MS212 Thermodynamics of Materials (소재열역학의이해)

Lecture Note: Chapter 5

Byungha Shin (신병하)Dept. of MSE, KAIST

1

2017 Spring Semester

CourseInformationSyllabus

Chap 1. Introduction and Definition of Terms (1 lecture)Chap 2. The First Law of Thermodynamics (2 lectures)Chap 3. The Second Law of Thermodynamics (4 lectures)Chap 4. The Statistical Interpretation of Entropy (2 lectures)Chap 5. Auxiliary Functions (2 lectures)Chap 6. Heat Capacity, Enthalpy, Entropy and the Third

Law of Thermodynamics (3 lectures)Chap 7. Phase Equilibrium in a One-Component System (3 lectures)Chap 8. The Behavior of Gases (3 lectures)Chap 9. The Behavior of Solutions (3 lectures)Chap 10. Gibbs Free Energy Composition and Phase Diagram (3 lectures)

Criterion for Equilibrium: U, SThe 1st and 2nd Law of Thermodynamics: 𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉

inconvenient choice of independent variables à new equation of state, similarly simple, with more practical choice of variables (such as P and T) without losing the meaning (criterion for equilibrium)

Criterion for equilibrium

𝑑𝑈 = 𝑇 𝑑𝑆 − 𝑑𝑆)** − 𝑃𝑑𝑉 − 𝛿𝑤-. ≤ 𝑇𝑑𝑆 − 𝑃𝑑𝑉 − 𝛿𝑤-.

𝛿𝑞 = 𝑇𝑑𝑆 − 𝑇𝑑𝑆)**; 𝛿𝑤 = 𝑃𝑑𝑉 + 𝛿𝑤-.work other than PV work such as electrical work

𝑑𝑈)𝑺,𝑽,𝒘𝒆𝒙;𝟎 ≤ 0 (energy minimum principle)

𝑑𝑆)𝑼,𝑽 ≥ 0 (entropy maximum principle)

(≥ 0)

(closed system, no work other than PV)

Legendre Transformation𝑧 = 𝑓(𝑥)

envelope of tangent lines (slope m + intercept at x=0, y)

𝑚 =𝑧 − 𝜓𝑥 − 0

𝑧 = 𝑓 𝑥 𝜓 = 𝑧 −𝑚𝑥 = 𝜓 𝑚 = 𝑧[𝑚]

equivalent

𝑧 = 𝑓 𝑥, 𝑦 𝜓 = 𝑧 − 𝑓.𝑥 − 𝑓I𝑦 = 𝜓(𝑓., 𝑓I)

= 𝑧[𝑓., 𝑓I]

𝑧

𝑧

𝑧 = 𝜓

𝑧 = 𝑧

𝑧

Auxiliary Functions𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉

𝑈J =𝜕𝑈𝜕𝑆 L

= 𝑇;𝑈L =𝜕𝑈𝜕𝑉 J

= −𝑃

partial derivative of U with respect to S

𝑑𝑈 =𝜕𝑈𝜕𝑆 L

𝑑𝑆 +𝜕𝑈𝜕𝑉 J

𝑑𝑉

𝑈 𝑃 = 𝑈 − 𝑈L𝑉 = 𝑈 + 𝑃𝑉 ≡ 𝐻𝑧 = 𝑓 𝑥 ↔ 𝑈 = 𝑈(𝑉)

𝜓 = 𝑧 − 𝑓.𝑥 ↔ 𝐻 = 𝑈 − 𝑈L𝑉

𝑧 = 𝑓 𝑥 ↔ 𝑈 = 𝑈(𝑆)𝜓 = 𝑧 − 𝑓.𝑥 ↔ 𝐹 = 𝑈 − 𝑈J𝑆

𝑈 𝑇 = 𝑈 − 𝑈J𝑆 = 𝑈 − 𝑇𝑆 ≡ 𝐴

𝑧 = 𝑓 𝑥, 𝑦 ↔ 𝑈 = 𝑈(𝑉, 𝑆)𝜓 = 𝑧 − 𝑓.𝑥 − 𝑓I𝑦 ↔ 𝐺 = 𝑈 − 𝑈L𝑉 − 𝑈J𝑆

𝑈 𝑇, 𝑃 = 𝑈 − 𝑈L𝑉 − 𝑈J𝑆 = 𝑈 + 𝑃𝑉 − 𝑇𝑆 ≡ 𝐺

Helmholtz Free Energy

Enthalpy

Gibbs Free Energy

Auxiliary Functions: Differential Form

𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉𝑑𝑈 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃𝑑 𝑈 + 𝑃𝑉 ≡ 𝑑𝐻 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃

