Thermodynamics and Kinetics of Solids 45 · Thermodynamics and Kinetics of Solids 45 15.10.01 uvib= 3 2 NA hn + 3NA hn e hn kT-1 Ê Ë Á Á ˆ ¯ ˜ ˜ 3N (5.128) Using the characteristic

Thermodynamics and Kinetics of Solids 45

15.10.01

uvib =32

NA hn +3NA hn

ehnkT -1

Ê

Ë

Á Á

ˆ

¯

˜ ˜

3N

(5.128)

Using the characteristic temperatureQ:=hnk

, the result is

uvib =32

Rq +3Rq

eq

T - 1(5.129)

The molar heat capacity (for solid state materials) isobtained by differentiation with regard to thetemperature:

cvib = 3R

qT

Ê Ë Á ˆ

¯ ˜

2eqT

eq

T - 1( )2 (5.130)

Checking the experimental finding that cv Æ 0 for T Æ 0K and Dulong-Petit’s law cv ª 3R for sufficiently hightemperatures (Fig. 5.5.):

limTÆ0

cvib = limTÆ0

3R qT

Ê Ë Á ˆ

¯ ˜

2

eqT

= limTÆ 0

3R ⋅2 qT

-q

T 2Ê

Ë Á Á

ˆ

¯ ˜ ˜

eqT -

q

T 2Ê

Ë Á Á

ˆ

¯ ˜ ˜

=1

eqT

= 0 (5.131)

Which is the functional relationship when we approach 0K?

Experimentally, a parabolic relationship is found (Fig.5.5). In a double logarithmic plot, a straight line with a

slope 3 is observed:

cp ~ T3 (5.132)

(as shown in Fig. 5.6 for diamond at low temperatures)At very low temperatures, 1 may be neglected compared

to eq

T in eqn. (5.130). Accordingly, the specific heat isgiven by

cvib = 3RqT

Ê Ë Á ˆ

¯ ˜

2e

q

T (5.133)

Einstein’s assumptions are therefore not correct andprovide the wrong temperature dependence.

Limiting Value at Very High Temperatures

limT Æ•

cvib = limT Æ•

3R qT

Ê

Ë Á

ˆ

¯ ˜

2e

qT

eqT -1

Ê

Ë

Á Á Á Á

ˆ

¯

˜ ˜ ˜ ˜

2 = R (5.134)

in agreement with the experiment.The wrong temperature dependence close to 0 Kelvin isdue to the assumption that neighbouring species of theconsidered particle are frozen-in at their lattice sites anddo not vibrate.

ii) Debey’s TheoryThe species are no longer (as in the case of Einstein)considered to be vibrating with a fixed frequency n. Afrequency spectrum as assumed. Analogously to thephotonic gas (black body radiation, see later), the number

Fig. 5.5. Temperature dependence of the molar heat capacity cv

of elemental solid bodies.

Fig. 5.6. Molar Heat Capacity cp of diamond at lowtemperatures cp vs T3.

46 Thermodynamics and Kinetics of Solids

15.10.01

of species N with vibrations in the frequency rangebetween n and n+ dn is

dN n( ) =4pVn2

c3 dn (5.135)

V: volume; c: wave velocity.Two kinds of waves are spreading out in a solid body,one longitudinal and two transversal ones with thevelocity wl and wt. Accordingly, it is assumed

dN n( ) =1

w l3 +

2

w t3

Ê

Ë

Á Á Á

ˆ

¯

˜ ˜ ˜ 4pVn2dn =

3

w 34pVn2dn (5.136)

w : average velocity.

dN(n)/dn increases stronger and stronger with increasingn ; for n Æ • a continuous spectrum occurs. An elasticcontinuum is only justified, however, as long as the wavelength of the elastic wave is much larger than the distanceof the lattice particles.Accordingly, a minimum wave length, i.e. a maximumfrequency, exists.Determination of nmax by integration of dN(n) byconsidering only the lowest 3N frequencies of the systemof N species with 3N degrees of freedom:

dN(n) = 3N0

nmax

Ú (5.137)

