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Thermodynamics and Kinetics of Solids
33
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III. Statistical Thermodynamics
5. Statistical Treatment of Thermodynamics
..., er-1
(r-states).
The number of particles of the energy state ei is Ni.
5.1. Statistics and Phenomenological Thermodynamics.
-
Calculation of the energetic state of each atomic or
molecular constituent by making use of
mechanics/quantum mechanics.
Because of the large number of species: Consideration
of probabilities, i.e. statistics. Determination of the
partition function (e.g. velocity).
In view of the very large number only one distribution
among all possible distributions is most probable (Fig.
5.1: comparision of the relative probability for throwing a
certain number of spots with 1, 2, 3 and many dices).
Number of microstates for the realization of a
macrostate (total number of spots)
The total number of particles is N:
W = N!
(5.1)
and under consideration of the exchange of dices without
new configurations
W=
N!
N 0 !N 1 !N2 !K
(5.2)
Determination of the distribution function for the
following model:
- N particles
- Particles may be distinguished.
- Particles are independent of each other (no mutual
influence).
- Each particle has one of the energetic states e0, e1, e2,
r- 1
 Ni = N
(5.3)
i= 0
The total energy of the system is E (fixed value)
r-1
 N ie i = E
(5.4)
i= 0
Of interest is the most probable distribution function, i.e.
the distribution for which
W=
N!
N 0 !N 1 !K N r-1 !
(5.5)
has a maximum.
We consider lnW (which has a maximum at the same
value as W):
r-1
lnW = ln N!-
 ln Ni !
(5.6)
i =0
Considering Stirling’s formula (ln n! = n ln n - n for large
numbers of n) we have
r-1
lnW = N lnN - N -
Â
r-1
N i lnN i +
i =0
 Ni
(5.7)
i= 0
Considering eq. (6.3) this results in
r-1
lnW = N lnN -
 Ni ln Ni
(5.8)
i =0
A maximum with regard to Ni is obtained from the
derivation with regard to
Ni under consideration of eqs. (5.3) and (5.4):
d ln W = -d
 N i lnN i = -  dN i -  lnN idN i = 0
(5.9)
Fig. 5.1. Comparison of the relative probabilities for throwing a
certain number of spots when using 2, 3 and many dices.
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 dN i = 0, sin ce dN = 0
(5.10)
 e i dNi = 0, sin ce dE = 0
(5.11)
Application of Lagrange’s multiplier method, i.e.
multiplication of eqs. (5.10) and (5.11) by l and m
(constant, but not fixed values) and adding them to eq.
(5.9):
 dNi +  lnN idN i + l  dN i + m e idNi = 0
(5.12)
(5.18)
 - ei e -e / kT
 e -e / kT
i
U=N
(5.19)
i
The denominator is named partition function Z
Z=
 e- e / kT
i
The counter of eq. (5.19) is kT2
(5.20)
dZ
.
dT
Accordingly we have
or
 dNi (1+ lnN i + l + me i ) = 0
(5.13)
Since the dN i may be arbitrarily chosen, the quantities in
parenthesis have to disappear:
1 + ln N i + l + me i = 0
-(1+l) -mei
e
(5.15)
According to eq. (5.3) we have
N=
1 dZ
dln Z
= NkT2
Z dT
dT
(5.21)
By knowlegde of Z the inner energy may be determined
and from that all other thermodynamic functions.
(5.14)
or
Ni = e
U = NkT 2
 Ni = e- (1+l )  e -me
i
(5.16)
Entropy
The entropy provides information about the direction of
irreversible processes and is a criterium for the presence
of equilibrium. From a statistic point of view there is a
transition to the most probable macrostate.
