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GEO 4860 Advanced petrology

GEO 4860 Advanced petrology. Part 1 : Thermodynamics and phase diagrams, mostly igneous petrology January 24 – March 16 Evaluation: Written exercises (20%) and mid-term exam (30%) Reidar Trønnes , Natural History Museum, UiO r.g.trønnes@nhm.uio.no

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GEO 4860 Advanced petrology

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  1. GEO 4860 Advancedpetrology Part 1: Thermodynamics and phase diagrams, mostly igneous petrology January 24 – March 16 Evaluation: Written exercises (20%) and mid-term exam (30%) ReidarTrønnes, Natural History Museum, UiO r.g.trønnes@nhm.uio.no www. nhm.uio.no/om-museet/seksjonene/forskning-samlinger/ansatte/rtronnes/index-eng.xml Part 2: Mostly metamorphic petrology, petrological and structural field relations, training in petrological research March 20 – June 14 Evaluation: Project report (20%) and final exam (30%) HåkonAustrheim, Dept. of Geosciences / Centre for Physics of Geological Processes, UiO h.o.austrheim@geo.uio.no www.mn.uio.no/geo/english/people/aca/tpg/hakonau/index.html

  2. Lectures and tutorials Tuesday and Friday 10-12 and 13-14 in GEO 114 (Skolestua) Examinations Mid-term exam: Friday, March 16, 10.15-13.15 Final exam: Wednesday, June 14, 14.30 – 17.30 Textbook Winter (2010): PrinciplesofIgneous and MetamorphicPetrology. 2. ed. Prentice Hall / Pearson. ppt-presentasjoner for hvert av kapitlene på: www.whitman.edu/geology/winter/.

  3. List ofparticipants: 12 Advantage for student – instructorcommunication Presentationof student status, M.Sc. thesisproject, supervisor, etc. Completedcourses(or courses in progress) Opticalmineralogy and petrography ? Isotope geochemistry ? Structuralgeology ?

  4. Mineral- and rock compositions Weight% versus atom% (or mol%) Verybasic features - Silicate minerals and commonrocks arenearly "fullyoxidized", withO as themost importantand oftentheonlyanion BUT: Fe occurs as a combination ofFe2+(as FeO) and Fe3+(as Fe2O3/ FeO1.5) - Oxide-, carbonate-, phosphate-, sulphate- and tungstate-mineralsarealso "fullyoxidized" - The contentsofhalogenides and sulphidesareverylow in commonsilicate rocks Chemical analyses of minerals and rocks arenormallypresented as oxides BUT: useful to recalculate to cation basis,especially for minerals (i.e. relatively simple stoichiometric compounds)

  5. Basic information: Valency/ oxidationstatefor major and trace elements - Thegreatmajorityof major and trace elements has onedominatingvalency in silicate minerals and rocksin planetary mantles and crusts. - Importantexception: Fe Importantquestion Choiceofstoichiometry / formulafor theoxides SiO2 Al2O3or AlO1.5 FeO Fe2O3or FeO1.5 CaO Na2Oor NaO0.5 K2Oor KO0.5

  6. Is 19 wt% Al2O3identical to 19 wt% AlO1.5 ? Yes! evenifAl2O3 has 2 cations+ 3 anions, whereas AlO1.5has 1 cation+ 1.5 O-ions Calculationof mineral formulasbasedonwt%: I prefer to useone-cation-oxides

  7. Calculationof mineral formula (structuralformula) for afeldsparanalysis Generell feltspat-formel: (K,Na,Ca)(Al,Si)4O8, i.e. 5 cations+ 8 O-ions Mineral formula, normalized to 8 O-ions = op*8/(O*29769) Si 2.947 Al 1.025 Fe3+ 0.035 Ca 0.029 Na 0.618 K 0.354 Scations5.008 Scharge16.00 Mineral formula, normalized to 5 cations = cp*5/18635 Si 2.943 Al 1.024 Fe3+ 0.035 Ca 0.029 Na 0.616 K 0.353 Scations5.000 Scharge15.97 oxygen-prop. op = 104*O*wt%/mw 21936 5723 194 109 1149 658 S 29769 mol. wt mw 60.08 50.98 79.84 56.08 30.99 47.10 cationprop. cp = 104*wt%/mw 10968 3815 129 109 2298 1316 S 18635 wt% 65.90 19.45 1.03 0.61 7.12 6.20 SiO2 AlO1.5 FeO1.5 CaO NaO0.5 KO0.5 4.007 4.002 1.001 0.998 Mineral formulanormalized to 5 cations: K0.35Na0.62Ca0.03(Fe0.04Al1.02Si2.94) O7.99 Mineral formulanormalized to 8 O-ions: K0.35Na0.62Ca0.03(Fe0.04Al1.03Si2.95) O8.00

