# Thermodynamic modeling course (LMV)

¡Hola!, this site will be dedicated to the last two days of the course and a place to download files.

### The last challenge

You've a learned quite a lot of things this week so I propose the following game.

After a long field campaign in the Antarctic a bunch a petrologist brought a collection of 10 unusual samples (samples LMV22-01 to LMV22-10 ) from two sites but with no contextual information (i.e., "xenoliths"). Back in Clermont-Ferrand they discovered that most of the samples contained significant amounts of an unknown element that they temporarily call X. Arduous chemical efforts for separation allowed to identify the element as trivalent having a molar mass of 100.

Further geochemical works identified another strategic element but in only half of the samples. The youngest petrologist got funded from a generous ERC project to constrain one of the two thermal gradients where the strategic element can be found for future explorations and (of course) revolutionize our current knowledge of the thermal state of the lithosphere.

You will start by looking at the petrological data (bulk rock chemistry, mineralogy and mineral chemistry of the 10 samples).

Site 1LMV22-01;LMV22-02;LMV22-03;LMV22-04;LMV22-05

Site 2 LMV22-06;LMV22-07;LMV22-08;LMV22-09;LMV22-10

No more information is given (but as you know the LMV is excellently equipped with all experimental techniques necessary to accomplish the project). Your challange is to discuss what kind of experimental data you would need to solve the puzzle.

Good luck!

Site 1 LMV23-01

Site 1 LMV23-06

### NEWS 3!!!

A python script in Jupyter to compute G (Clic here)

### NEWS 4 !!!

xxxxxxxxxxbolf     EoS = 2 |                                                                    MGO(3)X2O3(1)SIO2(3)G0 = -5900105 S0 = 280.3 V0 = 8.318 c1 = 600.0 c3 = -5100000 c5 = -4000.2  b1 = .2500E-4 b5 = -.2500E-3 b6 = 2007000. b7 = -300.55 b8 = 4   end
clerm    EoS = 2 |                                                                    MGO(1)X2O3(1)G0 = -2106030 S0 = 81.5 V0 = 3.978  c1 = 342.7 c2 = -.5037E-2 c3 = -2205100 c5 = -1678.1  b1 = .431E-4 b5 = -.431E-3 b6 = 1945000. b7 = -291.75 b8 = 4   end
ppx      EoS = 2 |                                                                   X2O3(1)G0 = -1560000 S0 = 49.9 V0 = 2.658  c1 = 160.0 c2 = .600E-2 c3 = -2560600 c5 = -599.2  b1 = .429E-4 b5 = -.429E-3 b6 = 2620000. b7 = -388 b8 = 4  end

### News!!

Experiments in simple systems (Clic here and here)

### News!!

Experiments done for nathalite (Clic here and here)

### Resources

• Here is a summary of the key expression needed to understand what is behind the scenes of a phase diagram (Clic here).

• A few notes about the simplest solid solutions (ideal and regular). (Clic here)

• An examle of olivine as a simple non-ideal binary solid solution implemented in python (Clic here)

### Other resources

• Oxygen fugacity based on Fe-Mg exchange in olivine-orthopyroxene in magnetite buffered assemblages. Clic here).