Chapter 1. Thermodynamics

Kinetic processes in materials

Chapter 1. Thermodynamics

Assoc.Prof. Nguyen Hong Hai

Hanoi University of Science and Technology

Textbook: Phase Transformations in Metals and Alloys. D.A. Porter and K.E. Easterling

Student DutiesClass attendance: ≥ 80%Homework: 100%Lab work: 100%

AssessmentMid-term grade: 0.3 (30%)Homework problemsIndependent laboratory project conducted by teamsFinal exam (multichoice and writing): 0.7 (70%)

Kinetic processes in Materials

Content

Rewiew of thermodynamics

Diffusion

Surfaces and interfaces

Phase transformation


Chapter 1. THERMODYNAMICS

The main use of thermodynamics in physical metallurgy is to

allow the prediction of whether an alloy is in equilibrium.

In considering phase transformation we are concerned with

changes towards equilibrium, and thermodynamics is

therefore very powerful tool.

However, the rate at which equilibrium is reached cannot be

determined by thermodynamics alone.



Figure 1a represents the physical space of materials.

The outermost circle is the performance circle,

which characterizes how a material behaves in

given environments.

The performance is realized by a set of properties,

in the second circle, relevant to those environments.

Those properties are achieved by an agglomeration

of phases with various structures and their

morphologies in the third circle.

The structures are obtained through proper processing in the fourth circle

with accurate chemical compositions in the fifth circle.


While the two outermost circles are environment-dependent, the three inner

circles can be directly related to the three key components in materials

science and engineering, crystallography, kinetics, and thermodynamics, as

shown in Figure 1b.

Thermodynamics and kinetics of materials are viewed as difficult subjects.

1.1. Equilibrum

System: mixture

of one or more

phases

Phase: a portion of the

system whose properties

and composition are

homogeneous and which

is physically distinct from

other part of the system.

A356


The main question of Thermodynamics is if a system is in equilibrium?

1.1. Equilibrum


Components: the

different elements

or chemical

compounds which

make up the system.

The composition of

a phase or the

system can be

described by giving

the relative amounts

of each component.

1.1. Equilibrum


T, 0C

1538

1394

G912

500

Fe 1 2 3 4 5 6 7

% C

l

l +

B 14870

l + l + Fe3C

11480

C

E F

+ Fe3C +

P S 7270 K

+ Fe3C

N

Fe3 C

Phase transformation: one or more phases in an alloy (the system)

change into a new phase or mixture of phases.

1.1. Equilibrum

-Al5FeSi - Al8Fe2Si

Stable Unstable

Desired

In reality


What determine the stability of phases?

At constant temperature and pressure the relative stability of a system is determined by its Gibbs free energy

G = H – TS (1.1)

H: enthalpy, the measure of the heat content of the system

H = E + PV (1.2)

E : internal energy of the system, P: the pressure and V: the volume.

For the condensed phases (solids and liquids) the PV term is usually very small in comparison with E, so H E.

T: absolute temperature

S: entropy of the system, the measure of the randomness of the system.

The intensive properties (independent on the size of the system): T and P

The extensive properties (directly proportional to the quantity of material in the system such as V,E,H,S,G)


1.1. Equilibrum

A system is said to be in equilibrum when

it is in the most stable state;

At constant temperature and pressure a

closed system (i.e. one of fixed mass and

composition) will be in stable equilibrum

if it has the lowest value of the Gibbs free

energy, or in mathematical term:

dG = 0 (1.3)

The state with the highest stability will be

that with low enthalpy and high entropy. Metastable (diamond)

Stable (graphite)

Unstable (dG 0)

dG = 0

dG = 0

AB

Arrangement of atoms

GibbsfreeenergyG


1.1. Equilibrum

Any transformation that results in a decrease in Gibbs free energy is

possible. Therefor a necessary criterion for any phase transformation is:

= G2 – G1 0 (1.4)

The answer to the question “How fast does a phase transformation

occur?” belong to the realm of kinetics

The usual way of measuring the size of the system is by the number of

moles of material it contains (the extensive properties are then molar

quantities, i.e. expressed in units per mole).

