1. Introduction to thermodynamics a. What is thermodynamics?

MSE 511 notes 2009 – J. R. Morris

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1. Introduction to thermodynamics

a. What is thermodynamics?

Thermodynamics is a description of material properties as a function of state. What do we mean by state?

A macroscopic state of a homogeneous material includes various physical properties. Some (but certainly

not all) of these properties are:

Temperature T

Pressure / stress P (σxy)

Energy U

Magnetic field H

Magnetism M

Composition x

What sort of properties are we interested in?

“Phase:” solid/liquid/gas

Phase fractions (such as what fraction is solid or liquid, or what fraction has a particular crystal

structure)

Heat capacity

Thermal expansion

Magnetism

Pressure

Let‟s introduce some notation and initial definitions:

Intensive variables: These are quantities that are independent of the sample size.

Examples: pressure, temperature, density, composition…

Extensive variables: Those quantities that are proportional to the system size.

Examples: energy (U), entropy (S), volume (V), number of atoms (N)

I will generally use capital letters for both intensive and extensive variables. However, we can usually

create intensive variables by normalizing extensive variables to values per atom, per mole (6.02×1023

atoms), or (occasionally) per volume. I will typically indicate these using small letters:

Energy/atom = U/N = u

Volume/atom = V/N = v

Note that I typically normalize per atom and not per mole. Example: the ideal gas law is often written as

PV = nRT

n = number of moles

R = “ideal gas constant” =8.315 J/(mol∙K)

We may also write this as


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PV = NkBT

N = number of atoms

kB = “Boltzmann‟s constant” = 1.381×10-23

J/K

Using Avagadro‟s number, NA = 6.022×1023

atoms/mole, we have

R = NA∙kB

An important note: The usage of the constants R and kB are because temperature is really an energy and

we have to convert from Kelvin to Joules.

Another important note: Temperature will be expressed almost always in Kelvin, not in Centigrade or

Fahrenheit.

b. First important concept: Energy and the first law of thermodynamics

If we consider a piece of material, how can we change its energy? For example: a piece of iron. The

following actions will change its energy:

Heat it

Bend it

Magnetize it

Oxidize it

Break a piece off

Anneal it

In general, the change in energy is the heat flow into the system, minus the work done by the system:

∆U = ∆Q - ∆W

∆Q = heat flow in

∆W = work done

This is simply conservation of energy, written differently. This is THE most important concept.


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In general, we will often write the differential change in energy as

dU = -PdV + đQ

where the first term is the work term, and the second term is the heat flow into the system. This is the

first law of thermodynamics, and is really a statement of conservation of energy.

c. The second law of thermodynamics

But what is this “đQ” thing? Why the funny notation? We understand the heat flow into the system, and

work done by the system. We also understand the pressure and volume (and energy) of the system. But

we don’t know what is meant by the “heat” of the system – this is undefined.

Ideally, we‟d like to describe the energy of a system in terms of state variables such as volume. Work

done is related to change in volume. What quantity is changing when heat flows in? The historic

approach is to assume that đQ is somehow related to temperature, but that there is some other quantity

coupled in. If we call this quantity the entropy S, then the first law becomes

dU = TdS – PdV.

This is not always correct, though we will use it often. If correct, then we would have

đQ = TdS.

In reality, we have

đQ ≤ TdS or ∆Q ≤ T∆S.

Thus, the correct version is

dU ≤ TdS – PdV.

Example of work: expansion of a gas

Piston has cross-section A.

External pressure = P

Force = pressure × area = PA

Work = force × distance = PA∆x

∆W = PA∆x

Change in volume = ∆V = A∆x

∆W = P∆V

Energy decreases; change is

∆U = -P∆V

(assuming no change in P and no heat flow)

Differential version:

dW = PdV dU = -PdV


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This is the second law of thermodynamics. The equality holds only for reversible processes, where the

system is essentially in equilibrium during the change. In reality, the entropy may increase even when

there is no volume or energy change – a spontaneous increase in randomness.

In equilibrium, the energy is a function of the entropy S and the volume V:

U = U(S,V)

We may also rearrange this to write the entropy in terms of the energy and volume:

S = S(U,V).

We will commonly use these with partial derivatives:

The advantage: we don‟t need to know the history of the system to know its properties.

But what is this thing we call entropy?

The problem is that thermodynamics describes the collective behavior of many particles, and generally it

is not possible (or useful) to describe the behavior of each of these. On average, in equilibrium, their

behavior is described by only a few quantities. For example, we don‟t care too much about the energy of

each individual particle, only the total energy of the system. One macrostate (such as the total energy)

represents many possible microstates (energies of each particular particle). For a given total energy,

energy may flow within the system to allow more possibilities.

How many ways can we distribute the energy? Let (U) be the number of different possible ways of

distributing energy U in the system. For example, one particle could have a lot of kinetic energy, while

the rest have none. Or all particles could have exactly the same kinetic energies. These are two very

unlikely cases, amongst many possibilities.

The entropy per particle is defined to be

S = kB ln (U).

This connection is the heart of thermodynamics (and is engraved on Boltzmann‟s tombstone). It is not

obvious, but it works. For a given energy, the system evolves to a state with the most number of possible

states (i.e. it moves to the most likely state). In other words, for fixed energy, the system maximizes the

entropy. At constant volume with no heat flow, this gives

∆S ≥ 0 or, more generally, TdS ≥ đQ.

This is the “usual” statement of the second law, that entropy can only increase (or stay the same) if the

system is closed (∆Q=0). Again, if the process is reversible, TdS = đQ.


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Example: A closed system with two subsystems:

Total energy: U = U1 + U2

Total number of possible states: (U) = 1 (U1) ∙ 2 (U2)

Note: number of possibilities multiply, not add.

Now, allow for a small amount of heat ∆U to flow from one part

of the system to another. Total energy is the same. What about the

number of possible states?

= 1 (U1)∙ 2 (U2)

At the maximum, this change will be zero. This gives

Multiplying both sides by kB then gives:

But, using the first law (with constant volume), this is simply

So: the condition that equilibrium is a maximum in entropy is equivalent to the temperatures being equal!

This is a “consistency” argument – what we defined ends up making some sense in the end. Similar

arguments under slightly different conditions are common.


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2. Example of entropy calculation from counting of possible states

How do we use the identity S = kB ln (U)? Let‟s do a relatively simple problem, relevant to alloys.

Suppose we have NA atoms of type A, and NB atoms of type B. How many ways can we arrange these on

a lattice with N=NA+NB sites?

A simple example: NA=2, NB=1. The ways we can arrange these are:

AAB

ABA

BAA

So, there are =3 ways of arranging these. If all have the same energy, then S=kB ln 3.

Now, a slightly more complicated example: NA=2, NB=2. The ways we can arrange these are:

AABB

ABAB

ABBA

BAAB

BABA

BBAA

So, there are =6 ways of arranging these. If all have the same energy, then S= kB ln 6.

More relevant is when there are many more atoms – NA and NB both above 105 or more! How do we

calculate ?

Let‟s count the ways to place the A atoms – once those are placed, the B atoms fill up remaining spaces.

There are N places to put the first A atom, N-1 places to put the second, etc., with N-NA+1 places to put

the last atom. This gives

N∙(N-1) ∙(N-2) ∙…∙(N-NA+1) =

Here, N! = N∙(N-1) ∙(N-2) ∙… However, this is not correct. This overcounts the possibilities, because it

is impossible to tell the difference between different A atoms. After they are placed, it is impossible to

tell which is the „first‟ and which is the „second‟. For NA=NB=2, this would give 12 arrangements, not 6.

Once the A atoms are placed, we must count the number of ways to re-arrange these on the same sites,

and divide this out. This number is simply NA!. So, the final result is

To calculate the entropy, we will use the fact that N, NA and NB are all large numbers. In this limit, we

may use Stirling’s approximation:

With some minor rearrangements, we may write the entropy per particle as


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where xA=NA/N and xB=NB/N. This is

known as the ideal entropy of mixing, and

plays an important role in alloys,

magnetism, and many other places. It is

an “ideal” limit, because we have assumed

that all arrangements have the same

energy, and that there are no correlations

between the particles. It is the most

random possibility. We graph this

function at right. It goes to zero when xA

or xB go to zero; interestingly and

importantly, it does this with an infinite

slope.

Self quiz: Is the entropy positive or negative? How can you tell from S = kB ln (U)? Does the

above equation for the ideal entropy of mixing satisfy this?

3. Basic applications of 1st & 2nd laws: heat capacity, Helmholtz & Gibbs free

energies

Now we have our basic tools: the first two laws of thermodynamics, and the statistical definition of

entropy. How do we use this? Again, let‟s ask a simple (but general question). Suppose we have a

system, and we want to increase its temperature by ∆T. How much heat must flow into the system? This

is the heat capacity of the system, defined as đQ/dT.

First, assume that the system is kept at constant volume. The first law may be written as

đQ = dU + PdV = dU for a constant volume process (dV=0).