Feeding into this expression eqn. (5.136), thisresults in

12pV

w 3n2dn

0

nmax

Ú = 4pVnmax3

w 3= 3N (5.138)

Solving this equation for w 3 and feeding this expressioninto eqn. (5.136), the result is

dN n( ) =9Nn2

nmax3 dn (5.139)

or

D n( ):=dN(n )

dn=

9Nn2

vmax3 (5.140)

D: "spectral distribution function" (or partition function)(Fig. 5.7. in the case of a 3-dimensional isotropic body).

The molar vibrational energy of the solid body isobtained by multiplying D(n) with the vibrational energy

of a single species uvib =12

hn +hn

ehnkT - 1

Ê

Ë Á

ˆ

¯ ˜ :

uvib =9NAn2

nmax3

12

hn +hn

ehnkT - 1

È

Î Í ˘

˚ ˙ 0

nmax

Ú dn

=98

NAhnmax +9hNAnmax

3n3

ehnkT - 10

nmax

Ú dn (5.141)

By making use of the definition

qD:=hnmax

k(5.142)

this expression may be written as

uvib =98

RqD +9hR

kh

kqD

Ê

Ë Á

ˆ

¯ ˜

3n3

ehnkT - 10

qD kh

Ú dn (5.143)

This results in the molar heat capacity by differentiationwith regard to T:

uvib = 9RT3

qD3

Ê

Ë Á

ˆ

¯ ˜

x4ex

ex -1( )20

qDT

Ú dx (5.144)

if x =hnkT

if being used for shortness.

Fig. 5.7. Partition Function in the case of Debey‘s model of asolid material.

Thermodynamics and Kinetics of Solids 47________________________________________________________________________________________________________________________

15.10.01

At low temperatures, x is very large; because of ex in thedenominator, the large values do not contribute to theintegral and the integration may run up to •.

Development of the expression x3

ex - 1 into a series:

x3

ex - 1=

x3e-x

1 - e-x = x3e-x + x3e-2x + x3e-3x + L (5.145)

Integration results in

x3

ex -10

•

Ú dx = 6.4950 (5.146)

Feeding this value into eqn. (5.143), this results under

consideration of x =hnkT

in the following expression

uvibT Æ0

=98

RqD + 9 ⋅6.4950RT 4

qD3 (5.147)

and accordingly in the following expression for the molarheat capacity

cvibT Æ0

= 233.8 ⋅RT3

qD3 (5.148)

cvib Æ 0 for T Æ 0. Temperature dependence ~ T3 (asobserved experimentally).

For high temperatures, x Æ 0. Development of thedenominator of eqn. (5.143) into a series and integrationgives

uvib =98

RqD + 9RTT

qD

Ê

Ë Á

ˆ

¯ ˜

3

+13

TqD

Ê

Ë Á

ˆ

¯ ˜

3

-18

TqD

Ê

Ë Á

ˆ

¯ ˜

4È

Î Í Í

+1

60T

qD

Ê

Ë Á

ˆ

¯ ˜

5

± L˘

˚ ˙ ˙

. (5.149)

Finally, the molar heat capacity is

cvib = 3R -

320

RqD2

T2 ± L (5.150)

(Dulong-Petit’s Law).

The dependence of the type of material on thetemperature dependence is expressed by Debey’stemperature. Since always the ratio qD / T appears, allmaterials should follow a single curve when the reducedtemperature T /qD is introduced (Fig. 5.8.). Table 5.2.shows the assumed Debey temperatures.

Fig. 5.8. Temperature dependence of the molar heat capacity ofsolids according to Debey’s theory.

Fig. 5.9. Temperature Dependence of the molar heat capacity ofsolar bodies with different composition.

Table. 5.2. Debye-temperatures of the elements and compoundsindicated in Figs. 5.8 and 5.9.