The number of moles of a given component in the system is given by the

mass of the component in grams divided by its atomic or molecular

weight.

The number of atoms or molecules within 1 mol of material is given by

Avogadro’s number (Na) and is 6.023x1023.


1.1. Equilibrum

Single component system: which contains a pure element or one

type of molecule that does not dissociate over the range of

temperature of interest.

In order to predict the phases that are stable or mixtures that are in

equilibrium at different temperatures it is necessary to calculate the

variation of G with T

1.2. Single component system


1.2.1. Gibbs Free Energy as a Function of Temperature

The variation of enthalpyH with T can be calculated by intergrating equation 1.5

The variation of entropywith temperature can also be derived from the specific heat CP, taking entropy at zero degrees Kelvin as zero

Specific heat: the quantity of heat (in jouls) required to raise the temperature of the substance by one degree Kelvin P

PT

HC

T

PdTCH298

T

P dTT

CS

0

(1.5)


Finally the variation of G with temperature (Fig 1.3) can be

obtained by combining Fig 1.2b and 1.2c using Equation 1.1

TS

H

G

H

T(K)

slope = -S

slope = CP

G

G = H – TS (1.1)

G decreases with increasing T at the rate given by –S

ST

G

P



Note:

1. At Tm both liquid and solid

phases have the same value of

G and can exist in equilibrium.

Tm is therefore the equilibrium

melting temperature at the

pressure concerned.

2. At Tm the heat supplied to the

system will not raise its

temperature but will be

supplying the latent heat of

melting (L) that required to

convert solid into liquid (line bc

in Fig. 1.4)

Fig. 1.4. Variation of entalpy (H) and free energy (G) with temperature for the solid and liquid phases of a pure metal.



The free energies of the liquid and solid at a temperature T are given by:

GL = HL - TSL

GS = HS - TSS

Therefore at a temperature T:

G = H - TS, [J.mol-1] (1.10)

H = HL – HS

S = SL - SS

If a liquid metal is undercooled by T below Tm before it solidifies,

solidification will be accompanied by a decrease of free energy G (J mol-1)

This free energy decrease provides the driving force for solidification.



Molar free energy

Temperature

GS

GL

G

T

TmT

At the equilibrium melting

temperature Tm the free

energies of solid and liquid are

equal, i.e. G = 0.

Consequently:

G = H - TmS = 0

And therefore at Tm

This is known as the entropy of fusion and for most metals is a constant R (8.3 J mol-1 K-1).

mm T

L

T

HS

(1.11)



Molar free energy

Temperature

GS

GL

G

T

TmT

Note: For small undercooling (T) the difference in the specific heats

of the liquid and solid (CLP – CS

P) can be ignored.

So, H and S therefore approximately independent of temperature.

Combining Equations 1.10 and 1.11 thus gives:

1.2.2. The driving Force for Solidification

Conclusion: G is proportional to the T (driving force for solidification

is proportional to the undercooling).

mT

LTLG

TST

TLG

m

G = H - TS (1.10)

mm T

L

T

HS

(1.11)


Molar free energy

Temperature

GS

GL

G

T

TmT

Metal Undercooling,

0 C

Metal Undercooling,

0 C

Metal Undercooling,

0 C

Hg 77 Sb 135 Mn 308

Ga 76 Ge 227 Ni 319

Sn 118 Ag 227 Co 330

Bi 90 Au 230 Fe 295

Pb 80 Cu 236 Pd 332

Al 195 Pb 80 Pt 370

The maximum undercooling, T, is obtained by metal atomisation,

which is needed for homogeneous nucleation. Tmax 1/3 Tm


1.2.2. The driving Force for Solidification

How much is maximum undercooling?

In practice the pure metals (single component system) are rarely applied.

In single component system all phases have the same composition, and

equilibrium simply involves pressure and temperature as variables

In alloys composition is also variable

Since pressure is usually fixed at 1atm, most attention will be given to

changes in composition and temperature

1.3. Binary Solutions


The Gibbs free energy of a binary solution of A and B atoms can be calculated from the energies of pure A and pure B

Assumed that A and B have the same crystal structure , can be mixed in any proportion to make a solid solution with the same crystal structure.

