Lecture 11 Statistical Thermodynamics.pdf

Statistical Thermodynamics

(Lecture 11)

1

^st

semester, 2021

Advanced Thermodynamics (M2794.007900) Song, Han Ho

(*) Some materials in this lecture note are borrowed from the textbook of Ashley H. Carter.

è

Establish the tie between macroscopic and molecular properties

è

Stochastic (probabilistic) approach à Focus on equilibrium states

è

Fundamental questions

l

What are the energy states available to the particles (atoms/molecules) comprising a macroscopic system?

l

How are these particles distributed among these states at equilibrium to satisfy the macroscopic constraints?

Introduction

Statistical Thermodynamics

Quantum Mechanics (accessible energy states)

E N

N N

j

j j j

j

=

=

å å

e

Statistical Mechanics (distribution among states)

Statistical Thermodynamics (microscopic)

Classical Thermodynamics

è

Let’s apply some elementary concepts of statistical thermodynamics to a coin-tossing experiment.

(Assumption) Coins are distinguishable.

Coin-Tossing Experiment

N₁: number of heads

N₂: number of tails (N₂ = N – N₁) w_k: thermodynamic probability

=number of microstates in macrostate k

s microstate of

number total

: or

where ^macro

1

W

= W W

=

å

= N

k k k

k w w

P

Statistical Thermodynamics

è

Continue on.

Macrostate (or configuration) is specified by the number of particles in each of the energy levels of the system. (= thermodynamic state in the classical theory)

Microstate is specified by the number of particles in each energy state. In general, there are more than one energy state (i.e. quantum state) for each energy level (or the energy level has degeneracy.)

First Postulate of Statistical Mechanics

“In equilibrium, each microstate can occur with an equal probability.”

Coin-Tossing Experiment

Statistical Thermodynamics

è

Continue on.

The average occupation number is

Let’s generalize the coin-tossing experiment with a larger number of coins.

How many ways are there to select from the N candidates N₁ heads and N-N₁ tails with distinguishable coins?

Coin-Tossing Experiment

( ) ( ) ( ) ( ) ( )

[ ]

heads) of

number average

(

2 1

0 4

1 6 2 4

3 1

16 4 1

.

. ₁

k

®

=

´ +

´ +

´ +

´ +

´

=

W =

=

=

å å

å å

N g e

P N w

N w

w N N

k

k jk k

k jk

k k

k jk j

)!

(

!

!

1

1k k

k

N N N

w N

= -

Statistical Thermodynamics

è

Continue on.

Calculate the maximum thermodynamic probability for different number of coins.

(use Stirling’s formula, for large n)

Coin-Tossing Experiment

! 6 2

! 2

! 4 4

max = =

= w N

! 70 4

! 4

! 8 8

max = =

= w N

300

max 10

! 500

! 500

! 1000 1000

»

=

= w N

n n n

n ! » ln - ln

The peak always occurs at N/2, but grow rapidly and becomes much sharper as N increases.

In other words, the most probable configuration is that of total randomness.

Statistical Thermodynamics

è

Continue on.

The “ordered regions” almost never occur. (w is extremely small compared to w_max) This leads to the very important conclusion!

i.e. The total number of microstates is very nearly equal to the maximum number.

For thermodynamic problem, the “outcomes” are the occupation numbers of each of n energy levels for the most probable macrostate, i.e. w_max(N₁, N₂, … , N_n). This most probable macrostate is the equilibrium state of the thermodynamic system.

In extending our formula of tossing coins from two levels to n levels,

Coin-Tossing Experiment

w

max

w

_k

»

= W å

Õ

=

=

= _n

j

n Nj

N N

N N w N

1 2

1 !

!

!

!...

!

!

Statistical Thermodynamics

è

The fundamental problem of statistical thermodynamics is to determined the most probable state or the equilibrium state, in given constraints.

à Method of Langrange multiplier

As a system proceeds toward a state of equilibrium, the entropy increases, and at equilibrium attains its maximum value. (Classical point of view)

The system tends to change spontaneously from states with low thermodynamic probability to states with high probability. (Statistical point of view)

Assembly of Distinguishable Particles

energy) of

ion (conservat

particles) of

ion (conservat

: s constraint

under the

) ,..., ,

(

Find

1 1

2 1 max

U N

N N

N N

N w

n j

j j n j

j

n

=

=

å å

=

=

e

Statistical Thermodynamics

è

Boltzmann made the connection between the classical concept of entropy and the thermodynamic probability.