The probability plays accordingly the same role as the
entropy. Therefore, a functional relationship is assumed:
S = S (W )
(5.22)
and by making use of eq. (5.15) follows
Ne -me i
Ni =
e -mei
(5.17)
Â
m has to have the inverse dimension of energy. Under
consideration of the average oscillation energy of a
particle, e = kT , we obtain
Ni = N
e
Derivation of this relationship:
2 independent systems of the same type of particles (1
and 2) are combined isothermally to the total system (1,
2) In this case, the entropies of the individual systems
(S1,2 = S1 + S2) are added up, while statistic weights are
being multiplied (W1,2 = W1 · W2):
S1, 2 = S ( W1, 2 ) = S( W1 ⋅ W 2 ) = S( W1 ) + S ( W2 )
(5.23)
-e i / kT
 e- e / kT
i
(5.18)
(more strict derivation of this equation by considering
quantum mechanics and going to classical mechanics).
This equation may be only fulfilled if
S = k* ln W
(5.24)
k* will be later identified as Boltzmann’s constant k.
Equation (5.18) represents the Boltzmann’s distribution
(Boltzmann’s e-relation).
5.2. The Various Statistics.
Inner Energy from Boltzmann’s distribution:
Because of E ≡ U in eq. (5.4) we have according eq.
Boltzmann’s statistics:
- particles which build up the system are independent
of each other and distinguishable
- any number of particles may occupy the same state.
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- location (3-dim. space)
- velocity or momentum (3-dim. momentum space)
Both are combined to the 6-dimensional phase space.
Quantum mechanical treatment of a particle in a cube
with the length a of each edge. Possible energies of the
particle:
e=
Fig. 5.2. Possibilities of realizing the total number 7 and 8 spots
with 2 distinguishable dices (B) and undistinguishable dices
(BE) as well as prohibition of the same number of spots.
Quantum statistics: It is impossible to determine
simultaneously exactly location and momentum of a
particle. The possibility to distinguish of particles is
therefore questionable.
Fig. 5.2.: Possibilities to throw dices with a total number
of 7 and 8 spots with 2 distinguishable dices (B), not
distinguishable dices (BE) and prohibition of the same
number of spots (FD).
Particles for which the sum of numbers of electrons,
protons and neutrons is even (H2, D+, D2, N2, 4He,
photons) have an integer spin.
Particles for which the sum of the numbers of electrons,
protons and neutrons is odd (e-, H+, 3He, NH +4 , NO) have
a half-numbered spin.
A system which consists of many particles is described in
the case of an integer-numbered spin by a symmetric and
in the case of a half-numbered spin by an antisymmetric
eigenfunction.
h2
n 2x + n 2y + n2z
8m a 2
(
)
(5.25)
By considering the relationship between energy and
momentum
e=
1
p 2x + p 2y + p2z
2m
(
)
(5.26)
eq. (5.25) results in the following possible components of
the momentum:
h
n
2a x
h
py =
ny
2a
h
pz =
nz
2a
px =
(5.27)
nx, ny and nz are the integer quantum numbers.
By using a cartesian coordinate system with the unit h/2a
of the axis px, py and pz, the states of the particle in the
cube are represented by the lattice points with integer
numbered coordinate values (Fig. 5.3.).
In the first case (integer-numbered spin) any number of
particles may be present in the same energy state, while
in the latter case (half-numbered spin) each energy state
may be only occupied by 1 particle (Pauli’s law).
The non-distinguishability results in the following
quantum statistics:
- Bose-Einstein statistics in the case of an integernumbered spin.
- Fermi-Dirac-Statistics in the case of a half-numbered
spin (Pauli’s law).
5.3. Momentum- and Phase space
Fig. 5.3. States of a particle in a cube with the length of a of
each edge in the momentum space.
Classical description of the state of a particle:
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Volume of each cell: h3/8a3. Since eq. (5.26) holds for
positive and negative momenta, all 8 octants of the
momentum space have to be taken into consideration.
Accordingly, a state of a species corresponds to a cell of
the volume h3/a3 in the full 3-dimensional momentum
space.
By considering the phase space, i.e. adding the physical
space to the momentum space, a state of the particle
corresponds in the 6-dimensional phase space toa cell of
the volume h3, since the species will occupy the volume
a3.
