Question

Consider again the aluminosilicate system treated in Problem 5.29. Calculate the slopes of all three phase boundaries for this system: kyanite andalusite, kyanite-sillimanite, and andalusite-sillimanite. Sketch the phase diagram, and calculate the temperature and pressure of the triple point.

Short answer

The temperature at the triple point is 690 K, and the pressure is 3.3 kbar.

Step-by-step solution

Given information

By using the relation we will calculate the slope of all three phase boundaries

$\frac{dP}{dT}=\frac{\Delta S}{\Delta V}$

Where,

$∆S$is the change in entropy

$∆V$is the change in volume

P is the pressure

T is the temperature

The slope of the phase for the kyanite and andalusite boundary is as follows:

$\frac{dP}{dT}=\frac{S_a-S_k}{V_a-V_k}$

Where,

S_a and V_a are the entropy and volume of the andalusite

S_k and V_k are the entropy and volume of the kyanite.

Substituting the values

$$\begin{gathered} \frac{dP}{dT}=\frac{93.22\mathrm{J}/\mathrm{K}-83.81\mathrm{J}/\mathrm{K}}{5.153\mathrm{J}/\mathrm{bar}-4.409\mathrm{J}/\mathrm{bar}} \\ =12.65bar/K \end{gathered}$$

Explanation

The slope of the phase for the andalusite and sillimanite boundary is as follows:

$\frac{dP}{dT}=\frac{S_a-S_s}{V_a-V_s}$

Where,

S_sand V_s are the entropy and volume of the sillimanite

Substituting the values

$$\begin{gathered} \frac{dP}{dT}=\frac{93.22\mathrm{J}/\mathrm{K}-96.11\mathrm{J}/\mathrm{K}}{5.153\mathrm{J}/\mathrm{bar}-4.990\mathrm{J}/\mathrm{bar}} \\ =-17.73bar/K \end{gathered}$$

The slope of the phase for the kyanite and sillimanite boundary is as follows:

$\frac{dP}{dT}=\frac{S_k-S_s}{V_k-V_s}$

Substituting the values

$$\begin{gathered} \frac{dP}{dT}=\frac{96.11\mathrm{J}/\mathrm{K}-83.81\mathrm{J}/\mathrm{K}}{4.99\mathrm{J}/\mathrm{bar}-4.409\mathrm{J}/\mathrm{bar}} \\ =21.17bar/K \end{gathered}$$

Explanation

The transition from kyanite to andalusite happens at 1 bar at T= 427K, giving up one point on the phase boundary, according to problem 5.29. Now you must come up with a term for this couple.

The transition pressure for kyanite and andalusite is as follows:

$$\begin{gathered} P=\frac{dP}{dT}·T+C \\ C=P-\frac{dP}{dT}·T \\ C=1\mathrm{bar}-(12.6\mathrm{bar}/\mathrm{K})(427\mathrm{K}) \\ C=-5.38kbar \end{gathered}$$

The transition pressure for andalusite and sillimanite is as follows:

$$\begin{gathered} C=P-\frac{dP}{dT}·T \\ C=1\text{bar}-(-17.73\mathrm{bar}/\mathrm{K})(875\mathrm{K}) \\ C=15.51kbar \end{gathered}$$

The transition pressure for kyanite and sillimanite is as follows:

$$\begin{gathered} C=P-\frac{dP}{dT}·T \\ C=1\mathrm{bar}-(21.17\mathrm{bar}/\mathrm{K})(532\mathrm{K}) \\ C=-11.26kbar \end{gathered}$$

Explanation

Calculate the temperature T and pressure p at the places where the kyanite andalusite ($k\leftrightarrow a$) line intersects the andalusite $\leftrightarrow$sillimanite ($a\leftrightarrow s$) line.

role="math" localid="1646933788841" $$\begin{gathered} P=(12.6\mathrm{bar}/\mathrm{K})T-5.38\mathrm{kbar}\text{for}k\leftrightarrow a \\ P=(-17.73\mathrm{bar}/\mathrm{K})T+15.51\mathrm{kbar}\text{for}a\leftrightarrow s \end{gathered}$$

