Transcript 8 Elementary statistical thermodynamics

8.2 Energy level and its degeneracy
2
1
0
1
2
3
3
4
5
Energy levels are said to be degenerate, if the same energy
level is obtained by more than one quantum mechanical state. They
are then called degenerate energy levels.
The number of quantum states at the same energy level is called
the degree of degeneracy.
N   ni
i
U   ni i
i
A molecular energy state is the sum of
an electronic (e), nuclear (n), vibrational
(v), rotational (r) and translational (t)
component, such that:
  t  r  v  e  n
The degree of freedom of movement
• Translation: x,y,z F=3
Rotation
• For linear molecules, F=2
• For non-linear molecules, F=3
Vibration
• A polyatomic molecule containing n
atoms has 3n degrees of freedom
totally. Three of these degrees of
freedom can be assigned to
translational motion of the center of
mass, two or three to rotational motion.
• 3n-5 for a linear molecule;
• 3n-6 for a nonlinear molecule
• CO2 has 3×3-5 = 4 degrees of freedom of
vibration; nonlinear molecule of H2O has 3×3-6
= 3 degrees of freedom of vibration.
8.2.1 Translational particle
The expression for the allowed translational energy levels of a
particle of mass m confined within a 3-dimensional box with
sides of length a, b, c is
2
2
x
2
n
2
y
2
2
z
2
n
h n
t 
(   )
8m a
b
c
Where h＝6.626×10-34J·s，nx, ny, nz are integrals called
quantum numbers. The number of them is 1,2,…∞ .
2
If a=b=c,
equation becomes
h
2
2
2
t 
(
n

n

n
)
x
y
z
3/ 2
8mV
all energy levels except ground energy level
are degenerate.
Example At 300K, 101.325 kPa, 1 mol
of H2 was added into a cubic box. Calculate
the energy level εt,0 at ground state, and the
energy difference between the first excited
state and ground state.
Solution Take the H2 at the condition as an ideal gas,
then the volume of it is
nRT 1 8.3145  300
V

 0.02462 m3
p
101325
The mass of hydrogen molecule is
m  M / L  2.0158 103 / 6.022 1023  3.347 1027 kg
 t,0
h2
3  (6.626 1034 )2
40


3


5.811

10
J
2/3
27
2/3
8mV
8  3.347 10  0.02462
h2
40
 t,1 

6

11.622

10
J
2/3
8mV
   t,1   t,0  (11.622  5.811) 1040  5.8111040 J
the energy difference is so small that the translational particles
are excited easily to populate on different excited states, and
that the energy changes of different energy levels can be think
of as a continuous change approximately.
8.2.2 Rigid rotator (diatomic)
The equation for rotational energy level of diatomic molecules is ：
r 
h2
8 I
2
J ( J  1)
J  0，
1，
2，

where J is rotational quantum number, I is the moment of
inertia (转动惯量)
m1m2
2
2
I (
) R0   R0
m1  m2
μ is the reduced mass (折合质量), The degree of degeneracy is
g r,J  2 J  1
8.2.3 One-dimensional harmonic oscillator
1
 v  ( v  )h
2
v  0,1, 2, 
Where v quantum number，when v=0，the energy
is called zero point energy.
 v,0
 v,2
1
 h
2
5
 h
2
3
 v,1  h
2
7
 v,3  h
2
One dimensional harmonic vibration is non-degenerate.
8.2.4 Electron and atomic nucleus
The differences between energy levels of electron
motion and nucleus motion are big enough to keep the
electrons and nuclei stay at their ground states.
Both degree of degeneracy, ge,0, for electron motion
at ground state and degree of degeneracy, gn,0, for
nucleus motion at ground state are different for
different substances, but they are constant for a given
substance.
8.5 Computations of the partition function
• 8.5.1 Some features of partition functions
• (1) at T=0, the partition function is equal to the
degeneracy of the ground state.
q   gi e

i
lim q  g 0
kT
T 0
i
• (2) When T is so high that for each term εi/kT=0,
lim q  
T 
• (3) factorization property If the energy is a sum
of those from independent modes of motion, then
q  qt qr qv qe qn
 i   t , i   r , i   v , i   e ,i   n ,i
gi  gt ,i g r ,i gv ,i g e ,i g n ,i
The partition functions for 5 mode motions are expressed as
qt   gt ,i e