2
2
ÊÁ a 8me ˆ˜
Ëh
¯
+
ny
ÊÁ a 8 me ˆ˜
Ëh
¯
2
+
n2z
ÊÁ a 8me ˆ˜
Ëh
¯
2
( 12
8me
2
) ( 12
p 2y
8me
2
+
) ( 12
p2z
8me
)
2
=1
(5.29)
This equation is the surface of a bowl with a radius
3
1
4 1
2
2 8me and the volume 3 p 8 ( 8me ) . Æ The number
of cells with energies < e is:
N(e) =
4 1
3
h 3 8 2 V 32 32
p (8m e ) 2 / 3 =
p 3m e
3 8
a
3
h
(6.30)
where (V = a3). The number of states with energies
between e and e + de is
dN ( e ) = D( e ) de =
dN ( e )
de
de
N (e ) = 1.03 ⋅10 28
which is a continuum in a first approach. If there is not
only 1 helium atom in the volume of 1 L, but 3 · 1022
atoms under atmospheric pressure, the number of states is
N (e ) = 3.3⋅10 5
5.4. DistributionFfunctions
Calculation of the number of quantum states with
translational energies < e:
Feeding eq. (5.27) into eq. (5.28) results in:
+
)
= 1 (5.28)
There exist as many different quantum states that belong
to the energy e as integer solutions nx, ny, nz are possible.
All those integer numbers nx, ny, nz, that result in a
value < 1 for the left hand side of eq. (5.28 belong to
quantum states with energy values < e.
p 2x
(
Even under such conditions, the number of quantum
states and accordingly the number of cells in the phase
space is still much larger than the number of species.
Eq. (5.25) may be rewritten in the following way:
n2x
of translation of a helium atom (m = 6.7 · 10-27 kg) in a
volume of 1 l at 300 K is under considering the
-21
translation of energy e = 32 kT 6.2 ⋅10 J :
(5.31)
D(e) is the density of states. Differentiation of eq. (5.30)
results in
V 3 1
(5.32)
dN ( e ) = D( e ) de = 4 2 p 3 m 2 e 2 de
h
The order of magnitude of the number of quantum states
Since the discrete energy levels are very close to each
other, we do not consider the occupation of the individual
levels but the occupation of the total number of energy
values between ei and ei + dei.
The number of energy levels between ei and e i + de i: Ai.
These are occupied by Ni species.
For the determination of the distribution function it has to
be calculated by how many microstates a macrostate may
be built up. That macrostate which may be generated by
the largest number of microstates is the most probable
one and characteristic for the system.
Bose-Einstein-Statistics.
The species may not be
distinguished and any
number of species may
occupy one state. The
energy levels are:
I, II, ..., Ai.
Fig. 5.4. Shows the
distribution of 2 species
over 3 cells. Dots are
used in that figure to
indicate that the species
may not be distinguished.
In general: Distribution of
Ni species over Ai cells:
Number of the possibilities of distributions:
Fig. 5.4. Illustration or the
derivation of Bose-EinsteinsStatistics.
Two
not
distinguishable species (dots)
are distributed over three cells
(I, II, III) of a group of energy
levels.
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A i ⋅ (A i +1)L( Ai + Ni - 1)
1⋅ 2LNi
d ln W = Â
i
Expansion of this expression by (Ai - 1)! results in the
following number of microstates
-Â
i
( N i + Ai - 1) !
N i !( A i - 1) !
i
Ni
dN i - Â ln NidN i = 0
Ni
(5.38)
or
The same holds for all energy intervals. Since each
distribution within one group may be combined with any
distribution in another group, the number of different
microstates is
W= P
Ni + Ai
dNi + Â ln( N i + A i )dN i
Ni + Ai
i
( N i + A i -1 )!
N i !( A i - 1)!