15.10.01

qD is directly proportional to nmax.On the other hand side,the vibrational frequency is proportional to the squareroot of the force constant D and inversely proportional tothe square root of the mass m. For the ductile and heavyPb, D is small and m is large. This results in a low valueof qD for Pb. The inverse holds for diamonds and a largevalue of qD (or nmax) is expected. So far, only solid bodiesof the elements have been considered. The treatment maybe extended, however, to solid, multiple-atomiccompounds.Rule of Neumann and Kopp"The molar heat capacities of solid compounds iscomposed additively of the molar heat capacities of theelements that make up the compound".

Accordingly, if the molar heat capacity measured athigh temperatures is divided by the number of atoms performula unit, approximately Dulong-Petit’s value 3R isobtained (Fig. 5.9. for 2- and 3-atomic compounds aswell as Pb and diamond). The data for compounds alsofall on to the curve shown in Fig. 5.8 if these are dividedby the number of atoms per formula unit.

At very low temperatures, also the electrons in metalsshow a contribution to the molar heat capacity.

5.8. The Electron Gas

Application of Fermi-Dirac’s Statistics.In the case of metals, free electrons are assumed. If thesewould be moving completely disordered like particles inan ideal gas, the monovalent metals should show a molarheat capacity of 4.5 R (3R for the vibrations of the latticeatoms, 3/2R or the translational motion of the electrons).

It has been observed, however, that the molar heatcapacity may be explained only by the vibration of thelattice constituents. The conducting electrons haveapparently no influence.

In a crystal of a metal of n atoms, the atomic orbitals areseparated into n molecular orbitals which form a band.

Fermi-Dirac’s Distribution Function:

NA

= f e( ) =1

ee-e FkT+1

(5.151)

Special cases:

i) T = 0 K

a) e << eF

Æ ee-e FkT << 1

and f (e) = 1

b) e > eF

Æ ee-e F

kT Æ •

and f (e) Æ 0

eF: "’Fermi Energy"

ii) T > 0 K

a ) e - eF >> kT (regime of sufficiently largedistance from Fermi’s energy)If e < eF we may consider e

e-e FkT << 1 . Under

consideration of 1

1 + xª 1- x , we come to the relation

f e( ) = 1- ee-e F

kT

For low energies, all levels are occupied. For higherenergies (however, still sufficiently far away from eF)Boltzmann’s distribution with (e - eF) as energy holdsfor the number of unoccupied energy levels. (In thecase of the probability 1 - e

e -e FkT that a state is

occupied, ee-e F

kT is the probability that a level is notoccupied).

In the case of e > eF, ee-e FkT >> 1 holds. Then, eqn.

(5.151) results in

f e( ) = e-e-e F

kT (5.152)


15.10.01

Accordingly, Boltzmann’s distribution holds with (e -eF) as energy for energies which are sufficientlyhigher than Fermi’s energy.

b) e = eF

Æ ee-e F

kT = 1 Æ f e( ) = 12 (5.153)

(independent of the temperature)

Fig. 5.10.: The dotted line holds for a temperature at

which kT =1

10eF (T ª 5000 K for metals).

Occupation Density N (e):

N e( ) = g ⋅ f e( )D e( ) (5.154)

D (e): density of states, g = 2.

Accordingly, eqn. 5.32 results in

N e( ) =4pV 2m( )

3 2

h3 ⋅e

12

ee-e F( )kT + 1

(5.155)

At 0 K and 110

eFR

K, the occupation densities are as

shown in Fig. 5.11. The shape of the curve isfundamentally different from the one expected accordingto Boltzmann’s statistics.

