Statistical Thermodynamics ?

Introduction To Statistical Thermodynamics -1

고려대학교 화공생명공학과

강정원

Statistical Thermodynamics ?

Link between microscopic properties and bulk properties

Microscopic Properties

T,P

U,H,A,G,S μ,Cp,…

Microscopic Properties

Statistical Thermodynamics

Molecular Properties Thermodynamic Properties

u_ij

r

) 1 ( +

=hcBJ J

E_J ^Eν^=(v+₂^1)hcv~

2 2 2

8mX h E_n = n

Potential Energy

Kinetic Energy

Two Types of Approach for Microscopic View

Classical Mechanics

– Based on Newton’s Law of Motion

Quantum Mechanics

– Based on Quantum Theory

) ,...,

,

2 ( ^{1 2 N}

i i

i U

m

H ⁼





−

i

i i

E m U

h ₂

2 2

8π

Hamiltonian

Schrodinger’s Wave Equation

Solutions

Using classical mechanics, values of

position and momentum can be found as a function of time.

Using quantum mechanics, values of

allowed energy levels can be found. (For simple cases)

Purpose of statistical thermodynamics

Assume that energies of individual molecules can be calculated.

How can we calculate overall properties (energy, pressure,…) of the whole system ?

Statistical Distribution

n : number of object

b : a property (can have 1,2,3,4, … discrete values)

b n_i

1 2 3 4 5 6

if we know “Distribution”

then we can calculate the average value of b

Normalized Distribution Function

 Probability Function





>=

<

>=

<

i

i i i

i i

P b

F b

F

P b b

) ( )

b (

P_i

b₁ b₂ b₃ b₄ b₅ b₆





=

=

=

i

i i

i

i i

i i i

i i

i

b P

b n

b n n

b b n

P

1 )

(

) (

) ( )

) (

b : energies of individual molecule,… (

F(b) : internal energy, entropy, …

Finding probability (distribution) function is the main task in statistical thermodynamics

The distribution of molecular states

Quantum theory says ,

– Each molecules can have only discrete values of energies

Evidence

– Black-body radiation – Planck distribution – Heat capacities

– Atomic and molecular spectra – Wave-Particle duality

Configurations…

Instantaneous configuration

– At any instance, there may be n_o molecules at ε₀ ^{, n}1 molecules at ε₁ ^{, n}2 molecules at ε₂ ^,

…  {n₀ , n₁ , n₂ …} configuration

ε

0

ε₂ ε

1

ε₃ ε₄ ε

5

{ 3,2,2,1,0,0}

Weight ….

Each configuration can be achieved in different ways …

Example1 : {3,0} configuration  1

Example2 : {2,1} configuration  3

ε

0

ε

1

ε

0

ε

1 ε

0

ε

1 ε

0

ε

1

Weight …

Weight (W) : number of ways that a

configuration can be achieved in different ways

General formula for the weight of {n₀ , n₁ , n₂ …} configuration

=

∏

=

i

ni

N n

n n W N

!

!

!...

!

!

!

3 2

1

Example1

{1,0,3,5,10,1} of 20 objects W = 9.31E8

Example 2

{0,1,5,0,8,0,3,2,1} of 20 objects W = 4.19 E10

Principle of equal a priori probability

All distributions of energy are equally probable

If E = 5 and N = 5 then

0 1 2 3 4 5

0 1 2 3 4 5

0 1 2 3 4 5

All configurations have equal probability, but possible number of way (weight) is different.

The dominating configuration

For large number of molecules and large number of energy levels, there is a dominating configuration.

The weight of the dominating configuration is much more larger than the other configurations.

Configurations W_i

{n_i}

The dominating configuration

0 1 2 3 4 5

0 1 2 3 4 5

0 1 2 3 4 5

W = 1 (5!/5!) W = 20 (5!/3!) W = 5 (5!/4!)

Difference in W becomes larger when N is increased !

Stirling’s Approximation

A useful formula when dealing with factorials of numbers.

N N

N

N ! = ln − ln

i i

i

i i i

i i

i

i

n n

N N

n n

n N

N N

n n N

n n W N









−

=

+

−

−

=

−

=

=

ln ln

ln ln

! ln

!

!... ln

!

! ln ! ln

3 2 1

The Boltzmann Distribution

Task : Find the dominating configuration for given N and total energy E.

 Find Max. W which satisfies ;





=

=

i

i i i

i

n E

n N

ε ⁰

0

=

=





i

i i i

i

dn dn ε

Method of Undetermined Multiplier

Maximum weight , W

Recall the method to find min, max of a function…

Method of undetermined multiplier :

– Constraints should be multiplied by a constant and added to the main variation equation.

ln 0

0 ln

 =



 



 ∂

= dni

W W d

Method of undetermined multipliers

ln 0 ln ln

 =





 + −



 



=  ∂

−

 +



 



=  ∂









i

i i

i

i

i i i

i i

i i

dn dn W

dn dn

dn dn W W

d

βε α

ε β

α

ln 0

=

−

 +



 



 ∂

i

dni

W α βε





∂

− ∂

∂

= ∂



 





∂

∂

−

=

j i

j j

i i

i i

n n n

n N N

n W

n n

N N

W

) ln ln (

ln

ln ln

ln

1 1 ln

ln ln

+

 =



 





∂

× ∂

 +



 





∂

= ∂

∂

∂ N

n N N N

n N N n

N N

i i

i



















 





∂

× ∂

 +



 





∂

= ∂

∂

∂

j

i i

j j

j j

i j

j i

j

j n

n n n n

n n n n

n

n 1 ln 1

) ln ln

(

N N n

n n

W _i

i i

ln )

1 (ln

) 1 ln (ln

−

= +

+ +

−

∂ =

∂

0

ln + + =

− i

i

N

n α βε

e i

N

n_{i α−βε}

=



 

−

−

=

=

=

j

j j

j

j

j

e e

e Ne

n N

α βε

α βε

1



= −

=

j i

i _j

i

e e N

P n _βε

βε

Boltzmann Distribution (Probability function for

energy distribution)

The Molecular Partition Function ( 분자 분배 함수 ₎

Boltzmann Distribution

Molecular Partition Function

Degeneracies : Same energy value but different states (g_j-fold degenerate)

q e e

e N

p n

i

j i

j i

i

βε βε

βε −

−

− =

=

= 

 ⁻

=

j

e j

q ^βε

 ⁻

=

j levels

j

e j

g

q ^βε

An Interpretation of The Partition Function

Assumption :

T 0 then q  1

T infinity then q  infinity

The molecular partition function gives an indication of the average number of states that are thermally accessible to a molecule at T.

kT /

= 1

β

An example : Two level system

Energy level can be 0 or ε

e kT

e

q =1+ ^−βε =1+ ^−ε/

Example

Energy levels : ε, 2ε, 3ε , ….

βε βε

βε

−

−

−

= − +

+ +

= e e e

q 1

... 1

1 ²