Example

Examples: Photon gas, an electromagnetic field in thermal equilibrium with its container

i

i

In this case h harmonic oscillatorQM we know that i ni

H = h

To describe the state of the field, we need to know how many n are in each oscillator

Photons are bosons n=0,1,2,3…..

1 2

1 1 2 2

, ,...,

... Q N,V,Tj

j j

n n n

n n nEe e

1 2

1 1 2 2

0 , ,...,

first we want to show that

Q N,V,Tj j

j j j j

n n n nj

n nn ne e e e

Cont.

1 1 1 2 2 2

3 3 3

0 1 2 0 1 2

0 1 2

Q N,V,T

e e e e e e

e e e

1 2 1 2 1 2 1 2 1 2

1 2 1 2 1 2 1 2

3 3 3

1 2 1

2 1 2

0 0 0 0 1 0 1

1 2 0 2 1

0 1 2

e e e e

e

e

e e e

e e e

1 2 3 1 2 3 1 2

1 2 3 1 2

1 2

3

3 0 0

1 0 1 1 2 3

1 2 3 1 2 3

1 2

1

1 2 3

1 2 3

0 0 1 00 0 0

0 0 0 1 2

0 2 1 0 2 2 1

0 2

0

0

0 ...

e e

e e

e e

ee

e e e

1 2

1 2 31 2 3

1 1 2 2 3 3

0 0 0

1 2

1 2 3

1 2 3 1 3 32

, ,...,

010 0 1 0

0 0 1 2 0 0 0 2 0

...

jn n n

n n n

e e

e

e

e e e

Cont

j

10 1

where we used1Q N,V,T

1 jj

j j

j xnj j x

nee

The average number of photons in a state j…

1 1 2 2

j

...

n

jjn

E

E

n nn en e

Qe

1 1 2 2 ...

ln , ,1

j j

n

j jn n n n

n neQ N V T

Q

ln , , 1 ln 1 j

j

Q N V T e

ln , ,

1

j

jj

Q N V T e

e

j

1average occupation number is n =

1jePlanck distribution

Monoatomic ideal gases

for the number of states to be N,

32 26

1 12

m

h

kT

To evaluate a monoatomic ideal gas we first see if it can

be represented by the Bolztmann statistics

2

32 2

for 10 , 10 , 300

61 for most gases Boltzmann Stat.

12

m gr a cm T K

kTm

h

,Q N,V,Twe can us

!e

Nq V T

N How do we calculate the molecular partition function q V,T ?

translation electronic nuclearq V,T

we can write thu

=

s

q q q

translation electronic nuclearH H H +H

sum over states sum over levels

q V,Tj

j k

kk

kT kTe e

Nuclear Contribution

for nuclear states ~106 ev to populate an excited state, T>1010 Kwe can consider only the ground state

Nuclear partition function qnuclear

degeneracy

sum over levels

1 2

1 2,nuclear

k

kk

kT kT kTq V T e e e

1

1

we arbitrarily define the ground state as 0

all other energies will be referenced to

12

1 2,nuclear kTq V T e

1

degeneracy is given by spin 2 1I

1

,nuclearq V T

For most chemical problems (not nuclear!), qnuclear is just a const.

Electronic Contribution

Electronic partition function qelectronic

degeneracy

sum over levels

1 2

1 2,electronic

k

kk

kT kT kTq V T e e e

1 we arbitrarily define the ground state as 0

12

1 2,electronic kTq V T e

for electronic states, 400-700 nm (Uv-Vis); 33000–15000 cm-1;

4–1.8 ev; 160 kT – 70 kT; 6.6x10-19J – 3x10-19 J

Most cases

@300K only ground state pop. T>104 K for pop. in excited state

1

,electronicq V T

What is the degeneracy 1 of an electronic energy level?

Electronic degeneracy

What is the degeneracy 1 of an electronic energy level?

We have to go back to atomic electronic structure and Spectroscopic Term Symbols 2S+1Lj (for light atoms with RS S-L coupling)