ÊA
ˆ
 lnÁË N ii +1˜¯ dNi = 0
(5.39)
i
with the boundary conditions
dN =
(5.33)
 dN i = 0
(5.40)
 e idN i = 0
(5.41)
i
and
dE =
Conditions that have to be fulfilled:
i
i) The total number of species is constant
N=
Application of Lagrange’s multiplier method:
 Ni
(5.34)
È ÊA
 dN i ÍlnÁ N i
i
Î Ë
i
i
˘
ˆ
+ 1˜ + a + be i ˙ = 0
¯
˚
(5.42)
ii) The total energy of the system is constant
E=
Â
This results in
Ni e i
(5.35)
i
That macrostate is the most stable one for which W or ln
W takes up a maximum under the boundary conditions
eqs. (5.34) and (5.35).
Equation (5.33) results in
ln W = Â ln( (N i + Ai )!) - Â ln( N i !) - Â ln(A i !)
i
i
(5.36)
i
Considering Stirling’s formula (ln n! = n ln n - n for large
numbers n):
ln W = Â ( N i + A i ) ln( N i + A i ) - Â ( N i + A i )
i
-
=
i
i
(5.37)
i
i
Ni
1
= -a -be i
Ai e
-1
(5.44)
or
Determination of a and b:
Making use of eq. (5.37a), the expression S = k ln W may
be written as:
È
 ÍÎ Ni ln
i
 ( Ni + A i ) ln( Ni + Ai ) -  N i lnN i -  A i lnA i
i
(5.43)
S=k
i
 N i lnN i +  Ni -  A i lnA i +  Ai
i
Ê Ai
ˆ
ln Á
+ 1˜ + a + be i = 0
N
Ë i ¯
i
Ni + Ai
N + Ai ˘
+ Ai ln i
Ni
Ai ˙˚
(5.45)
and according to eq. (5.44)
S=k
È
1
˘
 ÍÎ Ni ( lnB - be i ) - Ai ln ÊÁË1 - B ebe ˆ˜¯ ˙˚
i
(5.46)
i
(5.37a)
Maximum:
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1 be i
e <<1 (confirmation later!), the
B
following approximation holds under considering of ln
(1-x) = -x:
with B = e-a. If
È
Ai ˘
S = k ÍÂ N i ( ln B - bei ) + Â -be
˙
i
Î i
˚
i Be
and because of eq. (5.44) for Be
-bei
È
∂ lnB ˘
S = k N Í lnB - b
+ 1˙
∂b
Î
˚
Â
For the inner energie U of the system of N particles holds
U = Ne = N
(5.48)
e may be expressed by the distribution function (5.44):
E = Ne = Â N i e i = Â
i
-be i
Be
-1
(5.50)
 Ni = Â
i
i
Ê ∂U ˆ
∂ 2 ln B
Á
˜ =N
Ë ∂b ¯ v
∂b 2
(5.57)
Ê ∂S ˆ
∂2 ln B
Á ˜ = -kNb
Ë ∂b ¯ v
∂b 2
(5.58)
1
Ê ∂U ˆ
˜ =T= Æ Á
Ë ∂S ¯ v
kb
or b = -
Under considering of
N=
Be
Ai
-be i
-1
(5.51)
1
kT
N=Â
i
e=
i
 Ai e be i
=
∂
 A e be i
∂b i i
=
∂ È
Íln
∂b Î
i
˘
 A ie be ˙˚
i
(5.52)
 A ie be
i
- lnB
(5.60)
•
1
V 3
e 2 de
N = 4 2p 3 m 2
e
h
Be kT - 1
o
i
Ú
According to eq. (5.51) results in the limiting case
Be -bei >> 1
ln N = ln
Ai
Bee i / kT -1
Changing from summation to integration: The number of
states Ai has to be expressed as a function of e: Ai, i.e. the
number of energy levels between ei and ei + dei, is
identical with d N(e) in eq. (5.32):
 A ie be i
i
(5.59)
Accordingly, eq. (5.51) may be written in the following
way
e becomes in the case Be -bei >> 1 :
 Ai e i e be i
(5.56)
determine b:
(5.49)
Aiei
∂ ln B
∂b
Ê ∂U ˆ
˜ = T , eqs. (5.55) and (5.56) allow to
Because of Á
Ë ∂S ¯ v
With E = Ne ( e : average energy of each particle) eq.
