Thermodynamics and Kinetics of Solids 33________________________________________________________________________________________________________________________

01.08.97

III. Statistical Thermodynamics

5. Statistical Treatment of Thermodynamics

5.1. Statistics and Phenomenological Thermodynamics.

Calculation of the energetic state of each atomic ormolecular consti tuent by making use ofmechanics/quantum mechanics.

Because of the large number of species: Considerationof probabilities, i.e. statistics. Determination of thepartition function (e.g. velocity).

In view of the very large number only one distributionamong all possible distributions is most probable (Fig.5.1: comparision of the relative probability for throwing acertain number of spots with 1, 2, 3 and many dices).

Number of microstates for the realization of amacrostate (total number of spots)

W = N! (5.1)

and under consideration of the exchange of dices withoutnew configurations

W =

N!N0 !N1!N2 !K

(5.2)

Determination of the distribution function for thefollowing 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,

..., er-1

(r-states).- The number of particles of the energy state ei is Ni.

The total number of particles is N:

Nii= 0

r-1Â = N (5.3)

The total energy of the system is E (fixed value)

Nii= 0

r-1

 e i = E (5.4)

Of interest is the most probable distribution function, i.e.the distribution for which

W =

N!N0 !N1!KNr-1!

(5.5)

has a maximum.

We consider lnW (which has a maximum at the samevalue as W):

lnW = ln N!- ln Nii =0

r-1

 ! (5.6)

Considering Stirling’s formula (ln n! = n ln n - n for largenumbers of n) we have

lnW = N lnN - N - Nii =0

r-1

 lnNi + Nii= 0

r-1

 (5.7)

Considering eq. (6.3) this results in

lnW = N lnN - Nii =0

r-1

 ln Ni (5.8)

A maximum with regard to Ni is obtained from thederivation with regard to Ni under consideration of eqs. (5.3) and (5.4):

d ln W = -d Ni lnNi = - dNi - lnNidNi = 0 (5.9)

Fig. 5.1. Comparison of the relative probabilities for throwing acertain number of spots when using 2, 3 and many dices.

34 Thermodynamics and Kinetics of Solids________________________________________________________________________________________________________________________

01.08.97

dNi = 0, sin ce dN = 0 (5.10)

eidNi = 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 + lnNidNi + l dNi + m e idNi = 0 (5.12)

or

dNi 1+ lnNi + l + me i( ) = 0 (5.13)

Since the dNi may be arbitrarily chosen, the quantities inparenthesis have to disappear:

1 + lnNi + l + me i = 0 (5.14)

or

Ni = e- 1+l( )e-mei (5.15)

According to eq. (5.3) we have

N = Ni = e- 1+l( ) e-mei (5.16)

and by making use of eq. (5.15) follows

Ni =Ne-me i

e-meiÂ(5.17)

m has to have the inverse dimension of energy. Underconsideration of the average oscillation energy of aparticle, e = kT , we obtain

Ni = Ne-e i / kT

e- ei / kTÂ(5.18)

(more strict derivation of this equation by consideringquantum mechanics and going to classical mechanics).

Equation (5.18) represents the Boltzmann’s distribution(Boltzmann’s e-relation).

Inner Energy from Boltzmann’s distribution:Because of E ≡ U in eq. (5.4) we have according eq.

(5.18)

U = N- ei e-ei / kT

e-e i / kTÂ(5.19)

The denominator is named partition function Z

Z = e- ei / kTÂ (5.20)

The counter of eq. (5.19) is kT2 dZdT

.

Accordingly we have

U = NkT2 1Z

dZdT

= NkT2 dln ZdT

(5.21)

By knowlegde of Z the inner energy may be determinedand from that all other thermodynamic functions.

EntropyThe entropy provides information about the direction ofirreversible processes and is a criterium for the presenceof equilibrium. From a statistic point of view there is atransition to the most probable macrostate.

The probability plays accordingly the same role as theentropy. Therefore, a functional relationship is assumed:

S = S W( ) (5.22)

Derivation of this relationship:2 independent systems of the same type of particles (1and 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 arebeing multiplied (W1,2 = W1 · W2):

S1,2 = S W1,2( ) = S W1 ⋅ W2( ) = S W1( ) + S W2( ) (5.23)

This equation may be only fulfilled if

S = k* lnW (5.24)

k* will be later identified as Boltzmann’s constant k.

5.2. The Various Statistics.

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.