The entropy is an extensive property. Therefore, the combined entropy of the two subsystems is simply the sum of the entropies of each subsystem.

From the characteristic of the thermodynamic probability,

Then,

Finally,

Thermodynamic Probability and Entropy

) ( w f

S =

) (

) (

) (

or

_{total A B}

B A

total

S S f w f w f w

S = + = +

B A

total

w w

w =

) (

)

( w

_total

f w

_A

w

_B

f =

) (

) (

)

( w

_A

f w

_B

f w

_A

w

_B

f + =

Statistical Thermodynamics

è

Continue on.

The only function to satisfy the above condition is the logarithm.

Thermodynamic Probability and Entropy

w k

S = ln

) (

) ( )

(w_A f w_B f w_Aw_B

f + =

) 10

(1.38 constant

Boltzmann :

wherek ´ ^-23JK^-1

Statistical Thermodynamics

è

Schrӧdinger’s equation: the governing equation for the dynamical behavior of matter on the atomic scale (just as Newton’s law for macroscopic particles)

è

Time independent Schrӧdinger’s equation

è

The equation has solutions only for specific values of , termed the eigenvalues of the energy.

Quantum States and Energy Levels

( ) 0

8

2

2 2

÷÷ ø - =

çç ö è + æ

Ñ y p e f y

h

m

Time independent Schrӧdinger’s Equation

( ) f

e = kinetic energy + potential energy = 1 / 2 p

²

/ m +

particle) a

finding of

density y

probabilit :

|

| ( function

wave y

²

y =

z y

xy y

y y =

z y

x e e

e

e = + +

z y

x f f

f

f = + +

) (e -f

Statistical Thermodynamics

è

Apply Schrӧdinger’s equation to the translational kinetic energy states of a dilute monatomic ( ) particle in the absence of

external force fields (“particle in a box” à on the boundary)

è

Substituting into Schrӧdinger’s equation and separating variables,

with the boundary conditions:

Quantum States and Energy Levels

Statistical Thermodynamics

only

,

0 e e

_tr

f = =

= 0 y

x y

z a

b c

) ( ) ( )

( x

_tr

y

_tr

z

tr

y y

y y =

) ( )

( )

( x

_tr

y

_tr

z

tr

tr

e e e

e

e = = + +

) 8 (

) ( )

( 1

2 2 2

2

h x m dx

x d

x ^tr

tr tr

p e y

y ^÷÷_ø

çç ö è -æ

=

0 ) ( )

0

( = _tr a =

tr y

y

è

The solution to the equation,

è

Applying B.C.

è

Similarly for y and z:

è

Total translational energy:

Quantum States and Energy Levels

Statistical Thermodynamics

) 8 (

) ( )

( 1

2 2 2

2

h x m dx

x d

x ^tr

tr tr

p e y

y ^÷÷_ø

çç ö è -æ

=

0,

0 ) 0

( = Þ B =

ytr

÷÷ ø ö çç

è

æ +

= B

h

x x m

A

x ^tr

tr 2

2 ( )

sin 8 )

( p e

y

) 0 ( sin 8

0 )

( ₂

2 ÷÷ =

ø ö çç

è Þ æ

= h

x a m

A

a ^tr

tr

e y p

3,...) 2,

, 1 (

8 )

( )

( 8

2 2 2

2

2 ÷÷ø =

ö ççè

÷÷æ ø çç ö

è

=æ Þ

= _{x tr} ^x _x

tr n

a n m x h

h n

x

a p me p e

÷÷ ø ö çç

è æ

÷÷ø çç ö

è

= æ ₂

2 2

) 8

( b

n m

y h ^y

etr _÷÷

ø ö ççè

÷÷æ ø çç ö

è

=æ ₂

2 2

) 8

( c

n m

z h ^z

etr

úú û ù êê

ë é

÷÷ø ö ççè

+æ

÷÷ ø ö çç

è +æ

÷÷ø ö ççè

÷÷ æ ø çç ö

è

= æ +

+

= ₂

2 2

2 2

2 2

) 8 ( )

( )

( c

n b

n a

n m

z h y

x _{tr tr} ^{x y z}

tr e e

e e

è

The total solution:

