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Last class, we were talking about the evolution phase diagram based on thermodynamic principles.
We took up two parameters, which are basically the omega s and the omega l, which is nothing
but, the interaction parameter in the liquid between A and B and interaction parameter
in the solid between A and B, and used them to basically see how phase diagram evolve
from a isomorphous system. And you get isomorphous system, when both of them are close to 0,
and as you start varying one, keeping the other one fixed, how the phase diagram evolves,
so we will try to see that.
And today, I just had brought you image from one of the reference as you can see here,
this is from T.B. Massalski, published in Metall Trans in 1999, which gives you a flayer
of how really the phase diagram evolves; actually I was talking about that. You can see here,
on the left axis you have the omega s and on the top, which is the X axis is the omega
l, and you see how as you start varying these things, how various from phase diagrams come.
And you see, when you take both of them as 0 0, that is what ends of the view of the
isomorphous system. And if you keep omega l as 0 and keep decrease, or keep the omega
s as 0, keep decreasing the omega l, make omega l more and more negative, you see when
it is minus 10 and when it is minus 20, what you see? You start seeing pseudo eutectic
phase diagram evolves, so the liquid becomes more and more stable, so the liquidus curve
starts plunging down. And, as you start seeing, you see more and
more of liquid stability there. And on the other side, keeping the same, the omega s
as 0, and let us say, increase the omega l to positive side, again you see that, you
start seeing that, because the moment omega l is positive and omega s is 0, that means
solid is more stable than the liquid, and as a result solid stability regime increases
and you start seeing a congruent melting phase develops. And similar thing can also happen
when omega l is omega s is negative. For example, keeping omega l as constant at 0, and if you
make the omega s negative, this is what you see in the diagram below, where you see the
moment you have minus 15 for the omega s, you again start seeing for the same omega
l being kept as 0, so you see the concrete melting task evolving.
So, you can see that, we can have either this under the ideal conditions. You can have a
pseudo eutectic type, when the liquid is more stable; when the solid is more stable, you
start getting what is called the concrete melting, which is an indication of the formation
of the inter metallic. And if this becomes stronger, that means, if the omega s becomes
more and more negative, you would actually see that inter metallic starts actually evolving
inside the phase diagram. And this is as far as either the omega l being negative or omega
s being negative, but if you think if omega s is positive, what is going to happen? Keeping
the omega l as 0 and if you start making it positive, for example you see, this is plus
15 and everything is kilojoules per mole, all the numbers that we are talking are basically
the enthalpy unit because omega s is nothing but, the enthalpy unit, because it is an interaction
parameter and that multiplied by X A and X B will give you the actually enthalpy of mixing.
So, it will have the same unit for the enthalpy, which is nothing but kilojoules per mole.
So, you can see here, if you keep this as 0 and then make omega s as positive 15, you
start developing eutectic. And we saw yesterday how this eutectic actually develops because
of the miscibility gap, and the miscibility gap and the liquid of plunging down, side
downwards and both of them meeting, both the liquidus, solidus and the miscibility gap
meeting. Basically, what you see down is this, it is the miscibility gap and this is the
liquid of both of them merging to each other, and you will get that. Now, if you keep this
as constant and let us say make omega l more and more negative, you see that liquid is
becoming more and more stable with respect to the solid and as a result, the eutectic
temperature starts falling. You will see the same thing can be also seen,
if you want to increase the omega s keeping the omega l the same. For example, if you
keep the omega l as 0 and increase the omega s to plus 30, what you see here is that, the
terminal solubility vanished on both the sides and you will also see the liquid is coming
down. Like this we will be able to see, how various and similarly, if you keep this as
0 and make it this as plus 30 and make this more and more negative, similar thing happens.
So, as the liquid becomes more and more stable by making the omega l more and more negative,
you see that the liquidus line comes down, that means the eutectic becomes more and more
deep eutectic. Yesterday, I also gave you an idea of how
to quantitatively measure, how deeply the eutectic is with respect to basically, we
take the melting points of two pure metals and then find out what is the melting point
of mixture of the two pure metals at the eutectic composition, and take down what is the eutectic
temperature; and the difference between this two will give you a measure of the deeper
eutectics, and of course, it has to be normalized with respect to the melting point of the mixture
and that is what we have seen in the last class.