Thermodynamics and Kinetics of Solids 35________________________________________________________________________________________________________________________

01.08.97

Quantum statistics: It is impossible to determinesimultaneously exactly location and momentum of aparticle. The possibility to distinguish of particles istherefore questionable.

Fig. 5.2.: Possibilities to throw dices with a total numberof 7 and 8 spots with 2 distinguishable dices (B), notdistinguishable dices (BE) and prohibition of the samenumber 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, NH4

+ , NO) havea half-numbered spin.A system which consists of many particles is described inthe case of an integer-numbered spin by a symmetric andin the case of a half-numbered spin by an antisymmetriceigenfunction.

In the first case (integer-numbered spin) any number ofparticles may be present in the same energy state, whilein the latter case (half-numbered spin) each energy statemay be only occupied by 1 particle (Pauli’s law).

The non-distinguishability results in the followingquantum statistics:- Bose-Einstein statistics in the case of an integer-

numbered spin.- Fermi-Dirac-Statistics in the case of a half-numbered

spin (Pauli’s law).

5.3. Momentum- and Phase space

Classical description of the state of a particle:

- 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 cubewith the length a of each edge. Possible energies of theparticle:

e =h2

8ma2 nx2 + ny

2 + nz2( ) (5.25)

By considering the relationship between energy andmomentum

e =1

2mpx

2 + py2 + pz

2( ) (5.26)

eq. (5.25) results in the following possible components ofthe momentum:

px =h

2anx

py =h

2any

pz =h2a

nz

(5.27)

nx, ny and nz are the integer quantum numbers.

By using a cartesian coordinate system with the unit h/2aof the axis px, py and pz, the states of the particle in thecube are represented by the lattice points with integernumbered coordinate values (Fig. 5.3.).

Fig. 5.3. States of a particle in a cube with the length of a ofeach edge in the momentum space.

Fig. 5.2. Possibilities of realizing the total number 7 and 8 spotswith 2 distinguishable dices (B) and undistinguishable dices(BE) as well as prohibition of the same number of spots.

36 Thermodynamics and Kinetics of Solids________________________________________________________________________________________________________________________

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Volume of each cell: h3/8a3. Since eq. (5.26) holds forpositive and negative momenta, all 8 octants of themomentum space have to be taken into consideration.Accordingly, a state of a species corresponds to a cell ofthe volume h3/a3 in the full 3-dimensional momentumspace.

By considering the phase space, i.e. adding the physicalspace to the momentum space, a state of the particlecorresponds in the 6-dimensional phase space toa cell ofthe volume h3, since the species will occupy the volumea3.

Eq. (5.25) may be rewritten in the following way:

nx2

ah

8meÊ Ë Á ˆ

¯ ˜

2 +ny

2

ah

8meÊ Ë Á ˆ

¯ ˜

2 +nz

2

ah

8meÊ Ë Á ˆ

¯ ˜

2 = 1 (5.28)

There exist as many different quantum states that belongto the energy e as integer solutions nx, ny, nz are possible.

All those integer numbers nx, ny, nz, that result in avalue < 1 for the left hand side of eq. (5.28 belong toquantum states with energy values < e.

Calculation of the number of quantum states withtranslational energies < e:Feeding eq. (5.27) into eq. (5.28) results in:

px2

12 8me( )2 +

py2

12 8me( )2 +

pz2

12 8me( )2 = 1 (5.29)

This equation is the surface of a bowl with a radius12 8me and the volume 4

3 p 18 8me( )

32 . Æ The numberof cells with energies < e is:

N e( ) =43

p18

8me( )32 /

h3

a3 =8 2

3p

Vh3 m

32e

32 (6.30)

where (V = a3). The number of states with energiesbetween e and e + de is

dN e( ) = D e( )de =dN e( )

dede (5.31)

D(e) is the density of states. Differentiation of eq. (5.30)results in

dN e( ) = D e( )de = 4 2pVh3 m

32e

12de (5.32)

The order of magnitude of the number of quantum states

of translation of a helium atom (m = 6.7 · 10-27 kg) in avolume of 1 l at 300 K is under considering thetranslation of energy e = 3

2 kT 6.2 ⋅10-21J( ):

N e( ) = 1.03 ⋅1028

which is a continuum in a first approach. If there is notonly 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

Even under such conditions, the number of quantumstates and accordingly the number of cells in the phasespace is still much larger than the number of species.

