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Byeong-Joo Leewww.postech.ac.kr/~calphad
ByeongByeong--Joo LeeJoo Lee
POSTECH POSTECH -- [email protected]@postech.ac.kr
Fundamentals Fundamentals
Microscopic vs. Macroscopic View Point
State function vs. Process variable
First Law of Thermodynamics
Byeong-Joo Leewww.postech.ac.kr/~calphad
Special processes1. Constant-Volume Process: ∆U = qv2. Constant-Pressure Process: ∆H = qp3. Reversible Adiabatic Process: q = 0
4. Reversible Isothermal Process: ∆U = ∆H = 0
Second Law of thermodynamicsSecond Law of thermodynamics -- Reversible vs. IrreversibleReversible vs. Irreversible
Byeong-Joo Leewww.postech.ac.kr/~calphad
S = measurable quantity + un-measurable quantity
= q/T + Sirr= qrev/T
Second Law of thermodynamicsSecond Law of thermodynamics -- Maximum WorkMaximum Work
wqUU AB −=−
wdUq system δδ +=
irrsystem dST
qdS +=
δ
wdU + δ
Byeong-Joo Leewww.postech.ac.kr/~calphad
irr
system
system dST
wdUdS +
+=
δ
irrsystemsystem TdSdUTdSw −−=δ
systemsystem USTww ∆−∆=≤ max
Second Law of thermodynamicsSecond Law of thermodynamics -- Entropy as a Criterion of EquilibriumEntropy as a Criterion of Equilibrium
※ for an isolated system of constant U and constant V,
(adiabatically contained system of constant volume)
equilibrium is attained when the entropy of the system is maximum.
※ for a closed system which does no work other than work of
volume expansion,
dU = T dS – P dV (valid for reversible process)
U is thus the natural choice of dependent variable for S and V
Byeong-Joo Leewww.postech.ac.kr/~calphad
U is thus the natural choice of dependent variable for S and V
as the independent variables.
※ for a system of constant entropy and volume, equilibrium is attained
when the internal energy is minimized.
irrsystemsystem TdSdUTdSw −−=δ
irrsystemsystem TdSdUTdSPdV −−=
irrsystem TdSdU +=0
Second Law of thermodynamicsSecond Law of thermodynamics -- Condition for Thermodynamic EquilibriumCondition for Thermodynamic Equilibrium
※ Further development of Classical Thermodynamics results from the fact
that S and V are an inconvenient pair of independent variables.
+ need to include composition variables in any equation of state and
in any criterion of equilibrium
+ need to deal with non P-V work
(e.g., electric work performed by a galvanic cell)
Byeong-Joo Leewww.postech.ac.kr/~calphad
※ Condition for Thermodynamic Equilibrium of a Unary two phase system
βα dSdSdS systemisolated +=_
αβ
β
α
αα
β
β
α
αα
βα
µµdn
TTdV
T
P
T
PdU
TT
−−
−+
−=11
The same conclusion is obtained using minimum internal energy criterion.
New Thermodynamic FunctionsNew Thermodynamic Functions –– Reason for the necessityReason for the necessity
※ Further development of Classical Thermodynamics results from the fact
that S and V are an inconvenient pair of independent variables.
+ need to include composition variables in any equation of state and
in any criterion of equilibrium
+ need to deal with non P-V work
(e.g., electric work performed by a galvanic cell)
Byeong-Joo Leewww.postech.ac.kr/~calphad
dU = TdS – PdV
S, V are not easy to control. Need to find new state functions
which are easy to control and can be used to estimate equilibrium
→ F, G
dF ≡ dU – TdS – SdT
For a reversible process
dF = [TdS – PdV – δw’] – TdS – SdT = – SdT – PdV – δw’
dFT = – PdV – δw’ = – δwT.Total
▷Maximum work that the system can do by changing its state at Constant T, V = -∆F.
For a irreversible isothermal process
Helmholtz Free Energy Helmholtz Free Energy -- Work Function, F Work Function, F ≡ U ≡ U – STST
Byeong-Joo Leewww.postech.ac.kr/~calphad
For a irreversible isothermal process
FT = [q – w] – TS
TS = q + TSirr
w = PV + w’ = w’ For constant V
FT,V + w’ + TSirr = 0
▷ If cannot do a maximum work, it’s due to the creation of ∆sirr.
