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Byeong-Joo Leewww.postech.ac.kr/~calphad

ByeongByeong--Joo LeeJoo Lee

POSTECH POSTECH -- MSEMSE

[email protected]@postech.ac.kr

Equipartition TheoremEquipartition Theorem

kT2

1

L+++= '''''' εεεε LL ''''''/)''''''(/ ZZZeeZ kTkT === ∑∑ +++−− εεεε

VT

ZNkTU

∂∂

=ln2

βε

∂∂

−==><Z

N

U ln

translational kinetic energy :

rotational kinetic energy :

2

21 mv

21 ωI

The average energy of a particle per independent component of motion is


rotational kinetic energy :

vibrational energy :

kinetic energy for each independent component of motion has a form of

2

2

1 ωI2

212

21 kxmv +

2

ii pb

22

22

2

11 ff pbpbpb +++= Lε f

pbpbpbdpedpedpeP ff

2222

211

02

01

0

βββ −∞−∞−∞

∫∫∫= L

ii py 2/1β= ii

yb

i

pbKdyedpe iiii 2/1

0

2/1

0

22 −−∞−−∞== ∫∫ βββ

f

f

f KKKKKKZ LL 21

2/2/1

2

2/1

1

2/1 −−−− =⋅= ββββ

Equipartition TheoremEquipartition Theorem

kTffP

22

ln==

∂∂

−=><ββ

ε

RTu3

=

The average energy of a particle per independent component of motion is kT2

1

f

f KKKZ L21

2/−= β

※ for a monoatomic ideal gas :


RTu2

=

RTu2

5=

RTqu )3( +=

※ for a monoatomic ideal gas :

for diatomic gases :

for polyatomic molecules which are soft and vibrate easily

with many frequencies, say, q:

※ for liquids and solids, the equipartition principle does not work

Einstein and Debye Model for Heat Capacity Einstein and Debye Model for Heat Capacity –– Background Background


Einstein and Debye Model for Heat Capacity Einstein and Debye Model for Heat Capacity –– Concept Concept

3N independent (weakly interacting) but distinguishable

simple harmonic oscillators.

νε hii )(21+=

kTh

kTh

e

eP

/

2/

1 ν

ν

−

−

−= )1ln(

2

1ln / kThe

kT

hP νν −−−−=


for N simple harmonic vibrators

−

+=

∂∂

=12

1ln/

2

kTh

V e

hhN

T

PNkTU ν

νν

average energy per vibrator 12 / −

+=><kThe

hhν

ννε

Einstein and Debye Model for Heat Capacity Einstein and Debye Model for Heat Capacity –– number densitynumber density

Let dNv be the number of oscillators whose frequency lies

between v and v + dv

ννν dgdN )(=

where g(v), the number of vibrators per unit frequency band,

satisfy the condition

NdgdN 3)( == ∫∫ ννν


NdgdN 3)( == ∫∫ ννν

The energy of N particles of the crystal

νννν

ε νν dge

hhdNU

kTh)(

12 /∫∫

−

+=><=

ννν

ν

ν

dge

ekThk

T

UC

kTh

kTh

V

V )()1(

)/(2/

/2

∫ −=

∂∂

=

Einstein and Debye Model for Heat Capacity Einstein and Debye Model for Heat Capacity –– EinsteinEinstein

2/

/2

)1(

)/(3

−=

kTh

kTh

EV

E

E

e

ekThkNC ν

νν

All the 3N equivalent harmonic oscillators have the same frequency vE

ννν

ν

ν

dge

ekThk

T

UC

kTh

kTh

V

V )()1(

)/(2/

/2

∫ −=

∂∂

=


k

h EE

νθ =

2/

/2

)1(3 −

=T

T

EV

E

E

e

e

TR

cθ

θθ

Defining Einstein characteristic temperature

Einstein and Debye Model for Heat Capacity Einstein and Debye Model for Heat Capacity –– DebyeDebye

A crystal is a continuous medium supporting standing longitudinal

and transverse waves

2

3

9)( ν

νν

m

Ng =

ννννν

ν

ν

de

ekTh

Nk

C m

kTh

kTh

mV ∫ −=

2/

/232

)1(

)/)(/3(

3

ννν

ν

ν

dge

ekThk

T

UC

kTh

kTh

V

V )()1(

)/(2/

/2

∫ −=

∂∂

=


νν deNk kTh∫ −

=0 2/ )1(3

kT

hx

ν=

TkT

hx mm

Θ==

ν

dxe

ex

TR

c T

x

x

V ∫Θ

−Θ=

/

0 2

4

3 )1()/(

3

3

set

Einstein and Debye Model for Heat Capacity Einstein and Debye Model for Heat Capacity –– ComparisonComparison