The heat capacity at constant volume, CV, may then be written as

For a reversible process, we may write this as

However, for liquids & solids, it is more common to use constant pressure measurements. What is the

heat capacity at constant pressure? At constant pressure, we have

đQ = dU + PdV = d(U+PV) (using dP=0).

The quantity H=U+PV is called the enthalpy. Using this, the constant pressure heat capacity, CP, may

then be written as


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This is the mathematical statement that in a constant pressure process, the heat flow into the system is the

change in the enthalpy.

In general, for a reversible process,

dH = dU + VdP + PdV = TdS + VdP.

From this, we have

Now, let‟s consider a constant temperature process – again, this is a common condition. During a

reversible process, the work done by the system is dW = TdS – dU. For a constant temperature process,

we may write

-dW = d(U-TS) = dF

where F=U-TS is called the “Helmholtz free energy.” The reversible work done by a system during a

constant temperature process is the change in the Helmholtz free energy. If the system does work dW, the

free energy decreases.

For a non-reversible process, we have dF ≤ -dW. If no work is done (a constant volume process), then the

free energy will spontaneously decrease: ∆F ≤ 0. Thus, at a given temperature and volume, the system

will spontaneously move to a state of lowest Helmholtz free energy.

In general, dF = dU – TdS – SdT = – PdV – SdT, or

Note that because P>0 and S>0, the Helmholtz free energy F decreases with increasing volume

and with increasing temperature. From above, we can calculate the heat capacity CV from F:

In general, if we know F(V,T), we can calculate essentially all thermodynamic quantities, usually

from derivatives and other combinations. What about volume derivatives? The first derivative

gives –P, as noted above. The second derivative has a problem, though: Consider the quantity


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Is this quantity extensive or intensive? F and V are both extensive, so in fact this is neither – it

scales as 1/V. But this is not useful. Let‟s define an intensive quantity instead, by scaling it

with the volume:

What is this quantity B? To understand this, consider a small, isothermal change in volume. The

volume strain is ∆V/V, and produces a change in pressure ∆P. Note that if ∆V>0, then ∆P<0 –

an expansion reduces pressure. We know from linear elasticity that the pressure change is

proportional to the volume strain, and this constant is the bulk modulus. It is also common (less

in materials science) to discuss the isothermal compressibility, defined as

Thus, the curvature of F,

is proportional to the bulk modulus, and will be much larger

for a solid (which is hard to compress) than for a gas (which is much more compressible).

Self quiz: What is the compressibility and bulk modulus of an ideal gas? Evaluate this at T=273

K and P=101 kPa (roughly atmospheric pressure). Compare the bulk modulus of this with that of

steel (B=160×109 Pa) and with water (B=2.2×10

9 Pa).

Self quiz: Sketch the free energy vs. volume for a solid and for a gas. Note that a solid has a

well-defined volume at P=0, but a gas does not. Pay attention to the slope of the curve.

Now, what about a constant T, constant P process? Again, the second law gives

dU ≤ TdS – PdV

or dU – TdS + PdV ≤ 0.

For a constant temperature, constant pressure process, this gives

d(U-TS+PV) ≤ 0.

Defining the Gibb‟s free energy G=U – TS + PV = F + PV, this shows that G is a minimum under

constant temperature and pressure conditions. In equilibrium, the change is zero. Thus, the system

should evolve to minimize G at a given P and T. The differential form of G is

dG = VdP – SdT.

Therefore,


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Again, we can relate the second derivatives to physical quantities as well:

In this class, we will commonly use the Gibb‟s free energy, and simply call it the free energy.

Occasionally, we will also use the Helmholtz free energy.

There is a special property of G, when we consider how it scales with system size. If G(P,T,N) is

the free energy of a system with N particles, the free energy of a system with twice as many

particles is twice as much, for the same P and T. More generally,

G(P,T,N ) = N∙G(P,T,1) = Nµ(P,T)

where we have introduced the chemical potential µ. The chemical potential will play an

important role later on, when there is more than one type of atom. Note that the Helmholtz free

energy does not have an equivalent:

F(V,T,N) N∙F(V,T,1).

This can be seen by noting that there is a big difference between 1023

atoms in a liter, and one

atom in a liter!

4. Phase transformations and free energies

So, what is an example of a constant pressure, constant temperature process? At first, this seems

like nothing of interest could happen. Can we add heat to a system, and not get a change in

either pressure or temperature? This seems unlikely, but in fact is a common occurrence. The

most familiar case is ice in a cup of water. We know that if we have ice and water coexisting in

equilibrium, that the temperature of the system is equal to freezing. Adding some small amount

of heat will melt some of the ice, but not raise the

temperature.

This gives us the first result concerning phase

transformations: during a reversible, constant (P,T) process,

the system does not change its Gibbs free energy. However,

the system may change phase. Thus: phase transformations

only occur (in equilibrium) when the Gibb’s free energies are

equal. In the following figure, a sketch is shown of the

Gibb‟s free energies for a solid, liquid and gas as a function

of temperature. There are a number of important here:

The melting and boiling temperatures are shown,

where the Gibb‟s free energies are equal. It may be


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possible to cool a liquid below the melting point, or a gas below the boiling point, but the

system is then out of equilibrium.

The slopes are all negative. Why? The derivative (at constant pressure) is proportional

to –S, and entropy is always positive. Therefore, the slopes are negative. (They are also

curving downward, consistent with a positive heat capacity.)

Which phases have higher entropies? We immediately see that the slope of the liquid

line has a higher magnitude – and therefore has the higher entropy. This is intuitive – the

gas is very random, but high in energy. The solid phase has the lowest entropy.

Generally, at high temperatures, the Gibbs free energy is dominated by the entropy terms. At

low temperatures, the enthalpy dominates. The equilibrium phase is determined by balancing

these, minimizing G=H-PV.

Self-quiz: Carbon dioxide is an example of a substance that doesn‟t have a liquid phase at

atmospheric pressure: the solid phase (“dry ice”) transforms directly to a gas. Sketch the

graph of the Gibbs free energies, including the liquid phase. At some pressure, there will

be a temperature where all three phases can coexist. Sketch the Gibbs free energies vs. T

for this case.

For a phase transformation at constant pressure, there are some immediate important results.

Consider the case of melting: at Tmelt, we have Gliq=Gsol. This immediately implies

Hsol-TSsol = Hliq-TSliq.

However, we have already stated that the liquid entropy is higher than that of the solid phase:

∆S=Sliq-Ssol > 0.

Note that this immediately shows that we need to provide heat to melt the system, the latent heat:

L = ∆Q = Tmelt∆S.

However, to keep G constant, we must have

∆H = Hliq-Hsol = Tmelt∆S.

Thus, as the system moves to a higher energy, both the enthalpy and entropy increase. The

change in enthalpy is known as the latent heat L, and is equal to the heat of transformation.

Graphing this, we get something like the following:

Note the changes in entropy that occur at the phase

transformation. The enthalpy vs. temperature

would look very similar.

The discontinuities in entropy indicate a first-order

phase transformation (as they occur in a first

derivative in the free energy). There are other

phase transformations, such as second-order phase


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transitions, where the entropy does not change at the transition, but its derivative (the heat

capacity, a second derivative of the free energy) is discontinuous. Magnetism and

superconductivity are two common examples of these. We will focus primarily on first-order

transitions.

Can we get transformations as a function of pressure? Of course:

this is sketched below for transformations from low pressure

(gas) through the liquid and solid phases. Again, this is sketched

using the thermodynamic relationships: The derivative

The volume always decreases with pressure, and so does the

slope. In this graph, there are a number of assumptions that are

not always true:

The liquid is assumed to have a larger volume (lower density)

than the solid phase. This is common, but is not always true.

Ice is less dense than water, which is why it floats.

The liquid is assumed to have a much smaller volume (and a

much higher density) than the gas. At high temperatures and pressures, this is not true.

Self quiz: Sketch this graph for water, where the density of the liquid is higher than that of the

solid.

In general, it is often useful to map out the transitions as a function of temperature and pressure.

A portion of this is shown here for water. The

melting point (“mp” in the figure) and the normal

boiling point (“nbp”) are labeled for atmospheric

pressure. The triple point (point A) occurs where

all three phases can exist.

Besides its immediate usefulness, these diagrams

provide some important information on the

underlying thermodynamic quantities, particularly

the free energy. Let us consider the line

connecting the solid and liquid phases. If we pick

a point on this line, (T,P) and (T+∆T,P+∆P), we

must satisfy

Gsol(T, P) = Gliq(T, P)

Gsol(T+∆T, P+∆P) = Gliq(T+∆T, P+∆P)

Assume that ∆T, ∆P are small. Then we can

approximate the free energies by expanding in ∆T

Portion of the pressure-temperature phase

diagram for water, from

http://www.cbu.edu/~mcondren/water-

phase-diagram.jpg

http://www.cbu.edu/~mcondren/water-phase-diagram.jpg

http://www.cbu.edu/~mcondren/water-phase-diagram.jpg


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and ∆P. For the solid phase, this is:

Similarly, for the liquid, we have

Setting these expressions equal, and using Gsol(T, P) = Gliq(T, P), this becomes

Vsol∆P – Ssol∆T = Vliq∆P – Sliq∆T.

