Problem Set #3 Solutions

 

 

1. The Gibbs free energy of a phase over a limited range of pressure and temperature is

 

G = G0 + V0(P-P0) - S0(T-T0)

 

The pressure at room conditions (1 bar) is very small compared with the pressures of interest; we will assume P0=0.  We can also set T0=273 K for convenience, since this is close enough to room temperature for our purposes.  We then have, for each phase

 

G = G0 + V0P - S0T

 

where T is now measured in degrees centigrade. We find the phase diagram by determining the lines of equilibrium coexistence.  For example, setting Gibbs free energies of andalusite and kyanite equal to each other

 

G0an+V0anP - S0anT = G0ky + V0kyP - S0kyT

 

solving for T as a function of P, we find the equation of a straight line

 

T = DG/DS + (DV/DS)P

 

where DG = G0ky - G0an, DV = V0ky - V0an, and, DS = S0ky - S0an.  The first term (being careful to use consistent units!)

 

DG/DS = (-2760 kJ/mol + 1376 kJ/mol)/(83.76 J mol-1 K-1 - 93.22 J mol-1 K-1) = 146 C

 

Is the temperature of the phase transition at zero pressure.  The Clapeyron slope (watch the units!)

 

DV/DS = (4.409 x 10-5 m3 mol-1 - 5.153 x 10-5 m3 mol-1)/ (83.76 J mol-1 K-1 - 93.22 J mol-1 K-1) = 7.86 x 10-7 C/Pa = 78.6 C/kbar

 

Where we have converted to more convenient units in the last step.  Applying the same procedure for the other phase transitions, we find (T in C, P in kbar)

 

Ky-an:             T = 146 + 78.6P

Ky-si:              T = 299 + 47.0P

Si-an:               T = 800 - 56.4P

 

Plotting these lines gives us

 

which is not a phase diagram yet.  Notice that the three lines meet in a point, called the triple point, near 4.8 kbar and 520 C.  We have to decide which part of each line to retain, the part above the triple point, or the part below it. An easy way to do this is to examine the physical properties of the phases.  For example, the high temperature portion of ky=an is unstable since sillimanite has a higher entropy.  The high pressure portion of si=an must be unstable since kyanite has a smaller volume.  Finally, andalusite must be the stable low pressure phase since it has the largest volume: the low pressure portion of ky=si is then unstable.  Our final phase diagram then looks like this

 

 

b) We calculated the Clapeyron slopes already.  Notice that ky=an and ky=si have positive slopes because the change in volume and entropy have the same sign.  The other transition, an=si has a negative slope because si has a larger entropy, but a smaller volume than an.

 

c) Our phase diagram is similar to that determined by Bohlen et al. (1991).  In particular, their triple point (4.2 kbar, 530 C) differs from ours by only 0.6 kbar and 10 C.  The difference in temperature is within the quoted uncertainty of Bohlen et al.'s determination.  The difference in pressure is larger than the uncertainty.  This might be caused by several factors.  The measured values of G0, V0, and S0 that we used in our calculations also contain uncertainties.  These may be large enough to account for the difference in phase diagrams.  The Bohlen et al. phase diagram may also contain systematic errors which the authors have not taken into account.  They were very careful to try to eliminate the effects of kinetics, but this is difficult to do completely, especially at these relatively low temperatures.

 

d) The rock containing kyanite and silliminate should originate at depths greater than 4.8 kbar.  Using the equation of hydrostatic equilibrium (P=rgz where z is the depth), and assuming a density of 2500 kg m-3 for the crust, we find (units!)

 

z = (4.8 x 108 Pa)/(2500 kg m-3 x 9.8 m s-2) = 19,600 m = 19.6 km

 

the rock might be brought to the surface by a combination of tectonic processes and erosion.  For example, compression, which may occur at continent-continent collisions may thicken the crust.  Subsequent erosion may then expose portions of the crust which before were buried at great depth.  A tectonic environment where these processes are occurring today is in the Himalaya at the boundary between the Indian and Eurasian plates.

 

2. a) Lattice points are shown as small open circles; b) the unit cell by dark lines.  Notice that the unit cell is centered-rectangular.

c) Since all angles are unequal, the mineral is triclinic.