 t ,i
kT
;
i
qe   g e,i e
i
qr   g r ,i e

 r ,i
kT
; qv   g v ,i e
i

 e ,i
kT
;
i
qe   g e,i e
i

 e ,i
kT

 v ,i
kT
q  [ g t,i exp(
 t ,i
i
[ g v,i exp( 
i
[ g n.i exp(
i
 r,i
kT
)]  [ g r,i exp(
kT
)]  [ g e,i exp( 
 v,i
 n,i
kT
i
i
)]
 qt  qr  qv  qe  qn
kT
)] 
 e,i
kT
)] 
8.5.2 Zero-point energy
• zero-point energy is the energy at ground state or
the energy as the temperature is lowered to
absolute zero.
• Suppose some energy level of ground state is ε0,
and the value of energy at level i is εi, the energy
value of level i relative to ground state is
  i  0
0
i
• Taking the energy value at ground state as zero,
we can denote the partition function as q0.
q   gi e
0

 i0
0
kT
q  e kT q
0
i
q e
 t ,0 / kT
q e
 e ,0 / kT
0
t
0
e
 r ,0 / kT
qt ;
q e
qe ;
q e
0
r
0
n
qr ; q  e
 n ,0 / kT
0
v
qn
Since εt,0≈0, εr,0=0, at ordinary temperatures.
q  qt , q  qr
0
t
0
r
 v ,0 / kT
qv
• The vibrational energy at ground state is
 v,0  h
1
2
• therefore
q e
0
v
h /2 kT
qv
• the number of distribution in any levels does not
depend on the selection of zero-point energy.
N
N
N
  i / kT
 ( i0  0 )/ kT
  i0 / kT
ni  gi e
 0  0 / kT gi e
 0 gi e
q
q e
q
8.5.3 Translational partition function
Energy level for translation
2
2
x
2
n y2
h n
nz2
 i ,t 
(  2  2)
8m a
b
c
The partition function
qt   gt ,i exp(
i
 t ,i
kT
)
 h 2  nx2 n y2 n 2 

z
qt     exp 
 2  2  2  / kT 
c 
nx 1 n y 1 nz 1
 8m  a b









h2
h2
h2
2
2
2
  exp  
n  exp  
n  exp  
n
2 x  
2 y  
2 z 
nx 1
 8mkTa
 ny 1
 8mkTb
 nz 1
 8mkTc

 qt , x qt , y qt , z


h
2
qt , x   exp  
n
2 x 
nx 1
 8mkTa


2

h2
2
qt , z   exp  
n
2 z 
nz 1
 8mkTc



h2
2
  exp  
n
2 y 
n y 1
 8mkTb


qt , y
Take qt,x as an example

qt, x
2
x
2
2
n
h
  exp(
 )
8mkT a
nx 1

2
h
2
(设
 ）
2
8mkTa
  exp( n )
2
nx 1
2
x
For a gas
 2 at ordinary temperature α2<<1, the summation
converts into an integral.

qt,x   exp( n )dnx
0
2
2
x
From mathematic relations in Appendix


0
e
 x 2
1  12
dx  ( )
2 
1
1
1
2

mkT
2
qt,x 
( ) 2  (
)
a
2
2
h
q t,y In like
q t,z manner,
2 mkT 3 2
qt  (
)  a b c
2
h
2 mkT 3 2
(
) V
2
h
• Example Calculate the molecular partition
function q for He in a cubical box with sides 10cm
at 298K.
• Solution The volume of the box is V=0.001m3.
The mass of the He molecule is
0.004/(6.022×1023)=6.6466×10-27kg.
Substituting these numbers and the proper natural
constants, we have
 2  6.6466 10 1.38 10
q
34 2
(6.626

10
)

27
23
 298 


3/2
 0.001  7.820 1027
For ideal gas,
M
m
L
nRT NkT
V

p
p


8.2052 107 N  M
1 
kg

mol


qt 
 p 
 Pa 
3
2
T K 
5
2
8.5.4 Rotational partition function
The rotational energy of a linear molecule is given by
εr = J(J+1)h2/8π2I and each J level is 2J+1 degenerate.
 r  J ( J  1)
qr   g r ,i e
i

 r ,i
kT
h2
8 I
2
J  0，
1，
2，


h2 
  (2 J  1) exp  J ( J  1) 2

8

IkT
J 0



define the characteristic rotational temperature
h2
r  2
8 Ik
J ( J  1) r
qr   (2 J  1) exp(
)
T
J 0

Θr<<T at ordinary temperature, The summation can be
approximated by an integral

qr   (2 J  1) exp   J ( J  1)r / T  dJ
0
Let J(J+1)=x, hence J(2J+1)dJ=dx, then

qr   exp( xr / T )dx  T r  8 IkT h
0
2
2
For a homonuclear diatomic molecule, such as O2, it comes back
to the same state after only 180o rotation.
8 2 IkT
qr 