Determination of the Fermi energy (energy up to whichall energy levels are occupied at 0 K) from the knowledgeof the occupational density (eqn. 5.155):

The total number of species has to be equal to theintegral of the occupatiol density between 0 and •. Since

there are no species with energies e > e F at 0 K, theintegration may be restricted to the range from 0 up to eF.It may be also taken into account that f (e) = 1. Then, eqn.(5.155) results in

N = N e( )0

e F

Ú de = 4pV2mh2

Ê Ë Á ˆ

¯ ˜

32

e1

2 de0

eF

Ú

= 4pV2mh2

Ê Ë Á ˆ

¯ ˜

32

⋅23

eF32 (5.156)

Solving this eqn. for eF leads to

eF =3NpV

Ê Ë Á ˆ

¯ ˜

23 h2

8m(5.157)

Determination of the total energy of species at 0 K:The total energy is the integral of the occupation densitymultiplied with the respective energy:

U0 = eN e( )0

•

Ú de = 4pV2mh2

Ê Ë Á ˆ

¯ ˜

32

e32

0

e F

Ú de

= 4pV2mh2

Ê Ë Á ˆ

¯ ˜

32

⋅25

eF52 (5.158)

Substitution of eF by the expression given in eqn. (5.157):

U0 =35

NeF =3h2

40mN

3NpV

Ê Ë Á ˆ

¯ ˜

23

(5.159)

Æ Average energy at 0 K:

e 0 ≡U0N

=35

eF =3h2

40m3NpV

Ê Ë Á ˆ

¯ ˜

23

(5.160).Fig. 5.10. Fermi-Dirac-distribution;, full line holds for T = 0 K,the dotted line holds for T =

e F10K .

Fig. 5.11. Occupation density in multiples of 4pV(2m/h2)3/2, Thefull line holds for T = 0 K, the dotted line holds at T =

eF10R

K.


15.10.01

The average energy is 3/5 h of the value of the Fermienergy and may be calculated from the mass and thevolume that is available for the species.

For the calculation of eF, only knowledge of the molarvolume is required (Tab. 5.3.).

Example: Li. According to eqn. (5.160), the averageenergy per electron is 2.72 · 105 J/mol at 0 K.

According to Boltzmann’s statistics, the molar innerenergy of a monoatomic gas would be 3

2 RT . The energyof 2.72 · 105 J/mol would only be reached by theelectrons at 22000 K.

Calculation of the average energy at higher temperatures:Generally holds

U = 4pV2mh2

Ê Ë Á ˆ

¯ ˜

32 e

32

e e- eFkT +1

0

•

Ú de (5.161)

There exist only numeric solutions, e.g. as shown inFig.!5.12.

Boltzmann’s statistics is shown for comparison. Thisfigure shows why electrons in metals provide a verysmall contribution to the molar heat capacity at normaltemperatures.

The electrons have a very high energy which increases,however, much more slowly with increasing temperaturethan expected classically (∂U/∂T).Pauli Principle: Only electrons may be excited which areat the most kT below the Fermi energy. Since kT << eF

this is only a small fraction of the electrons whichcontribute to the molar heat capacity.

Contribution of the electrons to the molar heat capacity ofmetals:Differentiation of eqn. 5.161 after T and division by nresults in:

cel =p2

2R

kTeF

= gT

(5.162)

Classically is expected: 3/2R for the translationalmotion.According to Table 5.3, thevalue for g is in the case ofLi:

g Li = 0. 9 ⋅10-4 R K-1

(5.163)

For an experimental proveof the value cel, conditions are required in which the smallheat capacity of the electrons is not completely coveredby the vibrational energy contribution of the lattice (ei).Experiments should be therefore either performed at hightemperatures at which cel is very large, or at very lowtemperatures, at which the molar heat capacity of thelattice becomes very small because of the T3-relationship.

For low temperature holds:

c v = c vib + cel = 233.8RT3

qD3 +

p2

2R

kTeF

= AT3 + gT (5.164)

Fig. 5.12. Dependence of the average energy of the electron gason the temperature.

Table 5.3 Quantities related to Fermi’s energy of electrons in metals

12


15.10.01

Figure 5.13. shows a plot of cv/T against T2; Theintersection with the ordinate provides the g-value and theslope provides the A-value.