Now, if you see that, if the omega l becomes positive, so far we have seen whether the
omega s is positive and the omega l is either 0 or negative, and if you make the omega l,
start making the omega l positive, then you start seeing that development of the monotectic
phase diagram. You start seeing a miscibility gap inside the liquid, not only the solid
also will have miscibility gap, but also the liquid will have the miscibility gap and you
start developing either the monotectic or the syntectic kind of phase diagrams.
That is how you see all the phase diagrams can be, in fact though this is schematically
shown in principle, one can calculate all these phase diagrams simply by just varying
the two parameters. And whole calculation is based on what we have seen, the equilibrium
principle, which is nothing but, the chemical potential of A in alpha is equal to chemical
potential of A in liquid.
And if you use this as the starting point for your equilibrium, from that, you can find
out at each temperature, what is the liquidus composition, what is solidus composition,
and from that, we can come across all this. And in case there is third phase to be evolved,
for example, in case of eutectic or in case of monotectic, you have three phases coming
into picture, and you would see how to deal with those three phases. By considering third
phase, and that is what we are going to see today, what happens, how do I consider the
third phase? At the moment, we have been talking about
only two phases, omega s and omega l, and how do I consider that third phase. And that
is what we will see today, and that also will give you an idea of how to understand the
miscibility gap.
Yesterday, somebody was asking me, sir if this is the miscibility gap, how does this
gets closed? What is thermodynamic behind it? We will also see today, how what is thermodynamics
behind this, the miscibility gap getting closed at higher temperature and why does it kept
closed. To understand this, what we see is, how does the free energy curve looks like
at the temperature does.
For example, you take a system which is an isomorphous system and draw free energy curve
at the temperature, how does free energy curve look like? If I draw it, this is G verses
X we are talking and at this temperature, what is stable? Solid alpha is stable, this
is liquid and this is liquid plus alpha, so alpha is more stable, very easy. So, as a
result, I will see the free energy curve for alpha which looks like this and of course,
there will be a free energy curve for the liquid which is something like this, this
is liquid, this is alpha, Am I right. And this free energy curve is expressed in terms
of the free energy expression that we have given earlier, where G alpha is X A G A plus
X B G B plus delta H mixing plus R T X A L N X A plus X B L N l X B; so that expression
which we have given can be used to calculate this free energy. This is all fine, the moment
you see a miscibility gap here, how does this free energy curve is going to change?
The difference between this and this system is that, in this system the omega s is positive,
right. So, there is a positive interaction between the A and B; that means, A and B do
not attract each other, they repel each other. The moment this is positive, what happens
to delta H mixing, delta H mixing is positive. Because, delta H mixing is nothing but, omega
into X A X B regular solution model, as long as I assume that regular solution model is
varied. So, this is also positive, if this is positive, if that is positive then, you
would see that, if I try to draw, you will see that if I plot, that delta H mixing as
a function of composition, you will see that delta H mixing curve will be something like
this, this is delta H mixing, am I right. So, once it is like that, and you have that,
T delta H mixing curve minus T into delta S mixing, which is the second part of the
free energy of mixing will always be negative. The only difference is, depending on the temperature
of this, this value is going to depend; this value, how large absolutely it is, or how
small absolutely it is, depends on what is the T. Because, delta S mixing is only a function
of composition, it is a symmetric function we have seen that, X A L N X A plus X B L
N X B plus X B. So, as a result, this is going to be solely
dependent on the temperature and as a result, at room temperature if I am considering, where
the temperature is not too high, you would see that possibly this particular value, T
delta S mixing is lower than the delta H mixing. For a certain case, as long as delta S mixing
value, the maximum value of delta S mixing is smaller than the minimum value of T delta
S mixing, you would see that the delta G mixing which is nothing but, delta H mix minus T
into delta S mix. If this is positive, which what we have seen here, but its absolute value
if it is smaller than this negative value, then the delta G mixing is still negative.