2
 r
h
T
where σ is called the symmetry number. σ is the number of
indistinguishable orientations that a molecule can exhibit by
being rotated around symmetry axis. It is equal to unity for
heteronuclear diatomic molecules and is equal to 2 for
mononuclear diatomic molecules.
For HCl, σ = 1; and for Cl2, σ = 2.
8.5.5 Vibrational partition function
Vibrational energies for one dimensional oscillator are
1
 v  (v  )h
2
v  0,1, 2,   
Vibration is non-degenerate, g=1. The partition function is
qv   gv ,i e
i

v ,i
kT
 
1

  exp    v   h / kT 
2
v 0
 


Define the characteristic vibrational temperature
h
v  , v
k
v
3 v
5 v
qv  exp( )  exp( 
)  exp( 
)  
2T
2T
2T
v
v
2 v
 exp( )  [1  exp( )  exp(
)  ]
2T
T
T
• Characteristic vibrational temperatures are usually
several thousands of Kelvins except for very low
frequency vibrational modes.
 v,H  5986 K
2
 v,CO  3084 K
 v,O  2239 K
2
we cannot use integral instead of summation in the calculation
of vibrational partition function.
At low T，
v
T
to mathematics
 1 ，
exp(
v
T
)  1 ，according
when x  1， 1  x  x     
2
1
1 x
v /2T
1
e
1
v /2T
qv  e


 v /2T
v /T
v /2T
1 x 1 e
e
e
take the ground energy level as zero,
qv0  exp  v,0 / kT   qv 
1
h

1 exp( )
kT
For NO, the characteristic vibrational temperature is
2690K. At room temperature Θv/T is about 9;
the qv0  1 , indicating that the vibration is almost in the
ground state.
8.5.6 Electronic and nuclear partition function
qe  ge,0 exp(
 g e,0 exp(
 e,0
kT
 e,0
kT
)  ge,1 exp(
)[1 
ge,1
g e,0
exp(
 e,1
kT
)  
 e,1   e,0
kT
)  ]
( e,1   e,0 )  400 kJ  mol ,
-1
Energy difference is large, so electrons are generally at ground
state, all terms except first one in the summation expression is
negligible.
qe  ge,0 exp(
 e,0
kT
)
 e ,0 / kT
q e
0
e
qe  ge,0
• If the quantum number of total angular momentum
for electronic motion is j, the degeneracy is (2j+1).
Then the electronic partition function can be
written as
0
e
q  2 j 1
• A rare exception is halide atoms and NO molecule.
The difference between the ground state and the
first excited state of them are not so large, the
second term in the summation has to be
considered.
Nuclear motion
Nuclear motion is always in the ground state at ordinary
chemical and physical process because of large energy
difference between ground and first excited state. Its
partition function has the form of
qn  g n,0 exp(
 n,0
kT
)
q  gn,0  2I  1
0
n
where I is a quantum number of nuclear spin.
8.6 Thermodynamic energy and partition function
Independent particle system:
U   ni i
i
N
  i / kT
ni  gi e
;
q
N
  i / kT
U   gi e
i
i q
(8.48)
8.6 Thermodynamic energy and partition function
 q    
  i / kT  

      gi e

 T V  T  i
 V
1 
  i / kT   i 
  gi e
    2 
 k  T 
i
1
 2  gi e i / kT  i
kT i
 q 
kT     gi e  i / kT  i
 T V i
Substitute this equation into equation (8.48), we have
2
N 2  q 
2   ln q 
U  kT 
  NkT 

q
 T V
 T V
Substitute the factorization of partition function for q
  ln qt qr qv qe qn 
U  NkT 

T

V
2
Only qt is the function of volume, therefore
  ln qt 
2 d ln qr
2 d ln qv
U  NkT 
 NkT
  NkT
dT
dT
 T V
2 d ln qe
2 d ln qn
 NkT
 NkT
dT
dT
 Ut  U r  Uv  Ue  U n
2
If the ground energy is specified to be zero, then
0

0
2  ln q 
U  NkT 

 T V
0
 0 / kT
Substitute q =qe
into this equation, it follows that
U  U  N 0
0
It tells us that the thermodynamic energy depends
on the zero point energy. Nε0 is the total energy of
system when all particles are localized in ground
state. It (denoted as U0) can also be thought of as
the energy of system at 0K. Then,
U  U U0
0
• U0 can be expressed as the sum of different
energies
U  U U U U U
0
0
t
0
r
0
v
U  Ut
U  Ur
U 0
U 0
0
t
0
e
0
r
0
n
0
e
0
n
Nhv
U  Uv 
2
0
v
The calculation of
U t0 , U r0 , U v0
• (1) The calculation of
U t0
  ln qt 
U  U t  NkT 

