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Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 22
Statistical Thermodynamics
Calculation of Internal Energy from Molecular Properties
- Suppose the internal energy to system contains (N) from
particles
U U 0   n i ( i   0 )
(1)
i 0
where, U0 the energy at i= 0
U U 0  n 0 (0)  n1 (1   0 )  n 2 ( 2   0 )  .....
- The equation of partition function for the particle at i=0
Q   g i e  (i 0 )/ kT
(3)
i
- By differentiating this eq. with T we will find :
(   )
dQ
  g i e  (i  0 )/ kT i 2 0
dT
kT
i
(4)
By multiply the two sides in kT2
kT
kT
2
2
dQ
  g i e  (i  0 )/ kT ( i   0 )
dT
i
dQ  g i  (i 0 )/ kT  

  e
  ni ( i   0 ) 
dT  i ni

 i
(5)
(6)
- If we substituted Q value from equation (3) and U-U0 from
equation (1) into equation (6)
kT
2
dQ
Q

(U U 0 )
dT N A
[1]
(7)
(2)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 22
- where NA is the Avogadro’s number
- But k is the Boltzmann constant = R/NA
- R=k NA
- Substitute R value in equation (7)
RT 2 dQ
d ln Q
(U U 0 ) 
 RT 2 (
) (8)
Q dT
dT
- Equation (8) is the relation between the internal energy of
statistical thermodynamics and the partition function for the
particle.
- This is clear to us that we can calculate the thermodynamic
energy to any type of motion by the value of partition function
with temperature.
d ln q t q r qv
)
dT
d ln qv
d ln q r
 d ln q t
 RT 2  (
)(
)(
dT
dT
 dT
(U  U 0 )  RT 2 (

)

(9)
 (U U 0 )  (U U 0 )t  (U U 0 )r  (U U 0 )v
- The mathematics uses the partition function in three
dimensions to calculate (U-U 0)t they found:
(U U 0 )trans  RT
2
 2 mkT 3 
d
ln (
)V 
dT 
h

3
 RT
2
[2]
(11)
(10)
Dr.Eman Zakaria Hegazy Quantum Mechanics and Statistical Thermodynamics Lecture 22
And also they calculate (U-U0)rot they found as the same:
(U U 0 ) rot  RT

2
d ln q rot
dT
(12)
3
RT
2
- Also they calculate the internal energy for the vibrational
motion (U-U0)vib
(U U 0 )vib
(U U 0 )vib
d ln qvib RT 2 dqvib
 RT

dT
qvib dT
2
(x / T )e  x
x
 RT

RT
(1  e  x )2
(e x  1)
2
(13)
- This vibration for one molecule but the energy of multi
molecules is
(U U 0 )vib 
3n 6
 RT
i
xi
(14)
(e x i  1)
- The total internal energy is summation of equations (11),(12)
and (14)
3n 6
x
3
3
(U U 0 )  RT  RT   RT x i i
(15)
2
2
(e  1)
i
(U U 0 )  3RT 
3n 6
 RT
i
xi
(e  1)
(16)
xi
[3]