Einstein and Debye Model for Heat Capacity Einstein and Debye Model for Heat Capacity –– More about DebyeMore about Debye

Behavior of dxe

ex

TR

c T

x

x

V ∫Θ

−Θ=

/

0 2

4

3 )1()/(

3

3

at T → ∞ 2

4

)1( −x

x

e

ex→ x2 RcV 3=

34

5

4

3

Θ

=T

R

cV π: Debye’s T3 law

at T → ∞ and T → 0

at T → 0


53

Θ

=R

: Debye’s T lawat T → 0

Einstein and Debye Model for Heat Capacity Einstein and Debye Model for Heat Capacity –– More about CpMore about Cp

Tce ⋅= 'γ

for T << TF


Effusion: Langmuir EquationEffusion: Langmuir Equation

Question: The rate at which particles strike a unit surface of a container

per unit time, given the pressure and temperature of the gas

Application:

1. Estimate of the time needed for a totally clean surface to be

covered with a monolayer of atoms or molecules,


covered with a monolayer of atoms or molecules,

assuming that all the molecules that hit the surface stick to it

2. Calculate how many atoms will escape from a small hole

in a vessel per unit time, given the area of hole

(measure of vapor pressure)

3. How may particles may evaporate from a surface per unit time

Maxwell Distribution of Speed in Dilute GasesMaxwell Distribution of Speed in Dilute Gases

kTegZ

Nn /)()( εεε −= 2

2

2222

2

2

8)(

8r

mL

hnnn

mL

hzyx =++=ε

εεπ

πεε dh

mVdrrdg 2/1

3

2/32 )2(2

48

1)( ==

εε

επ

πεε dkTkT

Ndn

−

= exp1

2)( 2/1

2/32/3

2

2

=h

mkTVZ

π


π kTkT

dvkT

mvv

kT

mNdvvn

−

=2

exp2

4)(2

2

2/3

ππ

m

kTv

32 >=<2/1

0 8)(

=>=<∫∞

m

kT

N

dvvvnv

π

2/1

* 2

=m

kTv

2

2

1mv=ε

Number of atoms N* that collide with side walls within a time


A x - axis

τxvL =

τ


Number of atoms N* that collide with side walls within a time

][2

1,,

*

ixix

i

vofyprobabilitV

NAvN τ∑=

−

=∑Z

ikTmV

h

V

NAvN ix

i

2

3/2

2

,

* 8exp

2

1τ

τ


−

=∑Z

ikTmV

h

V

NAvN ix

i

2

3/2

2

,

* 8exp

2

1τ

kTmV

h3/2

22

8≡γ

im

kTv ix ⋅⋅

= γ2/1

,

2 2/3

2

2

=h

mkTVZ

π2/1

2

3/1 2

=h

mkTVZx

π


diiim

kT

VZ

N

A

N)exp(

2

2

1 22

0

2/1*

γγτ

−

= ∫∞

2/1*

2

1

=mkTV

NkT

A

N

πτ

2

22

0 2

1)exp(

γγ =−∫

∞diii

2/1

*

)2( mkT

P

A

N

πτ=


2/1

*

)2( mkT

P

A

N

πτ= 4

2/1

23

23

3

12*

10

1038.110022.6

102

101325)()sec( −

−−

−− ×

××

×

×

×=

TM

atmPcm

A

N

πτ

ττ

1523

* 101072.2 =×= P

A

NAssume 1015 atoms per cm2 in the surface monolayer.

For O2 (M=32 g/mol) at 300K and at 10-10 atm,

about 37sec is necessary for monolayer deposition.


2/1)2( mkT

P

A

N equilevapor

πτ=

To keep the surface clean for 1 hour, the pressure

should be 10-12 atm.