(5.48) results in
S = kN[ lnB - be + 1]
(5.55)
(5.47)
>> 1 :
È
˘
Í
˙
S = k ÍlnB Ni - b Ni e i + N ˙
123
Í
˙
i 24
1
4
3
N
ÍÎ
˙˚
E
Â
Accordingly, the following expression holds for the
entropy (5.49)
Be
e
kT
>>1 :
(5.53)
i
(5.61)
N = 4 2p
V 32 1 • 12 - e kT
m
e e
de
B Ú0
h3
N = 4 2p
2
V 32 1
3
( kT ) 2 2 u2 e -u du
3m
h
B
0
(5.62)
and eq. (5.52) may be rewritten:
or
∂
∂
( ln N + ln B) = ln B
e=
∂b
∂b
B is a function of b.
(5.54)
•
Ú
(5.63)
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1
The integral has the value
pÆ.
4
This results in
3
(2p mk T ) 2 V
B=
h3
N
(5.64)
(B ≡ e-a)
Application of the same procedure as for the BoseEinstein and Fermi-Dirac-Statistics. The cells may be
occupied without limitations; all species may be
distinguished from each other.
Number of possibilities to distribute Ni species over Ai
states:
A i Ni
Fermi-Dirac-Statistics
Again the species may not be distinguished, but in
addition Pauli’s law holds, i.e. each quantum state may
only be occupied by one species.
Number of microstates:
A i ( A i - 1)( Ai - 2 )L (A i - N i +1)
1⋅ 2⋅L ⋅ N i
W=
N!
i
A N 0 A N1 LA N
i L
N 0 !N 1 !L N i !L 0 1
= N!
Ai !
N i !( A i - N i )!
A Ni
’ N ii !
(5.68)
i
With the same boundary conditions and the same
procedure as before, this results in
For all energy intervals holds
i
N!
N 0 !N1 !L N i !L
Number of possibilities to realize the distribution
Expansion by (Ai - Ni)! results in the number of micro
states
W=’
Number of possibilities to distribute N species in groups
of N0, N1, N2, ... species each with the same propertieson
A0, A1, A2, …energy levels:
Ai !
N i !( A i - N i )!
(5.65)
Analogously results as above under the same boundary
conditions
Ni
1
= -a -be i
Ai e
Also for the Boltzmann-Statistics holds
b=-
Ni
1
= -a -be i
Ai e
+1
(5.69)
(5.66)
1
kT
(5.70)
3
B=
For
b=-
1
kT
and in the limiting case
for
B ⋅ e -be i >> 1
B = e -a the same value
holds as in the case of
Bose-Einstein’s
statistics.
(2p m kT )
3
h
2
V
N
(5.71)
(5.67)
Comparison of the Statistics
Table 5.1. B-values for the H2-molecule and conducting electrons in sodium at different temperatures
and pressures.
Boltzmann-Statistics
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Difference in the distribution functions: "1" in the
denominator.
-a e i
If e e kT >> 1 , the quantum statistics result in the
Boltzmann statistics.
z:= Â gi e -
For e e
>>1 the right hand side of the distribution
functions becomes very small, i.e. Ni / Ai (occupation
probability) becomes very small.
The number of quantum states is very much larger than
the number of species (holds, e.g., for a gas under
normal conditions).
ei
Since e i ≥ 0 , e kT takes up values between 1 and • .
For B e
ei
kT
kT
>>1 , B has to be sufficiently large: large mass,
high temperature, high dissolution. B-values for H2 and
e-: Table 1.
1. row: The same density as at 273 K and p = 1.013 bar is
assumed at all temperatures.
2. + 3. row: p = constant
For H2, the condition B>>1 is fulfilled except for
extremely low temperatures and high pressures.
Electrons: small mass, large concentration (in the case of
metals) Æ Fermi-Dirac-Statistics.
5.5. Partition Function and Thermodynamic Potential
Making use of the different statistics, thermodynamic
quantities are derived.