Assumed that 1 mol of homogeneous solid solution is made by mixing together XA mol of A vµ XB mol of B.

XA and XB are the mole fraction of A and B in the alloy.

Note: XA + XB = 1 (1.13)

1.3.1. The Gibbs Free Eneggy of Binary Solutions

The Gibbs free energy of the system before mixing is given by:

G1 = XAGA + XBGB J.mol-1 (1.14)0 XB 1A B

GA

GB

G1

Free energy per mole beformixing


After mixing the free energy of the

solid solution, G2 will be:

G2 = G1 + Gmix (1.15)

where Gmix is the change in Gibbs

energy caused by the mixing.

G1 = H1 - TS1

G2 = H2 - TS2

Hmix = H2 - H1

Smix = S2 - S1

Gmix = G2 – G1 = Hmix - TSmix (1.16)

Smix is the difference in entropy between the mixed and unmixed states.


1.3.1. The Gibbs Free Eneggy of Binary Solutions

where Hmix is the heat absorbed or evolved during mixing, representing the difference in internal energy (E) before and after mixing.

It can be called heat of mixing.

If Hmix = 0, the resultant solution is said to be ideal and the free energy

change on mixing is only due to the change in entropy:

Gmix = -T Smix (1.17)

Smix can be found as follows: S = k ln (1.18)

where k is Boltzmann’s constant

is a measure of randomness or the

number of distinguishable ways of arranging the atoms in

the solution.

NA = XA Na ; NB = XB Na

1.3.2. Ideal solution

NA and NB are number of A and B atoms, Na is Avogadro’s number

!!

!

BA

BA

NN

NN (1.19)


By substituting into Equations 1.18

and 1.19, using the Stirling’s

approximation (ln N! N lnN – N) and

the relationship Nak = R

where R is universal gas constant,

we have:

Smix = -R (XAlnXA + XBlnXB) (1.19’)

Gmix = RT (XAlnXA + XBlnXB) (1.20)

Note: Since XA and XB are less than unity, Smix is positive, i.e. there is an increase in entropy on mixing.

The free energy of mixing, Gmix :

Gmix is function of

composition and temperature


1.3.2. Ideal solution S = k ln(1.18)

!!

!

BA

BA

NN

NN (1.19)

G2 = G1 + Gmix

G2 = XAGA + XBGB + RT (XAlnXA + XBlnXB)

Note:

1. The actual free energy of the solution G will also depend on GA

and GB

2. As the temperature increases, GA

and GB decrease and the free energy curves assume a greater curvature

3. The free energy curves must end asymptotically at the vertical axes of the pure components


1.3.2. Ideal solution

For ideal solution a random arrangement of atoms is the equilibrum, or most stable arrangement

In alloy it is interest to know how the free energy of a given phase will

change when atoms are added or removed.

If a small quantity of A, dnA mol, is added at constant temperature and

pressure, the size of the system will increase by dnA and therefore the

total free energy of the system will also increase by a small amount dG’.

If dnA is small enough dG’ will be proportional to the amount of A added:

dG’ = A dnA (T, P, nB constant) (1.22)

The proportionality constant A is called the partial molar free energy of A

or the chemical potential of A in the phase:

1.3.3. Chemical Potential

Note: G’ refers to whole system, whereas

G denotes the molar free energy and is

therefore independent of the size of the

system.

BnPTA

An

G

,,

'

AnPTB

Bn

G

,,

'


Definition: chemical potential is the change of the free

energy of whole system (G’) while a very small amount of

a component is added.