è

Let

è

For , there are numerous combinations of à numerous quantum states with the same energy à the energy levels are degenerate

à degeneracy = number of quantum states with the same energy =

Quantum States and Energy Levels

Statistical Thermodynamics

úû ê ù

ë

é ÷

ø ç ö

è ú æ

û ê ù

ë

é ÷÷ø

çç ö è ú æ

û ê ù

ë

é ÷

ø ç ö

è

= æ

= c

z A n

b y A n

a x A n

z y

x _{tr tr x} ^x _y ^y _z ^z

tr

p p y p

y y

y ( ) ( ) ( ) sin sin sin

1 ,

, n energy whe slational

point tran -

zero finite

a is there

box) a

in particle (no

, , all for 0

0 ,

, of any If

= Þ

= Þ

=

z y x z

y x

n n n z

y x n

n

n

y

3 /

V

1

c b

a = = =

(

^{2 2 2}

)

3 / 2 2

8

^{x y z}

tr

n n n

mV

h ÷÷ + + ø

çç ö è

= æ e const

tr

=

e

ⁿ_x^,n_y^,n_z

g

tr

Quantum States and Energy Levels

Statistical Thermodynamics

6

=

Þ g

_tr

3

or 2 , 1 ,

,

_{y z}

=

x

n n

n

n

_x

n

_y

n

_z

1 2 3

1 3 2

2 1 3

2 3 1

3 2 1

3 1 2

) (or

14

_tr

2 2

2

n n fixed

n

_x

+

_y

+

_z

= e =

3 / 2 2 2

2 2

3 / 2 2

tr

8

) 14

8 ( mV

n h n

mV n h

z y

x

+ + =

e =

1 ,

,

_{y z}

=

x

n n

n 1

8 ) 3 8 (

_2/3

2 2 2

2 3

/ 2 2

tr

= + + = Þ =

Þ

_{x y z}

g

_tr

mV n h

n mV n

e h

For zero-point energy,

è

In quantum theory, the energy levels are discrete, but in classical physics, the energy levels are continuous. Why is the discrepancy?

Let’s consider translational energy mode.

At room temperature, the mean kinetic energy of a helium gas atom is,

Then, the nominal quantum number n_j for one-liter volume of helium gas is,

This implies that there are so many energy levels below this small energy value (6.2x10^-21J), and thus the energy levels are very closely spaced and may be treated as an energy continuum.

Density of Quantum States

Statistical Thermodynamics

( )

² ₂ ^{2/3 1/2 9}

3 / 2 2

10 7 . 8 2

8 ÷÷ » ´

ø çç ö

è

æ ´

=

÷÷ ® ø çç ö

è

= æ

h n mV

mV n

h _tr

j j

tr

e e

J

tr kT

21 23) 293 6.2 10 10

38 . 1 2 ( 3 2

3 _{- -}

´

=

´

´

´

= e =

è

Under the conditions that the quantum numbers are large and the

energy levels are very close together, we can regard the n’s and the ɛ’s as continuous functions.

Let’s evaluate the density of states g(ɛ), or degeneracy.

The possible combination (n_x, n_y, n_z) in a given ɛ_tr,corresponds to points on a sphere of radius R in (n_x, n_y, n_z) space.

Thus,

Density of Quantum States

Statistical Thermodynamics

2 2 tr

3 / 2 2

2

2 2 8

h R n mV

n n

n = _x + _y + _z = e º

e e e

e e e

e e

e e

e e

d d ) ( dN

) ( N .

T . O . H d d

) ( ) dN (

N

) ( N ) d (

N d

) ε ( g

»

úû- êë ù

é + +

=

- +

=

è

Continue on.

Define N(ɛ), as the number of states contained within the octant of the sphere of radius R.

Then,

To complete our discussion, we need to consider additional energy states regarding spin!

Density of Quantum States

Statistical Thermodynamics

2 3 tr 2 3 2

3 8

6 3

4 8

1 /

/

h V m R

) (

N e p p _÷ e

ø ç ö

è

= æ

×

=

e p e

e e e e

e m d

h d V

d ) ( d dN

) (

g = = 4 2₃ ^3/2 _tr^1/2

bosons zero

spin for

1

2 ) 4

( ₃ ^3/2 _tr^1/2

=

=

=

s

s m d

h d V

g g g

e p e

g e e