Maps of Phase Equilibirum - Binary Isomorphous Phase Diagrams

Sugar and water is a familiar and interesting system, but that phase diagram is probably more relevant to food scientists. Materials scientists and engineers typically work with phase diagrams for metal alloys, ceramic systems, and polymer mixtures. Let's look at perhaps the simplest of phase diagrams - the isomorphous (iso- means "same" and "morphous" means shape or structure) binary phase diagram.

In earlier sections (Section 6.5 and Section 10.3) we found that sometimes, when one atom substitutes for another in a lattice, there is only a small disruption in the host lattice and we can form a complete solid solution. This is a solution that in which one can change the concentration of the impurity from very small to very large values and never yield the formation of a new phase. Copper and nickel are one such pairs that form a complete solid solution: the nickel can substitute into the copper host lattice without leading to a significant disruption in structure, and the copper can do the same in the nickel. They are completely miscible. Because they are completely miscible, they have a single phase field in the lower-temperature region from pure nickel to pure copper (Figure 10.6.1). That is, the phase of pure nickel (which has an FCC structure) and the phase of pure copper (which also has an FCC structure) are so structurally similar (i.e. they are isomorphous) that they mix completely.

There's a lot of content here that is very new for most people, and lots of terminology. Remember the following:

1. Phase diagrams are maps of the phases present for at different thermodynamics conditions (temperature, composition, pressure, etc.)
2. We can use the features of the phase diagram extract information about which phases are present, the composition of those phases, and the relative amount of each phase. -3. We can use the information in 2. to create schematic sketches of the expected microstructure of a material.

We'll learn to do all of this below! As is tradition in most introductory MSE courses, we'll use the Cu-Ni phase diagram as a playground to learn about isomorphous phase diagram. Notably, there are really interesting materials in this phase diagram, including cupronickel alloys (used for coinage), monel (used in marine and chemical processing industries, as well as decorative articles), and constantan, extremely useful in electrical applications.

Phase Fields in Isomorphous Phase Diagrams

Let's look at the phase diagram in Figure 10.6.1 in detail:

• At lower temperature we observe this region of complete miscibility in Figure 10.6.1. We label it $\alpha$ because in science we like to use Greek letters to represent things; $\alpha$ simply communicates that the material has a specific crystal structure in this phase field. Note - $\alpha$ does not necessarily mean FCC, or BCC, or whatever. Greek letters will represented different crystal structures in different phase diagrams. <\br> If I hold my system at a temperature and composition that places us in the $\alpha$, I expect to only observe a solid solution.

• At higher temperature, we'd expect the alloy to melt, and so we see, at higher tempreature, a phase field of liquid solution $\ell_{\mathrm{s}}$. Note - the liquid phases of Ag and Cu are also miscible (like water and alcohol), although not all liquid phases need be miscible (like oil and water). <\br> If I hold my system at a temperature and composition that places us in the $\ell_{\mathrm{s}}$, I expect to only observe a liquid solution.

• At intermediate temperature we have a region of two-phase equilibrium, as denoted by the label $\ell_{\mathrm{s}}+\alpha$. In this region we expect both the liquid phase and the solid phase to be present in equilibrium. <\br> If I hold my system at a temperature and composition that places us in the $\ell_{\mathrm{s}}+ \alpha$, I expect to observe a both a liquid solution and and solid solution.

Phase Boundaries and Melting Points

There's clearly sharp boundaries between these phase fields, denoted by solid lines. We give those phase boundaries names:

1. The so-called liquidus separating the $\ell_{\mathrm{s}}$ and $\ell_{\mathrm{s}} + \alpha$ phase fields. Above this line, we only have a liquid phase.
2. The so-called solidus separating the $\alpha$ and $\ell_{\mathrm{s}} + \alpha$. Below this line we only have a solid phase.

One can think of these phase boundaries very similarly to the solubility limit in the sugar-water phase diagram on the previous page. We'll discuss this more below.

There are two other points of interest that one should recognize:

1. The point at $C = 0 \text{ wt% Ni}$ (pure copper) and $T = 1084 ^{\circ}\text{C}$: the point where we transition from pure, solid copper to pure, liquid copper upon heating. Or, the melting point of copper, labeled $T_{\mathrm{m}}(\text{Cu})$.
2. The point at $C = 100 \text{ wt% Ni}$ (pure nickel) and $T = 1452 ^{\circ}\text{C}$: the point where we transition from pure, solid nickel to pure, liquid copper upon heating. Or, the melting point of nickel, labeled $T_{\mathrm{m}}(\text{Ni})$.