Under a Constant T, V, an equilibrium is obtained when the system hasmaximum (w’+ ∆sirr) or minimum F.
Helmhomlz Free Energy Helmhomlz Free Energy -- ExampleExample
Equilibrium between condensed phase and gas phase.
Use Helmholtz Free Energy Criterion
to determine equilibrium amount of
gaseous phase at a given temperature,
Byeong-Joo Leewww.postech.ac.kr/~calphad
gaseous phase at a given temperature,
and how it changes with changing
temperature
dG ≡ dU + PdV + VdP – TdS – SdT
For a reversible process
dG = [TdS – PdV – δw’] + PdV + VdP – TdS – SdT = – SdT + VdP – δw’
dGT,P = – δw’
▷Maximum work that the system can do by changing its state at Constant T, P = -∆G.
For a irreversible isothermal process
Gibbs Free Energy Gibbs Free Energy -- Gibbs Function, G Gibbs Function, G ≡ U + PV ≡ U + PV – STST
Byeong-Joo Leewww.postech.ac.kr/~calphad
For a irreversible isothermal process
GT,P = [q – w] + PV – TS
TS = q + TSirr
w = PV + w’
GT,P + w’ + TSirr = 0
▷ If cannot do a maximum work, it’s due to the creation of ∆sirr.
Under a Constant T, P, an equilibrium is obtained when the system hasmaximum (w’+ ∆sirr) or minimum G.
Thermodynamic RelationsThermodynamic Relations -- For a closed systemFor a closed system
dU = TdS – PdV
dH = TdS + VdP
dF = – SdT – PdV
Byeong-Joo Leewww.postech.ac.kr/~calphad
dF = – SdT – PdV
dG = – SdT + VdP
Thermodynamic RelationsThermodynamic Relations -- For a multicomponent systemFor a multicomponent system
),,,,,( Lkji nnnPTGG =
L
LLLL
+
∂∂
+
∂∂
+
∂∂
+
∂∂
= j
nnPTj
i
nnPTinnTnnP
dnn
Gdn
n
GdP
P
GdT
T
GdG
kikjjiji ,,,,,,,,,,,,,,
Gµ≡
∂
▷
▷ Chemical Potential
Byeong-Joo Leewww.postech.ac.kr/~calphad
i
nnPTikj
nµ≡
∂ L,,,,
∑++−=n
iidnVdPSdTdG1
µ
∑−= iidnPdVw µδ
∑− iidnµ is the chemical work done by the system
▷ Chemical Potential
Gibbs Energy for a Unary System Gibbs Energy for a Unary System -- from from dG = dG = –SdT + VdPSdT + VdP
Gibbs Energy as a function of T and P
dPVdTSdG +−=
dPVdTSPTGPTGP
P
T
Too
oo∫∫ +−=− ),(),(
TCT )(∫
Byeong-Joo Leewww.postech.ac.kr/~calphad
dTT
TCSTS P
T
To
o
)()( ∫+=@ constant P
dTdTT
TCSTGTG
T
T
PT
Too
o o∫ ∫
+−=
)()()(
@ constant T dPVPGPGP
Po
o∫=− )()(
Gibbs Energy for a Unary System Gibbs Energy for a Unary System -- from from G = H G = H – STST
Gibbs Energy as a function of T and P
STHG −=
TSTSPTSHTHPTHPTG oooo ))1,(),(())1,(),((),( +−−+−=
STHTSHTSSHH oooo ∆−∆+−=+∆−+∆= )(
@ constant P
dTTCHTH P
T
o )()( ∫+= dTT
TCSTS P
T
o
)()( ∫+=
Byeong-Joo Leewww.postech.ac.kr/~calphad
dTTCHTH PT
oo
)()( ∫+= dTT
STST
oo
)( ∫+=
dTT
TCTdTTCTSHTG P
T
TP
T
Too
oo
)()()( ∫∫ −+−=
@ constant T
dPVT
VTdPV
P
STdP
P
HH
T
P
PT
P
PT
P
P ooo
+
∂∂
−=
+
∂∂
=
∂∂
=∆ ∫∫∫ dPTVP
Po
)1( α−= ∫
dPVdPT
VdP
P
SS
P
PT
P
PT
P
P ooo
α∫∫∫ −=
∂∂
−=
∂∂
=∆
dTdTT
TCSTGTG
T
T
PT
Too
o o∫ ∫
+−=
)()()(
Gibbs Energy for a Unary System Gibbs Energy for a Unary System -- Temperature DependencyTemperature Dependency
dTT
TCTdTTCTSHTG P
T
TP
T
Too
oo
)()()( ∫∫ −+−=
Byeong-Joo Leewww.postech.ac.kr/~calphad
Empirical Representation of Heat Capacities2−++= cTbTacP
서로 다른 출발점에서 유도된 위의 두 식은 같은 식인가?