Notes:

For binary solution at constant temperature and pressure

the separate contributions can be summed:

dG’ = A dnA + B dnB (1.24)

If the solution contains more than 2 components and T and P change Equation 1.24 must be added to give the general equation:

dG’ = -SdT + VdP + A dnA + B dnB + C dnC + …

o If A and B are added in the correct proportions (dnA : dnB = XA : XB, for example if XA = 2/3, XB = 1/3, dnA : dnB = 2) , the size of the system can be increased without changing A and B.

o So if Xamol A and XB mol B is added the free energy of the system will increase by the molar free energy. Therefore from (1.24):

G = A XA + B XB J mol-1 (1.25)

BnPTA

An

G

,,

'

AnPTB

Bn

G

,,

'



Comparison of Equation 1. 21 and 1.25

gives A and B for an ideal solution as:

A = GA + RT lnXA

B = GB + RT ln XB (1.26)

A and B can be obtained by extrapolating the tangent to the G curve to the sides of the molar free energy diagram

A

BA

XB

B

G

Molar free energy

G2 = XAGA + XBGB + RT (XAlnXA + XBlnXB) (1.21)

G = A XA + B XB (1.25)



In reality Hmix 0 (mixing is

endothermic (heat absorbed) or

exothermic (heat evolved).

It is assumed that the heat of mixing,

Hmix, is only due to the bond

energies between adjacent atoms.

Necessary conditions: volumes of

pure A and B are equal and do not

change during mixing so that the

interatomic distances and bond

energies are independent of

composition.

o The structure of a binary solid

solution is shown schematically in

Fig.

1.3.4. Regular Solutions

Three types of interatomic bonds

are present:

A-A bonds each with an energy AA

B-B bonds each with an energy BB

A-B bonds each with an energy AB


The internal energy of the solution, E, will

depend on the number of bonds of each type

PAA, PBB and PAB such that:

E = PAAAA + PBBBB + PABAB

Before mixing pure A and B contain only A-A and B-B bonds, so the change in internal energy on mixing is given by

Hmix = PAB (1.27)

where = AB – (AA + BB)/2 (1.28)

If = 0, Hmix = 0, the solution is ideal, the atoms are completely randomly arranged and

PAB = NazXAXB bonds mol-1 (1.29)

where Na: Avogadro’s number, z is the number of bonds per atom.

If < 0, the atoms in the solution will prefer to be surrounded by atoms of the opposite type and this will increase PAB: PAB > NazXAXB

If > 0, PAB will tend to be less than in a random solution: PAB < NazXAXB


1.3.4. Regular Solutions

In which case the following

equation is good approximation:

Hmix = NazXAXB = XAXB (1.30)

where = Na z (1.31)

Real solution that closely obey

Equation 1.30 are known as regular

solutions.

The variation of Hmix with

composition is parabolic and is

shown in Fig 1. 13 for > 0

A XB B

Hmix

per mol


1.3.4. Regular Solutions Hmix = PAB (1.27)

PAB = NazXAXB (1.29)

In alloys where the enthalpy of mixing is not zero ( and 0) the

assumption that a random arrangement of atoms is the equilibrum, or

most stable arrangement is not true, and the calculated value for Gmix

will not give the minimum free energy.

The actual arrangement of atoms will be a compromise that gives the

lowest internal energy consistent with sufficient entropy, or

randomness, to achieve the minimum free energy.

1.3.6. Real Solutions

= AB – (AA + BB)/2


Gmix = RT (XAlnXA + XBlnXB) (1.20)

In system with < 0 the

internal energy of system

is reduces by increasing

the number of A-B

bonds, i.e. by ordering

the atoms as shown in

figure a.

If > 0 the internal energy

can be reduced by

increasing the number of

A-A and B-B bonds, i.e. by

the clustering of the atoms

into A-rich and B-rich

groups (fig. b).

In systems where there is a size difference between the atoms the quasi-chemical model will underestimate the change in internal energy on mixing since no account is taken of the elastic strain fields which introduces a strain energy terminto Hmix. When the size difference is very large the interstitial solid solution are energrtically most favourable(Fig. c).

= AB – (AA + BB)/2


1.3.6. Real Solutions

If the atoms in a substitutional solid solution are completely randomly

arranged (each atom position is equivalent) PAB, the number of A-B bonds,

in such solution is given by Equation 1.29:

PAB = NazXAXB bonds mol-1

If (= Na z ) < 0 (or < 0) and the number of A-B bonds is greater than

this, the solution is said to contain short-range order (SRO).