The melting points provide an opportunity to navigate these phase diagrams and build some intuation. Let's say we're at $C = 0 \text{ wt% Ni}$ and $T = 1050 ^{\circ}\text{C}$. What phase field are we in? Well, it looks like in the $\alpha$ phase field, or we're solid. Now, we heat our specimen to $T = 1100 ^{\circ}\text{C}$ without changing composition. Now we're in the liquid phase field. Something happened... a phase transition where $\alpha \rightarrow \ell$. We melted our specimen, and our phase diagram - our map of phase equilibria - tells us this.

Example Analysis of the Isomorphous Binary Phase Diagram

We mentioned in Section 10.3 that there were really only four things we can do with a phase diagram. Let's walk through those things using the Cu-Ni phase diagram as an example, and then practice a bit. Those things were:

1. Identify phases present in the system.
2. Computer the chemical compositions of the phases, typically in weight percent or atomic percent.
3. Calculate the phase fractions - or how much of each phase in in the system.
4. Draw a schematic microstructure for the system.

Let's practice with item 1-3 first, and then address 4. We'll use a few points (A, B, D) on the section of the Cu-Ni phase diagram shown in Figure 10.6.2, and then I'll ask you to analyze the remaining point in Exercise 10.6.3.

Phase Identification.

Let's look at points A, B, C, and D in the phase diagram. These are at compositions $C = C_0 = 65 \text{ wt% Ni}$, where $C_0$ represents the overall composition of the mixture. This idea is very important. As we change temperature, we'll see we change the amount and compositions of phases but the overall composition of the entire material will stay at $C = C_0 = 65 \text{ wt% Ni}$. That is, at high temperature we might have a liquid, but 65% of its weight is due to the nickel atoms, and 35% is due to the copper. If we cool it to a solid phase without changing the composition, 65% of the weight of the material is still due to the nickel atoms, even though we have a different phase.

For the points of interest (A, B, C, and D), we have four different temperatures defined by the points: $1390 ^{\circ}\text{C}$, $1350 ^{\circ}\text{C}$, $1330 ^{\circ}\text{C}$, and $1280 ^{\circ}\text{C}$, respectively. Let's find the phases for each:

• Point A: We're within the liquid solution ($\ell_{\mathrm{s}}$) phase field. We only have one phase, and it is a liquid solution.
• Point B: We're within the two phase region, so we have both liquid solution ($\ell_{\mathrm{s}}$)and solid solution ($\alpha$) phases.
• Point C: We're still within the two phase region, so we have both liquid solution ($\ell_{\mathrm{s}}$)and solid solution ($\alpha$) phases. Now, there is a difference between this point and Point B in terms of the composition and fraction of the phases present. We'll get to that next, but we're only identifying the phases at this point
• Point D We're within the solid solution ($\alpha$) phase field. We just have one phase, the $\alpha$ phase, or FCC phase.

You can try yourself on nearly any binary phase diagram now! Select a system and Google it. Then, choose a temperature and pressure and you should be able to identify the phases expected at equilibrium. Don't be intimidated when you search out a new phase diagram, because they can seem very complex. They're all constructed in the same way, although there are further details we'll discuss in he more complicated ones... Regardless, you could give it a try with something like the Pu-Ga phase diagram, if you like.

Composition of Phases

Next, we want to know the composition of the phases. The composition is how much of each chemical component is present each of the phases. for example, at some specific temperature and overall composition, how much Ni do we have in the $\alpha$ phase, or how much Cu is in the liquid phase?

Let's briefly return to our example of the chocolate chip cookie. The cookie has two "phases" in this visual model: cookie dough and chocolate. Each of those phases have different sugar content: chocolate chips are about 50 wt% sucrose, while cookie dough is about 20 wt% sucrose. They clearly have different weight percentages of sugar.

Let's move to our metal alloy phase diagram, reproduced in Figure 10.6.3. If we want to analyze the composition of the four points, A-D, we can start with the easier ones based on what we know about the number of phases at each from the previous section. At Points A and D we only have a single phase:

• Point A: We only have one phase, and the overall composition of the alloy is $C = C_0 = 65 \text{ wt% Ni}$. The composition of the liquid phase is therefore $C_{\ell_{\mathrm{s}}} = 65 \text{ wt% Ni}$.
• Point D: We again only have one phase, and again the overall composition of this phase is $C = C_0 = 65 \text{ wt% Ni}$. So, the composition of the solid phase $\alpha$ is $C_{\alpha} = 65 \text{ wt% Ni}$.

