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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


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


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


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


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)


※ 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)


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



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,


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.


Gibbs Free Energy Gibbs Free Energy -- Gibbs Function, G Gibbs Function, G ≡ U + PV ≡ U + PV – STST



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


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


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 )(∫


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

)()( ∫+=


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

)()()( ∫∫ −+−=


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)


• 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


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


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

)(


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


495 K

1 mole of liquid xmoles of solid

(1-x) moles of liquid



POSTECH POSTECH -- MSEMSE

[email protected]@postech.ac.kr

Phase Diagram for HPhase Diagram for H22OO


Phase Diagram for FePhase Diagram for Fe


EquilibriumEquilibrium

• Thermal, Mechanical and Chemical Equilibrium

• Concept of Chemical Potential

In a one component system, iiG µ=


• 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


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)(, =


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


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


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


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


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


dPVSTHG diamondgraphite

P

→∆+∆−∆=∆ ∫1



POSTECH POSTECH -- MSEMSE


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+=


Definition of Mole fraction: xi

Definition of partial pressure: pi

Partial molar quantities:

Pxp ii =

K,,,,

'

kj nnPTi

in

QQ

∂∂

= 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

∂∂

=∂

∂ )/()/(


∑∑ 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+=


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,


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


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


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


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

QQ

≠

∂∂

=,,

'

kk

c

PT dnQdQ ∑=,'c

Solution Thermodynamics Solution Thermodynamics -- Partial Molar PropertyPartial Molar Property


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


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 +++=


)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=−=∆


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


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


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.


2 BBB NzxW =

BAAB xNzxW =

])2[(2

BBAAABBABBBAAABA EEExxExExNz

E −−++=−

Solution Thermodynamics Solution Thermodynamics -- Regular Solution ModelRegular Solution Model

ABBA

xs

m xxG Ω=∆


SnSn--In SnIn Sn--BiBi

Solution Thermodynamics Solution Thermodynamics -- SubSub--Regular Solution ModelRegular Solution Model

])([ 10

ABABABBA

xs

m xxxxG Ω−+Ω=∆


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


•• Extension into ternary and multiExtension into ternary and multi--component systemcomponent system

•• Sublattice ModelSublattice Model

•• Inherent InconsistencyInherent Inconsistency



POSTECH POSTECH –– MSEMSE


PropertyProperty of a Regular Solutionof a Regular Solution

BA

mixxs xxHG ⋅Ω=∆=∆


PropertyProperty of a Regular Solutionof a Regular Solution


Standard StatesStandard States


Standard StatesStandard States

WhichWhich standardstandard statesstates

shallshall wewe use?use?


Phase DiagramsPhase Diagrams -- Relation with Gibbs Energy of Solution PhasesRelation with Gibbs Energy of Solution Phases


Phase DiagramsPhase Diagrams -- Binary SystemsBinary Systems


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


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


byeongbyeong--joo lee joo lee - cmse.postech.ac.kr · second law of thermodynamics --entropy as...

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