Rearranging, this becomes

∆P∆V = ∆T∆S

or

where ∆V=Vliq – Vsol and ∆S=Sliq – Ssol. Note that from above, the change in entropy is related to

the latent heat: L=T∆S. In the limit that ∆P and ∆T go to zero, this becomes

This is known as the Clausius-Clapeyron equation. While we have derived this for solid and

liquid phases, it applies to all lines in the P-T phase diagram. Also note that the latent heat and

the volumes are typically both normalized as quantities per mass, per mole, or per atom – as long

as they are done consistently, this equation still holds.

Again, let‟s examine the sign of the slope. By definition, the latent heat is positive, as is the

temperature. Therefore, if Vsol < Vliq,(the solid phase is denser than the liquid), then the slope is

positive. Conversely, if Vliq < Vsol,(the liquid phase is denser than the solid), then the slope is

negative.

Self quiz: For the water phase diagram above, let = N/V be the number of atoms per

volume (the number density). Do the densities satisfy gas>liq > sol ? How can you tell

from the phase diagram?

The above description provides much information on phase

transformations as a function of temperature and pressure. However,

what happens if the volume is fixed? Again, imagine putting water in a


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sealed container, partially filled. After some time, the water will be in equilibrium with the

vapor. This is a constant volume process at some fixed temperature; therefore, equilibrium must

minimize the Helmholtz free energy F(V,T). How do we determine equilibrium?

Again, let‟s begin by sketching the Helmholtz free energy vs. volume. This is done here for the

liquid & gas phase. Note that we must satisfy .

Therefore, the slope should be negative. For a liquid, the density

(or volume/atom) in the limit that P=0 is well defined, so there is

a place where the slope is zero. We can even consider what

happens if the volume of the liquid is greater than this (negative

pressure) but this can not be in true equilibrium. This is indicated

by the dotted line. For the gas, as the volume continues to

increase, the pressure continually drops. For an ideal gas,

P=NkBT/V. Notice also that the graphs curve upward: again, the

second derivative is related to the bulk modulus.

How do we describe coexistence from this graph? Suppose the

average volume per atom is close to the arrow shown in the above figure. The free energy of the liquid is

not stable, so the free energy of the gas looks appropriate. However, the system can have a lower free

energy, by having some atoms in the liquid phase (at a lower

volume per atom), and some in the gas phase (at a higher

volume per atom), maintaining the same average free

energy. To find out the coexisting Helmholtz free energy,

we construct a common tangent line as shown here. The

free energy can be as low as this value. With the addition of

this common tangent line, F(V,T) is everywhere sloping

downward, and continuous. The places where the line

intersect the liquid and gas curves shows the coexistence

volumes (or, more commonly quoted, the densities at

coexistence). The slope of the line gives the coexistence

pressure Pcoexist (at the given temperature), according to .

Note what happens as we compress the gas (isothermally). At first, the pressure simply increases. At

some point, the liquid begins to condense. As we decrease the volume further, the pressure does not

change, but more liquid appears. This continues until the system is completely liquid. Then, the pressure

will again rise.

We can complete the circle: let fliq be the Helmholtz free energy per atom, and the volume per atom of the

liquid be vliq. (This can be per mole or per mass just as well.) Similarly, let fgas and vgas be the same

quantities for the gas phase. It is not hard to see from the above graph that

fliq+Pcoexist vliq = fgas+Pcoexist vgas .

However, using G = F + PV, this is simply a statement that the Gibbs free energy per atom are the same

at coexistence. Equivalently, at coexistence, both phases have the same chemical potential µ.


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Example: Application of the Clausius-Clapeyron equation to vapor pressure

If we have a liquid or solid in equilibrium with the gas, we can make use of the Clausius-Clapeyron to

calculate a relationship with the pressure. Let ∆Hvap be the heat of vaporization – how much energy it

takes to convert the phase to a vapor. From the Clausius-Clapeyron equation, assuming we are dealing

with a solid phase, we have

However, the volume per atom in the vapor is much greater than that of the solid – perhaps 100 times

larger. Therefore, we can neglect vsol. Using Pvvap=kBT this becomes

Rearranging, we have

We can integrate this to find

Note that the heat of vaporization is here measured per atom. If this were measured per mole, this

becomes

Thus, a plot of (ln P) vs. (1/T) tends to give a straight line, whose slope gives a measure of the vapor

pressure.

Note that this does not only apply to solid-vapor conditions, but can also apply to adsorption. As an

example, for hydrogen storage, the goal is to find light materials with a heat of adsorption of 20-40

kJ/mol. This provides for a large adsorption, but not so large that it is difficult to get out again.


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Practice problems:

1. Naphtalene melts at 353 K at 1 atm pressure with an enthalpy change of 19 kJ/mole. The volume

increase is 19×10-6

m3. What change in the melting temperature will be observed if the pressure

is raised by 100 atm? Note: 1 atm = 101.3 kPa.

2.

3. The specific heat of water at 25 C is cp=4181 J/(kg∙K), while that of ice at 0C (the melting

temperature) is cp=2110 J/(kg∙K). The latent heat (or heat of fusion) is 3.34×105 J/kg. How

much heat must be supplied to bring one mole of ice from -10C to 10C, assuming that the heat

capacities are independent of T? Under certain conditions, water can be cooled considerably

below the freezing temperature without freezing (supercooled). Formulate an expression for the

latent heat vs. T. At what temperature does this go to zero? (Adapted from Pauling’s book).

The vapor pressure vs. temperature of a liquid is given in the table. Calculate the enthalpy

change on vaporization at T=373 K.

T (K) 326.1 352.9 415.0 451.7

Pv (Pa) 131.6 657.9 13160 52632

4

6

8

10

12

0.002 0.0025 0.003 0.0035

ln(P)

1/T


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Problems: The phase diagram of iron (Fe) 4. At right is the phase diagram for Fe.

a. (This part is not a big emphasis for MSE511 in

2008). For each of the temperatures below, sketch the

Helmholtz free energy vs. volume for the , and

phases on a single graph (one graph per temperature, with three different free energy

curves per graph). Indicate key pieces of

information that indicate differences between the different temperatures, and explain how you

constructed the graph. Make certain to pay

attention to relative volumes of the different phases, and the slopes connecting the free energy

graphs.

i. T=1500 K

ii. T=750 K iii. T=500 K

b. At P=0 sketch the Gibbs free energy of the , and liquid phases vs. temperature. Indicate the transitions. How would the Helmholtz free energy graph be different?

5. The change in entropy when a metal melts is typically close to S≈R (per mole) or S≈kB (per

atom) (with an accuracy of about 15%). Estimate the change in volume that occurs when Fe melts. Approximately how much heat (in Joules) must be supplied to melt one mole of Fe?

T=1500 K

T=500 K


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5. Interfacial free energies; classical nucleation theory

In what has gone before, we have considered free energies that scale with the number of atoms (or with

the volume, number of moles, etc.). However, there are additional contributions that are very important

in materials, and in the dynamics of phase transformations. In particular, interfaces have their own

contribution to the free energy. Typically, these contributions are positive, and as a result microstructural

evolution tends to minimize the amount of surface, to reduce the free energy. In general, for crystals, the

interfacial free energy depends upon the orientation of the crystal. For example, this tendency leads to

well defined facets observed in many single crystals. This is beyond the current treatment, and we will

ignore this for now.

Why have we ignored this contribution so far? If we write down the free energy of a system, as indicated

above, the free energy has a volume contribution, plus an area contribution:

Here, V is the volume and A is the area, and the free energy per volume is gvol. The interfacial free energy

per area is indicated by . Let R be a characteristic size of our system, so that V~R3 and A~R

2. In this

case,

(This holds only for cubic systems, but the results are more general.) The average free energy per volume

is then

When R is very large, the term /R can be ignored. However, for small volumes, this can not be ignored,

and even for large systems can control the microstructure evolution (which is dominated by interfaces).

Let us demonstrate the applicability of this for the case of the initial stages of crystal formation from the

liquid. Typically, a liquid can be cooled somewhat below the equilibrium melting temperature. The

liquid is then termed metastable: it is not the thermodynamically stable phase (as it has a higher free

energy than the solid), but can exist for measurable periods of time. To form the crystal, a small

crystalline region must form and grow. If there are no external causes for this, such as surfaces, cracks, or

other defects, then the nucleation is considered homogeneous; it can happen anywhere in the system with

the same probability. If, on the other hand, it is helped along by a defect, impurity, surface, or other

external cause, it is termed heterogeneous.