At equilibrium, rate of evaporation is the same as

the rate of deposition.

ex) for liquid Al in vacuum at 1250 K,

mass loss due to a hole of 2 × 10-3 cm2

→ 1.7 × 10-9 g/s

→ Knudsen effusion method atm

mkTA

NP

evaporequil

7

2/1

1030.1

)2(

−×=

= πτ


2/1)2( mkTA

NP

evaporequil π

τ=

aVNVNtvVN

tv>⇒==

)/(2

1

)/(

1

)/( 222 πσπσπσλ

Knudsen effusion method:

(valid for Knudsen flow)

Mean free path

for ideal gas

P

kT22πσ

λ =

2/1

2

1

2

1

=T

T

P

P


ex) O2 gas (diameter = 3 × 10-10 m) at 300 K

at 1 atm, λ = 10-7 m

at 10-9 atm, λ = 100 m

λ / a > 1 Knudsen flow

λ / a < 0.01 Viscous flow

)(1002.1

)10103.1()103(2

300)1038.1(

)(

7

5)(

210

23

meterPP atmatm

−

−

− ×=

×××

××=

πλ

Diffusion in GasesDiffusion in Gases

∂∂

−=x

CDJm dx

x

CCC o

∂∂

+= **

*

6

1λ+><= xCvN

λ

∂∂

><−=−=x

CvNNJ

3

1 λ><= vD3

1


∂x3 3

2/18

>=<m

kTv

π P

kT22πσ

λ =

P

T

m

kD

2

2/3

2/1

2/31

3

2

σπ

= for L<<λ

Flux during PVD Flux during PVD –– Evaporation (Langmuir) or Diffusion Controlled ?Evaporation (Langmuir) or Diffusion Controlled ?

cmP

kT 5

210

2==

πσλ

Consider W evaporation at 2473K onto cold substrate along 3cm path.

in vacuum

atmPequil 101023.1 −×=

smatomsmkT

P

A

NJW ⋅×=== 216

2/1

*

/1085.4)2( πτ

cm410−=λin 0.1 atm N2 gas


smP

T

m

kD /1066.1

1

3

2 23

2

2/3

2/1

2/3

−×=

=σπ

cm10=λin 0.1 atm N2 gas

314 /106.3 matomkT

P

V

NC ×===

smatomsx

CDJW ⋅×=∆∆

−= 213 /102

Flux during PVD Flux during PVD –– GeneralGeneral

Langmuir Equation

kT

P

V

NC ==

2/1

*

)2( W

WW

mkT

P

A

NJ

πτ==

Diffusion Controlled

x

CDJW ∆∆

−=

PJ W=


DxkTmkT

PJ

W

WW

/)2( 2/1 ∆+=

π

DxkT

PJ

W

WW

/∆=

In General

Size Distribution of Molecules in PolymerSize Distribution of Molecules in Polymer

For a polymer with NAmers (segment) : NA (Avogadro number)

X = number of segments in molecules

Nx = number of molecules with size X

N = total number of molecules

nx = fraction of molecules with size X = NX/N

Nb = number of bonding between segments

P = 고분자도 (중합도, degree of polymerization) = N /N


P = 고분자도 (중합도, degree of polymerization) = Nb/NA

∑ −=X

Xb NXN )1(

AN

NP −=1

Size Distribution of Molecules in PolymerSize Distribution of Molecules in Polymer

∑ =X

Xn 1 ∑ =X

AX

N

NXn

∑−X

XX nn lnMaximize

With Constraints

∑ −

−

=X

X

Xe

en λ

λ

∑∑

−

−

=X

X

A

e

eX

N

Nλ

λ

∑−

− =λ

λ ee X ∑

−− =

λλ e

Xe X

▷

Peak Size ?


∑ −−

−= λ

λ

e

ee X

1∑ −

−

−=

2)1( λλ

e

eXe X

▷

λ−−=

eN

N A

1

1

AN

Ne −=− 1λ

21 )1( PPNN X

AX −= −

)1(1)1()()1/(

)( 1

1

1 PPN

N

N

Nee

ee

en X

A

X

A

XX

X −=

−=−=

−= −

−

−−−−−

−λλ

λλ

λ

21 )1( PXPN

XNW X

A

XX −== −

Entropy of Mixing in Polymer Solutions Entropy of Mixing in Polymer Solutions -- Flory Huggins TheoryFlory Huggins Theory

∑−=∆ iiM xxRS ln

n

n

n

iN

niNNZZ

N

niNZ

N

niNZniNw

−−=

−−

−−= −

−

+2

2

1 )1()1()(

N = N1 + nN2

Consider the number of ways for distribution of

segments of the (i+1)th n-mer


∏=

=Ω2

12!

1N

i

iwN

= ∏

=

2

12!

