6.5. Gaseous Systems Composed of Molecules with Internal Motion

Assumptions ( ideal Boltzmannian gas ) :

1. Molecules are free particles ( non-interacting).

2. Non-degeneracy (MB stat) :3 1n

22

mkT

1

1 1, , , 1, , ,

! !

N N

NZ N T V Q T V Z T V Q T VN N

internal1 1 11, , , translZ T V Q T V Q Q 3

Vj T

j T g e

= quantum # for internal DoF

internal

internal, 0transl

transl

H H H

H H

internal internaltransl translH H H HHe e e e e

Internal DoF j T g e

int , ln lnA N T N kT j N kT j intint ln

T

AkT j

N

intint

N

AS

T

int

1 jU N

j

int

int,

V

V N

UC

T

2 ln jN kT

T

lnln

jN k j T

T

2 ln jN k T

T T

elec nucl vib rotj T j T j T j T j TMolecules :

Homopolar molecules (A-A) : elec nucl rot vibj T j T j T j T

6.5.A. Monatomic Molecules

Let 4 5~ 10 10ionT K

k

( All atoms are neutral &

in electronic ground state )

Nuclear spin Hyperfine structure : T ~ 101 – 100 K.

Level-splitting treated as degeneracy : 2 1n ng S

Inert gases ( He, Ne, Ar, ... ) : Ground state L = S = 0 : 1eg

0/ /0

k T k Te e

j T g e

= 0 denotes ground state. 0 = 0.

0j T g

0

e ng g 2 1nS

L = 0; S 0 : 2 1 2 1nj T S S 2 1eg S

int , ln ln 2 1 ln 2 1nA N T N kT j N kT S S

intint ln

T

AkT j

N

intint

N

AS

T

int

1 jU N

j

int

int,

V

V N

UC

T

0 0

ln 2 1 ln 2 1nkT S S

ln 2 1 ln 2 1nN k S S

L = 0, S 0

2 1 2 1nj T S S

L 0, S 0

/2 1 J

L Sk T

elecJ L S

j T J e

kT 2 1L S

elecJ L S

j T J

12 1 2 1 2 1

21

2 1 2 1 2 12

L S L S S L S

L S S L L S L

2 1 2 1

2 1 2 1

L S L S

S L S L

kT 0 /02 1 k T

elecj T J e Ground state 0 = 0.02 1J

CV, int = 0 in both limits CV has a maximum.

Const of motion:L2, S2, J2, Jz

S-O: L-S coupling

6.5.B. Diatomic Molecules

Let 4 5~ 10 10dissocT K

k

( All atoms are neutral &

in electronic ground state )

Non-degenerate ground state ( most cases ) ge = 1 & jelec (T) = 1

j T g e

Degenerate ground state ( seldom ) :

1. Orbital angular momentum 0, but spin S = 0 :

In the absence of B, depends on |z|

doublet ( z = M ) is degenerate ( ge = 2 = j(T) )

CV = 0

2. = 0, S 0 : ge = 2S + 1 = j (T) CV = 0

3. 0 & S 0 :

Spin-orbit coupling Beff fine structure

0elecfine struct

j T g g e

E.g., NO ( 1/2, 3/2 ) ( splitting of doublets ) :

178Kk

/

0 1k T

elecj T g g e

0elecfine struct

j T g g e

1 elec

elecelec

jU N

j

/1

/0 1

k T

k T

g eN

g g e

,

elecV elec

V N

UC

T

// 1

1 22 / /0 1 0 1

1 k Tk T

k T k T

g eN g e

kT g g e g g e

2 /0 1

2/0 1

k T

k T

g g eN k

kT g g e

//

2

k Tk Te

eT kT

2

/ /1 0

0 1

1

1 1k T k T

N kkT g g

e eg g

1 12

1 0

0 1

2

/1

0

1 1 0

0k T

g gkT

kT g gN k

ge kT

kT g

CV has max. for some kT ~

Vibrational States

310 Kk

for diatomic gases

Full contribution for T 104 K

No contribution for T 102 K

Harmonic oscillations (small amplitude) :1

2n n

2 /

2/ 1

v

v

Tv

T

ek

T e

v k

1

2

0

n

vibn

j e

/2

1

e

e

ln vibvib

ju

1ln ln 1

2vibj e

1

2 1

e

e

1 1

2 1e 1

2n

vibvib

uc

T

2

21

ek

kT e

“vibrons”