5.4. DistributionFfunctions

Since the discrete energy levels are very close to eachother, we do not consider the occupation of the individuallevels but the occupation of the total number of energyvalues between ei and ei + dei.The number of energy levels between ei and ei + dei: Ai.These are occupied by Ni species.

For the determination of the distribution function it has tobe calculated by how many microstates a macrostate maybe built up. That macrostate which may be generated bythe largest number of microstates is the most probableone and characteristic for the system.

Bose-Einstein-Statistics.The species may not bedistinguished and anynumber of species mayoccupy one state. Theenergy levels are:I, II, ..., Ai.Fig. 5.4. Shows thedistribution of 2 speciesover 3 cells. Dots areused in that figure toindicate that the speciesmay not be distinguished.In general: Distribution ofNi species over Ai cells:Number of the possi-bilities of distributions:

Fig. 5.4. Illustration or thederivation of Bose-Einsteins-S t a t i s t i c s . T w o n o tdistinguishable species (dots)are distributed over three cells(I, II, III) of a group of energylevels.

Thermodynamics and Kinetics of Solids 37________________________________________________________________________________________________________________________

01.08.97

Ai ⋅ Ai +1( )L Ai + Ni - 1( )1 ⋅ 2LNi

Expansion of this expression by (Ai - 1)! results in thefollowing number of microstates

Ni + Ai - 1( )!Ni ! Ai - 1( )!

The same holds for all energy intervals. Since eachdistribution within one group may be combined with anydistribution in another group, the number of differentmicrostates is

W = Pi

Ni + Ai -1( )!Ni! Ai - 1( )!

(5.33)

Conditions that have to be fulfilled:

i) The total number of species is constant

N = Nii

 (5.34)

ii) The total energy of the system is constant

E = Nii

 e i (5.35)

That macrostate is the most stable one for which W or lnW takes up a maximum under the boundary conditionseqs. (5.34) and (5.35).

Equation (5.33) results in

ln W = lni

 Ni + Ai( )!( ) - ln Ni!( )i

 - ln Ai!( )i

 (5.36)

Considering Stirling’s formula (ln n! = n ln n - n for largenumbers n):

ln W =i

 Ni + Ai( )ln Ni + Ai( ) - Ni + Ai( )i

Â

- Ni lnNi +i Ni - Ai lnAi

iÂ

i + Ai

i (5.37)

= Ni + Ai( ) ln Ni + Ai( )-i

 Ni lnNi - Ai lnAii

ÂiÂ

(5.37a)

Maximum:

d ln W =Ni + AiNi + Aii

 dNi + lni

 Ni + Ai( )dNi

-NiNii

 dNi - ln NidNi = 0 (5.38)

or

lnAiNi

+1Ê

Ë Á

ˆ

¯ ˜

i dNi = 0 (5.39)

with the boundary conditions

dN = dNii = 0 (5.40)

anddE = e idNi

i = 0 (5.41)

Application of Lagrange’s multiplier method:

dNii

 lnAiNi

+ 1Ê

Ë Á

ˆ

¯ ˜ + a + be i

È

Î Í

˘

˚ ˙ = 0 (5.42)

This results in

lnAiNi

+ 1Ê

Ë Á

ˆ

¯ ˜ + a + be i = 0 (5.43)

or

NiAi

=1

e-a -be i - 1 (5.44)

Determination of a and b:

Making use of eq. (5.37a), the expression S = k ln W maybe written as:

S = k Ni lnNi + Ai

Ni+ Ai ln

Ni + AiAi

È

Î Í ˘

˚ ˙ i (5.45)

and according to eq. (5.44)

S = k Ni lnB - be i( ) - Ai ln 1-1B

ebe iÊ Ë Á ˆ

¯ ˜

È Î Í

˘ ˚ ˙

i (5.46)

38 Thermodynamics and Kinetics of Solids________________________________________________________________________________________________________________________

01.08.97

with B = e-a. If 1B

ebe i <<1 (confirmation later!), the

following approximation holds under considering of ln(1-x) = -x:

S = k Ni lnB - bei( ) +Ai

Be-be iiÂ

iÂ

È

Î Í

˘

˚ ˙ (5.47)

and because of eq. (5.44) for Be-bei >> 1 :

S = k lnB NiÂN

1 2 3 - b Nie i

iÂ

E1 2 4 3 4

+ N

È

Î

Í Í Í Í

˘

˚

˙ ˙ ˙ ˙

(5.48)

With E = Ne (e : average energy of each particle) eq.(5.48) results in

S = kN lnB - be + 1[ ] (5.49)

e may be expressed by the distribution function (5.44):