𝑑𝑈 − 𝑇𝑑𝑆 − 𝑆𝑑𝑇 = −𝑆𝑑𝑇 − 𝑃𝑑𝑉𝑑 𝑈 − 𝑇𝑆 ≡ 𝑑𝐴 = −𝑆𝑑𝑇 − 𝑃𝑑𝑉

differential form of 𝐻 = 𝐻(𝑆, 𝑃)

differential form of 𝐴 = 𝐴(𝑇, 𝑉)

𝑑𝑈 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃 − 𝑇𝑑𝑆 − 𝑆𝑑𝑇 = −𝑆𝑑𝑇 + 𝑉𝑑𝑃𝑑 𝑈 + 𝑃𝑉 − 𝑇𝑆 ≡ 𝑑𝐺 = −𝑆𝑑𝑇 + 𝑉𝑑𝑃

differential form of 𝐺 = 𝐺(𝑇, 𝑃)

𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉

𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉

Physical Meaning of Free Energy

Wmax = (Free Energy)initial – (Free Energy)final = – D(Free Energy)

State 1

maximum (when reversible) amount of work that you can get out

P

V

P1

P2V1 V2

1

2

isotherm

State 2

minimum (when reversible) amount of work that you’d have to put in

==

D(free energy)

In terms of useful work, Wuseful (not counting pushing the atmosphere out of the way, i.e, PV work)• At constant S and P, 𝑊XYZ[X\ = −∆𝐻 = − 𝑇∆𝑆 + 𝑉∆𝑃 −𝑊Z^ = 𝑊Z^• At constant T and V, 𝑊XYZ[X\ = −∆𝐴 = − −𝑆∆𝑇 − 𝑃∆𝑉 −𝑊Z^ = 𝑊Z^• At constant T and P, 𝑊XYZ[X\ = −∆𝐺 = −(−𝑆∆𝑇 − 𝑉∆𝑃 −𝑊Z^) = 𝑊Z^

Criterion for Equilibrium: H, A, G

𝑑𝐻 = 𝑑𝑈 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃 ≤ 𝑇𝑑𝑆 + 𝑉𝑑𝑃 − 𝛿𝑤-.

𝛿𝑞 = 𝑇𝑑𝑆 − 𝑇𝑑𝑆)**; 𝛿𝑤 = 𝑃𝑑𝑉 + 𝛿𝑤-.work other than PV work such as electrical work

(≥ 0)𝑑𝑈 = 𝑇 𝑑𝑆 − 𝑑𝑆)** − 𝑃𝑑𝑉 − 𝛿𝑤-. ≤ 𝑇𝑑𝑆 − 𝑃𝑑𝑉 − 𝛿𝑤-.

• 𝛿𝑤-. ≤ −𝑑𝐻)J,_:max useful work under const. S, P• 𝑑𝐻)J,_,`ab;c ≤ 0:enthalpy minimum principle

Similarly, • 𝛿𝑤 ≤ −𝑑𝐴)d:max work under const. T

• 𝑑𝐴)d,L,`ab;c ≤ 0: Helmholtz free energy minimum principle

• 𝛿𝑤-. ≤ −𝑑𝐺)d,_:max useful work under const. T, P• 𝑑𝐺)d,_,`ab;c ≤ 0:Gibbs free energy minimum principle

𝑑𝐴 ≤ −𝑆𝑑𝑇 − 𝑃𝑑𝑉 − 𝛿𝑤-. = −𝑆𝑑𝑇 − 𝛿𝑤total work including PV work

• 𝛿𝑤-. ≤ −𝑑𝐴)d,L:max useful work under const. T, V

𝑑𝐺 ≤ −𝑆𝑑𝑇 + 𝑉𝑑𝑃 − 𝛿𝑤-.

Open SystemExchange of heat: diathermal wall (O), adiabatic wall (X)Exchange of work: moveable wall (O), immoveable wall (X)Exchange of mass: permeable wall (O), impermeable wall (X)

Isolated system: heat X, work X, mass X (𝛿𝑞 = 0, 𝛿𝑞 = 0, 𝛿𝑛) = 0)Closed system: heat O, work O, mass X (𝛿𝑞 ≠ 0, 𝛿𝑞 ≠ 0, 𝛿𝑛) = 0)Open system: heat O, work O, mass O (𝛿𝑞 ≠ 0, 𝛿𝑞 ≠ 0, 𝛿𝑛) ≠ 0)

Closed system𝑈 = 𝑈(𝑆, 𝑉)

Open system with C components𝑈 = 𝑈(𝑆, 𝑉, 𝑛g, 𝑛h, … . , 𝑛), … . , 𝑛k)