So, the delta G mixing is still negative then, free energy curve will be having a curve downwards.
I have seen how, drawn here, it will be like that, the moment this value is higher than
this u t, that the free energy curve will look.
If I draw free energy curve, the free energy curve would look something like this. Assuming
that the G A, G B is 0, we can always assume the standard stage. The free energy of that,
the free energy would be 0, as an extreme case of course, it may not be 0. If I assume
that G A is equal to 0, G B equal to 0, that X A G A plus X B G B term vanishes for schematic
case, we can guess that assume it 0. So, once that vanishes, then I can write this as G
alpha, but at the same time you may say, sir why not it will go the free energy curve go
like this, when this is going to be higher than this, why should it half?
And in place implication like that, why not it goes like this?
That is where, we have seen in one of the pervious classes, that free energy expression
which is here, X l G alpha. Even if you assume that the pure method terms are not there and
simply put omega X A X B plus R T X A L N X A plus X B L N X B and differentiate this
and find out the dou G alpha half dou X B and find out what is the value of at X A equal
to 1 and X B equal to 1, you see that this is minus infinity. I hope all of you have
done this assignment and submitted to me. You see that it is minus infinity and as result,
let me say the slope of the free energy curve as a function of composition, irrespective
of what is the omega value. However, large positive or however large negative it is,
this is bound to be negative and it is minus infinity. And because, it is minus infinity
on both sides of the free energy diagram, in terms of the composition you see that the
free energy has to form because, the negative in the free energy curve should have a slope
which is minus infinity towards this end and towards this end, the only difference that
you see when you keep on increasing this positive. If this is negative, there is no problem at
all; if this is negative, what is going to happen? This curve is going to become more
and more deeper down.
You will see that this curve will go down more and more if omega is negative. Omega
being negative is not a problem at all the free energy curve is not going to change.
Much expecting that the shape remains the same, the absolutely values are going to change.
But, if this value is positive and depending on how large this value is, you would see
that this free energy curve could in a ultimate case, could even be like this or could even
become drop a little and then suppose become like this. If the delta omega value is highly
positive, then you see it drops by a small extent and then starts rising and this drop
is because of this. So, you would see in all the cases, it will never start from 0 and
starts rising. So, it has to have a drop and how small it
is, how how measurable it is depends on the sensitivity of your measuring instrument.
But, in principle, thermodynamically and in fundamentally it has to come down and that
is what you see here. And the moment you see a curve like this, I can stop talking in terms
of what is called common tangent because, this curve is one single curve. But, it as
an inflection, the slope of it is changing at this point. In fact, will come to this
point this later, they are called spinodal points and when we talk about spinodal transmission,
we will try to understand that in greater detail. But, the moment you see a free energy
curve like this, then I can start talking about.
If I start drawing a tangent, you will see at one stage, there can be tangent, which
is something like this, which is a tangent for the both the ends. And this particular
tangent, where where you see that 2 points there, these 2 points are nothing but, the
equilibrium compositions of the original face, which is now decomposing into 2 phases. Because
of this limitation of the omega is being positive that A and B are not able to deserve into
each other. So, there are forces to segregate ((y)) A deserve into B to some existence.
B deserves into A to some existence and they are not able to completely deserves to the
existence to which A deserves into B and B deserves into A can be obtain from this common
tangent. So, that means, basically what you see, these 2 points at this temperature are
nothing but, these 2 points from this free energy curve.
And that is why we call this, we have composition of alpha 1 and composition of alpha 2, incidentally
you call this as alpha 1 plus alpha 2, and you do not call them alpha and beta basically
because, both are same structure. It is actually one single phase which is decomposing into
two with the structure remaining the same, expecting the composition being different.
