Thermodynamics Of Materials

Thermodynamics of Materials

Naeem ul Haq Tariq

Thermodynamics and Phase Diagrams • Thermodynamics: Prediction about the equilibrium of an alloy • Kinetics: Rate at which equilibrium is reached • Phase: A Portion of the system whose properties and composition are homogenious and which is physically distinct from other parts of the system

Phase Transformation

One or more Phases in an alloy

Q: Why transformation occurs? Q: How is phase stability measured?

New Phase or mixture of new phases

Ans1: Because the initial state of the alloy is unstable relative to the final state. Ans2: By thermodynamics ( for transformations that occur at constant temperature and pressure, the relative stability of a system is determines by its Gibbs Free Energy )

G = H - TS Enthalpy: Measure of heat content of the system H = E + PV Internal Energy: (Kinetic energy) + (Potential Energy)

• Atomic vibrations in solids/liquids • From traslational and rotational energies for the atoms and molecules with in a liquid or gas

From interactions, or bonds between the atoms with in the system

Reaction

H = E + PV

Heat absorbed or evolved

volume Change of the system PV

Change in internal energy E

At constant volume : W = P dV = 0 dE = q – W =q , Increase or decrease in E => Heat absorbed or released At constant pressure: dH = q , Increase in enthalpy => Heat absorbed Decrease in enthalpy => Heat released For condensed phases: PV <<<< E => H ~ E

Entropy: Measure of randomness of the system Equilibrium: The state of a physical system evolves irreversibly towards a time-independent state in which we see no further macroscopic physical or chemical changes

• At lower temperature solids are stable • Diamond (metastable) => Graphite (stable) • Any transformation that results in a decrease in free energy is possible, i.e ∆G = G2-G1<0 • Higher the energy hump => Slow transformation

sive Properties: Independent of the size of the system, e.g T, P

nsive Properties Dependent of the size of the system,e.g V,E, H, S, G

Gibbs Free energy as a Function of Temperature Specific Heat: The quantity of heat(in joules) required to raise the temperature of the substance by one degree kelvin. At constant pressure it is given by; Cp = (δH / δΤ)p H=0 for a pure element in its most stable state at 298K

Variation of entropy with temperature can be given as ; Cp= (δΗ/δT)p Variation of H with T can calculated by: Variation of entropy with temperature is given by: Taking entropy at zero K as zero => Variation of G with temperature is shown in fig.

When temperature and pressure vary the change in Gibbs Free energy can be given as follows; dG = -SdT + VdP At constant P => (δG/δT )p= -S

As we can't measure entropy directly

For a phase transition of a pure substance betwen the liquid and vapour phases, the Clapeyron equation reduces to the ClausiusClapeyron equation under the approximations that the vapour is ideal and the liquid phase volume is negligible next to the vapour volume.

Binary Solutions • In single component systems all phases have the same compositions • Equilibrium involves P & T variables • In alloys composition is also a variable

To understand phase changes in alloys

n and T are important

Dependence of free energy on composition, temperature and pressure should be known

Gibbs Free Energy of Binary Solutions • The Gibbs free energy of a binary solution of A and B atoms can be calculated from the free energies of pure A and pure B Assumptions: A and B have the same crystal structure in there pure states Can be mixed in any proportion to make solid solution with the same crystal structure

•

Imagine that 1 mol of homogeneous solid solution is made by mixing together XA mol of A and XB mol of B such that XA + XB = 1 XA & XB are the mole fractions of A and B in the alloy respectively • In order to calculate the free energy of the alloy, mixing can be made in two steps:  Bring together XA mol & XB mol of A & B  Allow the A and B atoms to mix together to make a homogeneous solid solution

Before Mixing

After Mixing

Mixing

• After step 1 the free energy of the system is given by; G1 = XAGA+XBGB Jmol-1 • GA & GB are the molar free energies of pure A & B at the temperature & Pressure of the above experiment • G1 can be most conveniently represented on a molar free energy diagram • For all alloy compositions G1 lies on the straight line between GA & GB

• After step 2 the free energy of the solid solution G2 can be expressed as; G2 = G1 + ∆Gmix Since,

G1 = H1- TS1 & G2 = H2- TS2

Putting, ∆Η mix = H2 - H1 & ∆Smix = S2 - S1 

∆Gmix = ∆Η mix - T ∆Smix

∆Η mix is the heat absorbed or evolved during step 2, i.e it is the heat of solution , and ignoring volume changes during the process, it represents only the difference in internal energy (E) before and after mixing.