U of an open system is still a state function:

𝑑𝑈 =𝜕𝑈𝜕𝑆 L,lm

𝑑𝑆 +𝜕𝑈𝜕𝑉 J,lm

𝑑𝑉 +n𝜕𝑈𝜕𝑛) J,L,lopm

k

);g

𝑑𝑛)

= 𝑇𝑑𝑆 − 𝑃𝑑𝑉 +n𝜕𝑈𝜕𝑛) J,L,lopm

k

);g

𝑑𝑛)

Chemical Potential𝜕𝑈𝜕𝑛) J,L,lopm

≡ 𝜇)electrochemical potentialor, chemical potential if particles have no charge

Imagine a semi-permeable wall allowing exchange of only i species:

𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉 +n𝜇)𝑑𝑛)

�

�

𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉 + 𝜇)𝑑𝑛) = 𝑇𝑑𝑆 − 𝑃𝑑𝑉 − 𝛿𝑤-.

𝜇)𝑑𝑛) = −𝛿𝑤-.transfer of mass, dni (>0) into the system from the surrounding, internal energy of the system increases by µi dni, which is the same as chemical work done by the surrounding, –dwex.

𝑑𝐺 = −𝑆𝑑𝑇 − 𝑃𝑑𝑉 +n𝜇)𝑑𝑛)

�

�

;

In other thermodynamic potential functions of an open system, the chemical potential term should be added.

For example:

𝜇) =𝜕𝑈𝜕𝑛) J,L,lopm

=𝜕𝐻𝜕𝑛) J,_,lopm

=𝜕𝐴𝜕𝑛) d,L,lopm

=𝜕𝐺𝜕𝑛) d,_,lopm

Thermodynamic RelationsThermodynamic potential function

1st derivative (equation of state) 2nd derivative

𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉 𝑇 =𝜕𝑈𝜕𝑆 L

;−𝑃 =𝜕𝑈𝜕𝑉 J

𝜕h𝑈𝜕𝑆h L

;𝜕h𝑈𝜕𝑆𝜕𝑉 =

𝜕h𝑈𝜕𝑉𝜕𝑆 ;

𝜕h𝑈𝜕𝑉h J

𝜕𝑇𝜕𝑆 L

;𝜕𝑇𝜕𝑉 J

= −𝜕𝑃𝜕𝑆 L

;−𝜕𝑃𝜕𝑉 J

𝑑𝐻 = 𝑇𝑑𝑆 + 𝑉𝑑𝑃 𝑇 =𝜕𝐻𝜕𝑆 _

; 𝑉 =𝜕𝐻𝜕𝑃 J

𝜕𝑇𝜕𝑆 _

;𝜕𝑇𝜕𝑃 J

=𝜕𝑉𝜕𝑆 _

;𝜕𝑉𝜕𝑃 J

𝑑𝐴 = −𝑆𝑑𝑇 − 𝑃𝑑𝑉 −𝑆 =𝜕𝐴𝜕𝑇 L

;−𝑃 =𝜕𝐴𝜕𝑉 d

−𝜕𝑆𝜕𝑇 L

;−𝜕𝑆𝜕𝑉 d

= −𝜕𝑃𝜕𝑇 L

;−𝜕𝑃𝜕𝑉 d

𝑑𝐺 = −𝑆𝑑𝑇 + 𝑉𝑑𝑃 −𝑆 =𝜕𝐺𝜕𝑇 _

; 𝑉 =𝜕𝐺𝜕𝑃 d

−𝜕𝑆𝜕𝑇 _

;−𝜕𝑆𝜕𝑃 d

=𝜕𝑉𝜕𝑇 _

;𝜕𝑉𝜕𝑃 d

Among sixteen 2nd derivatives, how many of them are independent?

Thermodynamic Relations

𝜕𝑆𝜕𝑇 _

?

𝑑𝑆)_ =𝛿𝑞)_𝑇 ;

𝛿𝑞)_𝑑𝑇 = 𝐶_

𝜕𝑉𝜕𝑇 _

?

∴𝜕𝑆𝜕𝑇 _

=𝑪𝑷𝑇

𝜶 ≡1𝑉

𝜕𝑉𝜕𝑇 _

thermal expansion coefficient

𝜕𝑉𝜕𝑃 d

?

𝜿𝑻 ≡ −1𝑉

𝜕𝑉𝜕𝑃 d

isothermal compressibility

• Derivatives of thermodynamic potential functions can be described by a combination of materials properties (CP, a, kT) and state properties (T, P, V)

Thermodynamic RelationsSome useful formulae

𝑑𝑧 =𝜕𝑧𝜕𝑥 I

𝑑𝑥 +𝜕𝑧𝜕𝑦 .