Again, we will treat the simplest case of homogeneous nucleation. If the interfacial free energy does not

depend on the orientation of the interface, then the nucleus will likely be spherical. The free energy

required to form the nucleus is


19

Here, ∆gvol=gsol-gliq is the difference in free energies per volume of the crystal. Note that we have usually

used values per mole or per atom, but here it is more convenient to discuss values per volume. Also note

that for T<Tm, ∆gvol is less than 0 – the free energy is reduced by forming the stable solid phase. Thus, for

large radii, Gnuc is less than zero, and the free energy can be lowered as long as R increases. However, for

small R, the interfacial free energy dominates, and the free energy increases as R increases. This is

shown roughly in the picture below. The maximum free energy G* of a nucleus can be found by

optimizing with respect to the radius. The resulting value is

Generally, the rates of transformation depend on the barrier according to a Boltzmann factor, so the

nucleation rate scales as exp(-G*

nuc/kBT). A high interfacial free energy suppresses nucleation, while a

low interfacial free energy makes for easy nucleation.

We can get a better estimate of some of these quantities, assuming that the temperature is close to the

transition temperature. Near the transition,

The term ∆gvol(Tm) is zero, because the free energies are equal. Note that we can write this in terms of the

solid and liquid free energies (remember that these are per volume for now):

Where ∆svol is the change in entropy per volume at

the transition. However, because ∆gvol(Tm)=0,

we have

Where Lvol is the latent heat per volume of the system.

(Notice by definition, both ∆hvol and ∆svol are

negative.) Therefore,

This gives a simple estimate of the difference in free

energies driving the transition. This quantity is

commonly referred to as the “driving force” for the

transition, as it controls how much free energy is reduced during the transformation.


20

If this were the entire story, the rate of nucleation would continue to increase as the temperature drops.

However, this is not the case: if it were, it would be very difficult to form glasses. A more careful

treatment says that the nucleation rate (number of nuclei forming per volume, per time) is also

proportional to the diffusion rate; if

diffusion is slow, then the system is

slow to rearrange into a crystalline

arrangement. Diffusion typically

behaves roughly according to

D(T)=D0exp(-Ediff/kBT) where Ediff is

a positive energy associated with

the barrier to diffusion. The

quantity D(T)exp(-G*/kBT) is

therefore a product of a function

D(T) that decreases rapidly as T

decreases, multiplied by a function

exp(-G*/kBT) that is small near Tm

and increases rapidly as T

decreases. Thus, nucleation is slow

both at low T and near Tm. The

nucleation rate has a sharp peak at

some characteristic temperature. The graph at right has the characteristic shape of a “Time-Temperature-

Transformation” (“TTT”) diagram (with rotated axes)T

A more detailed treatment of nucleation would include other effects ignored here. Typical other topics

include:

Elastic terms: Typically, there is a change in volume at the transition, and this can result in an

additional elastic energy for forming the nucleus. In that case, not only will the energy of the

critical nucleus change, but sometimes also the shape of the nucleus.

Heterogeneous nucleation: If the interface between the nucleating phase and the wall of the

container is very low, then the energy of the nucleus can be much lower by forming a “bubble”

on the wall. This is particularly true if there is a crack in the container, so that there is little

interface between the nucleating phase and the metastable “parent” phase.

Practice problems:

1. Show that

What is the radius R

* of the critical nucleus?

2. Turnbull showed that for FCC metals, the interfacial free energy is approximately = ½ -1/3Lvol,

where is the number of atoms/volume in the solid phase. The rate of nucleation is proportional

to exp(-G*

nuc/kBT). An observable nucleation rate may occur when G*

nuc ≈50kBTm. Using this, and

estimating Lvol≈kBTm, estimate the maximum undercooling ΔT/Tm. Compare this to Turnbull‟s

observation that ΔT/Tm is typically no larger than 0.2.

3. (Adapted from Porter & Easterling, problem 4.4.) The nucleation rate is

*

0 0 nucN=f C exp(-G / )Bk T .


21

Calculate the homogeneous nucleation rate in liquid copper at undercoolings of |T-Tm| = 180, 200

and 220 K, using the following data: Lvol=1.88x109 J/m

3, Tm=1356 K, =0.177 J/m

2, f0=10

11/sec,

C0=6x1028

atoms/m3, and k=1.38x10

-23 J/K. For each temperature, estimate the time it will take

for a 10 cm3 sample to form one critical nucleus. Repeat, assuming that is 10% smaller.

4. For slow vapor deposition on a surface, monatomic layers may nucleate as atoms accumulate.

The atoms lower their free energy by an amount g (per atom) by forming a layer, but the edge of

the layer has a positive step free energy s per atom. Assuming that the area occupied by an atom

is a2, and the length of the edge is equal to the number of atoms on the edge times a, write down

the free energy of a circular island of radius R. What radius gives the maximum energy? This is

the size of a critical nucleus. What is the energy of this critical nucleus, in terms of only g and s

(not R)?


22

6. Binary phase diagrams

a. Why?

From the point of view of traditional metallurgy, binary phase diagrams are perhaps the most important

topic we can cover. Alloy behavior is largely governed by the phase diagrams, and by the underlying

thermodynamic properties. Even some aspects of non-equilibrium properties may be understood from the

point of view of equilibrium phase diagrams, coupled with information on diffusion, nucleation, and

related kinetic properties.

However, it is a mistake to assume that this is limited to metallurgy: the same concepts and principles

underlie many material systems, including liquids, gases, polymers, sauces, and also are critical for

applications for other materials.

To draw a familiar example, consider the important material system ice cream. In its simplest form, ice

cream is a solidified mixture of cream and sugar. However, typically, a significant amount of salt is

needed to make ice cream. Why? It is not necessary for the flavor. Those who have made ice cream

know that the primary use of salt is not in the ice cream, but in the ice bath used to freeze this. What does

this do? It lowers the melting point of water. This is why salt is also used to break up ice on roads and

sidewalks. The lower melting point is used to freeze the cream-sugar mixture.

Previously, we looked at pressure-temperature phase diagrams, to show how (for example) pressure

affected the melting point of water. How can we instead form a phase diagram showing how composition

affects the melting temperature? The main point of this section is to demonstrate such diagrams, and how

they are interpreted. Later we will discuss the thermodynamics of such materials.

At right is a very incomplete picture of the water-salt phase

diagram, with water forming at high temperatures and ice at low

temperatures. The temperature at which ice forms drops with

temperature. This looks very simple, but be careful: the

interpretation is not so simple! To understand why, consider

what happens if you start with the composition shown at the red

dot. At higher temperatures, the system is a uniform

(homogeneous) mixture: salt water. However, at the temperature

shown by the red dot, the system is inhomogeneous: ice

essentially has no salt in it – the energy (or enthalpy) is too large.

Instead, the system is composed of nearly pure ice, plus salt water

at the composition shown by the line at the same temperature. The average composition of salt is the

same, but the system is composed of both liquid and solid phases with different compositions.

The above picture describes what happens if the composition is fixed, but the temperature is dropped.

What if we fix the temperature, and fix the composition? Imagine starting with pure water – perhaps hot

water (coffee temperature). If a little salt (or sugar) is added, it dissolves – forming a homogeneous

mixture. However, as more is added, the system at some point stops dissolving more salt, and the

remaining salt collects on the bottom. Again, the system is now inhomogeneous, composed of salt water

plus essentially pure solid salt. This is shown in the next figure. The red point indicates the

homogeneous regime, where the system is simply salt water. The solid line indicates the solubility limit

of salt in water. It slopes upward – indicating that at higher temperatures, water can dissolve more salt.


23

This is consistent with our experience. If the composition is to

the right of this line, then the system is a mixture of solid salt

(with essentially no water in it), plus salt water (with the

composition given by the blue line).

So, we have two lines in the composition-temperature binary

phase diagram, one that indicates the ice-salt water transition,

one that indicates the salt water-solid salt transition. Can we

form a complete diagram?

The more complete diagram is shown at right. If there is no salt, the diagram reduces to the water-ice

transition vs. temperature that is well known. Off to the right

somewhere is the pure crystalline salt phase. The main part of

the graph has four major parts: the “water” region which is a

homogeneous salt-water mixture, plus three inhomogeneous

regions, forming various mixtures of salt water, salt, and ice.

This is an example of what is called a eutectic phase diagram,

showing a high temperature liquid, two solid phases, with one

point where all three phases meet. This point, where water, ice,

and crystalline salt can coexist, is the lowest temperature that

the liquid may form (in equilibrium), and is called the eutectic

point, giving the so-called eutectic temperature and the eutectic

composition. This type of phase diagram is common in materials,

and will form the basis of much of our discussion.

This type of eutectic phase diagram is somewhat special, because

there is essentially no solubility of salt in ice, or of water in

crystalline salt. What happens in the case where there is some

solubility? This is shown for the Ag-Cu phase diagram at right.

There are three homogeneous regions, labeled as liquid, Ag, and

Cu. In fact, the regions labeled Ag and Cu are both Ag-Cu solid

mixtures – crystalline (FCC) phases. However, one is primarily

Ag, one is primarily Cu. In addition, there are three

inhomogeneous regions, unlabeled: the Ag-liquid region, the Cu-

liquid region, and the Ag-Cu region (the largest region, at lower temperatures). The eutectic point is

located just under 800 C, with a composition near 30 mass-% Cu. In this phase diagram, the solubility of

Cu in solid Ag is largest at the eutectic point, and the solubility of Ag in solid Cu is near or at a maximum

at the same temperature (somewhat hard to tell with this phase diagram).

