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Q 5.39

Expert-verifiedFound in: Page 176

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

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.

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

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

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

Where,

$\u2206S$ is the change in entropy

$\u2206V$ 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

$\frac{dP}{dT}=\frac{93.22\mathrm{J}/\mathrm{K}-83.81\mathrm{J}/\mathrm{K}}{5.153\mathrm{J}/\mathrm{bar}-4.409\mathrm{J}/\mathrm{bar}}\phantom{\rule{0ex}{0ex}}=12.65bar/K$

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_{s }and V_{s} are the entropy and volume of the sillimanite

Substituting the values

$\frac{dP}{dT}=\frac{93.22\mathrm{J}/\mathrm{K}-96.11\mathrm{J}/\mathrm{K}}{5.153\mathrm{J}/\mathrm{bar}-4.990\mathrm{J}/\mathrm{bar}}\phantom{\rule{0ex}{0ex}}=-17.73bar/K$

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

$\frac{dP}{dT}=\frac{96.11\mathrm{J}/\mathrm{K}-83.81\mathrm{J}/\mathrm{K}}{4.99\mathrm{J}/\mathrm{bar}-4.409\mathrm{J}/\mathrm{bar}}\phantom{\rule{0ex}{0ex}}=21.17bar/K$

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:

$P=\frac{dP}{dT}\xb7T+C\phantom{\rule{0ex}{0ex}}C=P-\frac{dP}{dT}\xb7T\phantom{\rule{0ex}{0ex}}C=1\mathrm{bar}-(12.6\mathrm{bar}/\mathrm{K})(427\mathrm{K})\phantom{\rule{0ex}{0ex}}C=-5.38kbar$

The transition pressure for andalusite and sillimanite is as follows:

$C=P-\frac{dP}{dT}\xb7T\phantom{\rule{0ex}{0ex}}C=1\text{bar}-(-17.73\mathrm{bar}/\mathrm{K})(875\mathrm{K})\phantom{\rule{0ex}{0ex}}C=15.51kbar$

The transition pressure for kyanite and sillimanite is as follows:

$C=P-\frac{dP}{dT}\xb7T\phantom{\rule{0ex}{0ex}}C=1\mathrm{bar}-(21.17\mathrm{bar}/\mathrm{K})(532\mathrm{K})\phantom{\rule{0ex}{0ex}}C=-11.26kbar$

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" $P=(12.6\mathrm{bar}/\mathrm{K})T-5.38\mathrm{kbar}\text{for}k\leftrightarrow a\phantom{\rule{0ex}{0ex}}P=(-17.73\mathrm{bar}/\mathrm{K})T+15.51\mathrm{kbar}\text{for}a\leftrightarrow s\phantom{\rule{0ex}{0ex}}$

Equating both the values

$(12.6\mathrm{bar}/\mathrm{K})T-5.38\mathrm{kbar}=(-17.73\mathrm{bar}/\mathrm{K})T+15.51\mathrm{bar}\phantom{\rule{0ex}{0ex}}T=\frac{15.51\mathrm{kbar}+5.38\mathrm{kbar}}{12.6\mathrm{bar}/\mathrm{K}+17.73\mathrm{bar}/\mathrm{K}}\phantom{\rule{0ex}{0ex}}T=688.7K\phantom{\rule{0ex}{0ex}}T=690K$

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:

$P=(12.6\mathrm{bar}/\mathrm{K})(688.7\mathrm{K})-5.38\mathrm{kbar}\phantom{\rule{0ex}{0ex}}P=3.298kbar\phantom{\rule{0ex}{0ex}}P=3.3kbar$

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.

$P=(-17.73\mathrm{bar}/\mathrm{K})T+15.51\mathrm{kbar}\text{for}a\leftrightarrow s\phantom{\rule{0ex}{0ex}}P=(21.17\mathrm{bar}/\mathrm{K})T-11.26\mathrm{k}\text{bar}\text{for}k\leftrightarrow s$

Equating both the values

$(-17.73\mathrm{bar}/\mathrm{K})T+15.51\mathrm{kbar}=(21.17\mathrm{bar}/\mathrm{K})T-11.26\mathrm{kbar}\phantom{\rule{0ex}{0ex}}\mathrm{T}=\frac{15.51\mathrm{kbar}+11.26\mathrm{kbar}}{21.17\mathrm{bar}/\mathrm{K}+17.73\mathrm{bar}/\mathrm{K}}\phantom{\rule{0ex}{0ex}}\mathrm{T}=688.2\mathrm{K}\phantom{\rule{0ex}{0ex}}\mathrm{T}=690\mathrm{K}$

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" $P=(-17.73\mathrm{bar}/\mathrm{K})(688.2\mathrm{K})+15.51\mathrm{kbar}\phantom{\rule{0ex}{0ex}}P=3.30kbar$

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

$P=(12.6\mathrm{bar}/\mathrm{K})T-5.38\mathrm{kbar}\text{for}k\leftrightarrow a\phantom{\rule{0ex}{0ex}}P=(21.17\mathrm{bar}/\mathrm{K})T-11.26\mathrm{k}\text{bar}\text{for}k\leftrightarrow s\phantom{\rule{0ex}{0ex}}$

Equating both the values and solving for T

$(21.2\mathrm{bar}/\mathrm{K})T-11.26\mathrm{kbar}=(12.6\mathrm{bar}/\mathrm{K})T-5.38\mathrm{kbar}\phantom{\rule{0ex}{0ex}}T=\left(\frac{11.26\mathrm{kbar}-5.38\mathrm{kbar}}{21.17\mathrm{bar}/\mathrm{K}-12.6\mathrm{bar}/\mathrm{K}}\right)\phantom{\rule{0ex}{0ex}}T=686.1K\phantom{\rule{0ex}{0ex}}T=690K$

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

$P=(12.6\mathrm{bar}/\mathrm{K})(686.1\mathrm{K})-5.380\mathrm{kbar}\phantom{\rule{0ex}{0ex}}P=3.26kbar\phantom{\rule{0ex}{0ex}}P=3.3kbar$

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.

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