를 이용하여 위의 두 식이 동일한 것임을 증명하라.
Gibbs Energy for a Unary SystemGibbs Energy for a Unary System
dPVdTT
TCTdTTCTSHTG
P
P
PT
TP
T
Too
ooo∫∫∫ +−+−=
)()()(
• V(T,P) based on expansivity and compressibility
• Cp(T)
Byeong-Joo Leewww.postech.ac.kr/~calphad
• S298: by integrating Cp/T from 0 to 298 K and using 3rd law of thermodynamics
(the entropy of any homogeneous substance in complete internal
equilibrium may be taken as zero at 0 K)
• H298: from first principles calculations, but generally unknown
※ H298 becomes a reference value for GT
※ Introduction of Standard State
Criterion of Thermodynamic Equilibrium, Thermodynamic RelationsCriterion of Thermodynamic Equilibrium, Thermodynamic Relations
Helmholtz Free Energy, Gibbs Free Energy
Correlation btw Free energy minimum & equilibrium
Chemical Potential vs. Gibbs energy
Byeong-Joo Leewww.postech.ac.kr/~calphad
Chemical Potential vs. Gibbs energy
Term as a Chemical Work
Thermodynamic Relations: Importance & Applications
∑− iidnµ
Statistical ThermodynamicsStatistical Thermodynamics
Basic Concept of Statistical Thermodynamics
Application of Statistical Thermodynamics to Ideal Gas
Understanding Entropy through the Concept of the
Byeong-Joo Leewww.postech.ac.kr/~calphad
Statistical Thermodynamics
Heat capacity
Heat capacity at low temperature
Application of CriterionApplication of Criterion
1.1기압 하 Pb의 melting point 는 600K이다.
1기압 하 590K로 과냉된 액상 Pb가 응고하는 것은자발적인 반응이라는 것을(1) maximum-entropy criterion과(2) minimum-Gibbs-Energy criterion을 이용하여 보이시오.
moleJHmelting /4810=∆KmolJTC lp ⋅×−= − /101.34.32 3
)(
Byeong-Joo Leewww.postech.ac.kr/~calphad
KmolJTC lp ⋅×−= /101.34.32)(
KmolJTC sp ⋅×= − /1075.9 3
)(
2. 1번 문제에서의 Pb가 단열된 용기에 보관되어 있었다면용기 내부는 결국 어떠한 (평형)상태가 될 것인지 예측하시오.
Numerical ExampleNumerical Example
KJTC lSnp /102.97.34 3
)(,
−×−=
KJTC sSnp /10265.18 3
)(,
−×+=
• A quantity of supercooled liquid Tin is adiabatically contained at 495 K.