Fig. 5.13a shows the measured results for potassium:2.08 · 10-3 J / mol K2, A = 2.57 · 10-3 J / K4 mol. Thisresults in the Debey temperature qD = 91 K. The g-valueis 20 higher than in the case of the assumption of a freeelectron gas (Table 5.3). The experimental values aregenerally somewhat higher because there are alwayssome interactions between the electrons themselves andwith the lattice.

Fig. 5.13b shows the contributions by the vibrationcvib, electrons cel and entire molar heat capacity as derivedfrom Fig. 5.13a. At lower temperatures there is anessential contribution by the electrons. However, at 1 Kthere is already cvib > cel in the case of potassium. At 10K, cvib ª 123 cel.Pressure of the electron gas at 0 K:Determination of the pressure from the volumedependence of the inner energy U0:

p = -∂U0∂V0

Ê

Ë Á

ˆ

¯ ˜

s(5.165)

Differentiation of eqn. (5.160) with regard to V:

p0 = -N3h2

40m3Np

Ê Ë Á ˆ

¯ ˜

23 d V- 23( )

dV

=h2

20m3p

Ê Ë Á ˆ

¯ ˜

23 N

VÊ Ë Á ˆ

¯ ˜

53

(5.166)

Making use of eqn. (5.157), this results in:

p0 =25

NV

eF (5.167)

and using eqn. (4.160) leads to

p0 =23

U0V

(5.168)

or

p0V =23

U0 (5.169)

In the case of potassium, a value p0 = 1.7 · 104 bar is theresult. The reason why the electrons don’t escape thecrystal originates from the attraction between electronsand positive atomic nuclei.Further thermodynamic data of the electron gas at 0!K:

i) Entropy.All levels are occupied up to eF; all energy levels areempty above this energy. There is only one possibility torealize this state. Accordingly,

S0 = 0 (5.170)

ii) Free Energy.

F0 = U0 - TS0 = U0 (5.171)

iii) Enthalpy.

H0 = U0 + p0V = U0 +23

U0 =53

U0 (5.172)

iv) Gibbs-energy

G0 = H0 - TS0 = H0 =53

U0 (5.173)

Combination of this equation with eqn. (5.160), results in

G0 =53

U0 = NeF (5.174)

Fig. 5.13. Molar heat capacity of Potassium at extremely lowtemperatures. Determination of A and g, (a). Contributions ofthe lattice vibration, electrons and total molar heat capacity (b).


15.10.01

i.e., the Fermi energy is identical with theelectrochemical potential of a single electron at 0 K.

Because of the same electrochemical potential of acomponent in different phases in equilibrium, this meansthat the Fermi-energy of 2 or more materials in contactwith each other is are the same. As a result, contactvoltages are established at the interfaces.

5.9. Photonic Gas Heat Iradiation)

Example of the application of the Bose-Einstein statistics(photon has no spin). Determination of the radiationequilibrium, i.e. determination of the energy of theradiation within a frequency range between n and n + dn:The density of states between the frequencies n and n +dn and the occupation probability of the quantum stateshas to be determined.Density of states (eqn. 4.103):

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

32e

12de (5.175)

Since the photon has no mass, eqn. 5.175 may not beapplied directly. Therefore, the number of quantum stateswith a momentum between p and p + dp is beingconsidered:

e =p2

2m; de =

1m

p dp (5.176)

This results in

dN p( ) = D p( )dp =4pVh3 p2dp (5.177)

Making use of de Broglie’s relationship, the momentummay be replaced by l or n = c/l:

p =hn

c, dp =

hc

dn (5.178)

Instead of eqn. (4.177), the following relationship may beused

dN n( ) = D n( )dn =4pVn2

c3 dn (5.179)

The application of the Bose-Einstein distribution functionfor the occupation probability

f e( ) =1

e-aee

kT - 1(5.180)

is not allowed since one of the boundary conditions (N =ÂNi) does not hold for a photonic gas since many morephotons exist in a body with high wall temperature thanat low temperatures.