According to Boltzmann’s statistics, the ratio of the
number of species Ni with the energy ei relative to the
total number of species N is
ei
Ni
g i e - kT
=
ei
N
gi e - kT
(5.74)
("molecular partition function") results from eq. 5.73, as
may be easily shown by substitution,
e = kT2
ei
kT
i
When is that the case?
-a
ei
∂z / ∂T
∂ lnz
= kT2
z
∂T
(5.75)
The average energy may be determined from the
differentiation of the partition function with regard to the
temperature.
When we do not consider the occupation probability of
an energy state ei by a single species and not the average
energy of this single species, but a large number (n
moles) of species, then the energy (= inner energy) is
analogously for the entire system
Ê ∂ ln Z ˆ
U = kT2 Á
˜
Ë ∂T ¯ v
(5.76)
Z: "system partition function"
Eq. 5.76 allows to relate the statistical treatment to
phenomenological thermodynamics by knowledge of the
partition function:
i) Heat capacity
∂ Ê 2 ∂ ln Z ˆ
∂ Ê
∂ ln Z ˆ
Ê ∂U ˆ
Cv = Á
˜ =
Á kT
˜ =
-k
=
Á
Ë ∂T ¯ v ∂T Ë
∂T ¯ v ∂T Ë
∂ (1 / T) ˜¯ v
Ê ∂[ ∂ lnZ / ∂(1/ T )] ˆ
Ê ∂[ ∂ lnZ / ∂(1/ T )] ˆ
k
= - kÁ
= - 2 T 2Á
˜
˜ =
∂T
T
∂T
Ë
¯v
Ë
¯v
(5.72)
Â
i
=
gi: degree of degeneration of the i-th state (statistical
weight).
This results in the average energy e of one species:
 N ie i  e i g i e e= i
= i
N
 gi e ei
ei
k Ê ∂[ ∂ lnZ / ∂(1/ T )] ˆ
k Ê ∂ 2 ln Z ˆ
=- 2 Á
˜
2Á
T Ë
∂(1/ T )
T Ë ∂(1/ T )2 ˜¯ v
¯v
ii) Entropy
dS =
Cv
dT
T
(5.78)
kT
kT
(5.73)
Integration:
T
i
S - S0 =
With the abbreviation
(5.77)
Cv
dT
T
0
Ú
(5.79)
Making use of eq. 5.77, this results in
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Ê ∂ lnZ ˆ
p⋅ V = kT Á
˜
Ë ∂ lnV ¯ T
T
1 ∂ Ê 2 ∂ lnZ ˆ
S - S0 =
Á kT
˜ dT =
T ∂T Ë
∂T ¯ v
0
Ú
T
=
1È
v) Enthalpy
Ê ∂2 lnZ ˆ
Ê ∂ lnZ ˆ ˘
˜ ˙ dT =
2 ˜ + 2kT Á
∂T ¯ v
Ë ∂T ¯ v ˙˚
Ú T ÍÍÎkT 2 ÁË
0
T
T
Ê ∂2 ln Z ˆ
Ê ∂ lnZ ˆ
= kTÁ
dT
+
2k Á
˜ dT
2 ˜
∂T
Ë ∂T ¯ v
Ë
¯v
0
0
Ú
(5.89)
Ú
Ê ∂ lnZ ˆ
Ê ∂ lnZ ˆ
H = U + pV = kT 2 Á
˜ + kT Á
˜
Ë ∂T ¯ v
Ë ∂ lnV ¯ T
(5.80)
Partial integration Æ
ÈÊ ∂ ln Z ˆ
Ê ∂ ln Z ˆ ˘
H = kT ÍÁ
˜ +Á
˜ ˙
ÎË ∂T ¯ v Ë ∂ lnV ¯ T ˚
2
(5.90)
vi) Gibbs Energy
T
T
Ú
Ú
Ê ∂ ln Z ˆ
Ê ∂ ln Z ˆ
Ê ∂ ln Z ˆ
S - S 0 = kTÁ
˜ - kÁ
˜ dT + 2 k Á
˜ dT =
Ë ∂T ¯ v 0 Ë ∂T ¯ v
Ë ∂T ¯ v
0
T
Ê ∂ ln Z ˆ
= kTÁ
˜ + k lnZ
Ë ∂T ¯ v
0
(5.81)
Ê ∂ lnZ ˆ
G = F + p⋅ V = -kT lnZ + kT Á
˜
Ë ∂ lnV ¯ T
È
Ê ∂ ln Z ˆ ˘
G = -kT Íln Z - Á
˜
Ë ∂ ln V ¯ T ˙˚
Î
(5.91)
Making use of eq. 5.76, this results in
U
S - S 0 = + k lnZ - ( k ⋅ lnZ )T =0
T
Æ S 0 = ( k ln Z) T =0
(temperature independent)
