MSE 511 notes 2009 – J. R. Morris 1 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 (σ xy ) 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 (6.02×10 23 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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MSE 511 notes 2009 – J. R. Morris
1
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.
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
MSE 511 notes 2009 – J. R. Morris
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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.
MSE 511 notes 2009 – J. R. Morris
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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.
MSE 511 notes 2009 – J. R. Morris
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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
MSE 511 notes 2009 – J. R. Morris
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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
MSE 511 notes 2009 – J. R. Morris
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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
MSE 511 notes 2009 – J. R. Morris
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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.
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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-
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
MSE 511 notes 2009 – J. R. Morris
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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?
MSE 511 notes 2009 – J. R. Morris
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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
MSE 511 notes 2009 – J. R. Morris
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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.
MSE 511 notes 2009 – J. R. Morris
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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.
MSE 511 notes 2009 – J. R. Morris
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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.
MSE 511 notes 2009 – J. R. Morris
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MSE 511 notes 2009 – J. R. Morris
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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