Reference: Daniel V. Schroeder, *An Introduction to Thermal Physics*, (Addison-Wesley, 2000) – Problem 5.29.

We can apply a similar analysis to that for graphite and diamond to the more complex case of aluminum silicate , which has three crystalline forms: kyanite, andalusite and sillimanite. Each is stable under some range of temperature and pressure.

Using the data in Schroeder’s book, we can get some insight into the stability of these three forms by considering the Gibbs free energy. The thermodynamic potential for the Gibbs free energy is

First, if and are constant, we have

Ignoring compressibility, we can assume that is constant as the pressure is increased. The molar volumes for the three forms are for kyanite, for andalusite and for sillimanite. Therefore the Gibbs energies are

The constants of integration , and can be found by requiring the energies to be those quoted in Schroeder for . As with our calculations for calcium carbonate, the difference in Gibbs energy between a pressure of zero and 1 bar is negligible so we can just set the constants of integration to the values quoted in Schroeder for . We get

At , is the lowest and since the slope of the line in the plot of versus is also the smallest, will be less than both and at all pressures, so at a temperature of 298 K, kyanite is always the most stable form. The graph looks like this:

The red line is kyanite, green is andalusite and yellow is sillimanite. The latter two lines do intersect at a pressure of around 9 kbar, but kyanite is always the most stable.

At constant pressure, we can use 1 to investigate the stability of the three forms as temperature varies. We have

By subtracting the two forms of this equation for two crystal types, we can get a relation for the difference in Gibbs energy between two types:

The entropy of a substance can be written as

so the total entropy as a function of temperature depends on the heat capacity . If we take Schroeder’s word for it that at high temperatures depends only on , then even though the raw value of for a sample of solid increases with , the difference between the entropy values for two substances should remain relatively constant as we increase . We can therefore integrate 10 to get

The entropy values quoted in Schroeder’s book for 1 mole at and are , and so using the Gibbs energy values from above, we get (with energies in kJ):

If we plot these three lines, we get

Here, is red, is green and is yellow. To use this graph to determine the temperature range for which each structure is the most stable, we observe that the most stable form has the lowest value of at that temperature. For kyanite to be the most stable, for example, we must have and . Therefore, and . Looking at the graph, these conditions are true if both the red and green lines are below the axis, which occurs when is less than the point where the red line crosses the axis, which is where . From 15 this occurs at .

For andalusite to be stable, we must have and , which means and . Thus the red line must be above the axis and the yellow line must be below the axis. The minimum satisfying these conditions is that we found above. The maximum is where the yellow line crosses the axis, which occurs when , at .

For sillimanite to be stable, we must have and , which means and . Thus both the green and yellow lines must be above the axis, which is true for . To summarize: kyanite is stable for , andalusite is stable for and sillimanite is stable for .

How accurate is the assumption that (the difference in entropy between two states of the substance) is actually a constant over a wide range of temperature? From 11 for one state, we can integrate:

Thus between two states, and is obtained by subtracting one equation of this type from the other:

will be constant if the two states have equal heat capacities. From the data in Schroeder’s book, , and . Using a reference temperature of , the log factor over the range of temperatures over which the state changes occur varies from to . The largest difference in heat capacity is , so we can expect the last term in 21 to be in the vicinity of . The entropy differences used in 15 and following equations are between around 2 and 12 so actual value of could be in error by a fair bit. Of course, we’ve also used the heat capacities for and assumed that they don’t vary over the temperature range, so we’d really need to correct *that* error as well.

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