byeong-joo lee calphad byeong-joo lee postech - mse [email protected] basic review of...

Byeong-Joo Lee www.postech.ac.kr/~calphad

Byeong-Joo LeeByeong-Joo Lee

POSTECH - MSEPOSTECH - [email protected]@postech.ac.kr

BasicBasic ReviewReview of of

ThermodynamThermodynamicsics


Objective Objective

Understanding and Utilizing Thermodynamic

Laws State function Thermodynamic Laws Statistical thermodynamics Gibbs energy

Extension of Thermodynamics Multi-Phase System Multi-Component System Partial Molar Quantities

Utilization of Thermodynamics Phase Diagrams Defect Thermodynamics


1-1.1-1. Understanding and Utilizing Understanding and Utilizing Thermodynamic LawsThermodynamic Laws

Why define state function ?

Why Thermodynamic Laws are

important ?

Statistical vs. classical

thermodynamics

Why use Gibbs energy ?

Why define state function ?

Why Thermodynamic Laws are

important ?

Statistical vs. classical

thermodynamics

Why use Gibbs energy ?


Fundamentals Fundamentals State function vs. Process variable

First Law of Thermodynamics “The change of a body inside an adiabatic enclosure from a given initial state to a given final state involves the same amount of work by whatever means the process is carried out”

It was necessary to define some function which depends only on the internal state of a system

– Internal Energy. For adiabatic process: UB – UA = -w Generally: UB – UA = q - w

dU = δq – δw : as a state function

Special processes

1.Constant-Volume Process: ΔU ＝ qv

2.Constant-Pressure Process: ΔH ＝ qp

3.Reversible Adiabatic Process: q ＝ 04.Reversible Isothermal Process: ΔU ＝ ΔH ＝ 0

0dU


Second Law of thermodynamics Second Law of thermodynamics - - Reversible vs. IrreversibleReversible vs. Irreversible

△S = measurable quantity + un-measurable quantity = q/T + S△ irr = qrev/T


Second Law of thermodynamics Second Law of thermodynamics - Maximum - Maximum WorkWork

wqUU AB

wdUq system

irrsystem dST

qdS

irrsystem

system dST

wdUdS

irrsystemsystem TdSdUTdSw

systemsystem USTww max


Second Law of thermodynamics Second Law of thermodynamics - Entropy as a - Entropy as a Criterion of EquilibriumCriterion 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 as the independent variables.

※ for a system of constant entropy and volume, equilibrium is attained

when the internal energy is minimized.irrsystemsystem TdSdUTdSw

irrsystemsystem TdSdUTdSPdV

irrsystem TdSdU 0


New Thermodynamic Functions New Thermodynamic Functions – Reason for – Reason for the necessitythe 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 = -ΔF.

For a irreversible isothermal process △FT = [q – w] – T S△ T S = q + T S△ △ irr

w = P V + w’ = w’△ For constant V

△FT,V + w’ + T S△ irr = 0

▷ If cannot do a maximum work, it’s due to the creation of Δsirr. Under a Constant T, V, an equilibrium is obtained when the system has

maximum (w’+ Δsirr) or minimum F.

Helmholtz Free Energy Helmholtz Free Energy - Work Function, F ≡ - Work Function, F ≡ U – STU – ST


Equilibrium between condensed phase and gas phase.

Use Helmholtz Free Energy Criterion to determine equilibrium amount of gaseous phase at a given temperature, and how it changes with changing temperature

Helmholtz Free Energy Helmholtz Free Energy - Example of - Example of ApplicationApplication


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 = -ΔG.

For a irreversible isothermal process △GT,P = [q – w] + P V – T S△ △ T S = q + T S△ △ irr

w = P V + w’△

△GT,P + w’ + T S△ irr = 0

▷ If cannot do a maximum work, it’s due to the creation of Δsirr. Under a Constant T, P, an equilibrium is obtained when the system has

maximum (w’+ Δsirr) or minimum G.