The degree of the ordering can be quantified by defining a SRO

parameters such that:

where PAB (max) and PAB(random) refer to the maximum number of bonds

possible and the number of bonds for a random solution, respectively.

1.3.7. Ordered Phases

)((max)

)(

randomPP

randomPPs

ABAB

ABAB


a) Random A_B solution with a total of 100 atoms and XA = XB = 0.5, PAB ~ 100, S = 0.

b) Same alloy with short-range order PAB = 132, PAB(max) ~ 200, S = (132 -100)/(200 – 100) = 0.32

Example of orders

In solutions with compositions that are close to a simple ratio of A:B atoms another type of order as the long-range one (c) can be found. Now the atom sites are no longer equivalent but can be labelled as A-sites and B-sites.



Example: ordered substitutional structures in the Cu-Au system

a) high-temperature disordered structure; atoms Cu or Au can occupy any site.

b) At low temperature solution with XCu = XAu = 0,5, i.e. 50/50 Cu/Au mixture, form an ordered structure in which the Cu and Au are arranged in alternate layer; each atom position is no longer equivalent and the lattice is described as a CuAusuperlattice.

c) Alloy with the composition Cu3Au: another superlattice is found.



The entropy of mixing of structures with long – range order is extremly small and

with increasing temperature the degree of order decreases until some critical

temperature there is no long-range order at all.

This temperature is a maximum when the composition is ideal required for the

superlattice.

Note: The critical temperature for loss of long-range order increases with increasing

, or Hmix, and in many systems the ordered phase is stable up to the melting point

However, long-range order can still

be obtained when the composition

deviates from the ideal if some atom

sites are left vacant or if some atoms

sit on wrong sites.

In such cases it can be easier to

disrupt the order with increasing

temperature and the critical

temperature is lower.



Some most common ordered lattices



Often the configuration of atoms that has the minimum free energy after

mixing does not have the same crystal structure as either of the pure

components. In such cases the new structure is known as an intermediate

phase.

Intermediate phases are often based on an ideal atom ratio that results in a

minimum Gibss free energy.

For compositions that deviate from the ideal,

the free energy is higher giving a characteristic

“U” shape to the G curve.

The range of compositions over which the free

energy curve has a meaningful existence

depends on the structure of the phase and

the type of interatomic bonding.

1.3.8. Intermediate Phases


Ideal compositionA B

G X


Gmix

Ideal compositiona)A B

G

GA

XB b)A B

G

GA

GB

In other structure fluctuations in

composition can be tolerated by

some atoms occupying “wrong”

positions or by atom sites being left

vacant, and in these cases the

curvature of the G curve is much less

(b).

When small composition deviations

cause a rapid rise in G the phase is

referred to as an intermetallic compound

and is usually stoichiometric, i.e. has a

formula AmBn, where m and n are

integers.


X

TEM picture showing a Si-rich zone arrounding in the presence of Sr



Al5FeSi - phase, platelete form, very hamful, stable

-Al5FeSi

Al8Fe2Si - phase, Chinese script form, less hamful, but metastable

- Al8Fe2Si

1.4. Equilibrum in Heterogeneous Systems

o The stable forms of pure A and B at a given

temperature and pressure can be denoted as

and respectively

Assumed is fcc and is bcc. The molar free

energies of fcc A and bcc B are shown as point a

and b

The first step in drawing the free energy curve of

the fcc is, therefore, to convert the stable bcc

arrangement of B atoms into an unstable fcc

arrangement. This requires an increase in free

energy, bc

The free energy curve for the phase can now be

constructed as before by mixing fcc A and fcc B. -

Gmix for of composition X is given by the

distance de as usual

Note: A-rich alloys will have the lowest free energy as a homogeneous phase and B-rich alloys as phase.


It is usually the case that A and B do not have the

same crystal structure in their pure states at a given

temperature. In such cases two free energy curves

must be drawn, one for each structure.

