MS212 Thermodynamics of Materials (소재열역학의이해)

Lecture Note: Chapter 6

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)

Heat Capacity

𝑑𝑈 =𝜕𝑈𝜕𝑇 &

𝑑𝑇 +𝜕𝑈𝜕𝑉 )

𝑑𝑉 = 𝒄𝑽𝑑𝑇 +𝛼𝑇𝜅)

− 𝑃 𝑑𝑉

Heat capacity is one of key quantities in calculation of thermodynamic functions

𝑑𝐻 =𝜕𝐻𝜕𝑇 1

𝑑𝑇 +𝜕𝐻𝜕𝑃 )

𝑑𝑃 = 𝒄𝑷𝑑𝑇 + 𝑉 1 − 𝛼𝑇 𝑑𝑃

𝑑𝑆 =𝜕𝑆𝜕𝑇 &

𝑑𝑇 +𝜕𝑆𝜕𝑉 )

𝑑𝑉 =𝒄𝑽𝑇 𝑑𝑇 +

𝛼𝜅)𝑑𝑉

𝑑𝑆 =𝜕𝑆𝜕𝑇 1

𝑑𝑇 +𝜕𝑆𝜕𝑃 )

𝑑𝑃 =𝒄𝑷𝑇 𝑑𝑇 + 𝛼𝑉𝑑𝑃

At dV = 0; ∆𝑈 = 6 𝒄𝑽𝑑𝑇)7

)8;∆𝑆 = 6

𝒄𝑽𝑇 𝑑𝑇

)7

)8;

At dP = 0; ∆𝐻 = 6 𝒄𝑷𝑑𝑇)7

)8;∆𝑆 = 6

𝒄𝑷𝑇 𝑑𝑇

)7

)8;

Heat Capacity1819, Dulong & Petit: 𝑐& ≈ 3𝑅 (per mole)1965, Kopp: for a solid compound 𝑐& ≈ ∑𝜈A 3𝑅�

� , e.g. cV of Al2O3 ≈ 5*(3R)

• only works for high enough temperatures• cV of softer materials rise to 3R faster

3R (24.94 J/mol-K)

softer

Einstein Model for cV (T)Einstein model of a solid 에너지

변위

ε0

ε1

ε2

ε3

ε4

n=0

n=1

n=4

n=3

n=2

(a) (b)• Atoms vibrating around the equil. position with the same frequency, nE• N particles: N harmonic oscillators for each orthogonal directions (x, y,

z) à 3N harmonic oscillators• Each harmonic oscillator: 𝜀A = 𝑖 + E

Fℎ𝜈H

• At high temperature 𝑘𝑇 ≫ ℎ𝜈H, U of each harmonic oscillator = kTà total U’ = 3NkT; U (per mole) = 3RT cV = 3R

Einstein Model for cV (T)Probability of a harmonic oscillator at energy ei at T :

Boltzmann Distribution,𝑒LMNO∑ 𝑒LMNO�A

𝑈′ = 3Q 𝑛A𝜀A�

A= 3Q 𝑁

𝑒LTUV(AXEF)/[)

∑ 𝑒LTUV(AXEF)/[)�

A

𝑖 +12 ℎ𝜈H

�

A=32𝑁ℎ𝜈H +

3𝑁ℎ𝜈H

𝑒TUV[) − 1

At high temperature, 𝑒TUV/[)à 1 + ℎ𝜈H/𝑘𝑇: 𝑈′ ≈32𝑁ℎ𝜈H + 3𝑁𝑘𝑇 ≈ 3𝑁𝑘𝑇

At low temperature, 𝑒TUV/[) ≫ 1: 𝑈′ ≈ 3𝑁ℎ𝜈H12 + 𝑒

LTUV[)

𝐶& =𝜕𝑈′𝜕𝑇 &

= 3𝑁𝑘ℎ𝜈H𝑘𝑇

F 𝑒TUV/[)

(𝑒TUV/[) − 1)F

𝑐& = 3𝑅𝜃H𝑇

F 𝑒_V/)

(𝑒_V/) − 1)F

ℎ𝜈H/𝑘 ≡ 𝜃H(per mole; molar heat capacity)

(Einstein characteristic temperature)

Einstein Model for cV (T)

𝜈H =12𝜋

𝐾𝑚

�

Natural frequency of a mass m held by a spring with spring constant K

softer materials: K , qE , cV approaches 3R sooner

The Einstein model underestimates CV at low T

Debye ModelEinstein: same frequency for all the oscillatorsDebye: a range of frequencies available to the oscillators;

n due to collective motion

Maximum available frequency:

(http://en.wikipedia.org/wiki/Debye_model)

0 La

λmin

𝜈 =𝑐d𝜆 , 𝜆g =

2𝐿𝑛 ; 𝜆iAg = 2𝑎

𝜈ikl =𝑐d2𝑎 ≈

5×10p𝑚/𝑠0.5×10Ls𝑚 = 10Ep𝑠LE

Number of vibration states, dnv, between nand n + dn ?