Calculate the fraction of the Tin which spontaneously freezes. Given
J at Tm = 505 K505 K
7070=∆ Sn
mH
Byeong-Joo Leewww.postech.ac.kr/~calphad
495 K
1 mole of liquid xmoles of solid
(1-x) moles of liquid
Byeong-Joo Leewww.postech.ac.kr/~calphad
ByeongByeong--Joo LeeJoo Lee
POSTECH POSTECH -- MSEMSE
[email protected]@postech.ac.kr
Phase Diagram for HPhase Diagram for H22OO
Byeong-Joo Leewww.postech.ac.kr/~calphad
Phase Diagram for FePhase Diagram for Fe
Byeong-Joo Leewww.postech.ac.kr/~calphad
Phase Diagram for FePhase Diagram for Fe
Byeong-Joo Leewww.postech.ac.kr/~calphad
EquilibriumEquilibrium
• Thermal, Mechanical and Chemical Equilibrium
• Concept of Chemical Potential
In a one component system, iiG µ=
Byeong-Joo Leewww.postech.ac.kr/~calphad
• Temperature and Pressure dependence of Gibbs free energy
dTT
TCTdTTCTSHTG P
T
TP
T
Too
oo
)()()( ∫∫ −+−=
Temperature Dependence of Gibbs EnergyTemperature Dependence of Gibbs Energy
Byeong-Joo Leewww.postech.ac.kr/~calphad
Temperature Dependence of Gibbs Energy Temperature Dependence of Gibbs Energy -- for Hfor H22OO
KatJHH lsm 2736008)( =∆=∆ →
KJS KsOH /77.44298),(2=
KJS KlOH /08.70298),(2=
KJC lOHp /44.75)(, =
Byeong-Joo Leewww.postech.ac.kr/~calphad
KJC lOHp /44.75)(, 2=
KJC sOHp /38)(, 2=
Temperature & Pressure Dependence of Gibbs Energy Temperature & Pressure Dependence of Gibbs Energy
VT
H
V
S
dT
dP
eq ∆∆
=∆∆
=
vaporphasecondensedvapor VVVV =−=∆ − 2RT
HP
VT
H
V
S
dT
dP
veq
∆=
∆=
∆∆
=
Clausius-Clapeyron equation
For equilibrium between the vapor phase and a condensed phase
Byeong-Joo Leewww.postech.ac.kr/~calphad
dTRT
H
P
dP2
∆= dT
RT
HPd
2ln
∆= +
∆−=
RT
HPln
TCCHTCHH PPPT ⋅∆+∆−∆=−∆+∆=∆ ]298[)298( 298298
+∆
+
∆−∆= T
R
C
TR
HCP PP ln
1298ln 298 constant
constant
KJTTC vOHp /1033.0107.1030 253
)(, 2
−− ×+×+=
KJTTC vlp /1033.0107.1044.45 253
)(
−−→ ×+×+−=∆
KJC lOHp /44.75)(, 2=
∫ →∆+∆=∆T
vlpevapTevap dTCHH373
)(373,,
dTRT
HPd
2ln
∆=
Phase Diagram Phase Diagram -- for Hfor H22OO
Byeong-Joo Leewww.postech.ac.kr/~calphad
732.19log65.4900,2
)(log +−−= TT
atmP
VT
H
V
S
dT
dP
eq ∆∆
=∆∆
=
dTRT
Pd2
ln =
75.21862
10279.0log465.5997,2
)(log2
3 ++×+−−= −
TTT
TatmP
for S/L equilibrium
Equilibrium vapor pressures vs. TemperatureEquilibrium vapor pressures vs. Temperature
Byeong-Joo Leewww.postech.ac.kr/~calphad
Equilibrium vapor pressures vs. TemperatureEquilibrium vapor pressures vs. Temperature
Byeong-Joo Leewww.postech.ac.kr/~calphad
Gibbs Phase RuleGibbs Phase Rule
•Degree of Freedom
number of variables which can be independently varied
without upsetting the equilibrium
•F = p(1+c) – (p-1)(2+c) = c – p + 2
Byeong-Joo Leewww.postech.ac.kr/~calphad
Example Example -- Phase Transformation of Graphite to DiamondPhase Transformation of Graphite to Diamond
• Calculate graphite→diamond transformation pressure at 298 K, given
H298,gra – H298,dia = -1900 J
S298,gra = 5.74 J/K
S298,dia = 2.37 J/K
density of graphite at 298 K = 2.22 g/cm3