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


Thermodynamic Relations Thermodynamic Relations - For a closed - For a closed systemsystem

dU = TdS – PdV

dH = TdS + VdP

dF = – SdT – PdV

dG = – SdT + VdP


),,,,,( kji nnnPTGG

j

nnPTji

nnPTinnTnnP

dnn

Gdn

n

GdP

P

GdT

T

GdG

kikjjiji ,,,,,,,,,,,,,,

i

nnPTikj

n

G

,,,,

n

iidnVdPSdTdG1

iidnPdVw

iidn is the chemical work done by the system

▷

▷ Chemical Potential

Thermodynamic Relations Thermodynamic Relations - For a - For a multicomponent systemmulticomponent system


Gibbs EnergyGibbs Energy

▷ Effect of Temperature on G

▷ Effect of Pressure on G


Gibbs Energy for a Unary System Gibbs Energy for a Unary System - from - from dG dG = –SdT + VdP= –SdT + VdPGibbs Energy as a function of T and P

dPVdTSdG

dPVdTSPTGPTGP

P

T

Toooo

),(),(

dTT

TCSTS P

T

Too

)()( @ constant

P

dTdTT

TCSTGTG

T

T

PT

Tooo o

)()()(

@ constant T dPVPGPGP

Poo

)()(


Gibbs Energy as a function of T and P

STHG TSTSPTSHTHPTHPTG oooo ))1,(),(())1,(),((),(

STHTSHTSSHH oooo )( @ constant

P

dTTCHTH P

T

Too

)()( dTT

TCSTS P

T

Too

)()(

dTT

TCTdTTCTSHTG P

T

TP

T

Toooo

)()()(

@ constant T

dPVT

VTdPV

P

STdP

P

HH

T

P

PT

P

PT

P

P ooo

dPTVP

Po

)1(

dPVdPT

VdP

P

SS

P

PT

P

PT

P

P ooo

Gibbs Energy for a Unary System Gibbs Energy for a Unary System - from - from G G = H - ST= H - ST


Use Empirical Representation of Heat Capacities2 cTbTacP

dTdTT

TCSTGTG

T

T

PT

Tooo o

)()()(

dTT

TCTdTTCTSHTG P

T

TP

T

Toooo

)()()(

Are both of above expressions the same?

to verify that the above expressions are exactly the same.

Gibbs Energy for a Unary System Gibbs Energy for a Unary System - - Temperature DependencyTemperature Dependency


Molar volume of Fe = 7.1 cm3Expansivity = 0.3 × 10-4 K-1

∆H(1→100atm,298) = 17 cal ※ The same enthalpy increase is obtained by heating from 298 to 301 K at 1atm

Molar volume of Al = 10 cm3

Expansivity = 0.69 × 10-4 K-1

∆H(1→100atm,298) = 23.7 cal ※ The same enthalpy increase is obtained by heating from 298 to 302 K at 1atm

∆S(1→100atm,298) = -0.00052 e.u. for Fe -0.00167 e.u. for Al

※ The same entropy decrease is obtained by lowering the temperature from 298 by 0.27 and 0.09 K at 1 atm.

※ The molar enthalpies and entropies of condensed phases are relatively insensitive to pressure change

Gibbs Energy for a Unary System Gibbs Energy for a Unary System - Effect of - Effect of PressurePressure


dPVdTT

TCTdTTCTSHTG

P

P

PT

TP

T

Tooooo

)(

)()(

• 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

Gibbs Energy for a Unary SystemGibbs Energy for a Unary System


Statistical Thermodynamics Statistical 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


1. The melting point of Pb is 600K at 1 atm. Show that solidification of supercooled liquid Pb at 590K and 1 atm is a natural process, based on (1) maximum-entropy criterion (2) minimum-Gibbs-Energy criterion.

moleJH melting /4810 KmolJTC lp /101.34.32 3)( KmolJTC sp /1075.9 3

)(

2. Estimate what will eventually happen (equilibrium state) if the Pb in problem #1 is contained in an adiabatic container.

Numerical ExamplesNumerical Examples

KJTC lSnp /102.97.34 3)(,

KJTC sSnp /10265.18 3)(,

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 K7070 SnmH

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