Some terminology should be learned. The two lines bordering the liquid phase are called liquidus lines.

These lines separate the homogeneous liquid phase from the liquid-solid two-phase regions. The lines

that separate the liquid-solid two-phase regions from the solid phases are termed the solidus lines.

Suppose the temperature and composition is in the liquid-solid two-phase region, between the liquidus

and solidus points, such as shown by the circle on the Ag-Cu phase diagram. The two phases will have

the compositions given by the liquidus and solidus lines at that temperature, when in equilibrium. The

line separating a pure solid phase region from the region with two coexisting solid phases is the solvus


24

line. The two solvus lines show the compositions of the two phases, when they are in chemical

equilibrium.

An important issue is to determine the fraction of each phase, in a two-phase region. Suppose the

composition x is greater than the liquidus composition xL but less than the solidus composition xS (such as

shown by the point in the Ag-Cu figure). In otherwords, xL<x<xS. What is the fraction fL in the liquid

phase, what is the fraction fS in the solid phase? The simple expression that gives the answer to this is the

lever rule:

Note that these two expressions simply have the labels L and S reversed. This is called the “lever rule”

because if these represents fraction by weight, then the weight of the solid portion (proportional to fs)

times the length of the “arm” x-xS “balances” the weight of the liquid portion (proportional to fL) times the

length of its arm x-xL:

There is an important subtlety related to phase diagrams that must be mentioned. In the Ag-Cu figure

shown, the composition x is given by weight. This is very convenient for people making materials – the

x=50 wt-% composition means the same weights of Ag and Cu must be mixed. However, for showing

what crystal phases exist, it is often useful to know the composition by atomic fractions instead. For

example, salt crystals are really NaCl – the composition is 50 at-% Cl. Both conventions are used, and

the choice needs to be specified.

Given this, you should ask: are the fractions given by the lever rule by weight, or by atomic fraction? The

answer is simple: they are the same as the composition. If the compositions are all wt.-%, then the

fractions are by weight. On the other hand, if the compositions are atom-%, then the fractions are atomic

(or molar) fractions.

What other types of phase diagrams occur? The simplest is

the solid-solution phase diagram, such as that for Cu-Ni.

This is shown in the figure here (from www.copper.org).

There are two homogeneous regions, shown in blue: the

liquid region, and the solid phase, denoted by α. The α

phase is a nearly random mixture of Ni and Cu atoms on an

FCC lattice. On the phase diagram, there is a two-phase

region, shown in white, where the α phase coexists with the

liquid phase. The liquidus and solidus lines are labeled.

Suppose we have a composition x near 40 wt % Ni,

starting with the liquid phase, and cool. What transitions

occur? When the temperature reaches the liquidus line,

solid phase starts to form. As long as the temperature is

below the liquidus, but above the solidus, the system has

both phases present. The fraction in each phase is given by

http://www.copper.org/


25

the lever rule. Below the solidus, only the α phase is present. Note that this is different from the pure

system, where the liquid and solid phases are both present only at a single temperature (in equilibrium).

Now the melting process is “spread out” over a range of temperatures.

b. Phase diagrams and

microstructure:

Equilibrium phase diagrams are primarily intended to

show what phases are present, and in what fractions, as a

function of composition and temperature. They are

limited – real materials are often out of equilibrium, so

the phase diagram is not complete. Yet the phase

diagram can yield important suggestions not only of

equilibrium phase fractions, but also there are important

correlations with microstructure.

As a simple example, examine the cooling sequence

shown at right for the Ag-Cu system, at a composition

near x=75 wt-% Cu. At high temperatures, the system is liquid. When the temperature reaches the

liquidus, the Cu phase begins to form – probably with many nucleation points, forming separate Cu

regions. As the system is cooled further, these grow, and the liquid becomes richer in Ag. At the just

above the eutectic temperature, the liquid is essentially at the eutectic composition. Just below this

temperature, the remaining liquid all solidifies. The limited time for diffusion causes the remaining liquid

to solidify in a very fine microstructure, formed of a mixture of solid Ag and solid Cu phase. So the final

microstructure has large grains of Cu, called the „primary‟ phase (because it is first to form), and a fine-

grained “eutectic” microstructure of Cu and Ag. These two microstructures are called “microconstituent

phases.” One of the microconstituent phases, the eutectic phase, is itself a mixture of two different

materials: the Ag phase (with some Cu dissolved in it), and the Cu phase (with some Ag dissolved in it).

The other microconstituent is the Cu phase (again, with some Ag).

This is shown at right for a “hypereutectic” Al-Si alloy (image

from http://www.azom.com/details.asp?ArticleID=3604 ). The

large, dark particles are Si crystallites; the surrounding regions

are mostly Al, with fine Si phases. The details of such

formation may be dependent on the system, as well as on the

cooling rate. The amount of „primary‟ phase and of „eutectic‟

phase may be found by using the lever rule, at the eutectic

temperature (to find how much liquid solidified in a eutectic

manner). If the composition of the liquid begins at the eutectic,

only the eutectic microstructure will result.

http://www.azom.com/details.asp?ArticleID=3604


26

A different microstructure results if the path shown at right

occurs. As the liquid is cooled to point A, solid Ag (with

some Cu) starts to form. At point B, the system completes

solidification, forming grains of Ag. However, a new change

occurs as the system is cooled to point C. At (or below) this

temperature, the equilibrium is mostly solid Ag, but partially

forms Cu. This is a „precipitation‟ transition, requiring

significant diffusion of the Cu atoms. The diffusion is fastest

at the grain boundaries, and therefore the Cu precipitates will

likely form on grain boundaries or at grain boundary junctions.

This is shown schematically in the inset picture.

Thus, for the Ag-Cu phase diagram, there are roughly four different regimes, with characteristic

microstructures. At either very low or very high Cu contents, the microstructure will typically be grains

of the “primary” phase, with precipitates of the secondary phase. At intermediate compositions, the

microstructure will usually be a combination of grains of “primary” phase, plus characteristic “eutectic”

microstructure. In addition to these four, at the eutectic composition only the eutectic microstructure

occurs.

In general, at low temperatures, both eutectic microstructures and

precipitates can occur. The micrograph at right, from

http://pwatlas.mt.umist.ac.uk/internetmicroscope///micrographs/in

dex.html , shows the structure of a 50 wt% Sn – 50 wt % Pb alloy.

Dark regions are lead, forming both as large primary particles and

in the „eutectic‟ region. Light regions are tin, forming both in the

eutectic region and as precipitates within the primary particles.

c. The Iron-Carbon phase diagram

The iron-carbon phase diagram occupies a special place in metallurgy, due to its importance throughout

history. The relevant portions are shown below (with the picture taken from

http://www.sv.vt.edu/classes/MSE2094_NoteBook/96ClassProj/examples/kimcon.html ). In addition to

the liquid phase, the most important crystalline phases are the (“ferrite”) and (“austenite”) phases of

iron, and the hard, brittle line compound Fe3C (“cementite”). (This is called a “line compound” because

there is essentially only one composition, and therefore appears as a line of the phase diagram.) The most

important features of the phase diagram is the eutectic transition, at 4.3 wt-% carbon and 1130 C, and the

“eutectoid” transition at 0.83 wt-% carbon and 723 C. The eutectic liquid solidifies into a mixture of

Fe (with ~2 wt-% C) and Fe3C. This is called “cast iron” and is important because the carbon lowers the

melting temperature significantly compared with pure iron. This allowed for the early usage of iron. The

“eutectoid” is not a eutectic, as the high temperature is a solid phase ( iron), not a liquid, but looks like a

eutectic phase diagram. Below the eutectoid, the iron transforms into iron and Fe3C.

http://pwatlas.mt.umist.ac.uk/internetmicroscope/micrographs/index.html

http://pwatlas.mt.umist.ac.uk/internetmicroscope/micrographs/index.html

http://www.sv.vt.edu/classes/MSE2094_NoteBook/96ClassProj/examples/kimcon.html


27

In this diagram, there are three major composition regions, characterized by different microstructures.

The highest carbon region, from 2 to 6.5 wt-% carbon, is “cast iron.” In this region, the microstructure is

primary iron or primary Fe3C, surrounded by eutectic composition. The iron transforms further, as

discussed later. In the composition range below 2 wt-%, the system will form pure iron (with carbon

additions), before subsequent transformations, avoiding the eutectic microstructure. This is termed

“steel.” At the eutectoid composition, the iron phase forms a very fine scaled microstructure (with

length scales defined by diffusion lengths in the solid phase). This microstructure is called “pearlite” in

this system, due to its characteristic appearance. (Note that the phrases “eutectic” and “eutectoid” apply

to many different phase diagrams, but

“pearlite” implies Fe-C.) If the system is

above 0.83% carbon (but below 2%), it is

“hyper-eutectoid” or “high carbon” steel.

Below 0.83% carbon, this is “hypo-eutectoid”

or “low carbon” steel. The presence of the

Fe3C significantly strengthens the iron, but is

best in small quantities (and kept to small

length scales).