So, as a result, the whole thing is alpha and alpha splitting into two phases, one having
A rich composition, another having a B rich composition; and that is how you see that
it will have. We will know more about this as we go to spinodal and try to understand
why spinodal acceptation looks like that. Now, if you look at what should happen if
this is a free energy curve, if that is the free energy curve which is leading to this
situation of alpha 1 having this decomposition, alpha 2 having this composition at this particular
temperature. And mind you, I have associated that if I
am on this side, if I am on this side, what does it mean? You remember that, maybe I will
draw this in a greater detail for you to understand, why this splitting has to occur.
For example, imagine a case something like this and you have a common tangent, which
is something like this. And I Mind you, common tangent does not mean it has to be always
horizontal, the free energy curve could be like this, could also be like that; though
generally, for the most of the in most of text books where shown like this. If it is
not that, it has to be a horizontal. So, if you take a case of this composition
composition to the left of it, and draw a tangent which goes and hit somewhere here,
you would see that the whole free energy curve, the whole free energy curve is above this
tangent. And in comparison, if I take a composition here and draw a tangent to that, assume that
this is the tangent, this is the point. You see that if this is the tangent for that particular
composition I am taking, a alpha of that particular composition. You see that, beyond this particular
composition, you see that there are certain compositions which have lower free energy
than that corresponding to this tangent. That means, these compositions are stable
compositions because, their free energy is lower. So, they can in principle come out
of this particular composition. So, if I take alpha of this composition, in principle alpha
half alpha, that composition is not stable. Why because, in principle, I can even show
what is the free energy of this free energy of this and if I take 2 phase mixture of alpha
1 plus alpha 2 of the same composition, that 2 phase mixture has a free energy which is
given by this. Because this is the common tangent, if I draw that line, any 2 phase
mixture, whenever you draw a 2 phase mixture, the free energy of a mixture can be always
found out by the line joining two free energy curves per energy points. So, if this is per
energy of the alpha, one of the equilibrium composition and if this is the free energy
for the alpha 2 of the equilibrium composition, then the line joining this two will give you
the free energy. If I have a phase mixture of alpha 1 and alpha
2 with different amounts of alpha 1 and alpha 2, what is the free energy for such a mixture
can be always obtained by this line. It is simple rule of mixture and if that is the
case here, the 2 phase mixture here has a lower free energy dynamics and because 2 phase
mixture has a lower free energy than this, you would see instead of single phase existing,
for this particular alive composition it would like to exist tablet 2 phase mixture. And
that is what you see if you draw a tangent, you see that the phase here which is B rich
phase, having compositions of all this composition. Anyone of the composition can, in principle
enucleate from this particular ((alien)) because, this a line has a higher free energy than
the 2 phase picture, which one will actually enucleate we will see when we talk about what
is called spinodal. We will know more details about it, which
phase should actually form and in principle usually the word form is the one which have
the highest driving force. And the highest driving force you can obtain by the difference
between this and this will give you the highest driving force. As a result, that is the composition
which is went to enucleate. We will know more about it when we go spinodal, but at the moment,
you can understand that, if I take this particular a line composition, a composition of this,
you see that none of this particular compositions have free energy lower than this line. As
a result, none of them can enucleate from an alpha of this composition that means what?
The alpha of single phase alpha is stable at this particular composition, but if I take
this particular composition, single phase alpha is not stable, it wants to split into
two, which is alpha 1 plus alpha 2 and alpha 2 will enucleate out of that alpha.
And whatever alpha is left out is what we start calling it as alpha 1 because, the moment
you see that alpha 2 of this particular composition enucleate, then immediately there has to be
an equilibrium between that alpha 2 and alpha. For example, if you have an alpha from which
alpha 2 have enucleate, this alpha 2 as to be in equal with alpha. What particular composition
of that can be equilibrium with this alpha 2 if you try to look at. If this is alpha
2, the composition of alpha that can be in equilibrium with alpha 2 is only this composition
because, the 1 that can come as common tangent. When you draw a common tangent, you would
see where where this common tangent attaches the alpha; that is, this composition that
alone can be in equilibrium with this alpha 2. You cannot have this particular composition
in equilibrium with this alpha 2 because, I cannot have a common tangent between that
and this, it is not possible to have common tangent.
