1. Pressure-Volume Equations of State *Motivation: Describing the compression of materials on a quantitative level, Comparing strength of materials, to seismic signals, fundamental thermodynamics,… *What is the function that describes reduction in volume for an increase in pressure? V(P) = ? (In general V(P,T,etc), here just look at pressure effects) Does it have some physical basis? Is it intuitive? Is it easy to manipulate? Does it work?

2. Gas EOS GASES ideal & real gas laws V~1/P => PV = nRT (ideal gas law) finite molecular volume => Veff = V-nb P(v-b) = RT (Clausius EOS) attractive forces => Peff = P-a/v 2 (P+a/v 2 )*(v-b) = RT (VdW EOS)

3. constant compressibility (F=k*x)  V/V 0 = -1/K *  P K = -V 0 *  P/  V (bulk modulus) Integrate : P = -K*ln(V/V 0 ) => V = V 0 exp(-P/K) linear compressiblity (Murnaghan EOS), pressure-induced stiffening K = K 0 + K’ * P K 0 + K’*P = -V 0 *dP/dV => dP/(K 0 + K’P) = -dV/V 0 ln(K 0 + K’*P) 1/K’ = lnV/V 0 => (K 0 +K’*P) 1/K’ = V/V 0 V = V 0 (K 0 +K’*P) 1/K’ Solid (condensed matter) EOS

4. polynomial expansion of K => K = K 0 + K’P + K’’P + … this has the problem that K -> 0 at high compression, which is physically non-sensical semi-emprical (physically reasonable, not from first principles, agrees with data) carefully choose variables: Eulerian finite strain measure: f= ½ [(V 0 /V) 2/3 – 1] Condensed Matter EOS (cont’d)

5. B-M EOS Birch-Murnaghan EOS: expand strain energy in Taylor series: F = a + bf + cf 2 + df 3 + … look at 2 nd order: F = a + bf + cf 2 apply boundary conditions: when no strain (f=0) F = 0 (F(0) = 0) => a = 0 F = bf + cf 2 From Thermo P = -dF/dVThermo P = -(dF/df)(df/dV) evaulate both parts: (first part) dF/df = b + 2cf; (second part) df/dV = d(½ [(V 0 /V) 2/3 – 1])/dV = -(1+2f) 5/2 /3V 0= combining: P = -(b*df/dV + 2cf*df/dV) so P = b*(1+2f) 5/2 /3V 0 + 2cf*(1+2f) 5/2 /3V 0

6. P = b*(1+2f) 5/2 /3V 0 + 2cf*(1+2f) 5/2 /3V 0 apply boundary conditions: when f = 0 P = 0 (P(0) = 0) this means b = 0 and P = 2cf*(1+2f) 5/2 /3V 0 so what is the constant c? Find out by analytically evaluating K, then apply the boundary condition that when f=0 K = K 0 & V = V 0 remember K = -V(dP/dV) = -V *dP/df * df/dV = -V * (2c/3V 0 ) [f*5/2*2*(1+2f) 3/2 + (1+2f) 5/2 ] * (-(1+2f) 5/2 /3V 0 ) = 2cV/9V 0 2 * (1+2f) 5/2 * [5f(1+2f) 3/2 + (1+2f) 5/2 ] evaluate for f = 0 => K = K 0 = 2cV 0 /9V 0 2 = 2c/9V 0 and c = 9V 0 K 0 /2 so P = 3K 0 f(1+2f) 5/2 2 nd order B-M EOS (cont’d)

