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Thermodynamics of solids4. Statistical thermodynamics and the 3rd law

Kwangheon ParkKyung Hee University

Department of Nuclear Engineering

2

4.1. Introduction to statistical thermodynamics

Classical thermodynamics Statistical thermodynamics

SystemU

Q

W

In classical thermodynamics, the atoms or molecules are not considered individually.

In statistical thermodynamics, the states (energies) of individual atoms or molecules are considered, and the thermodynamic properties are obtained based on the information of the individual atoms.

dU Q W

revQdST

1

r

SYS i ii

U n

lnS k

4.1. Introduction to statistical TDs.

3

a b c d

Possible energy state of each atom: 1 , 2

Atoms at the position a, b, c, d in a system.

4.2. Microstate, macrostate, and entropy

A simple example:

Number of possible state= 24 = 16 :number of microstates

4.2. Microstate, macrostate, entropy

4

Macrostate

Energy of macrostate

I

II

III

IV

V

14

1 23

1 22 2

1 23

24

4! 14!

4! 43!1!

4! 62!2!

4! 41!3!

4! 14!

Energy per atomNum

ber o

f microstates

Homework 4.1.Draw the graph of the number of microstates vs. energy per atom when the number of atoms is 10. Assume 1=0, 2=1.


5

Number of microstates in a macrostate for a system containing large number of atoms

… r

n1 n2 n3 n4 … nr

Energy levels

Number of atoms

Number of microstates =1 2 3 4

1

! ! (macrostate J)! ! ! !... ! !

Tot TotJr

ri

i

N Nn n n n n n

Total number of microstates in the system: TotNr

The probability of macrostate J =Tot

JNr


1 21

...r

Tot r ii

N n n n n

Macrostate, J

6

The probability distribution for macrostates has an extremely sharp peak for systems with large numbers of particles and energy levels.

The entropy of an isolated system is a maximum at equilibrium, suggesting a connection between entropy and the number of microstates corresponding to a given macrostate.

lnS k where k is the Boltzmann constant (=R/Navog)


7

Lagrange MultipliersA formal procedure for determining a maximum point in a continuous function subject to one or more constraints.

: continuous function.

at maximum (or minimum)

When x, y, z are independent, changes dx, dy, dz are also independent. Hence at maximum,

‐‐‐(*)

But x, y, z are interrelated to each other,

Then eq.(*) is no longer valid! How can we get the condition for maximum (minimum)?‐‐‐‐‐ > Lagrange multiplier method.

Then, the conditions for maximum (or minimum) f become,

Lagrange multiplier, should be chosen.

( , , )f x y z

0f f fdf dx dy dzx y z

0f f fx y z

( , , ) 0g x y z

0g g gdx dy dzx y z

0f g f g f gdx dy dzx x y y z z

0f gx x

0f g

y y

0f gz z

4.3. Fundamentals for statistical thermodynamic calculation

4.3. Fundamentals for calculation

8

Stirling’s formularln ! lnN N N N

Important integrals

n=0, n=1, n=2, n=3,

2

0

n axx e dx

12 a

12a 3

14 a

2

12a

4.3. Fundamentals for calculation

Homework 4.2. Using the method of Lagrange multiplier method, find the maximum (or minimum) of the following functions. (1) f(x,y)=x2y with the constraint of x+y=12.(2) f(x,y,z)=2x2+4y2+9z2 with the constraint of x+y+z=31(3) f(x,y,z)=xyz with the constraint of 2 3 3 6x y z

Homework 4.3. Check Stirling’s formula by comparing the relative difference between the real result and the approximation of the following calculation; 5!, 20!, 100!

9

4.4. Conditions for equilibrium in statistical thermodynamics

… r

n1 n2 n3 n4 … nr

Energy levels

Number of atoms

Evaluation of entropy1 2

1...

r

Tot r ii

N n n n n

11

1

!ln ln ! ln ! ln ! ln !!

r rTot

Tot i Tot irii

ii

NS k k N n k N nn

1 1

ln ln ln lnr r

Tot Tot Tot i i i Tot Tot i ii i

k N N N n n n k N N n n

1

using r

Tot ii

N n

1 1 1 1ln ln ln ln

r r r rTot i

i Tot i i i ii i i ii Tot

N nk n N n n k n k nn N

1

lnr

i iTot

i Tot Tot

n nS kNN N

4.4. Conditions for equilibrium

10

1 1 1

ln ln ln lnr r r

i i iTot i i i i Tot

i i iTot Tot Tot

n n ndS kN d k d n k d n n n NN N N

dS=?