That means what? That means, what the moment alpha 2 comes out of alpha, the composition
of alpha surrounding this particular alpha 2 change. And it actually it becomes poorer
in B and richer in A. Why does it become poorer in B? Because, it is alpha 2, which is rich
in B is coming out of alpha. Other result, the alpha which is surrounding that will loss
B atoms because, all the B atoms are coming together and forming A alpha 2 phase. And
because alpha 2 phase is coming out the alpha, which is around that will lower its B atoms
and other result, the composition of alpha stood towards pure a side such that, that
particular alpha becomes e equilibrium with the alpha 2. So, that alpha which you see
it around this, alpha 2 will start calling it as alpha 1. So, now that alpha, which is
so far away from this alpha 1 and alpha 2, you may still have alpha.
Having this particular composition until the whole the reaction is complete. At the beginning
of the reaction, you will see the alpha 2 will come out and around this alpha 2, always
you will have alpha having a composition which is which is decided by this common tangent.
And that is what you will call as alpha 1 and this is what we call it as alpha 2; alpha
2 is B rich and alpha 1 is A rich and this is how you see this happening. And similarly,
if I choose a composition here for the alpha, again you will see that if I draw a tangent
there, you will see all this free energy point. That means free energy of any of this composition
is higher than this tangent, so other result, nothing can come out of that particular alpha.
So, single phase alpha is stable at that, everything we are talking is at a particular
temperature. See, because free energy composition diagram
is always drawn at a fixed temperature, we are talking at the temperature, let say that
temperature is this temperature. So, at this temperature to the left of this point, single
phase alpha is stable to the right of this point. Again, single phase alpha is stable,
that is what this free energy curve tells me and between this 2 point single phase,
alpha is not stable and that is why whether I choose this composition or choose this composition,
if I draw a tangent here, you will see all the composition to the left of this; tell
precipitate out of this. So, that means, from alpha alpha 1 will precipitate and everything
will be ((inverse)) of water. we are talk. So far, once alpha 1 comes out, alpha 1 is
a rich place. Other result the A atoms from alpha how to come together and form the A
rich alpha 1 phase, and other result A atoms in the alpha surrounding the alpha 1 will
deplete in the A atoms and that becomes alpha 2.
So, will have alpha 1 surrounded by alpha 2; in the previous case, it is alpha 2 surrounded
by alpha 1. So, depending on whether you have, taking a A rich composition or B rich composition,
this is what will happen. This keeps on happening until you reach the whole composition of alpha
1 and alpha 2 reach these two compositions, this is what the free energy concept tells
you. At that temperature that these are the only two compositions that can be equilibrium,
no other composition can be equilibrium. So slowly, you see that the alpha composition
changes and the alpha 2 composition remain. What it is, according to the phase diagram?
According to the p n composition, that the remaining whole alpha vanished and you will
have alpha 2 growing at the existence of alpha. And you would have alpha plus alpha 2 and
alpha 1 at the end of this. We will know more about it when we go to the
spinodal, but at the moment, what you need to understand is that this is how we will
know that this. Any composition that I choose higher here, here you see all this composition
have higher free energy than at 2 phase mixture; the 2 phase mixture composition given by this,
and as a result, a 2 phase mixture is more stable in this between these two points, and
in these two points are called bimodal points. We will talk about this again, when we come
to spinodal decomposition. So, what you see is that, you will see between this two is
the alpha 1 and alpha 2. And we call it as alpha 1 and alpha 2 because both of them have
the same structure; we do not call them as alpha and beta. So, the moment you will bring
in a beta, that means, we are talking about different. So, this is how you see this happens.
Now, what should happen if I increase the temperature, what should happen?
If I increase the temperature, this delta s mixing, is it a function of temperature?