7. P = 3K 0 f(1+2f) 5/2 substitute f= ½ [(V 0 /V) 2/3 – 1] P = 3K 0 /2 * [(V 0 /V) 2/3 – 1] * (1 + 2* ½ [(V 0 /V) 2/3 – 1]) 5/2 = 3K 0 /2 * [(V 0 /V) 2/3 – 1] * (1 +[(V 0 /V) 2/3 – 1]) 5/2 = 3K 0 /2 * [(V 0 /V) 2/3 – 1] * ((V 0 /V) 2/3 ) 5/2 = 3K 0 /2 * [(V 0 /V) 2/3 – 1] * (V 0 /V) 5/3 P = 3K 0 /2 * [(V 0 /V) 7/3 – (V 0 /V) 5/3 ] ( This is 2 nd order BM EOS!) K = -V(dP/dV) = K 0 (1+7f)(1+2f) 5/2 (after derivatives and a a lot of algebra) K’ = dK/dP = (dK/dV)*(dV/dP) = (dK/dV)/(dP/dV) = (12 +49f)/(3+21f) K 0 ’ = K’(f=0) = 4 2 nd order B-M EOS (cont’d)

8. 3 rd order B-M EOS F = a + bf + cf 2 + df 3 apply boundary conditions and use derivative relations P = -dF/dV & K = -V(dP/dV) to solve for coefficients (just like in 2 nd order B-M EOS) get another term, a lot more algebra & K’ not constrained to 4 P = 3K 0 /2 * [(V 0 /V) 7/3 – (V 0 /V) 5/3 ]*[1 + 3/4*(K 0 '-4) *((V/V 0 ) -2/3 - 1)] = 3K 0 f(1+2f) 5/2 * [1 + 3/4*(K 0 '-4) *((V/V 0 ) -2/3 - 1)] (this is the 3 rd order B-M EOS) and, in general, P = 3K 0 f(1+2f) 5/2 * [1 + x 1 f + x 2 f 2 + …]

9. F vs f define a Normalized Pressure: F = P/{3/2 * [(V 0 /V) 7/3 – (V 0 /V) 5/3 ]} (yes, it’s confusing that there is another variable named F) remember (f= ½ [(V 0 /V) 2/3 – 1]) = P/ 3f(1+2f) 5/2 F = K 0 (1 + f(3/2*K 0 '-6)) (3 rd order B-M EOS) *if you plot F vs f you get and equation of a line with a y-intercept of K 0 and a slope of K 0 *(3/2*K 0 '-6) *if K 0 ’ is 4, then the line has a slope of zero positive slope means K’>4 negative slope means K’<4

10. P-V vs F-f plots Ringwoodite Spinel (Mg 0.75,Fe 0.25 ) 2 SiO 4 in 4:1 ME

11. P-V vs F-f plots (cont’d) Ringwoodite Spinel (Mg 0.75,Fe 0.25 ) 2 SiO 4 in 4:1 ME

12. F-f tradeoffs (cont’d) K=168,K’=6.2 K=164,K’=3.9 K=183,K’=3.1

13. Trade-off between K & K’ Ringwoodite Spinel (Mg 0.75,Fe 0.25 ) 2 SiO 4 in 4:1 ME

14. References Birch, F., Finite Strain Isotherm and Velocities for Single-Crystal and Polycrystalline NaCl at High Pressures and 300° K, J. Geophys. Res. 83, 1257 - 1268 (1978). T. Duffy, Lecture Notes, Geology 501, Princeton Univ. W.A. Caldwell, Ph.D. thesis, UC Berkeley 2000

15. (for DAC experiments, which are generally isothermal, we look at the Helmholtz free energy F because its minimization is subject to the condition of constant T or V) F = U-TS => dF = dU – TdS -SdT = (TdS – PdV) –TdS –SdT = -SdT – PdV dF = -SdT – PdV P = -(dF/dV) T also K = -V(dP/dV) Thermodynamics refresher

16. df/dV = d(½ [(V 0 /V) 2/3 – 1])/dV = d(1/2 V 0 2/3 * V -2/3 – 1/2 ) = ½ * -2/3* V 0 2/3 *V -5/3 = -1/3 * V 0 2/3 *V -5/3 = -1/3 * (1/V 0 ) * (V 0 /V) 5/3 = -1/3 * (1/V 0 ) * ((1 + 2f) 3/2 ) 5/3 = -(1+2f) 5/2 /3V 0 nitty gritty f = ½ [(V 0 /V) 2/3 – 1] 2f + 1 = (V 0 /V) 2/3 (2f + 1) 3/2 = V 0 /V