2 /

2/ 1

v

v

Tv

V vibvib T

eC N c N k

T e

v k

2

2

1 v

vV vib

v

TC N kT

T

vT N k equipartition

value

vT 2

/v TvV vib

C N k eT

0 vib DoF frozen out

2 /

2/ 1

v

v

Tv

V vib T

eC N k

T e

Very high T anharmonic effects Cvib T ( Prob 3.29-30)

Mathematica

Nuclear Spin & Rotational States: Heteropolar Molecules

Heteropolar molecules ( AB ) : no exchange effects

interaction between nuclear spin & rotational states negligible.

2 1 2 1n A Bg S S From § 6.5.A :

Cnucl = 0

Molecule ~ rigid rotator with moment of inertia ( bond // z-axis )

, , 0diag I II 20I r A B

A B

m m

m m

= reduced mass

r0 = equilibrium bond length

2

12rot l l

I

0,1, 2,l

2

12rot l l

I

0,1, 2,l j T g e

2

0

2 1 exp 12rot

l

j T l l lI kT

0

2 1 exp 1r

l

l l lT

2

2r I k

<< T room if no H or D present.

HCl : ~ 15 KN2 , O2 , NO : ~ 2-3 KCl : ~ 1/3 K

H2 : ~ 85 KD2 : ~ 43 KHD : ~ 64 K

( 1) ( 1)1

2 !

ppk kk

k

BB f n f m f n f m O f

k

S I I

Euler–Maclaurin formula :

n

i m

f i

S n

m

d x f xI

Bk = Bernoulli numbersB1 = −1/2,  B2 = 1/6, B3 = 0,B4 = −1/30,  B5 = 0, B6 = 1/42,B7 = 0, B8 = −1/30, …

0 1rotj T ( only l = 0 term survives )

T >> r : 0

2 1 exp 1rrotj T d l l l l

T

( 1) ( 1)1

2 !

ppk kk

k

BB f n f m f n f m O f

k

S I I

0

exp rd x xT

1x l l

r

T

V rotC N k2 ln rot

rot

jU N k T

T

N k T

0

2 1 exp 1rrot

l

j T l l lT

Better approx : set 2 1 exp 1rf l l l lT

0 1

0k

f

f e

( 1)

20

12 1 exp 1 0

2 !kr k

rotk

Bj T d l l l l f

T k

( 1)

1

0!

kk

kr

T Bf

k

( 1)

1

0!

kkrot

kr

T Bj T f

k

B1 = −1/2,  B2 = 1/6, B3 = 0,B4 = −1/30,  B5 = 0, B6 = 1/42,B7 = 0, B8 = −1/30, …

2 1 exp 1rf l l l lT

2 31 1 4 1

3 15 315 315r r r

r

T

T T T

Mulholland’s formula( Evaluated using

Mathematica )

2 ln rotrot

jU N k T

T

2 3 4

2 3

8 59

3 45 945 14175r r r rN k T

T T T

,

rotV rot

N V

UC

T

2 3 41 16 59

145 945 4725

r r rN kT T T

0

2 1 exp 1rrot

l

j T l l lT

Crot calculated using jrot summed up to l = lmax .

Thick curve : lmax = 20Thin curve : lmax = 30

T << r :

max

0

2 1 exp 1l

rrot

l

j T l l lT

T 0 :

2 /1 3 r Trotj T e 2 ln rot

rot

jU N k T

T

2 /6 r T

rN k e

,

rotV rot

N V

UC

T

22 /12 r TrN k e

T

Mathematica

Cp

jint indep of V int lnA N kT j is also indep of V.

intint

,

0N T

AP

V

All contributions from the internal DoF are indep of V.