Equating both the values

$$\begin{gathered} (12.6\mathrm{bar}/\mathrm{K})T-5.38\mathrm{kbar}=(-17.73\mathrm{bar}/\mathrm{K})T+15.51\mathrm{bar} \\ T=\frac{15.51\mathrm{kbar}+5.38\mathrm{kbar}}{12.6\mathrm{bar}/\mathrm{K}+17.73\mathrm{bar}/\mathrm{K}} \\ T=688.7K \\ T=690K \end{gathered}$$

Explanation

As a result, the pressure at the place where the kyanite $\leftrightarrow$andalusite (k $\leftrightarrow$a) line intersects with the andalusite $\leftrightarrow$sillimanite (a $\leftrightarrow$s) line is:

$$\begin{gathered} P=(12.6\mathrm{bar}/\mathrm{K})(688.7\mathrm{K})-5.38\mathrm{kbar} \\ P=3.298kbar \\ P=3.3kbar \end{gathered}$$

Calculate the temperature T and pressure P at the places where the andalusite$\leftrightarrow$sillimanite (a$\leftrightarrow$s) line interacts with the kyanite$\leftrightarrow$sillimanite (k $\leftrightarrow$s) line.

$$\begin{gathered} P=(-17.73\mathrm{bar}/\mathrm{K})T+15.51\mathrm{kbar}\text{for}a\leftrightarrow s \\ P=(21.17\mathrm{bar}/\mathrm{K})T-11.26\mathrm{k}\text{bar}\text{for}k\leftrightarrow s \end{gathered}$$

Equating both the values

$$\begin{gathered} (-17.73\mathrm{bar}/\mathrm{K})T+15.51\mathrm{kbar}=(21.17\mathrm{bar}/\mathrm{K})T-11.26\mathrm{kbar} \\ \mathrm{T}=\frac{15.51\mathrm{kbar}+11.26\mathrm{kbar}}{21.17\mathrm{bar}/\mathrm{K}+17.73\mathrm{bar}/\mathrm{K}} \\ \mathrm{T}=688.2\mathrm{K} \\ \mathrm{T}=690\mathrm{K} \end{gathered}$$

As a result, the pressure at the place where andalusite$\leftrightarrow$sillimanite (a $\leftrightarrow$s) and kyanite $\leftrightarrow$sillimanite (k$\leftrightarrow$s) lines connect is as follows:

role="math" localid="1646934600818" $$\begin{gathered} P=(-17.73\mathrm{bar}/\mathrm{K})(688.2\mathrm{K})+15.51\mathrm{kbar} \\ P=3.30kbar \end{gathered}$$

Calculation

Calculating the temperature T and pressure P at k$\leftrightarrow$a line with k$\leftrightarrow$s line.

$$\begin{gathered} P=(12.6\mathrm{bar}/\mathrm{K})T-5.38\mathrm{kbar}\text{for}k\leftrightarrow a \\ P=(21.17\mathrm{bar}/\mathrm{K})T-11.26\mathrm{k}\text{bar}\text{for}k\leftrightarrow s \end{gathered}$$

Equating both the values and solving for T

$$\begin{gathered} (21.2\mathrm{bar}/\mathrm{K})T-11.26\mathrm{kbar}=(12.6\mathrm{bar}/\mathrm{K})T-5.38\mathrm{kbar} \\ T=\left(\frac{11.26\mathrm{kbar}-5.38\mathrm{kbar}}{21.17\mathrm{bar}/\mathrm{K}-12.6\mathrm{bar}/\mathrm{K}}\right) \\ T=686.1K \\ T=690K \end{gathered}$$

Therefore, the pressure of the intersection point of k $\leftrightarrow$a line with k$\leftrightarrow$s line will be:

$$\begin{gathered} P=(12.6\mathrm{bar}/\mathrm{K})(686.1\mathrm{K})-5.380\mathrm{kbar} \\ P=3.26kbar \\ P=3.3kbar \end{gathered}$$

Conclusion

The phase diagram for the three boundaries of kyanite, andalusite, and sillimanite is shown below.

The temperature at the triple point is 690 K, and the pressure is 3.3 kbar, according to the above diagram and calculations.