This means that a does not show up when Lagrange’smultiplicators are applied, i.e. a = 0. For photons, theoccupation probability is therefore

f e( ) =1

ehnkT - 1

(5.181)

with e = hn. Making use of eqn. 5.179 this results in

N n( ) =8pVc3

n2

ehnkT -1

(5.182)

The value 8 (instead of 4) results from the fact, that 2independent polarisation states exist for the light.

The spectral energy density follows from eqn. (5.182) bymultiplication of each single light quantum with theenergy hv:

U n( ) =8phV

c3n3

ehnkT - 1

(5.183)

Division by the volume results in Planck’s iradiation law,i.e. the spectral radiation density r (n) of the black bodyradiation:

r n( ) =U n( )

V=

8pc3

hn3

ehnkT -1

(5.184)

Rearranging this equation results in

r n( ) =8phc3

k3T3

h3 =hn / kT( )3

ehnkT -1

(5.185)

Substitution

x : =hn

kTÆ (5.186)

provides

r n( ) =8phc3

k3T3

h3 =x3

ex -1(5.187)


15.10.01

The expression x3 / (ex - 1) is shown in Fig. 5.14.The maximum may be determined by differentiation. Thefollowing transcendental equation is obtained

exmax 3- xmax( ) = 3 (5.188)

The graphical solution is xmax = 2.8214.According to eqn. 5.186 we get

hnmaxkT

= 2.8214 (5.189)

or

nmax = 5.877 ⋅101 0T K-1s-1 (5.190)

(≡ Wien’s distribution law, according to which theproduct of T and lmax is constant at the maximum of theradiation density/wave length).

The total iradiation density of the black body is

UV

=8phc3

n3

ehnkT -10

•

Ú dn

=p4

15(5.191)

Accordingly, we have

UV

=8

15p5 k

chÊ Ë Á ˆ

¯ ˜

3kT4 (5.192)

(≡ Stefan-Boltzmann’s relationship)

U = bVT 4 with b =8

15p5 k

chÊ

Ë Á Á

ˆ

¯ ˜ ˜

3k (5.193)

cv = 4bVT 3 (5.194)

S =cvT

0

T

Ú dT =43

bVT 3 (5.195)

F = U - TS = -13

bVT 4 = -13

U (5.196)

G = F + pV = -13

U + pV (5.197)

Because of m = 0, G = 0 and accordingly

pV =13

U (5.198)

According to eqn. 5.193 an iradiation pressure of

p =13

UV

=845

p5 kch

Ê Ë Á ˆ

¯ ˜

3kT4 (5.199)

is observed which increases with the 4th power of thetemperature.

5.10. Kinetic Gas Theory.

Maxwell’s Velocity Distribution.Which fraction of the species has a velocity between vand v + dv?The treatment is analogous to the energy distribution ofthe electrons. Considering the density of states D(e) andthe probability that a given state is occupied, this resultsin the energy distribution which also provides thevelocity distribution.Fraction of molecules with an energy between e and e +de:

N e( )N

de =D e( )Nede

D e( )Nede0

•

Ú(5.200)

Considering eqn. (5.32), i.e.

D e( ) = 4 2pVh3 m

32e

12de and eqn. 5.72, i.e.

N e = Ne -e

kT

eei kT

iÂ

= ae-ekT , results in

Fig. 5.14. Representation of the function x3 /ex – 1 as a functionof x.