In Figure 10.6.3, points B and C are in the region of two-phase equilibrium, and we therefore have two phases at each of these points. At each point we'll have two phases with differing compositions. The weighted average will be the overall composition. Let's see:

• Point B: Two phases! A liquid phase and a solid phase. So, let's look at the phase diagram for guidance remembering that the phase boundaries communicate maximum solubility of their respective phases. That is, the liquidus tells us the maximum amount of Ni can be pushed into the liquid phase before I hit the solubility limit. The solidus tells us how much Cu can be pushed into the solid phase before liquid starts forming. So, the liquidus and solidus phase boundaries tell us the compositions of the phases in the two-phase region!

Let's demonstrate in Figure 10.6.4. Let's say we start with pure Cu at a temperature of $1350\,^{\circ}\text{C}$ and start adding Ni until we reach Point B. At first, no problem. We continue to add Ni to the liquid solution. However, soon we reach the maximum solubility of Ni in the liquid at $C_{L} = 61 \text{ wt% Ni}$ or $C_{\alpha} = 39 \text{ wt% Cu}$ . What happens if we put more Ni in? No more Ni can go in the liquid, so we form a solid - similar to the examples with with the sugar and water, above.

What will the composition of that solid be? It is defined by the solidus line because the that line defines the maximum solubility of Cu into the $\alpha$ phase: $C_{\alpha} = 71.5 \text{ wt% Ni}$ or $C_{\alpha} = 28.5 \text{ wt% Cu}$ . (Note, exactly why the $\alpha$ phase incorporates this concentration of Ni when it forms is due to free energy of solutions, analyzed using a Gibbs free energy curve. We won't cover that here, but you can take a deeper dive here, if you like, or take a more advanced course on solution thermodynamics)

This means if I ever have a point in a two-phase region and I want to find the composition of the phases in the mixture, I can draw an isothermal (single temperature) line through the point that terminates at the liquidus and solidus. I then find the concentrations at those points. Those are the concentrations of or liquid and solid phases, respectively. We call this a tie line analysis. It's a very good thing to do when you first start analyzing phase equilibria in a two-phase phase field. This tie line, revealing the compositions of the liquid and solid phases at Point B is shown in Figure 10.6.4.

Phase Fractions

What if we want to find the fraction of our mixture that is comprised of each phases? Two of our points in Figure 10.6.3 are simple because they are single phase (Point A and Point D). Point B and Point C are more difficult.

In general, we denote some phase fraction as $P_i$, where $P$ is the fraction of the mixture that is the phase, and $i$ denotes the phase of interest (e.g. $\ell_{s}$ or $\alpha$). However, the most common way we see phase fractions reported is as weight fraction $W_i = \frac{m_i}{m_\text{tot}}$, where $m_i$ is the mass of the the phase $i$ and $m_{\text{tot}}$ is the total mass of the system. We'll use that analysis here, but note that you can also analyze phase fractions in mole percent or volume percent because the densities of different phases may differ.

• Point A: We have a single phase, just liquid. So our phase fraction for the liquid phase is simply $W_{\ell_s} = 1.0$. Since we have just two possible phases at this point,, we can also say that $W_{\alpha} = 0$.
• Point D: We have a single phase, just solid. So our phase fraction for the solid phase is simply $W_{\alpha} = 1.0$ and for the liquid phase it is $W_{\ell_{\mathrm{s}}} = 0$.
• Point B: We have two phases. How do we do do this? It turns out (click here for a proof, but we don't go into it in detail), that the equilibrium phase diagram is constructed in a very similar way to a teeter-totter, where we have a lever mounted on a fulcrum and two loads on either side, like shown in Figure 10.6.5.

This equilibrium model us the relationships between phase fractions (analogous to masses on a teeter-totter) and line segment lengths (Figure 10.6.5(b.)). This means that we can relate masses $m_{\ell_{\mathrm{s}}}$ and $m_{\alpha}$ and line segments $\overline{C_{\ell_{\mathrm{s}}}C_0}$ and $\overline{C_0 C_{\alpha}}$ in equilibrium the same way we'd do for a teeter-totter.

Note, the length of line segment $\overline{C_{\ell_{\mathrm{s}}} C_0}$ is $C_0-C_{\ell_{\mathrm{s}}}$ and the length of line segment $\overline{C_0 C_{\alpha}}$ is $C_{\alpha}-C_0$

Often, to simplify the notation we define the line segments as (for example) $R = (C_0-C_{\ell_{\mathrm{s}}})$ and $S = (C_\alpha-C_0)$. The selection of symbols "$R$" and "$S$" are arbitrary - you can use whatever you like.