At right is a micrograph from

http://www.msm.cam.ac.uk/phase-

trans/2006/old/old.html showing a “modern

hypereutectoid steel” with 1.85 % C (plus

minor additions). The large dark phase is the cementite phase (the first formed from the iron); the small

structure is the pearlite. [This image is originally from Munoz et al., Journal of Nuclear Materials 349, 1-

5 (2006)].

http://www.msm.cam.ac.uk/phase-trans/2006/old/old.html

http://www.msm.cam.ac.uk/phase-trans/2006/old/old.html


28

This is far from a full description of iron and steel, but serves as a very basic introduction. Important to

remember are why the different regions (cast iron, low-carbon steel, high-carbon steel) get their names,

and understand why the different microstructures form. There is a lot of terminology, but the most

important are the “eutectic” and “eutectoid” transformations that occur in many systems. The phrase

“pearlite” (as well as ferrite, austenite, and cementite) are important to those dealing with steel and iron

systems.

7. Thermodynamics of binary phase diagrams

a. Back to the first law…

To describe the thermodynamics of binary systems, we first need to return to the first law. Assuming a

reversible process, a small change was described as

dU = TdS – PdV.

Is this complete? No – we described many ways of changing the energy, not just entropy or volume

changes. In particular, what happens if we add more atoms? Can we do this without changing the

entropy and volume? The answer is yes – but the energy changes. Easier to understand is to consider the

Helmholtz free energy, F=U-TS. The differential form we wrote down is

dF = -PdV – SdT.

We can certainly imagine adding particles at constant volume and temperature – the density and pressure

changes, but this will change the free energy. Then this equation does not apply. Why not? Because

adding particles takes work – different from the form that we used before. A more complete description

of the work is

dW = -PdV + dN

where we have introduced the chemical potential . This is a measure of how much work it takes to add

an atom. If we have different types of atoms, then each type has its own chemical potential:

dW = -PdV + AdNA + BdNB + …

Then the first law becomes

dU = TdS - PdV+ AdNA + BdNB + …

The Helmholtz free energy is

dF = -SdT – PdV + AdNA + BdNB + …

A very special case is the Gibbs free energy. We had written this as

dG = VdP – SdT

Adding particles at constant pressure and temperature makes perfect sense – the system is simply bigger,

without changing the density. The volume and entropy just scale with the number of particles. The

generalized version is

dG = VdP – SdT + AdNA + BdNB + …


29

but there is another important result, from the way that the free energy G scales with the number of

particles. Specifically,

G = ANA + BNB + …

For a single component system, the chemical potential is just the Gibb‟s free energy per atom (or per

mole).

What happens when we mix two atom types A and B? From a thermodynamic point of view, it is useful

to define the partial molar free energies of the pure phases as

gA = GA/NA.

Here, GA is the Gibbs free energy of the pure A phase, and NA is the number of moles. Similarly,

gB=GB/NB.

When we combine the different types of atoms, we can write the total Gibbs free energy as

G = GA + GB + Gmix = NAgA + NBgB + Gmix

Where the term Gmix is the change in free energy from mixing. Again, the system will try to reach the

lowest Gibbs free energy – so if Gmix>0, then the A and B atoms will tend to separate. On the other hand,

if Gmix < 0, then the system will preferentially mix. It will be useful to do all of this on a molar basis, with

N=NA+NB being the total number of moles, and xA=NA/N being the atomic fraction of A atoms, and

xB=NB/N being the fraction of B atoms. Clearly, xA+xB=1. The molar free energy is then

g = xAgA + xBgB + gmix.

The terms gA are the “pure phase” free energies we considered in the first part – so we will not pay close

attention to those now. The mixing term is of most interest. Not surprisingly, there is an enthalpy

contribution and an entropy contribution:

gmix = hmix - Tsmix .

Of course, all of these quantities depend on temperature, pressure (which we will largely ignore), and

composition. The easiest to deal with is the entropy of mixing. If mixing is completely random, then the

entropy that we derived earlier is

This is the “ideal” entropy of mixing. (We wrote this per atom before; here, we write things per mole,

and use the ideal gas constant R instead of Boltzmann‟s constant kB). Note that because xA and xB are both

less than one, this term is always positive. If we write x=xB and xA=1-x, we can write this as

.

We sketched this before (around pg. 7). It is important to understand that the slope of this plot goes to

infinity as x → 0, and to minus infinity as x → 1. This means that a small change of composition can

dramatically increase the entropy. We will use this form throughout the rest of this, though it is only an

approximation. In reality, two atom types won‟t mix completely randomly – there will be some


30

preference for either some level of mixing (A atoms like to be near B atoms) or de-mixing (A-A and B-B

bonds are prefered over A-B bonds).

b. “Ideal solution” thermodynamics

Let‟s consider a simple case: Suppose the enthalpy of mixing is zero: there is no enthalpy change from

mixing A and B atoms. This is a limiting case, but often useful. It is best for good “solid solution”

systems such as Cu-Ni. In this case, there is little correlation between the atoms, and the ideal entropy of

mixing is a good approximation. Neglecting hmix and using the above form for smix, we obtain the ideal

solution model for a mixture:

g = xAgA + xBgB + RT( xA ln xA + xB ln xB ).

This ideal solution model is sufficient to understand many aspects of simple phase diagrams. Note that

the above expression is for a single phase, such as a single crystal phase, or a liquid phase. The entropy

of mixing term always favors mixing, so the free energy is lowered.

Let‟s sketch this, roughly, at different temperatures, for the Cu-Ni system. First, consider T=TmeltNi

– at

high temperature, where the system is beginning to solidify. At this temperature, gNiliq

=gNisol

– for pure

Ni, the liquid and solid (FCC) free energies have the same value at the melting temperature. However, it

is above the melting temperature of Cu, so gCuliq

< gCusol

. In fact, gliq

< gsol

for all compositions except

pure Ni. This is shown schematically below. For both liquid and solid phases, the free energy is lowered

by mixing.

Now consider a slightly lower temperature, such that TmCu

< T<TmNi

. For Ni, the free energy of the solid

phase must be lower than that of the liquid: gNiliq

>gNisol

. For Cu, this is reversed: gCuliq

<gCusol

. Therefore,

the two curves must cross. This is shown in the middle figure above. It would be easy but wrong to

assume that the transition occurs at the composition where these cross. Why? Because these curves are

single-phase free energies. The system can, in a certain composition range, lower its free energy by

changing to two phases in coexistence. In the middle figure, the free energy at the composition where

gliq=gsol is higher than the average free energies of the „common tangent‟ shown. The system has the


31

lowest free energy by separating into a liquid with composition xliq and a solid with composition xsol.

These are the liquidus and solidus compositions at this temperature. This is analogous to the curves we

drew earlier for the Helmholtz free energy as a function of volume.

In general, if we have a free energy g(x) (at some temperature and pressure), and construct the tangent

line going through this curve at some value of x, the tangent line can be traced back to x=0 and x=1. It

can be easily shown that the intersections at these point are equal to the chemical potentials for atoms A

and B, respectively. That is to say, the intersections give the values A and B. So, for the common

tangent shown above, which meets both the liquid and solid curves, we must have Culiq

= Cusol

and Niliq

= Nisol

, and these values can be obtained from where they meet the Cu and Ni axes, respectively. The

two phases, in equilibrium, have the same chemical potentials. This is a general condition for multiphase

equilibrium: all components must have the same chemical potential. If there are three phases in

equilibrium, all three curves have the same common tangent.

Finally, we consider the temperature T = TmCu

. At this point, gCusol

=gCuliq

; however, for all other

compositions, the solid phase has the lowest free energy. This is shown at right in the figure above.

We can rewrite the “ideal solution” model in terms of the chemical potentials. Recall that the free energy

(per mole) is generally given by

g = xAA + xBB.

Comparing this to the above expression for the ideal solution model shows that

Aid = gA + RT ln xA

Bid = gB + RT ln xB .

These simple expressions give some estimate of the behavior of real solutions, and can be reasonably

accurate for some systems.

c. “Regular solution” thermodynamics

We obtained the ideal solution model using Hmix = 0. However, this is often a poor approximation. How

do we do better? With some idea on how Hmix changed with concentration, then a better approximation

can be made.

To get some idea, consider first a pure crystal phase. If the interactions between atoms are restricted to

near neighbors, and there are z neighbors per atom, then there are ½zN bonds in the system. Why ½ ?

Because each bond is shared by two atoms – so without this, we are double counting. If the enthalpy of a

bond is h, then the total enthalpy is

H = ½ zhN

Now, what happens if we have two types of atoms, A and B? There are now three types of bonds: A-A,

A-B, and B-B bonds. Note that the different types of bonds have different enthalpies. If the A and B

atoms are totally separate, then there are no A-B bonds, and the enthalpy may be written as

H = ½ zhAANA + ½ zhBBNB.


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If we write this per mole, this is

h = H/N = ½ zhAA xA + ½ zhBB xB

To find out about the enthalpy if the A and B atoms mix, we need to think about how they mix. An

ordered arrangement will be different than a random arrangement. The ideal entropy of mixing was

calculated by assuming complete randomness, so let us proceed using that. If atoms are arranged

randomly, then an A atom will usually have xAz neighbors that are A atoms, and xBz neighbors of type B.