E = Ne = Nieii

 =Aiei

Be-be i - 1Â (5.50)

Under considering of

N = Nii =

AiBe-be i -1i

 (5.51)

e becomes in the case Be-bei >> 1 :

e =

Aieiebe i

iÂ

Aiebe i

iÂ

=

∂∂b

Aiebe i

iÂ

Aiebe i

iÂ

=∂

∂bln Aiebei

iÂ

È

Î Í

˘

˚ ˙ (5.52)

According to eq. (5.51) results in the limiting caseBe-bei >> 1

ln N = ln Aii

 ebei - lnB (5.53)

and eq. (5.52) may be rewritten:

e =∂

∂blnN + ln B( ) =

∂∂b

lnB (5.54)

B is a function of b.

Accordingly, the following expression holds for theentropy (5.49)

S = k N lnB - b∂ lnB

∂b+ 1

È

Î Í

˘

˚ ˙ (5.55)

For the inner energie U of the system of N particles holds

U = Ne = N∂ lnB

∂b(5.56)

Because of ∂U∂S

Ê Ë Á

ˆ ¯ ˜

v= T , eqs. (5.55) and (5.56) allow to

determine b:

∂U∂b

Ê

Ë Á

ˆ

¯ ˜

v= N

∂2 ln B∂b2 (5.57)

∂S∂b

Ê

Ë Á

ˆ

¯ ˜

v= -kNb

∂2 lnB∂b2 (5.58)

Æ∂U∂S

Ê Ë Á

ˆ ¯ ˜

v= T = -

1kb

or b = -1

kT(5.59)

Accordingly, eq. (5.51) may be written in the followingway

N =Ai

Bee i / kT -1i (5.60)

Changing from summation to integration: The number ofstates Ai has to be expressed as a function of e: Ai, i.e. thenumber of energy levels between ei and ei + dei, isidentical with d N(e) in eq. (5.32):

N = 4 2pVh3 m

32

e12de

Bee

kT - 1o

•

Ú (5.61)

Bee

kT >>1 :

N = 4 2pVh3 m

32 1B

e12

0

•

Ú e-e kTde (5.62)

or

N = 4 2pVh3 m

32

1B

kT( )32 2 u2e-u2

0

•

Ú du (5.63)

Thermodynamics and Kinetics of Solids 39________________________________________________________________________________________________________________________

01.08.97

The integral has the value 14

p Æ .

This results in

B =2p mk T( )

32

h3VN

(5.64)

(B ≡ e-a)

Fermi-Dirac-StatisticsAgain the species may not be distinguished, but inaddition Pauli’s law holds, i.e. each quantum state mayonly be occupied by one species.Number of microstates:

Ai Ai - 1( ) Ai - 2( )L Ai - Ni +1( )1⋅2 ⋅L ⋅ Ni

Expansion by (Ai - Ni)! results in the number of microstates

Ai!Ni! Ai - Ni( )!

For all energy intervals holds

W =Ai!

Ni! Ai - Ni( )!i’ (5.65)

Analogously results as above under the same boundaryconditions

NiAi

=1

e-a -be i + 1 (5.66)

For

b = -1

kT(5.67)

and in the limiting caseB ⋅ e-be i >> 1 f o rB = e-a the same valueholds as in the case ofBose-Einstein’sstatistics.

Boltzmann-Statistics

Application of the same procedure as for the Bose-Einstein and Fermi-Dirac-Statistics. The cells may beoccupied without limitations; all species may bedistinguished from each other.Number of possibilities to distribute Ni species over Ai

states:

AiNi

Number of possibilities to distribute N species in groupsof N0, N1, N2, ... species each with the same propertiesonA0, A1, A2, …energy levels:

N!N0 !N1!LNi!L

Number of possibilities to realize the distribution

W =

N!N0 !N1!LNi!L

A0N 0A1

N1LAiNi L

= N!Ai

Ni

Ni !i’ (5.68)

With the same boundary conditions and the sameprocedure as before, this results in

NiAi

=1

e-a -be i (5.69)

Also for the Boltzmann-Statistics holds

b = -1

kT(5.70)

B =2p m kT( )

32

h3VN

(5.71)

Comparison of the Statistics

Table 5.1. B-values for the H2-molecule and conducting electrons in sodium at different temperaturesand pressures.