P V

T P

U VC C P

V T

int0P VC C

P VC C nR

int 0U

V

DoF ; f transl: 3 ; 3rot: 2 ; 2vib: 1 ; 2

11

2PC

f Rn

rot frozen

vib frozen

5 7 9

2 2 2R

rv

Homopolar Molecules

High T ( classical region / phase space distinguishable p’cles ; MB ) :

2 22

1 1

2 2 sinrot p pI I

2

1rot

rotj T dp dp d d eh

max

20

rotI

d eh

max

2 hetero

homo

2

2 1 2 1

2 12

A Br

nucl rot

Ar

TS S hetero

j TT

S homo

2 lnV

jC N k T

T T

homo same as hetero with B = A

Low T ( quantum region / BE, FD ) :

nucl rot nucl rot symmetrized

,rot l

symm l even

anti symm l odd

( Particle exchange ~ spatial inversion )

1 2 1 2nucl m m m m

m m pairings 12 1 2 2 1

2 A A A AS S S S

m = m pairings 2 1AS

, , 1, , 1,A A A Am m S S S S

There’re

2 1A AS S anti-symmetric pairings (m m only )

2 1 2 1 2 1 1A A A A AS S S S S symmetric pairings (both ) &

Fermion nuclei ( anti-symm. ) :

, ,

, ,

nucl symm rot l odd

nucl rotnucl anti symm rot l even

2 1 1

2 1A A

nA A

S Sg

S S

with

2 1 1 2 1FD odd evennucl rot A A rot A A rotj S S j S S j

/

/

2 1 exp 1rroe odd

od

ven

even dt

l

j T l l lT

Boson nuclei ( symm. ) :

, ,

, ,

nucl symm rot l even

nucl rotnucl anti symm rot l odd

2 1 1

2 1A A

nA A

S Sg

S S

with

2 1 1 2 1BE even oddnucl rot A A rot A A rotj S S j S S j

where

/

/

2 1 exp 1rroe odd

od

ven

even dt

l

j T l l lT

2 1 1 2 1FD odd evennucl rot A A rot A A rotj S S j S S j

2 1 1 2 1BE even oddnucl rot A A rot A A rotj S S j S S j

S I 1

2even odd

rot rot rotj T j T j T 2 r

T

22 12

BE FD rotnucl rot nucl rot A

jj j S 22 1

2Ar

TS

Same as classical approach

High T ( classical region) :

Ortho : higher degeneracy

Para : lower degeneracyRatio of ortho-para components :

1 oddFD A rot

evenA rot

S jn

S j

1 even

BE A rotodd

A rot

S jn

S j

2 1 1 2 1FD odd evennucl rot A A rot A A rotj S S j S S j

2 1 1 2 1BE even oddnucl rot A A rot A A rotj S S j S S j

High T :1

2odd even

rot rot rotj j j 1FD BE A

A

Sn n

S

/

/

2 1 exp 1rroe odd

od

ven

even dt

l

j T l l lT

For H2 : SA = ½ n 3

For D2 : SA = 1 n 2

Low T : smallest l term dominates

1evenrotj T 2 /3 r Todd

rotj T e 2 /3 r

oddTrot

evenrot

je

j

/

/

2 1 exp 1rroe odd

od

ven

even dt

l

j T l l lT

1 oddFD A rot

evenA rot

S jn

S j

1 even

BE A rotodd

A rot

S jn

S j

2 /

00rFD T

Tn e

2 /

0

rBE T

Tn e

For H2 (FD) : wholly para as T 0

For D2 (BE) : wholly ortho as T 0

with l = 0

Disagree with exp. on H2

Reason (Dennison) :

Transition rate for nuclear spin-flip extremely low ( T ~ 1 yr )

ortho-para ratio not eqm values

Lab prep done in room T >> r

1FD BE A

A

Sn n

S

Let

/ 2 /lneven evenodd odrot

dC N k T jT T

11

2 1FD even odd

A AA

C S C S CS

11

2 1BE even odd

A AA

C S C S CS

Mathematica

FD : wholly para as T 0, C = Ceven .