1ln

N

i

ic wN

kS

iN

NZZN

ZN

ZniNw

−=

−

−=+1 )1()1()(

]ln)1()1ln()2([lnlnln 22

21

1 nnZnZNN

nNN

N

NN

k

Sc +−+−−++

−

−=

( )2211

22

11

22

11

lnln

lnlnlnln

φφ xxR

N

nNx

N

NxR

N

nNN

N

NNkSM

+−=

+

−=

+

−=∆

Elasticity of Rubber Elasticity of Rubber –– Scope Scope

Ref. Ref. –– Castellan, Gilbert W., Physical Chemistry 3rd Ed., Benjamin/Cummings,Castellan, Gilbert W., Physical Chemistry 3rd Ed., Benjamin/Cummings,

New York, 1983 (Chap. 29)New York, 1983 (Chap. 29)

fdLTdSdU +=

LTTT T

fT

L

U

L

ST

L

Uf

∂∂

+

∂∂

=

∂∂

−

∂∂

=

Ideal rubber:


T

f

T

f

L

=

∂∂

Ideal rubber:

0=

∂∂

LL

U

TL

STf

∂∂

−=

Polymer molecules themselves are not stretched, but the degree of

alignment is changed.

Position of one end of an N-mer with respect to the position of the other end

Elasticity of Rubber Elasticity of Rubber –– Statistics of Random Walk Statistics of Random Walk

LR N

L

N

R

LR

R PPNN

NNNP

!!

!),( =

+=l

xNN R

2

1

−=l

xNN L

2

1

N

lxNlxN

NNxP

−

+=

2

1

!2

/!

2

/

!),(

−

−+

+

+=−

Nl

x

l

xNl

Nl

x

l

xNlNxP 1ln

21ln

2),(ln


22 Nlx =

−=

−=

2

2

22

2

2exp

2

1

2exp),(

Nl

x

NlNl

xKNxP

π

−

+

+

=−NllNll

NxP 1ln2

1ln2

),(ln

2

2

11ln

−=

+Nl

x

Nl

x

Nl

x2

2

2),(ln

Nl

xNxP −=

Elasticity of Rubber Elasticity of Rubber –– Statistics of Random Walk Statistics of Random Walk

dxdydzNl

zyxl

NdxdydzNzyxP

++−

=−

2

2222/3

2

2

)(3exp

32),,,( π

dxdydzN

zPN

yPN

xPdxdydzNzyxP

=3

,3

,3

,),,,(

2222 Rzyx =++

RN − 22/3

3

dRRdxdydz 24π=


2/1N

Rl rms=

dRNl

RRl

NdRNRP

−

=−

2

22

2/3

2

2

3exp4

32),( ππ

2

2

24

0

2/3

22

2

3exp

324 NldR

Nl

RRl

NR =

−

= ∫∞

−

ππ

Elasticity of Rubber Elasticity of Rubber –– Deformation & EntropyDeformation & Entropy

LLL o ∆+=

00 zz α→

oLL /=α oLL /∆=ε εα =−1

When a solid with original dimensions x0, y0, z0 is stretched in the z direction

at constant volume

02/10

1xx

α→ 02/10

1yy

α→


Elasticity of Rubber Elasticity of Rubber –– Deformation & EntropyDeformation & Entropy

dxdydzNl

zyxNldxdydzPu

++−

=−

2

2222/3

2

2

)(3exp

3

2π

00 zz α→

When a solid with original dimensions x0, y0, z0 is stretched in the z direction

at constant volume

02/10

1xx

α→

02/10

1yy

α→

dxdydzzyx

NldxdydzP

++−

=

− 22222/3

2 )//(3exp

2 αααπ


( )2

2222

2

111

11

3Nl

zyx

kS i

−+

−+

−

−=∆

ααα

)ln(ln usus kSSS Ω−Ω=−=∆

dxdydzNl

zyxNldxdydzPs

++−

=2

2

2

)//(3exp

3

2 αααπ

Flory, Paul J., Principles of Polymer Chemistry, Cornell University Press, Ithaca,

NY, 1960.

Elasticity of Rubber Elasticity of Rubber –– ElasticityElasticity

−+−=∆ 32

2

2

αα

νkS

−+

−−=∆3

1

3

11

1

3

2

2

3 2αα

kS i

−=

∂∂

−=

∂∂

−=2

1

αα

να oToT L

kTS

L

T

L

STf

−=

−==

21 LLkTkTf oν

αν

σ


−=

−==2 LLVVA

o

oαασ

L

dLE

L

dL

V

kT

L

dL

L

L

L

L

V

kTdL

L

L

LV

kTd o

o

o

o

=

+=

+=

+=

22

2

3

22

221

αα

νννσ

+=

+=22

22

αα

ρα

αν

cM

RT

V

kTE

c

A

M

N

V

ρν=

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