𝑑𝑦

𝑧 = 𝑧(𝑥, 𝑦)when dy = 0

1 =𝜕𝑧𝜕𝑥 I

𝜕𝑥𝜕𝑧 I

;𝜕𝑧𝜕𝑥 I

=1𝜕𝑥𝜕𝑧 I

• inversion formula

• upstairs-downstairs-inside-out formula

𝑑𝑧 =𝜕𝑧𝜕𝑥 I

𝑑𝑥 +𝜕𝑧𝜕𝑦 .

𝑑𝑦

divide by dyand set dz =0

0 =𝜕𝑧𝜕𝑥 I

𝜕𝑥𝜕𝑦 {

+𝜕𝑧𝜕𝑦 .

𝜕𝑥𝜕𝑦 {

𝜕𝑦𝜕𝑧 .

𝜕𝑧𝜕𝑥 I

= −1

Thermodynamic RelationsGuidelines

If a partial derivative contains,1. a thermodynamic potential function, then bring it to numerator and

replace it with differential form expressed by state properties (first column of Table in Slide 11).

2. dX)S, then bring S inside the bracket and bring to numerator using upstairs-downstairs-inside-out formula, and if necessary, inversion formula. Then use 𝑑𝑆)_ = 𝐶_𝑑𝑇/𝑇 or 𝑑𝑆)L = 𝐶L𝑑𝑇/𝑇.

3. dS)X in numerator, but if X is not P nor V, then use Maxwell relations of thermodynamic potential function based on X and variable in the denominator.

4. dX)V, then bring V inside the bracket and bring it to numerator using }_}L d

}L}d _

}d}_ L

= −1 and, if necessary, inversion formula. Then it can be replaced with an expression containing a, kT.

Thermodynamic RelationsExample 1

𝐶_ − 𝐶L = 𝑃 +𝜕𝑈𝜕𝑉 d

𝜕𝑉𝜕𝑇 _

=𝑇𝑉𝛼h

𝜅d

𝜕𝑈𝜕𝑉 d

= 𝑇𝜕𝑆𝜕𝑉 d

− 𝑃 Guideline 1

= 𝑇𝜕𝑃𝜕𝑇 L

− 𝑃 Guideline 3

= −𝑇𝜕𝑃𝜕𝑉 d

𝜕𝑉𝜕𝑇 _

− 𝑃

= −𝑇

𝜕𝑉𝜕𝑇 _𝜕𝑉𝜕𝑃 d

− 𝑃 Guideline 4

= −𝑇𝛼𝑉−𝜅d𝑉

− 𝑃

𝜕𝑉𝜕𝑇 _

= 𝛼𝑉

Guideline 4

Thermodynamic RelationsExample 2

𝜅J ≡ −1𝑉

𝜕𝑉𝜕𝑃 J

Guideline 2 with =1𝑉

𝜕𝑆𝜕𝑃 L

𝜕𝑉𝜕𝑆 _

=1𝑉

𝜕𝑆𝜕𝑃 L𝜕𝑆𝜕𝑉 _

Guideline 4 with inversion formula

Guideline 2

Adiabatic compressibility

}L}_ J

}_}J L

}J}L _

= −1

Guideline 2 with inversion formula

=1𝑉

𝐶L𝑇

𝜕𝑇𝜕𝑃 L

𝐶_𝑇

𝜕𝑇𝜕𝑉 _

=1𝑉𝐶L𝐶_

𝜕𝑇𝜕𝑃 L

𝜕𝑉𝜕𝑇 _

= −1𝑉𝐶L𝐶_

𝜕𝑉𝜕𝑃 d

Guideline 4 with u-d-i-o formula

∴ 𝜅J ≡𝐶L𝐶_𝜅d

Gibbs-Helmholtz EquationG = H – TSG à S and H ? Yes

𝑆 = −𝜕𝐺𝜕𝑇 _

𝐻 = 𝐺 + 𝑇𝑆 = 𝐺 − 𝑇𝜕𝐺𝜕𝑇 _

−𝐻𝑇h = −

𝐺𝑇h +

1𝑇𝜕𝐺𝜕𝑇 _

=𝜕(𝐺 𝑇⁄ )𝜕𝑇 _

𝜕 1 𝑇⁄ = −𝜕𝑇/𝑇h

𝐻 =𝜕(𝐺 𝑇⁄ )𝜕(1 𝑇⁄ ) _

Similarly, A = U – TSA à S and U ?

𝑆 = −𝜕𝐴𝜕𝑇 L

𝑈 =𝜕(𝐴 𝑇⁄ )𝜕(1 𝑇⁄ ) L