  8. Graphicpresentationof mineral compositions Along a simple axis, systemO-Fe: 3 phases(minerals): FeO (wüstite), Fe2O3 (hematite) and FeOˑFe2O3 (Fe3O4, magnetitt) Mol% FeO: 1Fe + 1O = 2, 1/2 = 50% Fe Fe2O3: 2Fe + 3O = 5, 2/5 = 40% Fe Fe3O4: 3Fe + 4O = 7, 3/7 = 43% Fe Weight% FeO: 55.85 Fe + 16.00 O = 71.85, 78% Fe Fe2O3: 2ˑ55.85 Fe + 3ˑ16.00 O = 159.70, 70% Fe Fe3O4: 3ˑ55.85 Fe + 4ˑ16.00 O = 231.55, 72% Fe

  9. Graphicpresentationof mineral compositions In triangular diagram, system MgO-SiO2-Al2O3. Mineral endmembers (components) forsterite(Mg2SiO4), enstatite(MgSiO3), silica(SiO2), spinel(MgAl2O4) og pyrope(Mg3Al2Si3O12) Mol% Mg2SiO4: 2MgO + 1SiO2, 1/3 = 33.3% SiO2 MgSiO3: 1MgO + 1SiO2 = 2, 1/2 = 50.0% SiO2 SiO2: 100% SiO2 MgAl2O4: 1MgO + 1Al2O3 = 2, 1/2 = 50.0% Al2O3 Mg3Al2Si3O12: 3MgO + 1Al2O3 + 3SiO2 = 7, 3/7 = 42.9 % MgO 3/7 = 42.9% SiO2 1/7 = 14.3% Al2O3 Weight% Mg2SiO4: 2ˑ40.3044 + 60.0843 = 140.6931, 42.7% SiO2 MgSiO3: 40.3044 + 60.0843 = 100.3887, 59.9% SiO2 SiO2: 100% SiO2 MgAl2O4: 40.3044 + 101.9612 = 142.2656, 71.7% Al2O3 Mg3Al2Si3O12: 3ˑ40.3044 + 101.9612 + 3ˑ60.0843 = 403.1273, 30.0% MgO 44.7% SiO2 25.3% Al2O3 Al2O3 Sp Sp Py Py Fo MgO Fo En SiO2 En

  10. Definitionofthe terms phase, systemand components Phase: chemicallyhomogenoussubstance Examples: water, ice, steam, kyanite, sillimanite, quarts, granitic melt System: a collectionofphases under consideration Wedecidetheboundariesofour system. Try to findthe most convenient system in order to describe and understand thechemical and thermodynamicequilibria. Examples: - hand specimentof a basalt - a 10 m long by 2 m tall roadcutexposing lenses ofeclogite in gneis - an experimentalsamplecapsulewith 50 mol% MgSiO3 + 50 mol% Mg3Al2Si3O12 - theentireEarth’s mantle + core System components: chemicalentitiesnecessaryfor thecharacterization of a system(e.g. elements, oxides or more complexformulaunits) Phasecomponents: chemicalentitiesnecessary for thecharacterization of a phase(e.g. elements, oxides or more complexformulaunits)

  11. Considerationsw.r.t. choiceofcomponents Al-silicate system containingthephaseskyanite, sillimanite, andalusite. Polymorphswithchemicalcomposition: Al2SiO5 3 elements: Al, Si, O 2 ooxides: Al2O3, SiO2 Howmanycomponents? 1: Al2SiO5

  12. The ”olivine system” of solid solutionbetween forsteritt (Mg2SiO4) og fayalitt (Fe2SiO4) 4 elements: Mg, Fe, Si, O 3 oxides: MgO, FeO, SiO2 Howmanycomponents? 2: Mg2SiO4and Fe2SiO4 Howmanyphases? 1: olivine