Actually yes, and no, according this it is actually no. But, we if you carefully look
at this omega, omega is what interaction parameter; it tells you actually about how strong is
the A B bound; whether A and B attract each other or repel each other and this bound energy
are in principles, function of temperature. So, in principle, there are models that are
available where omega is consider as a temperature function and if you go to cal pad geranial’s
and cal pad, the book that I have told you half minutes burns out book, you will see
their omega is also taken as a function of temperature. But for the most of the basic
thermodynamic understanding, people consider this to be constant. If this is constant,
the only temperature function is this and this being the whole thing being negative,
because this whole thing is minus T delta s mixing so, this whole thing being negative.
So, you would see that this keeps on going more and more negative. As I increase the
temperature, as it becomes more and more negative, what you see, because this is fix delta s,
mixing is fix and this is becoming more and more negative; you see you will reach a stage
where this free energy curve changes from this shape to this shape. You would actually
see that, this free energy curve slowly starts becoming like this and then it further becomes
like this because, this pump start decreasing. Because, the negative value of the entropy
of mixing is becoming more and more, and you would see at some stage, the free energy curve
like this. And be ((hound)) the temperature, the free energy curve becomes more and more
negative again because, this is much more negative, you can see that just looking at
the T delta s mixing. So, we can understand how this free energy curve, which has a con
K con X mixed, is getting converted to one single type of free energy, which indicates
stability. We will also talk about how stability is related to this, when we come to the point
of spinodal. So, this is an indication of instability for free energy, we will see why
it so, when we come to spinodal decomposition. At the moment, you just leave it. So, the
moment you see that this free energy curve that means, there is any, for example, if
you take any point here, a composition here or here or here or here.
If I draw a tangent, what does that tell me? That is, all points are free energy points.
That means, at any given composition, free energy of all this composition is above the
top tangent. So, there is no chance for any other composition to precipitate out of alpha
of this composition. I can seen it from all the composition, anywhere I choose a composition,
draw a tangent, you would see all the whole free energy curve is above that of the tangent.
Because it is the above the tangent, there is no chance for this. That is why this free
energy curve indicates its single phase is stable and that is the reason why, above this
particular temperature, if you come to this temperature, you will see the free energy
curve becomes like this. And that basically means that, this has to merge.
You will also see, as you go to higher and higher temperature, this composition starts
coming closer and that is what you seen here. Also, as you go from this to anything like
this, for example, if you see there are too many curves here that may confuse you. For
example, if I take this free energy curve, draw a common tangent, you would see, may
be I will show you here.
If you see, this is a this decreasing like this, and draw again a common tangent, you
would see this points would be inside these two points; that means the equilibrium composition.
The bimodal points are shifting towards each other as the temperature increase, and to
an extent that you would see that, slowly they start coming closer and closer to an
extent that at some stage, both become margin to each other and become one point.
That is, this point that you see here, and that temperature in principle. How do I calculate
the temperature, can you tell me? Can I calculate the temperature; it should be very easy, what
is the temperature? Where it should happen, you simply equate this with this, simply equate
this with this you can find out what is the T and again that is not sufficient; something
else we have to do. Equate and put one boundary conditions X A equal to X B because, both
this functions being cemetery, this actually happens at 50.
So, equate this to this, and then take X A equal to 0.5 and X B equal to 0.5. And once
you take that, you would be able to find out this T, which is this. In fact, you would
see when you calculate this, later you would see this will become equal into that. We call
it as T star and T star is omega by 2 R, one can do this calculation and so, the higher
the omega value, is the higher the omega value is the higher will be the T star. That is
what yesterday when we are talking about, as I increase the omega more and more or in
other words, as I increase the size factor more and more, you would see that this temperature
increases, initially it was the miscibility gap was like this.
Then the miscibility gap has become like this, so this temperature has gone up. When I increase
the omega, this temperature goes up and that is what you see this happening. So, like this,
we can in principle find out how this miscibility gap closes that temperature and this miscibility
gap closes can be easily found out by equating this with this. And finally,… In fact, it
is not so simple. Also, we will see as we go along, when we go spinodal, there is also
another way to find out by taking the second differential of the free energy. How to do
that? We will see when we go to the spinodal at the moment you take it this way, this is
clear know, let us now look at, yes.