density of diamond at 298 K = 3.515 g/cm3
Byeong-Joo Leewww.postech.ac.kr/~calphad
dPVSTHG diamondgraphite
P
→∆+∆−∆=∆ ∫1
Byeong-Joo Leewww.postech.ac.kr/~calphad
ByeongByeong--Joo LeeJoo Lee
POSTECH POSTECH -- MSEMSE
[email protected]@postech.ac.kr
Thermodynamic Properties of Gases Thermodynamic Properties of Gases -- mixture of ideal gasesmixture of ideal gases
Mixture of Ideal Gases
Definition of Mole fraction: x
1 mole of ideal gas @ constant T:
1
212 ln),(),(
P
PRTTPGTPG =−
PRTddPP
RTVdPdG ln===
PRTTGTPG o ln)(),( += PRTGG o ln+=
Byeong-Joo Leewww.postech.ac.kr/~calphad
Definition of Mole fraction: xi
Definition of partial pressure: pi
Partial molar quantities:
Pxp ii =
K,,,,
'
kj nnPTi
in
∂∂
= iiG µ=
i
compT
i VP
G=
∂∂
,
iiQnQ ∑='
Thermodynamic Properties of Gases Thermodynamic Properties of Gases -- mixture of ideal gasesmixture of ideal gases
Heat of Mixing of Ideal Gases
i
o
i HH =
0' =−=∆ ∑∑ i
o
iii
mix HnHnH
PRTxRTGG ii
o
i lnln ++=
T
TG
T
TG i
o
i
∂∂
=∂
∂ )/()/(
Byeong-Joo Leewww.postech.ac.kr/~calphad
∑∑ ii
i
ii
i
Entropy of Mixing of Ideal Gases
Gibbs Free Energy of Mixing of Ideal Gases
mixmixmix STHG ''' ∆−∆=∆
ii
i
i
o
i
i
ii
i
mix xRTnGnGnG ln' ∑∑∑ =−=∆
ii
i
mix xRnS ln' ∑−=∆
fRTddG ln=
1→P
fas 0→P
For Equation of state α−=P
RTV
fRTdVdP ln= dPf
dα
−=
ln
Thermodynamic Properties of Gases Thermodynamic Properties of Gases -- Treatment of nonideal gasesTreatment of nonideal gases
Introduction of fugacity, f
fRTGG o ln+=
Byeong-Joo Leewww.postech.ac.kr/~calphad
fRTdVdP ln= dPRTP
fd
α−=
ln
RT
P
P
f
P
f
PPP
α−=
−
== 0
lnlnid
RTP
P
P
RT
PV
RT
Pe
P
f==−≅= − αα 1/
※ actual pressure of the gas is the geometric mean of the fugacity and the ideal P
※ The percentage error involved in assuming the fugacity to be equal to the
pressure is the same as the percentage departure from the ideal gas law
dPPRT
VdP
RTP
fd
−=−=
1ln
α
RT
PVZ ≡ dP
P
Z
P
fd
1ln
−=
dPP
Z
P
f PP
PPP
1ln
0
−=
∫
=
==
Thermodynamic Properties of Gases Thermodynamic Properties of Gases -- Treatment of nonideal gasesTreatment of nonideal gases
Alternatively,
Byeong-Joo Leewww.postech.ac.kr/~calphad
PdRTP
fdRTfRTddG lnlnln +
==
JRTP
fRTG 112971137376150lnln
150
=+−=+
=∆
Example) Difference between the Gibbs energy at P=150 atm and P=1 atm
for 1 mole of nitrogen at 0 oC
Solution Thermodynamics Solution Thermodynamics -- Mixture of Condensed PhasesMixture of Condensed Phases
Vaper A: oPA
Condensed
Vaper B: oPB
Condensed
+ →
Vaper A+ B: PA + PB
Condensed Phase A + B
Byeong-Joo Leewww.postech.ac.kr/~calphad
Condensed Phase A
Condensed Phase B
Condensed Phase A + B
condensed
A
ovapor
A
o GG = condensed
B
ovapor
B
o GG = condensed
A
vapor
A GG =condensed
B
vapor
B GG =
i
o
i
i
ii
i
mix GnGnG ∑∑ −=∆ '
i
o
ii
i p
pRTn ln∑⇒ for gas
o
iie kpr =)(
Solution Thermodynamics Solution Thermodynamics -- ideal vs. nonideal vs. non--ideal solutionideal solution
o
iii pxp =iiie kpxr =)(
o
iiii pxkp =o
