ξ1 and N1 (in zeroth order ln Ξ = N )
1 V
1 V
ln Ξ1 − N1 = ± (2s + 1) ∓ 5 3 exp [2βµ] ± 3 3 exp [2βµ]
22 λ
22 λ
1
1
V
= ± (2s + 1) ∓ 5 ± 3
exp [2βµ]
(28)
22
2 2 λ3
Hence,
ln Ξ = N ±
and therefore
Fermi:
Bose:
2s + 1 V
exp [2βµ]
5
2 2 λ3
(29)
pV
2s + 1 V
exp [2βµ]
=1±
5
N kB T
2 2 N λ3
(30)
pV
1 V
=1+ 3
exp [2βµ]
N kB T
2 2 N λ3
(31)
pV
1 V
=1− 5
exp [2βµ]
N kB T
2 2 N λ3
(32)
(c) (3P) Write down a first-order
h energy for an ideal Bose gas.
i For ideal gases we have E =
3
3
V
1 V
N
k
T
.
Hence,
E
=
k
T
exp
[βµ]
+
exp
[2βµ]
. Also E = 32 P V such that
3 λ3
B
2
2 B
λ3
22
E
=
=
=
=
=
1 V
3
N kB T 1 − 5
exp
[2βµ]
2
2 2 N λ3
3
1 V
kB T N − 5 3 exp [2βµ]
2
22 λ
V
1 V
3
kB T
exp [βµ] + 3 3 exp [2βµ] −
2
λ3
22 λ
3
V
2 V
kB T
exp [βµ] + 5 3 exp [2βµ] −
2
λ3
22 λ
V
3
1 V
kB T
exp
[βµ]
+
exp
[2βµ]
5
2
λ3
2 2 λ3
1 V
exp
[2βµ]
5
2 2 λ3
1 V
exp
[2βµ]
5
2 2 λ3
(33)
5. (2P) How does the first-order corrected pressure of a Fermi gas/Bose gas compare to a classical
gas (lower, higher, equal)? The corrected pressure, compared to the classical ideal gas, is larger
for a Fermi gas (plus sign in partition function) and smaller for a Bose gas (minus sign in partition
function).
9
(25P) Problem 4
1. (4P) Consider a system, divided into two sub-systems 1 and 2, each characterised by their
entropy, volume, and number of particles (S1 , V1 , N1 ) and (S2 , V2 , N2 ), respectively. The total
system is isolated and is at equilibrium. the two subsystems are also in equilibrium with each
other and can exchange energy and particles, nut the total energy and total number of particles
remains constant. The subsystems can also change their volume prvided the total volume is
fixed. Show that at equilibrium the two subsystems have the same pressure P1 = P2 , the same
temperature T1 = T2 , and the same chemical potential µ1 = µ2 .
For the total system we always have
S
= S1 + S2
V
= V1 + V2
N
= N1 + N2
E
= E1 + E2
(34)
since the total system is isolated. This means that also
dS
= dS1 + dS2 = 0
dV
= dV1 + dV2 = 0
dN
= dN1 + dN2 = 0
dE
= dE1 + dE2 = 0
(35)
Starting from the first law of thermodynamics we write
dE1
= T1 dS1 − P1 dV1 + µ1 dN1
(36)
= −T2 dS2 + P2 dV2 − µ2 dN2
(37)
and
−dE2
such that with
dE1 = −dE2
T1 dS1 − P1 dV1 + µ1 dN1
= −T2 dS2 + P2 dV2 − µ2 dN2
(38)
= −T2 (−dS1 ) − P2 (−dV1 ) + µ2 (−dN1 )
= T2 dS1 + P2 dV1 − µ2 dN1
which holds only for
T1
= T2
P1
= P2
µ1
= µ2
(39)
2. (10P) A cylinder contains an ideal gas in thermodnamic equilibrium at pressure P , Volume V ,
temperature T , internal energy E, and entropy S. The cylinder is surrounded by a very large
heat reservoir at the same temperature T. By moving a piston a small change in volume ±∆V
can be made.
The cylinder walls and the piston can be switched to
10
be either perfect thermal conductors or perfect thermal insulators. For each of the five processes
below, state whether the changes in the quantities P, V, T, E, S (after re-establishment of the
equilibrium) have been positive, negative or zero. “Slow” (“fast”) means during the volume
change, the speed of the piston is much less (greater) than the speed of the gas particles at
temperature T .
(a) slow volume increase, thermal conductor isothermal expansion∆T = 0; ∆E = 0; ∆S =
∆V
V > 0;∆P < 0
(b) slow volume increase, thermal insulator reversible, adiabatic expansion∆Q = 0, hence ∆S =
0 ; ∆E < 0 because of external work; ∆P < 0; ∆T < 0
(c) fast volume increase, thermal insulator adiabatic free expansion, irreversible∆S > 0;∆P <
0;∆T = 0; ∆E = 0
(d) fast volume increase thermal conductor isothermal free expansion, irreversible∆S > 0;∆P <
0;∆T = 0; ∆E = 0;
(e) fast volume decrease, thermal conductor isothermal (free) compression, ∆P > 0;∆T = 0;
∆E = 0; ∆S < 0
3. (7P) Draw (and label) a thermodynamic cycle consisting of three processes involving an ideal
gas in a pressure-volume diagram.
(a) Leg 1-2: quasi-static adiabatic compression
Leg 2-3: isobaric cooling
Leg 3-1: quasi-static isothermal expansion
(b) Determine the sign of the heat and work transfers and change in internal energy for each
leg and for the cycle as a whole. Leg 1-2: quasi-static adiabatic compression
3
Q12 = 0, W12 = (P2 V2 − P1 V1 ), because P2 < P1 and V2 < V1 then W12 < 0.
2
On the other hand, we know: E12 = Q12 − W12 =⇒ E12 > 0
(c) Leg 2-3: isobaric cooling
3
W23 = P (V3 − V2 ) and E23 = (P3 V3 − P2 V2 ), because P3 = P2 and V3 < V2 then W32 < 0
2
and E32 < 0.
Then regarding to E32 = Q32 − W32 =⇒ Q32 < 0
(d) Leg 3-1: quasi-static isothermal expansion
E31 = 0,
Q31 = W31 .
For work done by system we have: W 31 =
W 31 =
V
´1
V3
V
´3
P dV and for ideal gas we could rewrite:
V1
N kB T
V1
dV = N kB T ln(
), because V3 < V1 then W31 > 0 and then Q31 > 0.
V
V3
11
4. (4P) Consider an ideal monoatomic gas with a mass of m=28g/mol and a heat capacity of
J
J
cV = 4 mol·K
. For simplicity, in this task, use a gas constant of R = 3 mol·K
.
(a) How much heat is required to heat 1kg of that gas from 210K to 420K at constant pressure?
For constant pressure we need cp = cV + R
Q = n (cv + R) ∆T
with n =
1000g
28g/mol ,
hence
1000g
J
(4 + 3)
· 210K
28g/mol
mol · K
1000
J
=
mol
· 210K
4
mol · K
= 52500J
Q =
(b) How much has the internal energy of that gas increased?
∆E = ncv · ∆T
∆E
4J
1000g
·
· 210K
28g/mol mol · K
1000
J
=
mol
· 210K
7
mol · K
= 30000J
=
(c) How much external work was done?
∆E
W
= Q−W
= Q − ∆E
=
52500J − 30000J
=
22500J
(d) The gas is heated by a heat pump, what is the efficiency of that heat pump? Effficiency of
a heat pump ηH
ηH
=
=
=
12
Thigh
Thigh − Tlow
420K
420K − 210K
2