𝑛 𝜈 ~𝜈𝟑3𝑁~𝑣ikl𝟑Total # of vibrational states b/w 0 – nD:

𝑑𝑛w = 𝑑𝜈𝜈x

p3𝑁 =

3𝜈F𝑑𝜈𝜈xp

3𝑁

Debye frequency, nD

Total # of vibrational states b/w 0 – n:

Debye Model

In Einstein model, 𝑈y = 3𝑁ℎ𝜈H12 +

1𝑒TUV/[) − 1

total # of oscillators with fixed nE avg. energy of an oscillator

In Debye model,

𝑈′ = 63𝜈F𝑑𝜈𝜈xp

3𝑁ℎ𝜈12 +

1𝑒TU/[) − 1

Uz

{

# of oscillators with n energy of an oscillator with nℎ𝜈x𝑘 ≡ 𝜃x= 6

ℎp9𝑁ℎ𝜈p𝑑𝜈(𝜃x𝑘)p

12 +

1𝑒TU/[) − 1

Uz

{ (Debye temperature)

𝐶& =𝑑𝑈′𝑑𝑇 = 6

9𝑁ℎ}𝜈p𝑑𝜈(𝜃x𝑘)p

ℎ𝜈𝑘𝑇F 𝑒

TU/[)

𝑒TU/[) − 1 F

Uz

{=9𝑁ℎp

𝑘F𝜃xp6 𝜈F𝑑𝜈

ℎ𝜈𝑘𝑇

F 𝑒TU/[)

1 − 𝑒TU/[) F

Uz

{

Debye Model

𝑐& =9𝑁~ℎp

𝑘F𝜃xp6 𝜈F𝑑𝜈

ℎ𝜈𝑘𝑇

F 𝑒TU/[)

1 − 𝑒TU/[) F

Uz

{=9𝑅ℎp

𝑘p𝜃xp6 𝜈F𝑑𝜈

ℎ𝜈𝑘𝑇

F 𝑒TU/[)

1 − 𝑒TU/[) F

Uz

{

(per mole)

= 9𝑅𝑇𝜃x

p6

𝑥}𝑒l

1 − 𝑒l F 𝑑𝑥_z/)

{

letting 𝑥 =ℎ𝜈𝑘𝑇

• At high temperature, 𝜃x/𝑇 ≪ 1 (𝜃x/𝑇 → 0): 𝑐& ≈ 3𝑅

• At low temperature, 𝜃x/𝑇 ≫ 1 (𝜃x/𝑇 → ∞):

6𝑥}𝑒l

1 − 𝑒l F 𝑑𝑥�

{=4𝜋15

𝑐& ≈ 1943𝑇𝜃x

pJ/(mol � K)

Electronic Contribution to Heat Capacity

Fermi-Dirac distribution: 𝑓(𝜀) =1

1 + 𝑒LN�LN[)

Δ𝑈�� ∝ (𝑘𝑇)F

𝑐�� =𝜕Δ𝑈��𝜕𝑇 &,�

∝ 𝑇 =𝜋F𝑘F

2𝜀�𝑻 𝑐& = 𝐴𝑇p + 𝐵𝑇

(per mole of electrons)

f(ε)

0.0

0.5

1.0

εεF

T1 T2 T3 T1>T2>T3=0 K

𝑘𝑇

usually very small, 10-2 – 10-3

At low temperature: 𝑐& = 3𝑅 + 𝐵𝑇At high temperature:

3D lattice vibration

electroniccontribution

Heat Capacity at constant P

𝑐1 = 𝑐& +𝑉𝛼F

𝜅)𝑇 (slide 15, Lecture Note 4)

Empirically, 𝑐1 = 𝑎 + 𝑏𝑇 + 𝑐𝑇LF (for𝑇 ≫ 𝜃x)

Enthalpy (T, composition)