(5.82)
(5.83)
Relationship between the entropy S and the statistical
weight W because the entropy adds up while the
probability is multiplied when several systems are
combined, the assumption is made
S ~ lnW or S = k* ln W
(5.92)
Accordingly, we have
U
S = + k lnZ
T
It has been
(5.84)
r-1
lnW = N lnN -
 Ni ln Ni
oder
ÈÊ ∂ lnZ ˆ
˘
S = k ÍÁ
˜ + lnZ ˙
ÎË ∂ lnT ¯ v
˚
(5.93)
0
(5.85)
Making use of the partition function (without
degeneration)
ei
N i e - kT
=
N
z
iii) Free Energy
F = U - TS
(5.86)
In view of eq. (5.84) this results in
ÊU
ˆ
F = U - T Á + k ln Z˜ = -kT ln Z
ËT
¯
(5.87)
(5.94)
this results in
È
S = k* ÍN lnN - N
ÍÎ
Â
i
e-
ei
Ê e- ei kT ˆ ˘
ln Á N
˜˙
z
z ¯ ˙˚
Ë
kT
(5.95)
ei
ei
ei
e - kT
e - kT
e - kT e i ˘
˙
= k Nln N - N Â
ln N + N Â
ln z + NÂ
z
z
z kT ˙˚
ÍÎ
i
i
i
È
*Í
iv) p·V
Ê ∂F ˆ
Ê ∂ lnZ ˆ
p = -Á
˜ = kT Á
˜
Ë ∂V ¯ T
Ë ∂V ¯ T
(5.88)
È
ei
N
ei e= k Nln N - N ln N + N ln z +
Â
kT i
z
ÍÎ
*Í
kT
˘
˙
˙˚
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42
Thermodynamics and Kinetics of Solids
________________________________________________________________________________________________________________________
Making use of eqs. (5.73) and (5.74) this results in
Ne ˘
U˘
È
È
S = k* ÍN lnz +
= k* ÍN lnz +
˙
Î
Î
kT ˚
kT ˙˚
(5.96)
Comparison with eq. (5.84) results in
k* = k
(5.97)
and
Z = zN
(5.98)
Relationship between the molecular partition function
and the system partition function.
According to eq. 5.98, the system partition function
(without degeneration) may be written as
Z =Âe
i
i
- Ne
kT
=Âe
- EkTi
(5.99)
i
Ei: energy eigen value of the i-th quantum state of the
macro system of N species.
Determination of Z from z:
i) Boltzmann: The system consists of N not interacting
distinguishable species (their exchange provides a
new state).
Example: Crystal of N species, which may be
distinguished from each other because of the
localization at specific lattice sites. Exchange of 2
such species provides a new state
E i = N ei ; Z = zN
(5.100)
All species of the system are equal to each other and
have the same energy eigen values.
ii) B o s e - E i n s t e i n : The system consists of N not
interacting and not distinguishable species (their
exchange provides no new state)
Example: Ideal Gas of free molecules. The exchange
of 2 species provides not a new state.
Under the assumption that the number of states is much
larger than the number of species, it may be assumed that
each quantum state is only occupied by 1 species.
01.08.97