1 =2j +1

212

10 1

12

1

32

1H 2 1 2

2

He 2 0 1 1

3F 2

2

1

1 42

4

electronic

electronic

electronic

q

q

q

S

S

P

Fraction of population in excited states

12For heavy atoms,

12

1 2sometimes we have to consider

kT ,

,

electronic kTq V T e

st

to know whether we have to consider an additional electronic

state, we calculate the fraction of atoms in the 1 excited state

2

12

2

12

1 2

kT

kT

ef

e

312 1

7652

19.80.0259

19.80.00259

He 19.8

3

1

3 10

3

S

ef

ev

e

1

2

212

2

0.050.0259

0.050.00259

F 0.05

2

4 2

0.068f

ev P

e

e

for 300

25.9

T K

kT mev

Translation Contribution by levelTranslation partition function qtranslation

A single particle translation energy is given by 2

2 2 2

2,

8

x y zn n n x y z

hn n n

ma

sum over level

degeneracy

s

,translationk

kk

kTq V T e

2 2 2x y z

3

2 3

2

where we consider again a sphere

of radius R= n n n

8 and

4

k

ma d

h

because the between 2 consecutive states 0

we can consider a continuos change in energy

3

2 3

20 0

8,

4translation kT kTm

q V T e d a e dh

3

2

2

2,translation mkT

q V T Vh

Translation Contribution by stateTranslation partition function qtranslation

sum over states

,

,

,translation

n n nx y z

n n nx y z

kTq V T e

22 2 2

2 2

2

23

88

,

,

x y z h n

transla

hn n n

tion kma

T

n

k maT

n n nx y z

q V T ee

2 2

2

2

2

11 32

3

8

8

1,

2

h n

translation kT ma

h

kT ma

q V T e dn

because the between 2 consecutive states 0 we can

consider a continuos change in energy

n

3

2

2

2,translation kT m

q V T Vh

Q of ideal gas

combining all these information we obtain Q N,V,T

123

2

2 2 11

1Q V,

!

2N, T kT

elec ele ucc

N

n

mkTV

hNe

2

and for the thermodynamic properties...

, , - ln , ,

lnQ N,V,TE

lnQ N,V,Tp

lnQ N,V,TS ln , ,

lnQ N,V,T, ln

A N V T kT Q N V T

kTT

kTV

k Q N V T kTT

T P kTN

12

1 2

3

2

2 1

21A= ln

!kT

elec elec uc

N

n

mkTVkT

he

N

1ln ln ln 1 ln

N!

N

eN N N N N N

1nuclear contributions is constantusually omittedln nucNkT

123

2

1 22

2 A= ln ln kT

elec elec

mkT eVNkT NkT e

Nh

12

2

2

1

3

2

2 kTelec elec e

m

N

T

h

ekV 54@ room T 4 10 1

3

2

2

the only significant contributionis from the translations A=

2ln

mkT eVNkT

Nh

Helmholtz energy, A N,V,T - ln , ,kT Q N V T

< E>for an ideal monoatomic gas

12

1 2

3

2

2

2

1

!ln

E =

2elec elec

kT

N

N

mkTV

h

kT

e

T

12

2 12 at low T, the electronicterm is neglegible

3

2E =

kTelec

electronic

Nke

TN

q

12

2

1

3

22 2

3

2

2

2

1

!

lnln

E =

2ln

elec

elec

N

kTN

N

N

eT

kT kTT T

mkV

h

kTT

Internal energy <E>

2 lnQ N,V,T

kTT

<p> for a monoatomic gas

12

1 2

3

2

2

1

!p

2ln elec elec

k

N

T

N

mkTV

hkT

e

V

Pressure <p>

all contributions from translationp

NkT

V

12

1 2

3

2

2

1

!

2ln

lnp

elec eleckT

N

mkTe

hVkTN kTN

V V

lnQ N,V,T

kTV

S for a monoatomic gas

Entropy S(N,V,T)

12

12

3

2

1 22

2 12

2ln ln

1 3

2

kTelec elec

kTelec

electronic

mkT eVS Nk Nk e

Nh

N eNkT

T q

translation electronicS S S

5322

. 2

2lntrans

mkT e VS Nk

Nh

12

12

2 12. 1 2ln

kTeleckT

elec elec elecelectronic

eS Nk e

kTq

lnQ N,V,Tln , ,k Q N V T kT

T

for a monoatomic ideal gas

Chemical potential (T,P)

12

1 2

3

2

2

1

!n

=

2l

, -

elec eleckT

N

N

mk

T p kT

TeV

h

N

12

1 2

3

2

2

ln

2ln

=+

elec eleckT

N N NkT

N

mkTN V e

hkT

N

12

1 2

3

2

2 02

= ln 1 ln elec eleckTkT

N mkTkT N V e

N h

lnQ N,V,T

lnkTN

Reference

3

2

2 .

2ln ln eleckT

mkT qk

T V

Nh

3

.

2

2

2, ln ln lnelec kT

mkTkT

hqT P kT kT p

, lnoT p T kT p

12

1 2

3

2

2

2

ln

elec eleckT

kT

mkTV

he

N