40 Thermodynamics and Kinetics of Solids________________________________________________________________________________________________________________________

01.08.97

Difference in the distribution functions: "1" in thedenominator.If e-ae

e ikT >> 1 , the quantum statistics result in the

Boltzmann statistics.

When is that the case?

For e-aee i

kT >>1 the right hand side of the distributionfunctions becomes very small, i.e. Ni / Ai (occupationprobability) becomes very small.The number of quantum states is very much larger thanthe number of species (holds, e.g., for a gas undernormal conditions).Since e i ≥ 0 , e

e ikT takes up values between 1 and • .

ForB eei

kT >>1 , B has to be sufficiently large: large mass,high temperature, high dissolution. B-values for H2 ande-: Table 1.1. row: The same density as at 273 K and p = 1.013 bar isassumed at all temperatures.2. + 3. row: p = constantFor H2, the condition B>>1 is fulfilled except forextremely low temperatures and high pressures.Electrons: small mass, large concentration (in the case ofmetals) Æ Fermi-Dirac-Statistics.

5.5. Partition Function and Thermodynamic Potential

Making use of the different statistics, thermodynamicquantities are derived.

According to Boltzmann’s statistics, the ratio of thenumber of species Ni with the energy ei relative to thetotal number of species N is

NiN

=gi e-e i kT

gi e-e ikT

iÂ

(5.72)

gi: degree of degeneration of the i-th state (statisticalweight).

This results in the average energy e of one species:

e =

Nie ii

ÂN

=

e i g i e-eikT

iÂ

gi e- eikT

iÂ

(5.73)

With the abbreviation

z:= gi e-e ikT

i (5.74)

("molecular partition function") results from eq. 5.73, asmay be easily shown by substitution,

e = kT2 ∂z / ∂Tz

= kT2 ∂ lnz∂T

(5.75)

The average energy may be determined from thedifferentiation of the partition function with regard to thetemperature.

When we do not consider the occupation probability ofan energy state ei by a single species and not the averageenergy of this single species, but a large number (nmoles) of species, then the energy (= inner energy) isanalogously for the entire system

U = kT2 ∂ lnZ∂T

Ê Ë Á

ˆ ¯ ˜

v(5.76)

Z: "system partition function"

Eq. 5.76 allows to relate the statistical treatment tophenomenological thermodynamics by knowledge of thepartition function:

i) Heat capacity

Cv =∂U∂T

Ê Ë Á

ˆ ¯ ˜

v=

∂∂T

kT2 ∂ lnZ∂T

Ê Ë Á

ˆ ¯ ˜

v=

∂∂T

-k∂ ln Z

∂ 1 / T( )Ê

Ë Á

ˆ

¯ ˜

v=

= - k∂ ∂ lnZ / ∂ 1/ T( )[ ]

∂TÊ

Ë Á

ˆ

¯ ˜

v= -

kT2 T2 ∂ ∂ lnZ / ∂ 1/ T( )[ ]

∂TÊ

Ë Á

ˆ

¯ ˜

v=

=k

T2∂ ∂ lnZ / ∂ 1/ T( )[ ]

∂ 1/ T( )Ê

Ë Á

ˆ

¯ ˜

v= -

kT2

∂2 ln Z∂ 1/ T( )2

Ê

Ë Á

ˆ

¯ ˜

v

(5.77)

ii) Entropy

dS =CvT

dT (5.78)

Integration:

S - S0 =CvT0

T

Ú dT (5.79)

Making use of eq. 5.77, this results in

Thermodynamics and Kinetics of Solids 41________________________________________________________________________________________________________________________

01.08.97

S - S0 =1T

0

T

Ú∂

∂TkT2 ∂ lnZ

∂TÊ Ë Á

ˆ ¯ ˜

vdT =

=1T

0

T

Ú kT2 ∂2 lnZ∂T2

Ê

Ë Á

ˆ

¯ ˜

v+ 2kT

∂ lnZ∂T

Ê Ë Á

ˆ ¯ ˜

v

È

Î Í Í

˘

˚ ˙ ˙

dT =

= kT∂2 ln Z

∂T2Ê

Ë Á

ˆ

¯ ˜

vdT + 2k

0

T

Ú∂ lnZ

∂TÊ Ë Á

ˆ ¯ ˜

v0

T

Ú dT (5.80)