1

1 1ln lnr

i i i i Tot i i Toti i Tot

k n dn n dn N dn n dNn N

1 1 1

ln lnr r r

ii Tot i i Tot

i i i Tot

nk n N dn dn dNN

1 1Since 1

r ri

Tot ii i Tot

ndN dnN

1

lnr

ii

i Tot

ndS k dnN


11

The constrained maximum in the entropy function

Adiabatic, close wall

10

r

Tot ii

dN dn

1

0r

i ii

dU dn

Find the condition when the entropy of the system is the maximum at given constraints.

Maximum:1

ln 0r

ii

i Tot

ndS k dnN

Constraints:1

0r

Tot ii

dN dn

1

0r

i ii

dU dn

Using the Lagrange multipliers method, 0TotdS dN dU

1 1 1 1

ln ln 0r r r r

i ii i i i i i

i i i iTot Tot

n nk dn dn dn k dnN N

ln 0 where =1, 2, 3, ..., rii

Tot

nk iN


12

ln 0 where =1, 2, 3, ..., rii

Tot

nk iN

where =1, 2, 3, ..., ri

i k k

Tot

n e e iN

Evaluation of the Lagrange multipliers, and

1 1 11

i ir r ri k k k k

i i iTot

n e e e eN

1

1 1 where : partition functioni

kr

k

i

e ZZ

e

1

irk

i

Z e

Then, i

ki

Tot

n eN Z

1 1

ln lni

r r ki

i ii iTot

n edS k dn k dnN Z

1 1 1 1ln ln

r r r ri

i i i i i ii i i i

dS k Z dn dn k Z dn dn dUk

Since dUdST

1T


13

1

i i

i

kT kTi

rTot kT

i

n e eN Z

e

At equilibrium (maximum entropy condition),

i

Tot

nN

ii kT

Tot

n eN


14

4.5. Calculation of the macroscopic properties from the partition function

1

irkT

i

Z e

Partition function, Z:

1 1 1

1 1

ln ln ln

1 1 ln ln

ir r rkT

i ii i i

i i iTot

r r

i i i Toti i

n eS k n k n k n ZN Z kT

n k Z n U kN ZT T

And, F U TS lnTotF N kT Z

And, V

FST

lnln lnTot Tot TotV V

ZS N kT Z N k Z N kT T

4.5. Calculation from partition function

15

lnAnd, ln lnTot Tot TotV

ZU F TS N kT Z T N k Z N kTT

2 lnTot

V

ZU N kTT

22

2

ln lnAnd, 2V Tot TotV V V

U Z ZC N kT N kTT T T

T

FPV

G F PV

H U PV

PP

HCT

4.5. Calculation from partition function

16

A model with 2 energy levels

4.6. Application of the algorithm


17

Einstein model of a crystal

Assume that a crystal is composed of a system of atoms which vibrate as harmonic oscillators all with the same frequency, . And, each oscillator has three degrees of freedom with regard to its direction of vibration. Thus a system of N0 oscillators in a three‐dimensional crystal corresponds to 3N0 linear oscillators.

12i i h

where i is an integer.

3 i iU n 1

2 322 1 ....

ii h

h h h hkT kT kT kT kT kT

i i

Z e e e e e e

2 3 1Since 1 ... if 11

x x x xx

2

1

hkT

hkT

eZe

2

22

ln ln2V Tot TotV V

Z ZC N kT N kTT T

2

23

1

hkT

V Tot hkT

h eC N kkT

e

Because 1 atom corresponds to 3 oscillators in 3‐d space.

Since 1 atom acts as 3 oscillator.

2

2Or, 3 where

1

E

E

TE

V Tot E

T

e hC N kT k

e


18

Homework 4.4. Based on the Einstein model, derive internal energy from the partition function. Then obtain the specific heat capacity by differentiation of the internal energy. Also find the entropy. Consider 1 mole of solid. At Show that CV becomes 3R (Dulong‐Petit law).