𝜕𝐻𝜕𝑇 1

= 𝑐1 𝐻 𝑇F − 𝐻 𝑇E = 6 𝑐1𝑑𝑇)7

)8

𝐻 𝑇F − 𝐻 𝑇E = 6 𝑐1,�𝑑𝑇)��

)8+ ∆𝐻�� + 6 𝑐1,M𝑑𝑇

)7

)��(if there is a phase transformation a à b at Ttr between T1 and T2)

Single component

Multi-components𝑎𝐴 + 𝑏𝐵 = 𝑐𝐶 + 𝑑𝐷

∆𝐻 𝑇 = 𝑐𝐻� + 𝑑𝐻x − 𝑎𝐻~ − 𝑏𝐻  =Q𝜈A𝐻A

�

�

(enthalpy of reaction: > 0, endothermic; < 0, exothermic)

𝜕∆𝐻(𝑇)𝜕𝑇 1

=Q𝜈A𝜕∆𝐻A(𝑇)𝜕𝑇 1

�

�

= Q𝜈A𝑐1,A

�

�

= 𝑐𝑐1,� + 𝑑𝑐1,x − 𝑎𝑐1,~ − 𝑏𝑐1, 

∆𝐻 𝑇F − ∆𝐻 𝑇E = 6 ∆𝑐1𝑑𝑇)7

)8Kirchhoff’s law

= ∆𝑐1

Enthalpy (T, composition)

𝐴(¡) = 𝐴(�)

∆𝐻 𝑇F = ∆𝐻 𝑇E + 6 (𝑐1,� − 𝑐1,¡)𝑑𝑇)7

)8

∆𝐻 𝑇E = 𝐻~ � 𝑇E − 𝐻~ ¡ 𝑇E∆𝐻 𝑇F

∆𝐻 𝑇E

∆𝐻 𝑇F = 𝐻~ � 𝑇F − 𝐻~ ¡ 𝑇F

∆𝐻 𝑇E + 6 𝑐1,�𝑑𝑇)7

)8− ∆𝐻 𝑇F − 6 𝑐1,¡𝑑𝑇

)7

)8= 0

1

23

4

Rearranging the equation above, 1 2 3 4

consistent with the fact that H is a state function

Hess’ law

Enthalpy at Standard State• H does not have an absolute value, only changes in H can be measured.• Convention: assign zero to the enthalpy of element i in its stable state

at T = 298 K, P = 1 atm (standard state), 𝐻A{ 298K = 0

e.g., 𝐻£¤¥¦§¨©ª{ 298K = 0, 𝐻«7(£){ 298K = 0, 𝐻~�(¬){ 298K = 0

HF ¬ +12OF ¬ = HFO(�)

∆𝐻 = 𝐻¯F°(�) − 𝐻¯F ¬ −12𝐻°F(¬) ≡ ∆𝐻𝒇,¯F°(�)Enthalpy of reaction:

Enthalpy of formation (b/c H2O forms from its constituent “elements” by reaction)

If the reaction occurs at 298 K,

∆𝐻 298K = 𝐻²,¯F°{ = 𝐻¯F°{ − 𝐻¯F{ −12𝐻°F

{ = 𝐻¯F°{

Standard enthalpy of formation of H2O

Standard enthalpy of H2O

𝐻¯F°(�){ 298K ≠ 0

Enthalpy of Reaction: ExampleM+

12OF = MO

T1

T2

T3

∆𝐻 298K = 𝐻²,µ°{ = 𝐻µ°{ (< 0)

∆𝐻 𝑇E = ∆𝐻 298K + 6 ∆𝑐1𝑑𝑇)8

Fs·∆𝐻 𝑇E

∆𝐻 𝑇F

∆𝐻 𝑇p

∆𝐻i = 𝐻𝒍 − 𝐻𝒔

∆𝑐1 = 𝑐1,µ° − 𝑐1,µ −12 𝑐1,°F

∆𝐻 𝑇F = ∆𝐻 298K + 6 ∆𝑐1𝑑𝑇)º,»

Fs·

−∆𝐻i,µ + 6 ∆𝑐1𝑑𝑇)7

)º,»

∆𝐻 𝑇p = ∆𝐻 298K + 6 ∆𝑐1𝑑𝑇)º,»

Fs·− ∆𝐻i,µ + 6 ∆𝑐1𝑑𝑇

)º,»¼

)º,»

+ ∆𝐻i,µ° + 6 ∆𝑐1𝑑𝑇)½

)º,»¼

Kirchhoff’s law:

Hess’ law:∆𝐻 𝑇p = 6 (𝑐1,µ � +

12 𝑐1,°F ¬ )𝑑𝑇

)º,»