Ideal Solutions ∆Η mix = 0

=0

∆Gmix = ∆Η mix - T ∆Smix ∆Gmix = - T ∆Smix In statistical thermodynamics,

S = k ln ω (2)

(1)

S

S

th

S

config

• In case of thermal entropy, ω is the number of ways in which the thermal energy of the solid can be divided among the atoms or The total numbers of ways in which vibrations can be set up in the solid • In solutions, additional randomness exist due to the different ways in which the atoms can be arranged leading to the configurational entropy • If there is no volume change or heat exchange during mixing then the only contribution to ∆ Smix is the change in configurational entropy

• If there is no volume change or heat change during mixing then the only contribution to ∆Smix is the change in configurational entropy • Before mixing , A & B atoms are held separately in the system and there is only one distinguishable way in which the atoms can be arranged  S1= k ln 1 = 0 

∆Smix = S2

• Assuming that A & B mix to form a substitutional solid solution • All configurations of A & B atoms are equally probable • The number of distinguishable ways of arranging the atoms on atom sites is  ω = (NA+ NB) ! / NA ! NB !

(3)

NA & NB are number of A & B atoms • Since we are dealing with 1 mol of solution i.e. Na atoms ( Avogadro’s number),  NA = X A Na

&

NB = X BNa

• Using 1, 2 & Stirling approximatiom i.e,

Ln N ! = N ln N - N

Na k = R

( 1 )

( 2



∆Smix = - R (X A

ln X A + X B ln X B)

Note: Since X A & X B < 1 => ∆Smix

>0

From equation (1);

∆Gmix = - T ∆Smix 

∆Gmix = RT(X A

ln X A + X B ln X B)

• The actual free energy of the solution G will also depend on GA & GA

G1 = XAGA+XBGB G2 = G1 + ∆Gmix ∆Gmix = RT(X A

ln X A + X B ln X B)

G = G2 = XAGA+XBGB +RT(X A ln X A + X B ln X B) • As the temperature increases, GA&GB decrease and the free energy curves assume a greater curvature.

GB G1 GA G A

G2

Chemical Potential • In alloys it is of 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 to a large amount of a phase at constant T & P , the size of the system will be increase by dnA • Therefore, the total free energy of the system will also increase by a small amount dG’ • If dnA is a small enough dG’ will be proportional to the amount of A added  dG’ = µ A dnA ( T, P , nB constant )

• The proportionality constant is called the partial molar free energy of A or alternatively the chemical potential of A in the phase. ∀ µ A depends on the composition of the , and therefore dnA must be so small that the composition is not significantly altered

µ A = (δG’ / δnA)

T,P,n(B)

G’ is the free energy of the whole system not molar

For binary system; dG’ = µ A dnA + µ B dnB If T & P changes are also allowed, dG’ = -SdT + VdP + µ A dnA + µ B dnB +…..

• If 1 mol of the original phase contained X A & X B mol of A & B, the size of the system can be increased without altering its composition if A & B are added in the correct proportions, i.e. such that dnA : dnB = X A : X B For example: If the phase contains twice as many A as B atoms (X A = 2/3, X B = 1/3), the composition can be maintained constant by adding two A atoms for every B atoms,In this way the size of the system can be increased by one mole without changing µ A & µ Β • To do this X A mol A and X B mol B must be added and the free energy of the system will increase by molar free energy G

• So, dG’ = µ G= µ

A A

dnA + µ

XA + µ

B

B

dnB gives;

XB

Compairing with; G = G2 = XAGA+XBGB +RT(X A ln X A + X B ln X B)

µ

A

= GA+ RT ln X A

µ

A

= GA+ RT ln X A

Molar Free Energy

GA

GA

Regular Solutions • In practice ∆Η mix = 0 • So, the simple model for for an ideal solution can be extented to include the ∆Η mix term by using quasi-chemical approach Assumption:

• The heat of mixing, ∆Ηmix is only due the bond energies between the adjacent atoms • For this assumption to be valid, it is necessary that the volumes of pure A & B are equal and do not change during mixing so that the interatomic distances and bond energies are independent of the compositions

• A-A bonds each with an energy ε AA • B-B bonds each with an energy ε BB • A-B bonds each with an energy ε AB  Consider zero energy to be the state wherethe atoms are separated to infinity ε AA,ε BB, ε AB are negative quantities and become more negative as the bond become more stronger  The internal energy of the solution E will depend on the number of bonds of each type, PAA, PBB, PAB such that : E = PAA ε AA + PBB ε BB + PAB ε AB

• Before mixing pure A and B contain only A-A, & B-B bonds respectively • By considering the relationship between PAA, PBB and PAB in the solution it can be shown that the change in internal energy on mixing is given by,

∆Η mix = PAB ε Where, ε = ε AB - (ε AA+ ε BB)/2

∀ε

is the difference between th A-B bond energy and the average of the A-A and B-B bond energies

• If ε

= 0, ∆Η mix = 0 and the solution is ideal • In this case the atoms are completely randomly arranged and entropy of mixing is given by: ∆Smix = - R (X A ln X A + X B ln X B)

• In such solutions it can be shown that: PAB = Na z X A X B bonds / mol ……………………..(1) • If ε < 0, the atoms in the solution will prefer to be surrounded by atoms of the opposite type and this will increase PAB • If ε > 0, PAB tend to be less than in a random solution • If ε is not too different from zero, (1) is still good approximation in which case

