CHAPTER 6: Ideal Diatomic Gas
[q q q q ]
Q ~ rot vib tr el
N
for N indistinguishable molecules
N!
⎡ 2π ( m1 + m2 ) kT ⎤
qtr = ⎢
⎥ V
2
h
⎣
⎦
We will exam this more closely later
3/2
Actually, we also need to consider nuclear spin and its
ramifications for the rotational partition function.
Assuming a heteronuclear diatomic
qrot = 1 + 3e
−2θ r /T
+ 5e
−6θ r /T
+ ...
high T replace sum by ∫ → qrot
8π 2 IkT
= =
θr
h2
2
⎤
T ⎡ 1 θr 1 ⎛ θr ⎞
qrot (T ) = ⎢1 +
+
+ ...⎥
θ r ⎢⎣ 3 T 15 ⎜⎝ T ⎟⎠
⎥⎦
Erot = NkT + ...
T
The extra terms result
from doing a better sum
→ integral
transformation
CV ,rot = Nk + ...
T
q
=
homonuclear diatomic rot 2θ
r
keeping only the leading term
symm # (# indistinguishable
orientations)
The total wave function is either antisymmetric (fermions) or symmetric
(bosons) wrt interchange of the nuclei
Ψtot = Ψtr Ψrot Ψvib Ψel Ψnuc
consider H2: nuclear spins = 1/2
(fermions)
nuclear spin functions aβ − β a
S
αβ + β a
αα
ββ
T
para
ortho
nuclear spin rot
S (‐) J even (+)
T (+) J odd (‐)
nuclei of spin I
statistical weight 3
2I + 1 spin states per nucleus
qrot ,nucl ≠ qrot qnucl
if θ r << T
1
1 ∞
T
−θ r J ( J +1) / T
2
J
1
e
dJ
≈
≈
≈
+
=
(
)
∑
∑
∑
∫
0
2
2
2θ r
J even
J odd
J
qrot ,nucl
( 2 I + 1)
~
2θ r
2
= qrot (T )qnucl
T
2θ r
( 2 I + 1)
need θ r / T < 0.2
to be valid 2
1 +
In the above and following discussion, we assume a Σ g electronic state
Half‐integral spin
I ( 2 I + 1)
( I + 1)(2 I + 1)
antisymm nuclear spin function – even J
symm nuclear spin function – odd J
( I +1)( 2I +1) +I ( 2I +1) =( 2I +1)
Integral spin
I ( 2 I + 1)
( I + 1)(2 I + 1)
antisymm nuclear spin function – odd J
symm nuclear spin function – even J
Integral spin
qrot ,nucl = ( I + 1)( 2 I + 1) ∑ ( 2 J + 1)e −θr J ( J +1) / T
J even
+ I ( 2 I + 1) ∑ ( 2 J + 1) e
θ r J ( J +1) / T
J odd
prefactors switched for nuclei of ½ integral spin
2
H2
θ r = 85.3 ( K )
qrot ,mol = ∑ ( 2 J + 1) e
largest θr value
−θ r J ( J +1) / T
J even
para
H2 is 100% para at T = 0
→ 25% para at high T
+ 3∑ ( 2 J + 1) e
−θ r J ( J +1) / T
J odd
ortho
If at equilibrium
Measurements of Cv vs. T: disagreement between theory and experiment
problem O ↔ P interconversion slow
so high temperature distribution locked in
as sample is cooled
for most homonuclear diatomics don’t have to worry
about this other than the factor of 2 symmetry #)
Due to rotational
levels being
much closer
together
qe = ωe1e De / kT + ωe 2 e −ε 2 / kT + ...
De
using free atoms as the zero
of energy – generally need only retain the 1st term
8π 2 IkT − β hv /2
⎛ 2π mkT ⎞
De / kT
− β hv −1
q=⎜
V
e
1
e
ω
e
−
(
)
e1
⎟
2
σ h2
⎝ h
⎠
3/2
D
E
hv / kT
5 hv
= +
+ hv / kT
− e
NkT 2 2kT ( e
− 1) kT
2
Cv 5 ⎛ hv ⎞
e
= +⎜
⎟
Nk 2 ⎝ kT ⎠ ( e hv / kT − 1)2
hv / kT
E=
assumes no low‐lying
electronically states
5
Nhv
hv ⎞
⎛
NkT + hv / kT
− ⎜ De − ⎟ N
2
e
−1 ⎝
2 ⎠
does not contribute
to heat capacity
pV = NkT
<E>
+AnQ
T
⎡⎛ 2π mkT ⎞3/ 2 Ve5 / 2 ⎤
⎛ 8π 2 IkTe ⎞
S
= An ⎢ ⎜
⎥ + An ⎜
⎟
⎟
2
2
Nk
h
N
h
σ
⎠
⎢⎣⎝
⎥⎦
⎝
⎠
hv / kT
+ hv / kT
− An (1 − e − hv / kT ) + An ωel
e
−1
S=
μ0
8π 2 IkT
hv
⎛ 2π mkT ⎞
= − An ⎜
−
A
+
kT
n
⎟
2
kT
σ h2
2kT
⎝ h
⎠
D
+ An (1 − e − hv / kT ) − e − An ωel
kT
3/ 2
For general polyatomic molecules we don’t have a closed
form for the rot. energy levels. So it would appear that we
can’t simply extend the procedure followed for diatomics.