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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
Byeong-Joo Leewww.postech.ac.kr/~calphad
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 :
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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
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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 −−−−=
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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
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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
∫ −=
∂∂
=
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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
∫ −=
∂∂
=
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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
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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
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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
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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,
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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
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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
Effusion: Langmuir EquationEffusion: Langmuir Equation
A x - axis
τxvL =
τ
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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
Effusion: Langmuir EquationEffusion: Langmuir Equation
−
=∑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
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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=
Effusion: Langmuir EquationEffusion: Langmuir Equation
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.
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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
Effusion: Langmuir EquationEffusion: Langmuir Equation
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
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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
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∂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
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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=
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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
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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 ?
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∑ −−
−= 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
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∏=
=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:
Byeong-Joo Leewww.postech.ac.kr/~calphad
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
Byeong-Joo Leewww.postech.ac.kr/~calphad
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=
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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→
Byeong-Joo Leewww.postech.ac.kr/~calphad
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
Byeong-Joo Leewww.postech.ac.kr/~calphad
( )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
Byeong-Joo Leewww.postech.ac.kr/~calphad
−=
−==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=