∆Η mix = Ω

X A X B , where Ω = Na z ε

……….(2)

• Real solutions that closely obey equation (2) are known as regular solutions • The variation of ∆Η mix with composition is parabolic and is shown in fig. for Ω >0 • Tangents at X A = 0 and X A = 0 are related to Ω • The free energy change on mixing a regular solution is given as:

∆Gmix = Ω

X A X B + RT(X A ln X A + X B ln X B)

∆ Hmix

- T ∆ Smix

• For exothermic solutions ∆Η mix < 0 and mixing results in a free energy decrease at all temperatures • For ∆Η mix > 0 , the situation is more complecated, At high temperature: T∆Smix > ∆Η mix For all compositions and free energy curve has a positive curvature at all points. At low temperature: T∆Smix < ∆Η mix Free energy develops a negative curvature in the middle

>

Pringles

• Differentiating; ∆Smix = - R (X A ln X A + X B ln X B) 0 the T∆Smix curve becomes • As X A or X B vertical whereas the slope of ∆Hmix curve tends to a finitevalue Ω, this means that, except at absolute zero, ∆Gmix always decreases on addition of a small amount of solute • The actual free energy of the alloy depends on the values chosen for G A & G A and is given by;

G = XAGA+XBGB +Ω X A X B + RT(X

A

ln X A + X B ln X B)

Using, X A X B = X A X 2B + X 2A X B

G = XAGA+XBGB + Ω X A X B + RT(X G= µ

A

XA + µ

B

A

ln X A + X B ln X B)

XB

µ A = GA+ Ω (1− XA)2 + RT ln X A µ B = GB+ Ω (1− XB)2 + RT ln X B

Activity • For any solution chemical potential expression can be written by defining activity of a component ‘a’  µ A = GA+ RT ln a A

 µ

A

= GA+ RT ln a A

Comparing with

µ A = GA+ Ω (1− XA)2 + RT ln X A 2

 ln (a A / X A ) = Ω (1 - X A )2/ RT  ln (a B / X B ) = Ω (1 - X B )2/ RT Assuming A & B have the same crystal structure, the relationship between a and X can be represented as shown in fig. For an ideal solution a A = X A If ∆Hmix< 0 , activity of the components in the solution will be less in an ideal solution (line 2) If ∆Hmix> 0, vice versa

•

The ratio (a A / X A ) is usually referred to as , γ Α the activity coefficient of A, that is γ Α = (a A / X A )

•

For the dilute solution of B in A Letting in above equation X B 0

•

γ Α = (a A / X A ) ~ 1

(Raoult’s law)

γ Β = (a B / X B) ~ constant (Henry’s law)

• The activity of a component is just another means of describing the state of the component in a solution • Activity and chemical potential are simply a measure of the tendency of an atom to leave a solution

Real Solutions • When ε and Ω = 0, the assumption that a random arrangement of atoms is the equilibrium, or most stable arrangement is not true

• 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

• In the systems with ε < 0 the internal energy of the system is reduced by increasing the number of A-B bonds, by ordering the atoms as shown

• In the systems with ε > 0 the internal energy of the system is reduced by increasing the number of A-A & B-B bonds, by clustering of the atoms into A-rich and B-rich groups

• However, the degree of ordering or clustering will decrease as temperature increases due to increasing importance of entropy • In systems where there is a size difference between the atoms , the quasichemical model will underestimate the change in internal energyon mixing • Since no account is taken of the elastic strain fields which introduces a strain energy term to

∆Η mix

• When the size difference is large this effect can dominate over the chemical term

• When the size difference between the atoms is very large then interstitial solid solutions are energetically most favourable • New mathematical models are needed to describe these solutions

• In the systems where there is strong chemical bonding between the atoms there is a tendency for the formation of intermetallic phases • These are distinct from the solutions based on pure components • Since they have a different crystal structure and may also be highly ordered

Ordered Phases • If atoms in a substitutional solid solution are completely randomly arranged each atom position is equivalent and probability that any given site in the lattice will contain an A atom will be equal to the fraction of A atoms in the solution XA ,similarly XB for B atoms • In such solutions number of A-B bonds (PAB) is given by PAB = Na z X A X B

bonds / mol

• If Ω < 0, and number of A-B bonds is greater than this, the solution is said to contain SRO • The degree of ordering can be quantified by defining a SRO parameter S S = PAB - PAB (random) / PAB (max)- PAB (random) PAB (max)

= max. no. of bonds possible

PAB (random) = no. of bonds for random solution

• In solutions with compositions that are very close to a simple ratio of A : B atoms another type of order can be found known as long range order

Now the atom sites are no longer equivalent but can be labelled as A-rich and B-rich, such solution can be considered to be a different (ordered)phase separate from the random or nearly random solution

Q

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