15.5 Electronic Excitation

The electronic partition function is

where g0 and g1, are, respectively, the degeneracies of the ground state and the first excited state.

E1 is the energy separation of the two lowest states.

Introducing

For most gases, the higher electronic states are not excited (θe ~ 120, 000k for hydrogen).

therefore,

At practical temperature, electronic excitation makes no contribution to the internal energy or heat capacity!

0ln2

VT

ZNkTU

0

V

V T

UC

15.6 The total heat capacity

For a diatomic molecule system

Since

Discussing the relationship of T and Cv (p. 288-289)

Heat capacity for diatomic molecules

• Example I (problem 15.7) Consider a diatomic gas near room temperature. Show that the entropy is

• Solution: For diatomic molecules

,2

2ln

2

7 25

23

2

rot

T

h

mkNkS

)!(0

1ln1

121

0

contributenotdoes

e

eNk

Te

Nk

S

S

etemperaturroomAt

SSSSS

T

TT

vib

excit

excitrotvibtrans

• For translational motion, the molecules are treated as non-distinguishable assemblies

23

2

23

2

2ln

2

5

1lnln

2

freedomofdegreesthree2

3

h

mkT

N

VNkNkS

NZNkT

US

h

mkTVZ

NkTU

tr

tr

• For rotational motion (they are distinguishable in terms of kinetic energy)

rotsystem

rotsystem

rot

rot

T

h

mk

N

VNkNkS

TNkNk

h

mkT

N

VNkNkS

TNkNk

ZNkT

US

TZNkTU

2

2ln

2

7

2ln

2ln

2

5

2ln

assemblyhabledistinguisforln

moleculerhomonucleatodue2

1

2

252/3

2

23

2

• Example II (problem 15-8) For a kilomole of nitrogen (N2) at standard temperature and pressure, compute (a) the internal energy U; (b) the Helmholtz function F; and (c) the entropy S.

• Solution: calculate the characteristic temperature

first!

29.23352 Nforkandk

rvib

J

kkmolJkmol

eNk

eNk

eNkNkTU

Tbecause

NkTe

NkNkTU

UUUU

vib

r

Tvib

rotvibtrans

7

113

25.11

25.11

2983352

1089.1

227210314.80.1

1

33522272

1

33523352

2

12985.2

1

1

2

1

2

5

1

1

2

1

2

3

4.16

11015.5ln11002.6

10195.3ln

10195.3

103.274.22

1086.2734.22

10626.6

29810381.11065.424.22

2

1lnln

&ln

1lnln

626

33

33

23

2023

23

19

23

682

123263

23

2

NkT

NkTNkTF

Jkm

h

mkTVZ

NZNkTF

rotationvibassuchparticleshableDistinguisZNkTF

particleshabledistinguisnonNZNkTF

t

t

tt

J

NkT

NkTNkTNkT

FFFF

NkTF

ZZ

TforT

ZZNkTF

eZ

e

e

e

eZ

ZNkTF

tvibrot

rot

rotrot

rotrot

rotrotrot

vib

vib

vibvib

72326

24.11

24.11

62.5

2983352

29823352

10817.32981038.11002.641.15

41.15

4.1662.563.4

63.4

63.4ln72.1029.2*2

298

2ln

62.51

1ln62.5ln

11

ln

Chapter 16: The Heat Capacity of a Solid

16.1 Introduction 1. This is another example that classical kinetic theory

cannot provide answers that agree with experimental observations.

2. Dulong and Petit observed in 1819 that the specific heat capacity at constant volume of all elementary solids is approximately 2.49*104

J .kilomole-1 K-1 i.e.

3R.3. Dulong and Petit’s result can be explained by the

principle of equipartition of energy via treating every atom of the solid as a linear oscillator with six degrees of freedom.

4. Extensive studies show that the specific heat capacity of solid varies with temperature, becomes zero as the temperature approaches zero.

5. Specific heat capacities of certain substances such as boron, carbon and silicon are found to be much smaller than 3R at room temperature.

6. The discrepancy between experimental results and theoretical prediction leads to the development of new theories.

16.2 Einstein’s Theory of The Heat Capacity of a Solid

• The crystal lattice structure of a solid comprising N atoms can be treated as an assembly of 3N distinguishable one-dimensional oscillators!

• The assumption is based on that each atom is free to move in three dimensions!

From chapter 15:the internal energy for N linear

oscillators is U= Nkθ(1/2 + 1/(eθ/T

-1)) with θ = hv/k

The internal energy of a solid is thus

Here θ is the Einstein temperature and is denoted by θE.

)

1

1

2

1(3

T

E E

e

NkU

The heat capacity:

Te

NkNk

T

UC

T

EE

vv

E

)

1

32

3(

Case 1: when T >> θE

This result is the same as Dulong & Petit’s

Case 2:

As discussed earlier, the increase of is out powered by the increase of

As a result, when

T << θE

If an element has a large θE , the ratio will be large even for temperatures well above absolute zero

When is large, is small

Since

A large θE value means a big

On the other hand

To achieve a large , we need a large k or a small u (reduced mass), which corresponds to a light element or elements that produce very hard crystals.

k

hvE

u

kv

21

• The essential behavior of the specific heat capacity of solid is incorporated in the ratio of θE/T.

• For example, the heat capacity of diamond approaches 3Nk only at extremely high temperatures as θE = 1450 k for diamond.

• Different elements at different temperatures will poses the same specific heat capacity if the ratio θE/T is the same.

• Careful measurements of heat capacity show that Einstein’s model gives results which are slightly below experimental values in the transition range of