1. Gibbs Phase Rule The number of variables which are required to describe the state of a system: p+f=c+2 f=c-p+2 –Where p=# of phases, c= # of components, f= degrees of freedom –The degrees of freedom correspond to the number of intensive variables that can be changed without changing the number of phases in the system

2. Variance and f f=c-p+2 Consider a one component (unary) diagram If considering presence of 1 phase (the liquid, solid, OR gas) it is divariant 2 phases = univariant 3 phases = invariant

3. Free Energy Gibbs realized that for a reaction, a certain amount of energy goes to an increase in entropy of a system. G = H –TS or  G 0 R =  H 0 R – T  S 0 R Gibbs Free Energy (G) is a state variable, measured in KJ/mol Tabulated values of  G 0 R are in Appendix

4. Now, how does free energy change with T and P? From  G=  H-T  S: T and P changes affect free energy and can drive reactions!!

5. Volume Changes (Equation of State) Volume is related to energy changes: Mineral volume changes as a function of T: , coefficient of thermal expansion Mineral volume changes as a function of P: , coefficient of isothermal expansion For Minerals:

6. Volume Changes (Equation of State) Gases and liquids undergo significant volume changes with T and P changes Number of empirically based EOS solns.. For metamorphic environments: –Redlich and Kwong equation: V-bar denotes a molar quatity, a Rw and b RK are constants

7. Phase Relations Rule: At equilibrium, reactants and products have the same Gibbs Energy –For 2+ things at equilibrium, can investigate the P-T relationships  different minerals change with T-P differently… For  G R =  S R dT +  V R dP, at equilibrium,  G  rearranging: Clausius-Clapeyron equation Remember that a line on a phase diagram describes equilibrium,  G R =0!!

8.  V for solids stays nearly constant as P, T change,  V for liquids and gases DOES NOT Solid-solid reactions linear  S and V nearly constant,  S/  V constant  + slope in diagram For metamorphic reactions involving liquids or gases, volume changes are significant,  V terms large and a function of T and P (and often complex functions) – slope is not linear and can change sign (change slope + to –)  S R change with T or P? V = Vº(1-  P)

10. Example – Diamond-graphite To get C from graphite to diamond at 25ºC requires 1600 MPa of pressure, let’s calculate what P it requires at 1000ºC: graphitediamond  (K -1 ) 1.05E-057.50E-06  (MPa -1 ) 3.08E-052.27E-06 Sº (J/mol K) 5.742.38 Vº (cm3/mol) 5.29823.417

11. Clausius-Clapyron Example

13. Phase diagram Need to represent how mineral reactions at equilibrium vary with P and T