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?

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)

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

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)

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

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)

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)

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 + …]

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

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

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

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

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

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

(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

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