Therefore, the enthalpy of one mole of A atoms will be

hA = ½ z (xAhAA + xBhAB).

Similarly, the enthalpy due to one mole of B atoms will be

hB = ½ z (xAhAB + xBhBB).

Therefore, the total enthalpy will be

H = ½ z [(xAhAA + xBhAB)NA + (xAhAB + xBhBB) NB]

H = ½ zN [(xAhAA + xBhAB) xA + (xAhAB + xBhBB) xB]

H = ½ zN [ hAAxA2 + 2hABxAxB + hBBxB

2 ]

Again, writing per molecule, this is

h = ½ z [ hAAxA2 + 2hABxAxB + hBBxB

2 ]

So, we‟ve written down expressions for the unmixed enthalpy, and the mixed enthalpy. What is most

useful is the change in enthalpy from mixing:

hmix = ½ z [hAAxA2 + 2hABxAxB + hBBxB

2 – hAAxA – hBB xB]

hmix = ½ z [2hABxAxB – hAAxA (1-xA) – hBxB (1-xB)]

hmix = ½ z [2hABxAxB – hAAxAxB – hBBxAxB]

Finally, we may write this as:

hmix = ΩmixxAxB.

Here, we have introduced the notation

Ωmix = ½ z [2hAB – hAA – hBB].

The term in brackets has a simple interpretation. If there is originally an A-A bond and a B-B bond, these

may be broken and combined to form 2 A-B bonds. The change in enthalpy is then 2hAB-hAA-hBB, exactly

the expression above.


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There is a different approach to argue this form for hmix. In the limit that xA=0 or xB=0, the system is pure,

so the enthalpy of mixing must be zero. The simplest expression satisfying this is hmix = ΩmixxAxB. Other

terms may be added to generate different approximations, but this is the simplest term. In the usual

approximation, Ωmix is assumed to be independent of temperature, but this may also be changed.

Using this form of hmix, combined with the ideal entropy of mixing, we may write the Gibbs free energy

of mixing as

gmix = Ωmix xAxB + RT (xA ln xA + xB ln xB).

Use of this form is known as the regular solution model. In the limit that Ωmix=0, it reduces to the ideal

solution model.

Let‟s examine some behaviors of this expression. First, examine the plot of hmix shown in red here

(assuming a positive Ωmix). It looks similar to the

entropy of mixing, shown in green. However, the rise in

hmix near x=0 is not as steep as that for smix . If hmix>0,

then the enthalpy is minimized by keeping the system

separate. This is referred to as a “positive heat of

mixing.”

What we are really interested in is

gmix = hmix - Tsmix.

At high temperatures, this will behave like –Tsmix – the

enthalpy term may be neglected. In that case, gmix is

negative everywhere, with a minimum at x=0.5. At very low temperatures, the entropy term may be

neglected, and the behavior is dominated by the enthalpy. However, some care with this must be taken –

at sufficiently small x (or for x sufficiently close to 1), the infinite slope of smix(x) will win out at all

temperatures except T=0. Consider the

plot shown here. At moderate

compositions, the enthalpy dominates,

but at smaller compositions the entropy

dominates. The system can find its

lowest free energy by separating into

two separate compositions, near x=0.1

and x=0.9 in the graph shown.

As the temperature increases, the

entropy becomes more important. The

free energy will start to look more like

the figure at right. Now the lowest free

energies occur near x=0.25 and x=0.75


34

– the mutual solubility increases.

At some critical temperature, there will no longer be two

minima at the bottom, but just one. The free energy at the

critical point is shown at the bottom. At this point, the

curvature of gmix(x) at x=0.5 changes from negative

(curving downward) to positive (curving upward). This

can be shown to occur when RTmix =2Ωmix .

What sort of phase diagram occurs for this? This is a

typical “phase separating” system. At high temperatures,

the system is mixed; however, at some temperature, the

system will separate (or “segregate” into two separate

compositions, within the same nominal phase (FCC,

liquid,…). Note that other phases should also be

considered: for example, if this is the behavior of a

crystalline system, the free energy of the liquid phase

should also be considered. In a strongly segregating

system, there may be several “common tangents” with

different phases that must be constructed to form the entire

phase diagram.

What about the chemical potentials? For a regular solution,

we have

g = xAgA + xBgB + gmix = xAgA + xBgB + Ωmix xAxB + RT (xA ln xA + xB ln xB).

To write this as g = xAA + xBB, we use (xA+xB) = 1 to rewrite the enthalpy of mixing term as

hmix= Ωmix xAxB = Ωmix xAxB (xA+xB)

= Ωmix [ xBxA2 + xAxB

2]

= Ωmix [ xB(1-xB)2 + xA(1-xA)

2].

Then, we may rewrite g as

g = xA [gA + RT ln xA + Ωmix(1-xA)2] + xB [gB + RT ln xB + Ωmix(1-xB)

2]

Comparing that with the previous form gives

A = gA + RT ln xA + Ωmix(1-xA)2

B = gB + RT ln xB + Ωmix(1-xB)2


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It is conventional to define the activity of a substance A using

A = gA + RT ln aA.

This is done in analogy with the ideal solution model: Aideal

= gA + RT ln xA. Thus, for an ideal solution,

the activity is the same as the composition. For a regular solution, we have

aAregular

= xA exp[ Ωmix(1-xA)2/RT ]

aBregular

= xB exp[ Ωmix(1-xB)2/RT ]

It is also common to define the activity coefficient as A=aA/xA; again, for an ideal solution, this is simply

A=1. For a regular solution, we have

Aregular

= exp[ Ωmix(1-xA)2/RT ]

Bregular

= exp[ Ωmix(1-xB)2/RT ]

Thus, the activity and the activity coefficient show how much the system deviates from ideality. If a

system is more “active,” i.e. >1, this implies that its chemical potential is higher than would be expected

simply from its concentration. This is the case when the enthalpy of mixing is positive. On the other

hand, if <1 (for a system with a negative heat of mixing), then it tends to be less reactive (or less

“active”) than would be expected for an ideal solution.

To examine this, consider the regular solution activity coefficients when x=xB≪1, and xA=1-x is nearly

one. In that case,

Aregular

exp[ Ωmixx2/RT ] 1 (for Ωmixx

2/RT ≪ 1)

Bregular

= exp[ Ωmix(1-x)2/RT ] exp[ Ωmix/RT ] constant (for x ≪ 1)

This says what we know intuitively: If a small amount of material B is dissolved in a large amount of A,

the chemical properties of element A are nearly unchanged, and the changes are simply due to the fact that

A is slightly diluted. If B does not want to mix with A (Ωmix>0), then the dissolved B elements are

somewhat more reactive (to try to stop mixing). On the other hand, if B is happy in A, then it will be less

reactive once dissolved.

These are well known expressions: the high-concentration result that A1 when xA1 is called “Raoult‟s

Law”; the low concentration result that Aconstant when xA≪1 is called “Henry‟s Law”. An important

note for Henry‟s law is that if Ωmix>0, then the constant is greater

than 1; similarly, if Ωmix<0, then the constant is less than one. The

ideal solution result is that the constant is identically equal to 1.

It is common to plot the activity vs. concentration. Again, for an

ideal solution, the activity is equal to the concentration. The limits

of small x and of small (1-x) give Henry‟s law and Raoult‟s law.

For example, the plot below shows the activities of the liquid Ag-

Cu system as a function of concentration. It is apparent that


36

Raoult‟s law holds only for small concentrations. The enthalpy of mixing can be seen to be positive – the

activities are significantly larger than x for most concentrations.

A different situation occurs for Al-Si. The activity plots below show that again, there are significant

deviations from an “ideal” solution behavior in the liquid. However, now the activities are smaller than

the concentrations, indicating a negative enthalpy of mixing.

Practice problems: 1. Si melts at T=1687 K, with a change in entropy of 3.25 kB per atom.

A. What is the change in enthalpy on melting? B. Si liquid is denser than crystalline Si. Will the melting temperature increase or decrease if

pressure is increased?

C. Derive the approximate expression Gsol(T)-Gliq(T)≈S(T-Tm), near the melting temperature. D. A small amount of Al is added to the liquid, with an atomic concentration x. Write the

chemical potential Si of Si in the liquid at Tm, in terms of Gliq(Tm) for pure liquid Si. Assume

that the Al mixes ideally. E. Using the results of parts C and D, estimate the liquidus temperature for x=10 at-% Al.

Assume that there is no Al

solubility in crystalline Si, so the chemical potential for Si in the

solid phase does not depend on

concentration.

F. Using the measured liquid activities given in the plot above,

explain whether the assumption of

ideal mixing is reasonable for part D. Is the enthalpy of mixing in the

liquid phase positive or negative?


37

2. Below is the phase diagram of Al-Zn. Solid Al () has an FCC phase, and solid Zn () has an HCP phase.

A. Is Zn more soluble in FCC Al, or is Al more soluble in HCP Zn? Is the enthalpy of mixing

positive or negative in the FCC phase? Explain.