Let us see, this is all alpha, starting with alpha we are talking of a situation where
a free energy curve is like this. We are talking of a composition like this, it is the starting
composition. Let it be 20 percent or 30 percent B, that is alpha and in this alpha now, alpha
2 having this particular composition as enucleated. This is this is alpha 2 which is enucleated
and that as how much this as much more be content is in it. And how does this enculturation
occur? Where another question is, you say sir, alpha is continuing only 30 percent B
and if it 30 percent B, then how can in a 30 percent B, which is uniformly 30 percent
B. Suddenly, you will have a nucleus of the alpha
2 which as 70 percent B coming out. This is where you need to understand that every liquid
or solid, you at any given temperature, there always are vibrations. There are always atoms
moving and there are always clusters of varying composition though, over all the composition
is 30 percent. You can have a small region which is few nanometer regions, for even the
less than nanometer region, where the atoms are more concentrated in terms of either A
or B. But, the question is, according to thermodynamic mix, atom in given temperature. If this alpha
is not stable, then those clusters which have A B content of this would get stabilize suddenly
and that particular clusters will start growing; that is what we call it as R star, this how
you will liquid to solid. Cross machine, you may ask, sir liquid is
completely random, how does suddenly solid enucleate? Because, there is a short range
order inside the liquid and inside the short range order, you see some clusters having
a particular cluster structure. They may be 100 atom clusters, 10 atom clusters, and 1000
atom clusters, different different sizes do exist, and once a particular size of the cluster
has the value of the R star for that particular temperature, that particular cluster start
growing. So, let us assume that there are clusters.
So, that B has enucleated. How it has enucleated, let us not bother about it now. Once it is
enucleated, this is alpha 2, now this particular alpha 2, can it be in equilibrium with this
alpha? It is not possible, because I cannot draw a common tangent between this two, impossible,
yes, so this. It is inside alpha 2, because this alpha 2
is stable, it has a low free energy with respect to that alpha, with respect to that alpha,
because it as a low free energy. So, it tries to come out, but once it comes out, because
this cannot be the only way this alpha 2 can be stable, either to dissolve that into the
matrix or change the alpha around its composition such that it can exist with that. this way
cannot If you want to be friendly with somebody, you change your attitude or change your friend.
So that, you are in a in kind of harmony with the other friend, exactly the same way here.
So, if you want the alpha 2 to be to be alive, it can be alive only if alpha 1. Alpha around
it changes its composition such that, you can have a common tangent between those 2,
that alpha alone can be in equal with alpha 2.
So, as a result, because, already from the alpha B atoms have come together to form alpha
2, then obviously, this alpha has given out some B atoms. Because it has given out some
B atoms, its B atoms will decrease. So, in small domain, in a small domain like this,
the B atoms have decreased and when they have decreased, that means, the pre composition
of alpha has that actually started moving in this direction only. At least, a small
reason it might not have happen in the whole alpha, but at least just around alpha 2, it
will happen. And that particular small reason around this, I will call it as alpha 1 and
that composition will reach this particular composition so that this alpha 2 can be alive,
otherwise, alpha 2 cannot be alive. It has to dissolve; that this is the major problem,
that is why we will see alpha 1. And now, what should happen in future? This
alpha 1 alpha 2 is there, then alpha is also there. So, finally, at this temperature, in
principle, what is stable is only alpha 1 plus alpha 2 and what is the driving force
for that? The driving force is this minus this, is the delta G that is the delta G for
the alpha. To give you alpha 1 plus alpha 2, that is the driving force. So, driving
force is available, the only thing that you need is, you need diffusion so that diffusion
takes place. And what should what you will see is, this alpha slowly vanishes, the alpha
having this composition because, it is not stable. It slowly starts giving out the B
atoms more and more so that, alpha 2 starts growing.