Ceven

Codd

6.5.C. Polyatomic Molecules

DoF ; f ( linear molecules )transl: 3 ; 3rot: 3 (2) ; 3 (2) vib: 3n 6 (5) ; 2 [3n 6 (5) ]

1 classicalnucl rot nucl rotj T g j

Large moment of inertia r << T of interest.

= # of indistinct config in 1 rot. = 2 for H2O = 3 for NH3

= 12 for CH4 & C6H6

n - atom molecule:

1/23

21

2classicalrot i

i

kTj I

Ii = principal moments of inertia

(Prob. 6.27)

2 lnrot rotC N k T jT T

3

2classical

rotC N k

/ 2

/1

i

i

T

vib Ti

ej T

e

§

3.8 :

ii k

i = normal freq

2 /

2/ 1

i

i

Ti

Ti

eN k

T e

2 lnvib vibC N k T jT T

Low i ~ 103 K E.g., CO2 :

1 = 2 = 960 K, 3 = 1990 K, 4 = 3510 K

Usually, high i > Tdisassoc.

6.6. Chemical Equilibrium

A B X YA B X Y Chemical reaction : i = stoichiometric coeff.

Let N = # of reactions occured.

with N > 0 ( < 0 ) meaning reaction direction is ().

0A A AN N N

0B B BN N N

0X X XN N N

0Y Y YN N N

0 denotes initial value

For a closed, iosothermal, reaction chamber kept at constant pressure,

the natural thermodynamic potential is the Gibbs free energy G( T, P, { Ni } ).

A A B B X X Y YG N , , j i

ii P T N

G

N

0

A A B B X X Y YG N 0

At equilibrium : 0G A A B B X X Y Y

If C is a catalyst, then CC appears at both side of the reaction eq.

The equilibrium relation is therefore unaffected.

N > 0 ( < 0 ) for ()

Hemholtz Free Energy for Ideal Gas

, , ln ln ln 1i i i i i i i i iA N V T kT Z N kT N N N N Z

1!

NZ

Z NN

2

int2 i

p

m

31 i

i ii

VZ e j T

int = energy due to internal degrees of freedom.

ji (T) = partition function due to inti .

i = ground state energy of the i th atomic species.

22i

im kT

3

ln 1 lni ii i i i

NN kT N kT j T

V

1Z e

3

,

ln lnii i i i i

i V T

AkT n kT j T

N

ii

Nn

V

A A B B X X Y Y 3ln lni i i i ikT n kT j T

Let

00

ln ii i

nkT

n

30 0ln lni i i ikT n kT j T n0 = standard # density.

Eqm. cond.

00

Pn

kT with P0 =1 atm.For

gas,For solution, n0 = 1 mole / liter

Eqm. cond. :

0 0 0 0 ln 0

A B

X YA A B B X X Y Y

A BkT

X Y

0 ln[ ]i kT i 0

[ ] ini

n

0X Y

A B

X Ye

A B

00 0 0 0X X Y Y A A B B

K T

= equilibrium constant

where

Law of mass action

Internal Combustion

4 2 2 22 2CH O CO H O

Natural gas combustion :

4 2 22 3 2 4CH O CO H O 2 22 2O CO CO

2 22 2CO O CO i.e.

X Y

A B

X YK T

A B

2

22

2

COK T

CO O

2 2

1COR

CO K T O

Combustion :

101500 10K K 1/25210R O

Exhaust : 40600 10K K

Rapid cooling Actual (non-eqm) exhuast R value is closer to the combustion one.

Can be reduced by raising [O2 ] via reducing [CH4 ] & using catalyst (Pt,Pd) at exhaust.

1/220210R O