15.4 Rotational modes of diatomic
molecules
The moment of inertia,
where μ is the reduced mass
r0 is the equilibrium value of the distance
between the nuclei.
From quantum mechanics, the allowed angular
momentum states are
where
l = 0, 1, 2, 3… …
From classical mechanics, the rotational energy
equals
with w is angular velocity.
The angular momentum
therefore, the energy
Define a characteristic temperature for rotation
θrot can be found from infrared spectroscopy
experiments, in which the energies required to
excite the molecules to higher rotational states
are measured.
Different from vibrational motion, the energy
levels of the above equation are degenerate.
for level
Now, one can get the partition function
For
, virtually all the molecules are in
the few lowest rotational states.
As a result, the series of
can be
truncated with negligible errors after the first
two or three terms!
For all diatomic gases, except hydrogen, the
rotational characteristics temperature is of the
order of 10 k (Kelvin degree).
At ordinary temperature,
Therefore, a great many closely spaced energy
states are excited. The sum of
may be
replaced by an integral.
Define:
Note that the above result is too large for
homonuclear molecules such as H2, O2 and N2
by a factor of 2… why?
The slight modification has no effect on the
thermodynamics properties of the system such
as the internal energy and the heat capacity!
Using
  ln Z 
U  NkT 

 T V
2
 U 
CV , rot  
  Nk
 T v
At low temperature
Keeping the first two terms
(Note:
again
)
Using the relationship
2   ln Z 
2
U  NkT 
  NkT  3  e
 T V
2 rot
T
  rot 
2 2 
T 

(for
And
 U 
CV ,rot  
  6 Nk rote
 T V
2 ro t
T
)
  rot 
2 2 
T 

(for
)
Characteristic Temperatures
Characteristic temperature
of vibration of diatomic
molecules
Substance
θvib(K)
H2
6140
O2
2239
N2
3352
HCl
4150
CO
3080
NO
2690
Cl2
810
Characteristic temperature
of rotation of diatomic
molecules
Substance
θrot(K)
H2
85.4
O2
2.1
N2
2.9
HCl
15.2
CO
2.8
NO
2.4
Cl2
0.36
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,
  ln Z 
U  NkT 
 0
 T V
2
 U 
CV  
 0
 T V
At practical temperature, electronic excitation
makes no contribution to the external energy
or heat capacity!
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
3
 7
 2mk  2 T 5 2  
S  Nk   ln  2 
 ,
 2
 h  2 rot  
• Solution: For diatomic molecules
S  S trans  S vib  S rot  S excit
At room temperature
S excit  0
1
1 


Nk   

 e  T
T
2
e  1   Nk ln 

S vib 
 1  e  T
T

 0 (does not contribute!)




• For translational motion, the molecules are
treated as non-distinguishable assemblies
3
U  NkT
2
three degrees of freedom 
3
 2mkT  2
Z V

2
 h

U
S tr   Nk ln Z  ln N  1
T
3


5
V   2mkT  2 
S tr  Nk  Nk ln 

 
2

2
 N  h


 
• For rotational motion (they are distinguishable
in terms of kinetic energy)
U  NkT
Z
1

 due to homonuclea r molecule 
2

T
2 rot
U
 NkT ln Z for distinguis hable assembly 
T
T
 Nk  Nk ln
2 rot
S
 
5
 V
S system  Nk  Nk ln
2
N

 
V
7

S system  Nk  Nk ln 
2
N


  2mkT  3 2  
T




 Nk  Nk ln

  h 2   
2 rot

 

  2mk 3 / 2 T 5 2 




  h 2  2 rot 



• 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!
 vib  3352k and  r  2.9k for N 2
U  U trans  U vib  U rot
1
3
1 

  NkT
U  NkT  Nk   vib   

2
 2 e T 1 
because  r T
U
1

5
1

NkT  Nk   vib   3352

2
 2 e 298  1 
1
3352 

 Nk  2.5  298   3352  11.25 
2
e
1 

3352 

 Nk  2272  11.25 
e
1 

 1.0kmol  8.314 103 J  kmol1  k 1  2272
 1.89 10 7 J
F   NkT ln Z  ln N  1 non  distinguis hable particles 
F   NkT ln Z Distinguis hable particles such as vib & rotation 
Ft   NkT ln Z t  ln N  1
 2mkT 
Zt  V 

2
 h

3
2
 2  4.65  10 1.38110
 22.4m 
2
 68
6
.
626

10

 26
3


 22.427.3  10 
 22.4 273.86  10
19
3
2
20
3
 23
Jk  298 


1
3
2
2
3
2
 3.195 1033
 3.195  1033 
6

Ft   NkT  ln

1


NkT
ln
5
.
15

10
1
26

 6.02 10

  NkT  16.4
 
 
Fvib   NkT ln Z vib
Z vib 
e
3352
1 e
2298
 3352
298
e 5.62

1  e 11.24
1 

ln Z vib  5.62  ln 1  11.24   5.62
 e

T
Fvib   NkT ln Z rot Z rot 
for T  rot
 rot
298
 102.72 ln Z rot  4.63
2.9
Frot  4.63 NkT
Z rot 
F  Frot  Fvib  Ft
 4.63 NkT  5.62 NkT  16.4 NkT
 15.41NkT
 15.41 6.02 10 26  1.38  10  23  298
 3.817  107 J