)½− ∆𝐻i,µ + 6 𝑐1,µ ¡ +

12 𝑐1,°F ¬ 𝑑𝑇

Fs·

)º,»

+ ∆𝐻(298K)

initial

final

+6 𝑐1,µ° ¡ 𝑑𝑇)º,»¼

Fs·+ ∆𝐻i,µ° + 6 𝑐1,µ° � 𝑑𝑇

)7

)º,»¼

Entropy (T, composition)

𝜕𝑆𝜕𝑇 𝑷

=𝑐1𝑇 𝑆 𝑇F, 𝑃 − 𝑆 𝑇E, 𝑃 = 6 𝑐1𝑑 ln 𝑇

)7

)8

𝑆 𝑇F, 𝑃 − 𝑆 𝑇E, 𝑃 = 6 𝑐1,�𝑑 ln 𝑇)��

)8+∆𝐻��𝑇��

+ 6 𝑐1,M𝑑 ln 𝑇)7

)��(if there is a phase transformation a à b at Ttr between T1 and T2)

Single component

𝜕𝑆𝜕𝑇 𝑽

=𝑐&𝑇

𝑆 𝑇F, 𝑉 − 𝑆 𝑇E, 𝑉 = 6 𝑐&𝑑 ln 𝑇)7

)8

𝑆 𝑇F, 𝑉 − 𝑆 𝑇E, 𝑉 = 6 𝑐&,�𝑑 ln 𝑇)��

)8+∆𝑈��𝑇��

+ 6 𝑐&,M𝑑 ln 𝑇)7

)��

• At constant P

• At constant V

Multi-components

∆𝑆 𝑇F, 𝑃 − ∆𝑆 𝑇E, 𝑃 = 6 ∆𝑐1𝑑 ln 𝑇)7

)8;∆𝑆 𝑇F, 𝑉 − ∆𝑆 𝑇E, 𝑉 = 6 ∆𝑐&𝑑 ln 𝑇

)7

)8Kirchhoff’s law

Absolute Value of Entropy

𝑆 𝑇 = 𝑆 0K + 6 𝐶1𝑑 ln 𝑇)

{, 𝑆 0K =?

• 1906, Nernst

∆𝐺 𝑇 = ∆𝐻 𝑇 − 𝑇∆𝑆 𝑇

Chemical rnx: pure liquid à pure solid

lim)→{

𝜕∆𝐻𝜕𝑇 1

= 0; lim)→{

𝜕∆𝐺𝜕𝑇 1

= 0

lim)→{

∆𝑐1 = 0

lim)→{

∆𝑆 = 0

okay because cP,i à 0 as T à 0 K (Debye theory)

Si à 0 as T à 0 K ?

The 3rd Law of Thermodynamics

Planck: “The entropy of any homogeneous substance, which is in complete internal equilibrium, may be taken to be zero at 0 K.”

Meaning of S(0K) = 0

𝑆 0K = 𝑆©§ + 𝑆ÂÃÄÅ = 𝑘 lnΩ©§ + 𝑘 lnΩÂÃÄÅ = 0

Ω©§ = 1: internal equilibrium (all at the ground state)ΩÂÃÄÅ = 1: internal equilibrium (homogeneous)

If not in complete internal equilibrium? • S(0K) may not be zero: residual entropy or frozen-in entropy• Why? Atomic mobility frozen not to achieve the homogeneous

internal equilibrium state

Residual Entropy• Example 1: glass

• super-cooled liquid• liquid-state “frozen into” the solid-state• atoms not in their equilibrium positions

at ground state

• Example 2: solid solution

ΩÂÃÄÅ =𝑁!

𝑛~! 𝑛 !𝑆¤ªÉ = 𝑘 lnΩÂÃÄÅ = −𝑁𝑘(𝑋Ë ln𝑋Ë + 𝑋Ì ln𝑋Ì)

diffusion required

random mixture ordered: ground state when A and B “like” each other

phase separated:ground state when A and B “hate” each other

ΩÂÃÄÅ = 1

ΩÂÃÄÅ = 1

Residual Entropy• Example 3: isotopic mixture

• Example 4: frozen-in point defects such as vacancies

e.g., O16, O17

, O18….

𝑆¤ªÉ = −𝑁𝑘(𝑋EÍ ln𝑋EÍ + 𝑋EÎ ln𝑋EÎ + 𝑋E· ln𝑋E·)

At high T, nv out of (N+nv) lattice sites à removal of vacancies at 0 K requires __________ .

ΩÂÃÄÅ =(𝑁 + 𝑛&)!𝑁! 𝑛&!