Partial integration Æ

S - S0 = kT∂ ln Z

∂TÊ Ë Á

ˆ ¯ ˜

v- k

∂ ln Z∂T

Ê Ë Á

ˆ ¯ ˜

vdT + 2 k

0

T

Ú∂ ln Z∂T

Ê Ë Á

ˆ ¯ ˜

v0

T

Ú dT =

= kT∂ ln Z

∂TÊ Ë Á

ˆ ¯ ˜

v+ k lnZ

T0

(5.81)

Making use of eq. 5.76, this results in

S - S0 =UT

+ k lnZ - k ⋅ lnZ( )T=0 (5.82)

Æ S0 = kln Z( )T=0 (temperature independent) (5.83)

Accordingly, we have

S =UT

+ k lnZ (5.84)

oder

S = k∂ lnZ∂ lnT

Ê Ë Á

ˆ ¯ ˜

v+ lnZ

È

Î Í

˘

˚ ˙ (5.85)

iii) Free Energy

F = U - TS (5.86)

In view of eq. (5.84) this results in

F = U - TUT

+ kln ZÊ Ë Á ˆ

¯ ˜ = -kT lnZ (5.87)

iv) p·V

p = -∂F∂V

Ê Ë Á

ˆ ¯ ˜

T= kT

∂ lnZ∂V

Ê Ë Á

ˆ ¯ ˜

T(5.88)

p ⋅ V = kT∂ lnZ∂ lnV

Ê Ë Á

ˆ ¯ ˜

T(5.89)

v) Enthalpy

H = U + pV = kT2 ∂ lnZ∂T

Ê Ë Á

ˆ ¯ ˜

v+ kT

∂ lnZ∂ lnV

Ê Ë Á

ˆ ¯ ˜

T

H = kT∂ ln Z

∂TÊ Ë Á

ˆ ¯ ˜

v+

∂ ln Z∂ lnV

Ê Ë Á

ˆ ¯ ˜

T

È

Î Í

˘

˚ ˙

2

(5.90)

vi) Gibbs Energy

G = F + p ⋅ V = -kT lnZ + kT∂ lnZ∂ lnV

Ê Ë Á

ˆ ¯ ˜

T

G = -kT ln Z -∂ ln Z∂ lnV

Ê Ë Á

ˆ ¯ ˜

T

È

Î Í

˘

˚ ˙ (5.91)

Relationship between the entropy S and the statisticalweight W because the entropy adds up while theprobability is multiplied when several systems arecombined, the assumption is made

S ~ lnW or S = k* lnW (5.92)

It has been

lnW = N lnN - Ni ln Ni0

r-1

 (5.93)

Making use of the partition function (withoutdegeneration)

NiN

=e-e i kT

z(5.94)

this results in

S = k* N lnN - Ne- ei kT

zln N

e- ei kT

zÊ

Ë Á

ˆ

¯ ˜

iÂ

È

Î Í Í

˘

˚ ˙ ˙

(5.95)

= k* Nln N - Ne -ei kT

zln N + N

e-ei kT

zlnz + N

e-ei kT

zeikTi

ÂiÂ

iÂ

È

Î Í Í

˘

˚ ˙ ˙

= k* Nln N - N ln N + N lnz +NkT

eie-ei kT

ziÂ

È

Î Í Í

˘

˚ ˙ ˙

42 Thermodynamics and Kinetics of Solids________________________________________________________________________________________________________________________

01.08.97

Making use of eqs. (5.73) and (5.74) this results in

S = k* N lnz +Ne kT

È Î Í

˘ ˚ ˙ = k* N lnz +

UkT

È Î Í

˘ ˚ ˙ (5.96)

Comparison with eq. (5.84) results in

k* = k (5.97)

and

Z = zN (5.98)

Relationship between the molecular partition functionand the system partition function.

According to eq. 5.98, the system partition function(without degeneration) may be written as

Z = e- NeikT

i = e- Ei

kT

i (5.99)

Ei: energy eigen value of the i-th quantum state of themacro system of N species.Determination of Z from z:

i) Boltzmann: The system consists of N not interactingdistinguishable species (their exchange provides anew state).Example: Crystal of N species, which may bedistinguished from each other because of thelocalization at specific lattice sites. Exchange of 2such species provides a new state

Ei = N e i ; Z = zN (5.100)

All species of the system are equal to each other andhave the same energy eigen values.

ii) Bose-Eins te in : The system consists of N notinteracting and not distinguishable species (theirexchange provides no new state)Example: Ideal Gas of free molecules. The exchangeof 2 species provides not a new state.

Under the assumption that the number of states is muchlarger than the number of species, it may be assumed thateach quantum state is only occupied by 1 species.