ii
ie
i pxr
rp
)('=iiie kpxr =)('
Ideal Solution
Nonideal Solution
Byeong-Joo Leewww.postech.ac.kr/~calphad
iiiiii
ie
ir )(
iiie )(
Solution Thermodynamics Solution Thermodynamics -- Thermodynamic ActivityThermodynamic Activity
o
i
ii
f
fa =
Thermodynamic Activity of a Component in Solution
o
i
ii
p
pa =
1
212 ln),(),(
P
PRTTPGTPG =−
→ ix⇒ for ideal solution
Byeong-Joo Leewww.postech.ac.kr/~calphad
Draw a composition-activity curve for an ideal and non-
ideal solution
Henrian vs. Raoultian
),,,,,('' 21 cnnnPTQQ L=
i
nnPTi
c
innTnnP
dnn
QdP
P
QdT
T
QdQ
ijkjkj ≠=
∂∂
+
∂∂
+
∂∂
= ∑,,1,,,,,,
''''
LL
▷ Partial Molar Quantity
ij nnPTi
in
≠
∂∂
=,,
'
kk
c
PT dnQdQ ∑=,'c
Solution Thermodynamics Solution Thermodynamics -- Partial Molar PropertyPartial Molar Property
Byeong-Joo Leewww.postech.ac.kr/~calphad
kk
k
PT dnQdQ ∑=
=1
,'
kk
c
k
nQQ ∑=
=1
'
01
=∑=
kk
c
k
Qdn
01
=∑=
kk
c
k
Qdx
GibbsGibbs--Duhem EquationDuhem Equation▷Molar Properties of Mixture
kk
c
k
dXQdQ ∑=
=1
k
c
k
k QXQ ∑=
=1
''' QQQ o
mix −=∆
kk
oc
k
o nQQ ∑=
=1
'
Solution Thermodynamics Solution Thermodynamics -- Partial Molar Quantity of MixingPartial Molar Quantity of Mixing
definition of solution and mechanical mixing
Byeong-Joo Leewww.postech.ac.kr/~calphad
Qo
kk
c
k
kk
o
k
c
k
mix nQnQQQ ∆=−=∆ ∑∑== 11
)('
is a pure state value per molewhere
왜 partial molar quantity를 사용해야 하는가?
Solution Thermodynamics Solution Thermodynamics -- Partial Molar QuantitiesPartial Molar Quantities
)lnln(' BBAA
phaseref
B
o
B
phaseref
A
o
A ananRTGnGnG +++∆= →→
)lnln( BBAAB
o
BA
o
Am axaxRTGxGxG +++=
)lnln(' BBAAB
o
BA
o
A ananRTGnGnG +++=
Byeong-Joo Leewww.postech.ac.kr/~calphad
)lnln( BBAABBAAm axaxRTGxGxG +++=
)lnln()lnln( BBAABBAAB
o
BA
o
Am xxRTxxxxRTGxGxG γγ +++++=
ABBABBAAB
o
BA
o
Am LxxxxxxRTGxGxG ++++= )lnln(
2
22 )1(dX
dQXQQ −+= example)
21XaXH mix =∆
• Partial molar & Molar Gibbs energy
Solution Thermodynamics Solution Thermodynamics -- Partial Molar QuantitiesPartial Molar Quantities
Evaluation of Partial Molar Properties in 1-2 Binary System
• Partial Molar Properties from Total Properties
)lnln( BBAAB
o
BA
o
Am axaxRTGxGxG +++=ii
o
i
M
i aRTGGG ln=−=∆
Byeong-Joo Leewww.postech.ac.kr/~calphad
Gibbs energy of mixing vs. Gibbs energy of formation
• Graphical Determination of Partial Molar Properties: Tangential Intercepts
• Evaluation of a PMP of one component from measured values of a PMP
of the other0
2211 =∆+∆ QdXQdX 2
1
21 Qd
X
XQd ∆−=∆
2
2
2
1
2
02
1
2
01
2
2
2
2
dXdX
Qd
X
XQd
X
XQ
X
X
X
X
∆−=∆−=∆ ∫∫ ==
2
12 aXH =∆example)
kkk xRTaRT γlnln =
Solution Thermodynamics Solution Thermodynamics -- NonNon--Ideal SolutionIdeal Solution
▷Activity Coefficient
2
)ln()/(
T
H
T
R
T
TG M
ii
M
i ∆−=
∂∂
=∂
∆∂ γ
R∂ )ln( γ
Byeong-Joo Leewww.postech.ac.kr/~calphad
kkk xax =→ )1lim(
k
o
kkk xrax =→ )0lim(
▷ Behavior of Dilute Solutions
M
ii H
T
R∆=
∂∂
)/1(
)ln( γ
1.Gibbs energy of formation과 Gibbs energy of mixing의차이는 무엇인가?