  13. The phaserule For a system in equilibrium: P + F = C + m F:Numberofdegreesoffreedom (variance): numberof variables thatcanchange independentlyofeachother (e.g. p-T-X, pressure-temperature-composition) C:The smallest numberofchemicalcomponentsrequired to characterize all ofthephases P:Numberofphases present in equilibrium at any given location m:Numberofexternal variables thatinfluencethe system, commonly m=2 (p and T)

  14. Phaserule: P + F = C + 2 System: Al2SiO5 C = 1 (componentAl2SiO5is common to all 3 phases) P = 3  3 + F = 1 + 2 F = 0 (invariant point) P = 2  2 + F = 3 F = 1 (univariantline) P = 1  1 + F = 3 F = 2 (divariantfield)

  15. Olivinegroup thevery first melt olivine Liquidus Solidus Forsterite:veryhighmeltingpoint System forsterite-fayalite:full solid solution and liquidsolution Bulk composition

  16. System: (Mg,Fe)2SiO5or Mg2SiO4 – Fe2SiO4 C = 2 bothofthephasescomprisedifferent proportionsofthecomponents forsterite and fayalite This is a T-X-diagram at constant (fixed) pressure(1 bar)  Phaserule: P + F = C + 1 P + F = 2 + 1 = 3 F = 3 – P • P = 1  F = 2 • divariantfields, canvaryboth T and X • i.e. for a fixedT (e.g. 1200 C) can • olivinevary from Fo100to Fo0 P = 2  F = 1 - univariantcurves. IfT is fixed, the compositionsofthetwophasesarealsofixed - changing T  changingXsol and Xmelt Three phasesare never present in this diagram. Therefore, therearenobinary invariant points. BUTthemeltingpoints for pure forsterite and pure fayaliteareunary invariant points (in thetwo one-component systems Fo and Fa.

  17. The lever rule: Measuringthemassproportionoftwophases Bulk compositionFo20 Equilibrium at 1660 C 99.99% olivine(Fo20) + 0.01% melt (Fo51)

  18. Vektstang-regelen: Måling av mengdeforholdet mellom to faser Bulk compositionFo20 Equilibrium at 1700 C olivine(Fo84) + melt (Fo56) 7 41 100*41/48 = 85.4% 100*7/48 = 14.6%

  19. Vektstang-regelen: Måling av mengdeforholdet mellom to faser Bulk compositionFo20 Equilibrium at 1800 C olivine(Fo93) + melt (Fo77) 5 22 100*5/27 = 18.5% 100*22/27 = 81.5%

  20. Teaching Phase Equilibria http://serc.carleton.edu/research_education/equilibria/index.html

  21. One-component system Two-component system Eutecticpoint, 23% NaCl

  22. Simple experimenti freezer(s) withadjustableT Make a salt solution with2-3% NaCl Putplastic bags withthesolution in thefreezer(s) at twodifferent T(–2 and –18 ºC) –2ºC • - takeoutthenextday • pickoutthe pieces ofice • rinse in clean water • taste theice • taste thesolutions • estimatethemassproportion • solid/liquid –18ºC Whatwillyouobserve: –2ºC ? –18 ºC ?

  23. Phaserelations: basalt – peridotite, binary systems

  24. Meltingrelations for natural rocks – Multicomponent systems Not a simple binarysystem: A ternarysystem is muchbetter, but still only a simplifiedmodel Eutektic point basalt peridotite

  25. Secondlawofthermodynamics: Changeofinternalenergy (heat content) and entropy dQ = TdS(for a reversibel process) Thirdlaw: Scrystal = 0 at T = 0 K First law: (internalenergyof an isolatedsystem is constant) dE = dQ – dW = TdS – pdV Work: W = pV W = Fs (Nm = J) p = F/s2 (N/m2 = Pa) W = ps3 = pV (m3N/m2 = Nm) Our system: e.g. a crystal Contribution1: addedenergy(dE1) mayincreasetheinternalenergy (by heating, dQ) dQgoesintothe system -thereforepositive dE1 dE1 = dQ = TdS heating Contribution2: addedenergy(dE2) enablethe system to performworkon thesurroundings( = volumeincreaseagainstconstantpressure) dWgoesoutofthesystem -thereforenegative dE2 dE2= -dW = -pdV volumeincrease

  26. First law: The internalenergy in an isolatedsystem is constant dE = dQ – dW = TdS – pdV Gibbsfreeenergy:energyin addition totheinternalenergy Definition: G = E – (TS – pV) = E + pV – TS = H – TS At chemicalequilibrium: DG = Gprod – Greact = 0