ET

19

Debye model of a crystal

Atoms in a crystal are not harmonic oscillators with the same frequency, . They are oscillators coupled with other atoms. The oscillators can have many natural frequencies; however, the total number of frequencies should be 3 NTot. The total number of possible frequencies of the oscillators can have in a crystal, can be expressed as:

33 where is sound wave traveling speed

4 3

cVc

33 the number of total oscillators

4 3 = 3Max m Tot

V Nc

3 23 3 the number per unit range of frequency

3 9 or, =Tot Tot

m m

N N

33 3

3 0 0

where

9 91

1

m m

m Dm

xTotToth x

m DkT

hhx x

kT kT T

N h T x dxU d N kTe

e

Debye temperature:mD

hk


20

At high temperature region (T>D),3 33

2

0 09 9 3

1m mx x

Tot Tot TotxD D

T x dx TU N kT N kT x dx N kTe

(heat capacity per mole)3 3 V TotV

UC N k RT

At low temperature region (T~0K),3 3 33 4 4

0

39 91 15 5

mx

Tot Tot TotxD D D

T x dx T TU N kT N kT N kTe

34

(heat capacity per mole)12

5VV D

U TC RT


21

Monatomic gas model

Z

xlyl

zl21

2mv


22

And,



0PV N kT nRT

24

Random mixing of atoms in solid

0 0( ) [( ) , )N n A nB Solution N n A nB

, ,ln ln lnm A B A B A B A BS S S S k

And, 1A B

0,

0

!( )!n!A B

NN n

0,

0

0 0 0 0 0 0

0 0 0 0

0 0 00 0

00

0 0

0

!ln ln( )!n!

ln ( ) ln( ) ( ) ln

ln ( ) ln( ) ln

ln ( ) ll n( ) ln

( ) ln ( )

n ln

ln

m A B

n N

NS k kN n

k N N N N n N n N n n n n

k N N N n N n n n

k N N N n N n n n

N nnk

n N

n N n kN N

0 00

0 0 0 0

0

( ) ( )ln ln

ln lnm A A B B

N n N nn nNN N N N

S kN X X X X

0

0 0

where, A BN n nX X

N N


25

Homework 4.5. A system containg 500 particles and 15 energy levels is in the following macrostate:

(14, 18, 27, 38, 51, 78, 67, 54, 32, 27, 23, 20, 19, 17, 15)This system experiences a process in which the number of particles in each energy level changes by the following amounts:

(0, 0, ‐1, ‐1, ‐2, 0, +1, +1, +2, +2, +1, 0, ‐1, ‐1, ‐1)

Homework 4.6. Use the Einstein model to compute the change in internal energy of crystal when it is heated reversibly at one atmosphere pressure from 90 to 210K. Assume θE=250K.

Homework 4.7. The low temperature molar specific heat of diamond varies with temperature as,

where the Debye temperature D=1860K. What is the entropy change of 1g of diamond when it is heated at constant volume from 4K to 300K?

33 -1 -11.94 10 J mol KV

D

TC


26

4.7. The 3rd law of thermodynamics

Nernst’s statement:The entropy change in a process, between a pair of equilibrium states, associated with a change in the external parameters tends to zero as the temperature approaches absolute zero. (Here, external parameters are P, T, V,…)

Planck’s statement:The entropy of all perfect crystals is the same at the absolute zero, and may be taken as zero.

G H T S

CorrectWrong


27

A microscopic view points

0 0 12

3

4

1

i i

i

kT kTi

rTot kT

i

n e eN Z

e

At 0K,0 1Tot

nN

At 0K, all atoms are in the ground state. Or, all atoms are in the identical state. Hence, the number of microstates is 1 (=1).

if 0iikT

At 0K

ln 0S k


28

Some consequences of the 3rd law

Thermal expansion coefficient:

1

P

VV T

P T

V ST P 1

T

SV P

Thermal expansion coefficient is zero at 0K.

Heat capacity:

lnVV V

S SC TT T

As 0K, 0, ln - T S T

CV is zero at 0K.

Homework 4.8. Show that CP becomes 0 at 0 K.

29

4.8 Empirical predictions of entropy

298

298 0 298 0

o o o PCS S S dTT

An empirical relation between So298 the molar weight (MW) of various substances:

2983 Jln M 2 mol K

oWS R b

where b is a constant. The value of the term b depends upon the nature of chemical bonding within the phase

4.8. Empirical prediction of entropy

31

Trouton’s rule, Richards’ rule

Trouton’s rule: An empirical relation between the energy of vaporization of a crystal and the boiling temperature. The unit of Lv = J/mol.


33

Homework 4.9. Show that So3 for CdSb, CuAl2, and AuSn are 5.7, 5.3, and 5.7 J/g‐atom‐K.

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