And as alpha gives out more and more B atoms, what happens? The alpha becomes poorer in
B and so that whole of alpha, so after sometime you would see a situation like this that the
alpha 2s have grown. These are all alpha 2s because, the B atoms are coming, continues
and flowing into the alpha 2 from the alpha, and as they flow into this, you would see
alpha 2 increases and the remaining alpha completely becomes alpha 1. So, you will have
alpha 1 matrix and alpha 2 precipitate in this type and this can be seen in very easily
if you look at the gradient between this and this. Why does there is gradient come, for
example, you can see the moment you see a situation, I see a case where everything is
alpha, everything is alpha with a certain composition usually.
Suddenly, you see a beta. I am in, alpha 2 comes out; if the alpha 2comes out, alpha
2 have much higher composition then the alpha has in it. So, as result, this is, if I call
this as alpha 2 which is coming out, the remaining is alpha, but this alpha 2 cannot be in equilibrium
with this alpha. So, what is that, it can be in equilibrium with? So, it is this value.
So, you will see that alpha 2 will be in equilibrium alpha having a composition which is much lower
than this alpha. So, you will see that, this will be actually, if I draw the concentration
profile, it looks like, this is the alpha, this is the alpha 1, this is the alpha 2.
So, you see a concentration profile something like this, just listen to alpha 2, alpha will
have a lower composition. But again, from the alpha 2, alpha will have
the same composition with which you have stared. Now, what happen is, if you see, there is
a gradient c ((or)), B atoms are more in this. Everything I am talking in terms of percentage
B, as a function of distance. So, initially everything is alpha and inside the alpha 2
has come out and alpha 2 has higher B content. And because of this, alpha 2 has to be alive;
it can be in equilibrium with alpha having a lower B content than what it is. That what
you see this is the al starting alpha, this is alpha composition which is lower than the
thing. So, now, you see there is a gradient, because there is a gradient, you would start
seeing the B atoms are flowing from this alpha towards this. Then what should happen? Slowly,
this will grow; this will start growing more and more.
But will this change, will this composition change? This composition is equilibrium composition
of alpha 1 in equilibrium with the alpha 2, so that cannot change. Then, as this grows,
something else should happen. What should happen? Amount of alpha amount of we are not
talking about amount, now we are talking about in terms of concentration, anything should
happen. Of course, amount of alpha will decrease because, the alpha 2 is increasing, alpha
will decrease; anything else should happen to the composition of alpha will go down.
So, as a result, you see that this curve or this course, you would see a situation. Well,
it becomes well this B content in the alpha has to come down, because it is giving away.
It is like reservoir, giving the B atoms to the alpha 2. As you keep on giving more and
more atoms to alpha 2, the amount of B atoms in alpha should come down. So, this particular
value comes down and as this comes down, you will see a situation where finally, the whole
alpha around this alpha 2 will have alpha 1 composition. Then, will call this whole
thing as alpha 2 alpha 1, this as alpha 2. So, now, it is alpha 2 and alpha 1. Of course,
kinetically if I look at it, because if this is happening as the composition is decreasing,
the gradient is coming down. That means d c by d x is coming down; you know that J is
minus B into d c by d x, if the d c by d x coming down, then what should happen? The
cluster should, so the reaction is slow down, so the growth of alpha 2 will slow down with
time; that is fine. But thermodynamically, finally, it should
come to this and that is how you see that remaining alpha will be called now, not as
a alpha, it will called as alpha 1. So, that is how you see one single alpha has become
alpha 1 plus alpha 2. When the same thing can happen in a different fashion in spinodal,
we will know how it will happen, when we come to spinodal. You can see that, this is how
prating take place, and as you increase the temperature, this how you see this gets closed
up. I think with this we will close for today. I want to say the fact, another concept, which
will start in the next class. Obviously, that not balance has to be maintained. So, because
the mass balance has to be maintained, the only thing is, we can write an expression
now.
The fraction of alpha 1 into the composition of alpha 1 plus the fraction of alpha 2 into
the composition of alpha 2 should be equivalent to mi C naught which I have started. So, the
amounts of alpha 1 and alpha 2 will be adjusted in such way that, these compositions will
give you a particular amount. So, we can, this is how actually our lever rule comes.
How to do get lever rule? It is only this.