B. Estimate the relative (atomic) percentages of and phase formed by the initial solidification of the liquid with the eutectic composition.

C. Starting from pure phase with 40 atomic % Zn at T=400 ºC, what are the phases that form as the system is cooled? Sketch the microstructure at T=300 ºC and at T=200 ºC.

D. Sketch the free energy of the andphases as a function of composition. Make separate graphs for T=350 ºC, T=300 ºC, and T=250 ºC. Indicate the tie-line compositions.

E. (This part uses some material from the following section.)

With 40 atomic % Zn, suppose you wanted to form as much phase as possible from the

melt, at T=350 ºC. Would it be better to slowly cool the liquid, or quench it? Why?

385 ºC

282 ºC


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8. A brief introduction to solidification dynamics

Above, we used equilibrium phase diagrams to determine fractions of phases, and to make some

arguments about the microstructure that could result. However, experimentally there are sometimes

significant deviations from this, particularly during rapid solidification (including casting). This is due to

nonequilibrium effects in the process. Perhaps it is a surprise that a significant amount can be learned

from the equilibrium phase diagram – again, this will be our „map‟ of behavior. We will focus primarily

on eutectic solidification, which commonly occurs, or at least has features in common with many systems

of interest. The example of Ag-Cu has been given before, but also the cast iron region of the Fe-C phase

diagram is also an example; other systems that are relevant include Al-Si (which has a simple eutectic

phase diagram) and Al-Cu (in the Al-rich region). We note that this will be a qualitative description only;

the time required to develop a quantitative description is insufficient.

To begin: under what conditions should the microstructure be what we argued above? Those assume that

the system is close to equilibrium, in particular chemical equilibrium. This implies that the chemical

potentials are equal through all phases. Within a single phase, this implies a constant concentration

throughout the phase. If chemical diffusion occurs sufficiently fast, then the above arguments should be

reasonable. What is “sufficiently fast”? Roughly, this means that changes in microstructure (such as the

motion of solidification fronts) are slow compared to the time to reach chemical equilibrium. Thus, the

microstructures we have argued above are the “fast diffusion” limits – the diffusion is assumed to be so

fast that the system is always in chemical equilibrium.

Here, we will present qualitative results for two other limits. First, it is important to recognize an

experimental fact: diffusion in the liquid is typically much faster than in the solid, by factors of at least 10

and commonly much more. Thus, the “fast diffusion” limit is better for the liquid than for the solid.

Thus, another easy approximation is to assume very fast diffusion in the liquid, so the liquid has a single

composition, but also assume that diffusion is very slow in the solid – essentially, so that the composition

of the solid phase never changes, even when there are significant composition changes throughout the

solid phase. Finally, we will briefly consider what happens when solidification is so rapid that the

solidification front is sufficiently fast that the concentration in the liquid near the interface is different

than it is far from the interface. So, the three limits are:

Fast diffusion in both solid and liquid phases (chemical equilibrium everywhere)

Fast diffusion in the liquid, no diffusion in the solid (liquid composition is the same everywhere,

but solid composition varies in space).

Slow diffusion in the liquid, no diffusion in the solid.

The first case we have treated already, so we begin here with the second case. The limits of fast diffusion

in the liquid and slow diffusion in the solid are convenient, because in this limit, the actual values of the

diffusion don‟t matter. Easiest to imagine is a directional solidification set up, shown below. One end is

kept slightly cooler, so that the solid region grows from left to right. This temperature change is assumed

to be small, so that the system may be considered to be at a single temperature T. The figure shows the

situation near the beginning of the solidification process. If the system is in equilibrium, and the average

composition is x, then the solid phase has the solidus composition xsol(T) and the liquid phase has a

composition xliq(T). The phase fractions of the solid and liquid portions are given by the lever rule.


39

Now suppose that the temperature is decreased slightly, so that the solid region grows. The new growth

is in equilibrium with the liquid at the new temperature. However, the already solidified region does not

change composition. Moreover, the liquid

phase is in equilibrium with the newly

solidified region. This can be seen here. The

newly solidified region at T2 has a different

composition than the first part. The key is that

as solidification progresses, the composition of

the solid and liquid regions change.

To make this more definite, consider the phase

diagram below. Solidification begins at T=T1.

The initially solidified phase has the solidus composition. As the system is cooled to T2, solidification

proceeds, but the composition of the solid phase changes through the system. The initially solidified

portion has less solute than the region solidified later. Where is this solute? It is in the liquid – so the

liquid is solute rich, compared to the equilibrium condition. This process continues as the temperature is

cooled.

Note that there is still a lever-rule relationship between the average concentration in the solid, <xsol>, and

the liquid concentration xliq. The average of the entire system is x. However, we can write the average

concentration in terms of the liquid and solid average

concentrations, and the fractions of the phase that are liquid

and solid:

x = <xsol>fsol + xliq fliq.

Using this, and fsol+fliq=1, we can derive the lever rule. Note

that because we have assumed that xliq>xsol, <xsol> must be less

than x as long as fliq is greater than zero. Once <xsol> = x, the

whole system must be solid.

At a temperature T3 where the equilibrium diagram gives

xsol(T3)=x, equilibrium should be a single, solid phase.

However, because the initially solidified regions have

concentrations less than x, the average concentration in the

solid region <xsol> is less than x. Thus, there is some liquid left. There is more solute in the liquid – so

there is still a liquid region, rich in solute. Thus, the system is not completely solidified at T3. The

average composition of the solid phase is always less than the total concentration. This means that the

system continues to solidify, with the liquid becoming richer in solute, down to the eutectic temperature.

At this point, the liquid has the eutectic composition, and further cooling results in eutectic solidification.


40

There are two main points here: first, the liquid does not fully solidify at the equilibrium temperature, but

can persist to lower temperatures. Secondly, some liquid solidifies with a eutectic microstructure – so

there is a two-phase region, even when the equilibrium state is single phase.

To examine this more thoroughly, let‟s plot the composition of the solid region as a function of position.

The first part has composition xsol(T1), some a little

further on has composition xsol(T2), etc. This is

shown in the figure at right.

How can this be made more rigorous? Assume that

the liquidus and solidus lines are straight lines, with

xsol=kxliq. The partition coefficient k is less than one.

Thus, the initially solidified region has a composition

xsol(T1)=kx (where x is the initial composition of the

liquid). If the fraction solidified increases by dfsol, then this region rejects some solute into the liquid.

This amount of rejection is equal to (xliq-xsol)dfsol. The change in the liquid phase composition is equal to

the fraction of liquid times the change in composition, fliqdxliq=(1-fsol)dxliq. Using the relation above, we

have

(1-k)xliq dfsol =fliqdxliq

Using fliq+fsol=1, we have dfsol= -dfliq, so

.

Integrating this, we get

ln fliq(k-1)

= ln Cxliq

where C is a constant of integration, to be set by boundary conditions. In this case, the condition is that

when xliq=x, then fliq=1. From this, we get C=1/x, or

fliq(k-1)

= xliq/x or fliq=(x/xliq)[1/ (1-k)]

.

Similarly, using xsol = kxliq and fsol=1-fliq,

k(1-fsol)(k-1)

= xsol/x.

These are the Scheil equations or non-equilibrium Lever rule results. Given the values of xliq and xsol from

the phase diagram, and the known composition, the fraction of the liquid and solid phases can be

determined. Again, the value of k is approximately k=xsol/xliq. To compare, in equilibrium, the fraction in

the liquid phase is

(in equilibrium).

Note that k-1 is less than zero. This means that when fliq is small (the system is mostly solid), fliq(k-1)

is

large, so the non-equilibrium result predicts that xliq is large as well. However, there is a limit: once xliq

reaches the eutectic composition, the remaining liquid can solidify rapidly. This is in agreement with the

arguments we made above.


41


42

Thermodynamic formulas:

First law:

dU = dQ – dW = dQ – PdV + i i dNi (Ni = number of atoms OR number of moles of type i )

dQ TdS

Other energies:

Enthalpy: H = U+PV

dH TdS + VdP

Helmholtz free energy: F = U – TS dF = –SdT –PdV

Gibb’s free energy: G=H+PV=i i dNi

g = G/N = xAA + xBB

dG = VdP – SdT+ i i dNi

Entropy of mixing: (ideal mixing)

Smix= –kB [ xA ln xA + xB ln xB] (per atom)

Or

Smix= –R [xA ln xA + xB ln xB] (per mole)

Gibb’s free energy for solutions, regular solution model (per particle):

G = xAGA+ xBGB + mixxAxB - TSmix (per mole or per atom)

Clausius-Clapeyron equation:

dP/dT = S/v

(v=volume/atom or volume/mole; S=entropy/atom or entropy/mole)

Constants:

kB = Boltzmann’s constant = 1.381x10-23 J/K R = 8.316 J/K/mole h = Planck’s constant = 6.626x10-34 J·s NA= Avogadro’s number = 6.022x1023 atoms/mole

Documents

1. Introduction to thermodynamics a. What is thermodynamics?