; 𝑆¤ªÉ = −𝑘 𝑛Ï ln𝑛Ï

𝑁 + 𝑛Ï+ 𝑁 ln

𝑁𝑁 + 𝑛Ï

• Example 5CO CO CO COCO CO CO COCO CO CO CO

due to limited mobility

CO OC CO COCO CO OC COOC CO CO CO

ground stateΩÂÃÄÅ = 1 ΩÂÃÄÅ = 2�

𝑆¤ªÉ = 𝑘 ln 2� = 𝑁𝑘 ln 2 = 𝑅 ln 2 = 5.832J/(mol � 𝐾)

𝑁 = 𝑁~(per mole)

Experimental Verification of the 3rd Law∆𝑆ÐÏ = ∆𝑆Ð + ∆𝑆ÐÐ + ∆𝑆ÐÐÐ

If the 3rd law to be true, ∆𝑆ÐÏ = 0à ∆𝑆ÐÐ = −(∆𝑆Ð + ∆𝑆ÐÐÐ)

∆𝑆Ð = 6𝑐1(�)𝑇 𝑑𝑇

)��

{, ∆𝑆ÐÐÐ = 6

𝑐1(M)𝑇 𝑑𝑇

{

)��∆𝑆ÐÐ =

∆𝐻��𝑇��

,

𝑆 𝑇 = 6𝑐1𝑇 𝑑𝑇

)

{= 𝑆 298K + 6 𝑐1𝑑 ln 𝑇

)

{

Third law entropy

tabulated tabulated

third law entropy change

third law entropy change

experimental entropy change

For sulfur, a = orthorhombic, b = monoclinic, Ttr = 368.5 K,∆𝑆Ð = 36.86J/ mol � 𝐾 , ∆𝑆ÐÐ = 1.09J/ mol � 𝐾 , ∆𝑆ÐÐÐ = −37.8J/ mol � 𝐾 ,

∆𝑆ÐÏ ≈ 0J/ mol � 𝐾 (within experimental uncertainty)

or, = 𝑆 298K + 6 𝑐1(¡)𝑑 ln 𝑇)º

{+ ∆𝑆i + 6 𝑐1(�)𝑑 ln 𝑇

)

)º

Empirical formula

For elemental metalsS à L à V

Richard’s rule Trouton’s rule

FCC:∆𝐻Õ = 9.61𝑇Õ ± 0.01J/mol

BCC:∆𝐻Õ = 8.25𝑇Õ ± 0.19J/mol

∆𝑆Õ =∆𝐻Õ𝑇Õ

∆𝑆Õ ≈ 𝑅 ∆𝑆Õ ≈ 10𝑅

Pressure Dependence of H and SIn condensed phase, negligible!

𝜕𝐻𝜕𝑃 )

= 𝑇𝜕𝑆𝜕𝑃 )

+ 𝑉 = −𝑇𝜕𝑉𝜕𝑇 1

+ 𝑉 = −𝛼𝑇 + 1 𝑉

DH at T ≈ 1000 K ? • Typically, for a condensed matter, a ≈ 10-5–10-6 K-1, V ≈ 10-5 m3/mol• ∆𝐻 ≈ 𝑉(1 − 𝛼𝑇)∆𝑃 ≈ (10LF𝐿/mol)(∆𝑃atm) Ú

{.{·F{ÛÜ�ÝÞßßàá�â

≈ 0.1𝑅 � ∆𝑃

• For DP ≈ 100 atm, DH ≈10R• How much DT to cause DH ≈10R ? ∆𝐻 ≈ 𝑐1∆𝑇 ≈ 3𝑅∆𝑇 à ∆𝑇 = 3K!

𝜕𝑆𝜕𝑃 )

= −𝜕𝑉𝜕𝑃 )

= −𝛼𝑉

DS at T ≈ 1000 K ? • ∆𝑆 ≈ −𝛼𝑉∆𝑃 ≈ −(10LÛ𝐾LE)(10LF𝐿/mol)(∆𝑃atm) Ú

{.{·F{ÛÜ�ÝÞßßàá�â• For DP ≈ 100 atm, DS ≈ 10-4R• How much DT to cause DS ≈10-4R ? ∆𝑆 ≈ ãä

)∆𝑇 ≈ pÚ

)∆𝑇 à ∆𝑇 ≈ −0.3K!

𝐻 𝑇, 𝑃 ≈ 𝐻 𝑇 ; 𝑆 𝑇, 𝑃 ≈ 𝑆 𝑇