The weight fraction of the liquid solution is $W_{\ell_{\mathrm{s}}} = \frac{m_{\ell_{\mathrm{s}}}}{m_{\alpha} + m_{\ell_{\mathrm{s}}}}$ and the weight fraction for the $\alpha$ phase is $W_{\alpha} = \frac{m_{\alpha}}{m_{\alpha} + m_{\ell_{\mathrm{s}}}}$, and so

Or, using $R$ and $S$:

The algebra is a bit messy to follow, perhaps, but the result is incredibly simple! It says the that weight fraction of the liquid solution $W_{\ell_{\mathrm{s}}}$ can be computed using the length of the blue segment ($C_{\alpha}-C_0$) and dividing it by the total width of the two-phase region at that point ($C_{\alpha}-C_{\ell_{\mathrm{s}}}$). This is such a useful expression we give it a name. Eq. 10.6.4 is the lever rule.

Lever Rule Caveat

Eq. 10.6.4 is incredibly useful for analyzing phase fractions. And while it's general form will always be the same (one line segment length over another), it will not always look exactly the same depending on the phase diagram you are working with. We'll see examples of this in the next chapter, but the important point is that you will need to construct your own expression for the lever rule when analyzing different phase diagrams..

Example at Point B

Let's first compute the weight fractions of liquid solution $W_{\ell_{\mathrm{s}}}$ and solid solution $W_{\alpha}$ at Point B in Figure 10.6.5. Remember, when we have two phases in equilibrium the the total weight fraction of the two phases must sum to unity!

This is a very useful equation because it tells us that if we solve for one weight fraction, we know the other immediately! We'll demonstrate for Point B. First, let's use Eq. 10.6.4 to compute the liquid solution weight fraction:

Our liquid solution phase should comprise 62% of the material, by weight. That means if we have 1 kg of the Ni-Cu alloy at $1350\,^{\circ}\text{C}$ and $C_0 =65 \text{\,wt%}$, then 620 g of it will be liquid solution.

How much is solid solution $\alpha$? We can use Eq. 10.6.5, which tells us:

$$0.62 + W_{\alpha} = 1$$

and so

$$W_{\alpha} = 0.38$$

Alternatively, you can also solve for $W_{\alpha}$ in general using Eq. 10.6.4 and Eq. 10.6.5.

Which gives you the weight fraction for the solid solution $w_{\alpha}$ in a form similar to that of Eq. 10.6.4.

Extrema and Intuition

Let's check our intuition at two other compositions at $1350\,^{\circ}\textrm{C}$ in Figure 10.6.5: a composition near the liquidus line ($\textrm{B}_{\ell_{\mathrm{s}}}$, $C_0 = 61 \text{ wt% Ni}$) and a composition near the solidus line ($B_{\alpha}$, $C_{\alpha}= 71.5 \text{ wt% Ni}$)

The phase diagram tells us that as we increase the wt% of Ni along $1350\,^{\circ}\textrm{C}$, we'll have liquid solution until the solubility limit until we meet the liquidus, in which $\alpha$ will form. Intuitively, then, at $\textrm{B}_{\ell_{\mathrm{s}}}$ we expect that we'd be nearly all liquid solution ($W_{\ell_{\mathrm{s}}}$) because we're right at the solubility limit. We find this is true if we use Eq. 10.6.4, since $S >> R$:

$$W_{\ell_{\textrm{s}}} = \frac{S}{R+S} = \frac{S}{S} = 1$$

It's all liquid solution!

Similarly, at $B_{\alpha}$ we have $R >> S$ and we'd expect $W_{\alpha} = 1$. How can we show that? Let's simply use Eq. 10.6.7.

$$W_{\alpha} = \frac{R}{R+S} = \frac{R}{R} = 1$$

It's all $\alpha$ phase!

Summary

Whew! That feels like a lot, and can be confusing at first. If you feel like you understand it at this point, great, if not, that's also fine. The thing about phase diagrams is that their analysis using tie lines and lever rules is very algorithmic. If you practice, you really can't get the analysis of the compositions and phase fractions wrong. So, practice following the following algorithm:

1. Determine the location of the point of interest on the phase diagram. Is there one phase present or two?
2. If there are two phases, draw a tie line between the adjoining phase boundaries. (Here, the liquidus and the solidus.) The composition at the point the tie line intersects the liquidus gives you the composition of the liquid phase. The composition at the point the tie line intersects the solidus gives you the composition of the solid phase.
3. Apply the lever rule to find phase fractions.

This will get a bit more complicated with more complex phase diagrams, but this is essentially the way to analyze phase diagrams.