 T V
0
t
2
3/2


 2 mkT 
  ln 
 V
2
3
h


2 

 NkT
＝ NkT


T
2



V
The calculation of
U r0
  ln qr 
U U r  NkT 

 T V
T
d ln
 r
2
 NkT
＝NkT
dT
The degree of freedom of rotation for diatomic or
0
r
2
linear molecules is 2, the contribution to the energy of
every degree is also ½ RT for a mole substance.
The calculation of U
0
v
1
d ln
0
v / T
d
ln
q
1 e
v
U v0  NkT 2
 NkT 2
U
dT
dT U
1
 Nk v v /T
e
1
0
v
0
v
Usually, Θv is far greater than T, the quantum effect
of vibration is very obvious. When Θv/T>>1, U v0  0
Showing that the vibration does not have contribution
to thermodynamic energy relative to ground state.
• If the temperature is very high or theΘv is very
small, thenΘv/T<<1, the exponential function can
be expressed as
v /T
e
U  Nk v
0
v
1
v /T
e
1
 1
 Nk v
v
T
1
1
v
T
 NkT
1
• For monatomic gaseous molecules we do
not need to consider the rotation and
vibration, and the electronic and nuclear
motions are supposed to be in their ground
states. The molar thermodynamic energy is
3
U m  RT  U 0,m
2
• For diatomic gaseous molecules vibration and
rotation must be considered. If only lowest
vibrational levels are occupied, the molar
thermodynamic energy is
5
U m  RT  U 0,m
2
(U  0)
0
v
• If all vibrational energies are equally accessible,
the molar thermodynamic energy for vibration is
U  RT
0
v
• The molar thermodynamic energy for diatomic
molecules is then
7
U m  RT  U 0,m
2
(U  RT )
0
v
8.7 Heat capacity and partition function
• The molar heat capacity, CV,m, can be derived
from the partition function.
   2   ln q   
CV ,m    RT 
 
 T   T V  V
 U m 
CV ,m  

 T V
Replace q with
0  0 / kT
qq e
     ln q0   
 
CV ,m    RT 2 
 T  

T
V  
  
V
We can see from above equations that heat capacity does not
depends on the selection of zero point of energy.
• Electrons and nucleus are in ground state

   2   ln qt0   



CV ,m    RT
 T   

T
V  

  
V
 CV ,t  CV ,r  CV ,v


T
 2   ln qr0   
 RT 
 

 T V  T
 2   ln qv0  
 RT 
 

T


V 
The calculation of CV,t, CV,r and CV,v
• (1) The calculation of CV,t
2 mkT 3 2
qt  (
) V
2
h
     ln q0   
CV ,m    RT 2  t   
 T 

T
V  
  
V
3
 R
2
• (2) The calculation of CV,r
8 2 IkT
T
qr 

2
h
r
CV ,r
(linear molecules)
0


 

ln
q

2
r

 RT 
 R
T 
 T V 
If the temperature is very low, only the lowest rotation state is
occupied and then rotation does not contribute to the heat
capacity.
The calculation of CV,v
q e
0
v
CV ,v
 v,0
kT
qv 
1 e
 v
T
0


d
2   ln qv 

 RT 
 
dT 
 T V V
 v 
 R
 e
 T 
2
CV , v
1
v
T

e

v
T

 1

2
Generally, Θv/T>>1, equation becomes
 v 
 R
 e
 T 
2
CV ,v
v
T
e


v
2
T
 v 

 1  R 
 e

 T 
2
v
T
e


v
T



  v   v T
 R
0
 e
 T 
2
It shows that under general conditions, the contribution to heat
capacity of vibration is approximately zero.
2
When temperature is high enough,
e
v
T
 v 
 1 