Consider alloy X0

Futher reductions in free energy can be

achieved if the A and B atoms interchange

between the and phases until the

compositions e and e are reached (Fig b).

The free energy of the system Ge is now a

minimum and there is no desire for futher

change. Consequently the system is in

equilibrium and e and e are the equilibrium

compositions of the and phases.



If atoms are arranged as a homogeneous

phase, the free energy will be lowest as , i.e.

G0 per mole.

However, the system can lower its free

energy if the atoms separate into two phases

with compositons 1 and 1, for example (Fig.

a). The free energy of the system will then

reduced to G1.



This results is quite general and applies to

any alloy with an overal composition between

e and e: only the relative amounts of the

phases change, as given by the lever rule.

When the alloy composition lies outside this

range, however, the minimum free energy lies

on the G ang G curves and the equilibrium

state of the alloy is a homogeneous single

phase.

From Fig. b) it can be seen that equilibrium

between two phases requires that the

tangents to each G curve at equilibrium

compositions lie on a common line.

In the other words each component must

have the same chemical potential in the two

phases, i.e. for heterogeneous equilibrium:

A = A

, B = B

In the free energy curves that have been

drawn so far the surfaces, grain

boundaries and interphase interfaces have

been ignored.

Interphase interfaces can become extremly

important in the early stages of phase

transformations when one phase, , say,

can be present as very fine particles in the

other phase.

P

Atmos-phericpressure

If the phase is acted on by a pressure 1 atm, the phase is subjected

to an extra pressure P due to the curvature of the / interpace.

If is the / interfacial energy and the particles are spherical with

radius r, P is given appoximately by:

1.5. The Influence of Interfaces on Equilibrium

rP

2


P

Therefore the cuvre on the molar free energy-composition diagram in

Fig. 1.20b will be raised by an amount:

By definition, the Gibbs free energy contains a “PV” term: G = H – TS and H =

E + PV so an increase of pressure P therefore causes an increase in free

energy G:

G = V P

Vm is the molar volume of the phase

X

GGr

G

G

XXr XB

This free energy increase due

to interfacial energy is known

as a capillarity effect of the

Gibbs-Thomson effect.

r

VG m

2



An important practical

consequence of the Gibbs-

Thomson effect is that the

solubility of in is sentitive to the

size of the particles.

From the common tangent

construction it can be seen that the

concentration of solute B in in

equilibrium with across a curved

interface (Xr) is greater than X, the

equilibrium concentration for a

planar interface.

Taking the following typical values: = 200 mJ m2, Vm = 10-5 m3, R = 8.31 J mol-1 K-1, T = 500 K gives:

For r = 10 nm Xr / X 1.1

For particles visible in the light microscope

(r > 1 m) capillarity effects are very small

nmrX

X r 11


X

GGr

G

G

XXr XB


The thermodynamics can be used to

calculate the driving force for a

transformation: G = G2 – G1, but it can

not say how fast a transformation will

proceed.

The study of how fast processes occur

belongs to the science of kinetics.

1.6. The Kinetics of Phase Transformations

Before the free energy of the atom can

decrease from G1 to G2 the atom must

pass through a so-called transition or

activated states with a free energy Ga

above G1

Ga

G

Finalstate

Innitialstate

Activatedstate

G

G1

G2


As a result of the random thermal motion of the atoms the energy of

any particular atom will vary with time and occasionally it may be

sufficient for the atom to reach the activated state. This process is

known as thermal activation.

The rate at which a transformation

occurs will depend on the frequency with

which atoms reach the activated state.

Putting Ga = Ha - TSa and changing from

atomic to molar quantities enables this

equation to be wrotten as Arrhenius rate

equation

The probability of an atom reaching the activated state is given by

exp (-Ga/kT), where k is Boltzmann’s constant (R/Na) and Ga is

known as the activation free energy barrier.

kT

Grate

a

exp

Arrhenius rate equation

kT

Hrate

a

exp


1.6. The Kinetics of Phase Transformations

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Chapter 1. Thermodynamics