15.10.01

N e( )N

de =g a e

12e-e / kTde

g a e12e-e / kTde

0

•

Ú(5.201)

The integral has the value 12

p kT( )32 . This expression

leads to

N e( )N

de = f e( )de = 2p1

pkTÊ

Ë Á

ˆ

¯ ˜

32e

12 e-ekT de (5.202)

As in the case of Fermi-Dirac statistics, the term e1

2 isdetermining the relationship at low energies. Theexponential function becomes more and more importantfor higher temperatures (Fig. 5.15.). In contrast to theFermi-Dirac distribution, the position of the maximummoves to e = 0 for T = 0.Differentiation of f(e) (eqn. (5.202)) with regard to e

shows

emax =12

kT (5.203)

Velocity Distribution:Substitution

e =12

mn2 , de = mndn (5.204)

Equation (5.202) then results in

N n( )N

dn = f n( )dn =m

2pkTÊ

Ë Á

ˆ

¯ ˜

324pn2 ⋅e

-mv22 kT dn (5.205)

The velocity distribution curve is given in Fig. 5.16.Because of u2, a parabolic relationship is obtained.

The maximum decreases with increasing temperature.u ≠ umax because of non-symmetry.Without derivation:

umax =2kTm

(5.206)

u =8kTpm

(5.207)

u2 = 3kTm

(5.208)

Average kinetic energy:

e kin =12

mu2 =32

kT (5.209)

Number of collisions with a wall:Fig. 5.17: Number of species arriving from the directionux, uy and uz at the area A within the period of time dt:

dN uxuyuz( ) = NA ux dtN ux( )

Ndux

N uy( )N

duyN uz( )

Nduz

(5.210)

Considering that species will arrive from any direction,integration over all negative ux, positive and negative uy

and uz ...:

Number of collisions / time unit

Zw =dNAdt

= NkT

2pmÊ Ë Á ˆ

¯ ˜

12

=p

2pmkT(5.211)

Fig. 5.15. Energy distribution according to Boltzmann’sstatistics at two temperatures.

Fig. 5.16. Velocity Distribution for nitrogen at differenttemperatures.


15.10.01

5.11. Law of Equal Distribution of Energy

The energy of a species is composed additively of severalcontributions:

e i = h + h' +h' ' + L (5.212)

The individual terms are dependent on variouscoordinates u, u', u'', ... , e.g. the velocity u , themomentum p, the angular velocity w , the rotationalmomentum etc.In the 1-dimensional case holds:

Translational energy e trans =12

mu2 =1

2mp2 (5.213)

Rotational energy erot =12

Iw2 (5.214)

Vibrational energy evib =12

mu2 +12

Dx2 (5.215)

According to Boltzmann’s statistics, the number Ni ofmolecules that have a given energy ei (without assumingdegeneration) is given by

Ni =Nebei

ebe i

iÂ

=Neb h+ h' +h' '+L( )

eb h+h' +h' '+L( )

iÂ

=Neb h+h' + h'' +L( )ebh

eb h+h' + h' ' +L( )

iÂ ebhÂ

(5.216)

(The modification of the denominator corresponds to theseparation of the partition function into various factors.)

Summation of all possible values of h', h'', ... Æ NumberNj of species, which have the energy h of the type ofmotion which corresponds to the coordinate u:

N j = Nih' ,h' ' , ...Â =

N eb h' +h' '+L( )ebh{eb h' + h' ' +L( )Â ebhÂ

=Nebh

ebhÂ(5.217)

All other components (h ', h '', ... ) of the energy ei arewithout any influence.

In the following, only the contribution of the componenth to the energy e are considered.It will be assumed that the energy is proportional to thesquare of the coordinate u (eqns. (5.213) – (5.215)).

Æ For all degrees of freedom that contribute to theenergy by a square law relationship (h = au2),

h = -1

2b(5.218)

is independent of the nature of the coordinate u andaccordingly the type of motion.

e kin =12

mu2 =12

m ux2 + uy

2 + uz2( ) =

32

kT (5.219)

Since ux2 = uy

2 = uz2 , for each square term holds

h =12

kT (5.220)

Fig. 5.17. Illustration for the derivation of the pressure of a gasonto a wall.

Documents

Thermodynamics and Kinetics of Solids 45 · Thermodynamics and Kinetics of Solids 45 15.10.01 uvib= 3 2 NA hn + 3NA hn e hn kT-1 Ê Ë Á Á ˆ ¯ ˜ ˜ 3N (5.128) Using the characteristic