2. Solution에서 한 성분이 Henrian 또는 Raoultian 거동을한다는 것을 무엇을 의미하는가? Molar Gibbs energy가 다음과 같이 표현되는 A-B 2원
ExampleExample
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Molar Gibbs energy가 다음과 같이 표현되는 A-B 2원Solution phase에서 각 성분은 dilute 영역에서는 Henrian 거동을, rich 영역에서는 Raoultian 거동을 보인다는 것을증명하시오.
LxxxxxxRTGxGxG BABBAAB
o
BA
o
Am ++++= lnln
ABABBBBBAAAABA EWEWEWE ++=−
2
21
AAA NzxW =
2
21
BBB NzxW =
Solution Thermodynamics Solution Thermodynamics -- QuasiQuasi--Chemical Model, Guggenheim, 1935.Chemical Model, Guggenheim, 1935.
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2 BBB NzxW =
BAAB xNzxW =
])2[(2
BBAAABBABBBAAABA EEExxExExNz
E −−++=−
Solution Thermodynamics Solution Thermodynamics -- Regular Solution ModelRegular Solution Model
ABBA
xs
m xxG Ω=∆
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SnSn--In SnIn Sn--BiBi
Solution Thermodynamics Solution Thermodynamics -- SubSub--Regular Solution ModelRegular Solution Model
])([ 10
ABABABBA
xs
m xxxxG Ω−+Ω=∆
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SnSn--Zn FeZn Fe--NiNi
•• Composition and temperature dependence of Composition and temperature dependence of ΩΩ
ABBABBAAB
o
BA
o
Am xxxxxxRTGxGxG Ω++++= )lnln(
Solution Thermodynamics Solution Thermodynamics -- Regular Solution ModelRegular Solution Model
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•• Extension into ternary and multiExtension into ternary and multi--component systemcomponent system
•• Sublattice ModelSublattice Model
•• Inherent InconsistencyInherent Inconsistency
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ByeongByeong--Joo LeeJoo Lee
POSTECH POSTECH –– MSEMSE
[email protected]@postech.ac.kr
PropertyProperty of a Regular Solutionof a Regular Solution
BA
mixxs xxHG ⋅Ω=∆=∆
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PropertyProperty of a Regular Solutionof a Regular Solution
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Standard StatesStandard States
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Standard StatesStandard States
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Standard StatesStandard States
WhichWhich standardstandard statesstates
shallshall wewe use?use?
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Byeong-Joo Leewww.postech.ac.kr/~calphad
Byeong-Joo Leewww.postech.ac.kr/~calphad
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Phase DiagramsPhase Diagrams -- Relation with Gibbs Energy of Solution PhasesRelation with Gibbs Energy of Solution Phases
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Phase DiagramsPhase Diagrams -- Binary SystemsBinary Systems
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Phase EquilibriumPhase Equilibrium
αβ
β
α
αα
β
β
α
αα
βα
µµk
k
kkisosys dn
TTdV
T
P
T
PdU
TTdS ∑
−−
−+
−= ''11
' ,
2)2)(1()1( +−=+−−+= pccpcpf
1. Conditions for equilibrium
2. Gibbs Phase Rule
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LLo
T xnxnxn += εε
LLLLLo xxxfxfxfxfxfx +−=−+=+= )()1( εεεεεεε
εε
xx
xxf
L
oL
−−
=ε
ε
xx
xxf
L
oL
−−
=
3. How to interpret Binary and Ternary Phase Diagrams
▷ Lever-Rule
※ Example: Deposition of Silicon SiH4 + 2Cl2 = Si + 4HCl
Driving force of CVD Deposition Driving force of CVD Deposition -- from N.M. Hwang, SNUfrom N.M. Hwang, SNU
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