T 

 v 
 R
 e
 T 
2
CV ,v
v
2
T
v
 v 
T

Re
R


 T 
In gases, all three translational modes are active and their
contribution to molar heat capacity is
3
CV  R
2
The number of active rotational modes for most linear
molecules at normal temperature is 2
CV  2  12 R  R
In most cases, vibration has no contribution to the heat capacity,
CV  CV ,t
5
 CV ,v  0  R
2
8.8 Entropy and partition function
8.8.1 Entropy and microstate
Boltzmann formula
S  k ln 
k = 1.38062×10-23 J K-1
As the temperature is lowered, the Ω, and hence the S of the
system decreases. In the limit T→0, Ω=1, so lnΩ=0, because
only one configuration is compatible with E=0. It follows
that S→0 as T→0, which is compatible with the third law of
thermodynamics.
ln   ln WD  ln Wmax
D
S  k ln Wmax
• For example
48
10
 0.01
50
10
ln1048
 0.96  1
50
ln10
When N approaches infinity,
ln Wmax
1
ln 
8.8.2 Entropy and partition function
• For a non-localized system, the most probable distribution
ni
number is
g
WD  
i
i
ni !
ln WD   (ni ln gi  ln ni !)
i
• Using Stirling equation ln N!=N ln N - N and Boltzmann
distribution expression
•
N
  i / kT
ni  gi e
q
• We have,
ln WB   (ni ln gi  ni ln ni  ni )
i
ni i
N
  (ni ln gi  ni ln  ni ln gi 
 ni )
q
kT
i
q U
 N ln 
N
N kT
q U
S  k ln WB  Nk ln   Nk
(non-localised system)
N T
or
q0 U 0
S  Nk ln 
 Nk
N T
(non-localised system)
• For localized system
S  k ln WB  Nk ln q 
U
T
(localised system)
or
0
U
S  Nk ln q 0 
T
(localised system)
• Entropy does not depend on the selection of zero point
energy .
• Factorizing the partition function into different
modes of motions and using
U  U U U U U
0
0
t
0
r
0
v
0
e
• We can give
S  St  Sr  Sv  Se  Sn
0
n
For identical particle system, entropies for every mode of
motion can be expressed as
qt0 U t0
St  Nk ln 
 Nk
N T
0
r
0
r
q U
Sr  Nk ln 
N T
qe0 U e0
Se  Nk ln 
N T
0
v
0
v
0
e
0
e
q U
Sv  Nk ln 
N T
q U
St  Nk ln 
N T
8.8.3 Calculation of statistical entropy
• At normal condition electronic and nuclear
motions are in ground state, and in general
physical and chemical process the contribution to
the entropy by two modes of motion keeps
constant. Therefore only translational, rotational
and vibrational entropies are involved in
computation of statistical entropy.
S  St  Sr  Sv
8.8.3 Calculation of statistical entropy
(1) Calculation of St
2 mkT 3 2
q  qt  (
) V
2
h
3
U  NkT
2
0
t
0
t
qt0 U t0
St  Nk ln 
 Nk
N T
St
2 mkT 

 Nk ln
Nh3
3/2
V
5
 Nk
2
• For ideal gases, the Sackur–Tetrode equation is
used to calculate the molar translational entropy.
S m ,t
5
3

1
 R  ln  M / kg  mol   ln(T / K)  ln( p / Pa)  20.723
2
2

(2) Calculation of Sr
• For linear molecules
qr0  qr  T / r
U r0  NkT
• When all rotational energy levels are accessible
qr0 U r0
Sr  Nk ln 
N T
• We obtain
Sr  Nk ln(T / r )  Nk
S m ,r  R ln
T
 r
R
(3) Calculation of Sv
• Substitute

q  1 e
0
v
v / T

1

and U  Nk r e
0
v
v / T

1
1
• Into the following equation
Sv  Nk ln q  U / T
0
v
0
v

v /T


 v / T

 Nk ln 1  e
Sm,v  R ln 1  e
1
1
 Nk vT
 R vT
1
1
e
e
v / T
 v /T

1

1
1
1
residual entropy
• in some the experimental entropy is less than the
calculated value. One explanation to this
discrepancy is that the experimental system does
not reach a real state of equilibrium. In other
words, some disorder is present in the solid even
at T = 0 K. In this case, the entropy at T = 0 is then
greater than zero. This difference in entropy is
called the residual entropy.
8.9 Other thermodynamic functions and partition
functions
• 1
A, G, H and q
q U


A  U  TS  U  T  Nk ln   Nk 
N T


q
  NkT ln  NkT   kT  N ln q  N ln N  N 
N
  kT  N ln q   N ln N  N    kT  N ln q  ln N !
qN
A   kT ln
N!
A  kT ln q N
(for identical particles)
(localized system)
 A 
  ln q 
G  A  pV
p  
  NkT 


V

V

T

T
  ln q 
N
G  kT ln  q / N !  NkTV 
 (non-localized system)
 V T
  ln q 
G  kT ln q  NkTV 
 (localized system)
 V T
H  U  pV
N
  ln q 
  ln q 
 NkT 
  NkTV 

 T V
 V T
2