  27. Gibbsfreeenergy:energyin addition totheinternalenergy Definition: G = E – (TS – pV) = E + pV – TS = H – TS At chemicalequilibrium: DG = Gprod – Greact = 0 Lowestenergylevel: greateststability Simple meltingreaction: SiO2 = SiO2 solid melt (e.g. tridymite)

  28. The aluminium silicates: Al2SiO5 Sillimanite: Al[6]Al[4]Si O5 Orthorombic Kyanite: Al[6]Al[6]Si O5 Triclinic c c p Al[6] T Al[4] c Andalusite: Al[6]Al[5]Si O5 Orthorombic Al[5]

  29. ReactionAl2SiO5 = Al2SiO5 andalusitekyanite p DV = Vky – Vand positive? or negative? < 0 kyanite DS = Sky – Sand < 0 dp p1+dp, T1+dT p1 Considertwopointsonthephaseboundary (equilibriumbetweenkyanite and andalusite) DG = 0 p1,T1 andalusite DG = DE + p1DV – T1DS = 0 DG = DE + (p1+dp)DV – (T1+dT)DS = 0 dT T1 T Subtractionofupperequation from thelowerone: dpDV = dTDS dp/dT = DS/DV Clayperon-equation negative DV andDS givespositivedp/dT-slope

  30. Aluminium silicates: Al2SiO5 Kyanite: Al[6] Al[6] Si O5 Triclinic, D: 3600 kg/m3 Sillimanite: Al[6] Al[4] Si O5 Orthorombic, 3250 kg/m3 Andalusite: Al[6] Al[5] Si O5 Orthorombic, 3180 kg/m3 p Al2SiO5 = Al2SiO5 andalusitesillimannite DV = Vsill – Vand DS = Ssill – Sand positive? or negative? T < 0 > 0 dp/dT = DS/DV positive? or negative?

  31. General features regardingmelting and crystallisation What is melting? Solid/orderedcrystallatticebreaksdown (”dissolves”) What is required? (Reaction: crystal → melt) - heating to Tm - heat offusion, DHm= DE + pDV> 0 WhataboutDV? Mostly: DV > 0 possibleexception at veryhighp Highp → DV may be negative,butDEis alwayspositive Equilibrium: DGsm = DH ̶̶ TDS = 0, d.v.s. DS > 0 alwayspositive DSm Gibbsfreeenergy:energyin addition to theinternalenergy Definition: G = E – (TS – pV) = E + pV– TS = H – TS At equilibrium: DG = Gprod – Greact = 0 First law (Internalenergy in isolated system is constant) dE= dQ – dW = TdS – pdV

  32. Basic question(try to reasononlyintuitively): AssumingthatDVm>0, howdoes Tmchangewithincreasingp? Why? Tmincreaseswithincreasing p becausethevolumeincrease due to melting (DVm> 0) becomes more unfavourable as p increases. Increasing Tm is required to compensate for theunfavourablep-V-effect.

  33. Clapeyron-equationfor melting p Forsterite ReactionMg2SiO4 = Mg2SiO4 forsteritemelt Melt DV = Vsm – Vfast > 0 DS > 0 dp p1 dT DG = DE + p1DV – T1DS = 0 DG = DE + (p1+dp)DV – (T1+dT)DS = 0 T T1 dpDV = dTDS dp/dT = DS/DV T Melt positive DV andDS: positivedp/dT-slope Forsterite p

  34. Correctmeltingcurvebasedon more recentexperiments (Kanzaki1990; Zhang et al 1993) Older and somewhatwrongphase diagram for SiO2 from Perkins (2011, Mineralogy) Extremelyhigh T, difficultexperiments p-Tmeltingcurvesarenot linear, buthave mostly increasingdp/dTwithincreasing p and T. Why? Howcanweanswerthisquestion?

  35. ConsidertheClapeyron-slopeofthemeltingcurve: dp/dT = DSm/DVm Meltingreaction: SiO2(solid) =SiO2(melt) DSm= Smelt-Ssolid DVm= Vmelt-Vsolid Remember: DSm is alwayspostitive- for simplicitywecan assumethatit is nearlyconstant WhataboutDVm? Whatis most compressible: mineral or melt ? DVm→0 in thehighest p-range ofb-quartz and coesite dp/dT = DSm/DVm→ ∞

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