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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) = ?

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## Pressure-Volume Equations of State

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**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/v2 (P+a/v2)*(v-b) = RT (VdW EOS)**Solid (condensed matter) EOS**constant compressibility (F=k*x) DV/V0 = -1/K * DP K = -V0* dP/dV (bulk modulus) Integrate: P = -K*ln(V/V0) => V = V0exp(-P/K) linear compressiblity (Murnaghan EOS), pressure-induced stiffening K = K0 + K’ * P K0 + K’*P = -V0*dP/dV => dP/(K0 + K’P) = -dV/V0 ln(K0 + K’*P)1/K’ = lnV/V0 => (K0 +K’*P)1/K’ = V/V0 V = V0 (K0 +K’*P)1/K’**Condensed Matter EOS (cont’d)**polynomial expansion of K => K = K0 + 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 [(V0/V)2/3 – 1]**B-M EOS**Birch-Murnaghan EOS: expand strain energy in Taylor series: F = a + bf + cf2 + df3 + … look at 2nd order: F = a + bf + cf2 apply boundary conditions: when no strain (f=0) F = 0 (F(0) = 0) => a = 0 F = bf + cf2 From Thermo P = -dF/dV P = -(dF/df)(df/dV) evaulate both parts: (first part)dF/df = b + 2cf; (second part)df/dV = d(½ [(V0/V)2/3 – 1])/dV = -(1+2f)5/2/3V0 combining: P = -(b*df/dV + 2cf*df/dV) soP = b*(1+2f)5/2/3V0+ 2cf*(1+2f)5/2/3V0**2nd order B-M EOS (cont’d)**P = b*(1+2f)5/2/3V0 + 2cf*(1+2f)5/2/3V0 apply boundary conditions: when f = 0 P = 0 (P(0) = 0) this means b = 0 and P = 2cf*(1+2f)5/2/3V0 so what is the constant c? Find out by analytically evaluating K, then apply the boundary condition that when f=0 K = K0 & V = V0 remember K = -V(dP/dV) = -V *dP/df * df/dV = -V * (2c/3V0) [f*5/2*2*(1+2f)3/2 + (1+2f)5/2] * (-(1+2f)5/2/3V0) = 2cV/9V02 * (1+2f)5/2 * [5f(1+2f)3/2 + (1+2f)5/2] evaluate for f = 0 =>K = K0 = 2cV0/9V02 = 2c/9V0 and c = 9V0K0/2 so P = 3K0f(1+2f)5/2**2nd order B-M EOS (cont’d)**P = 3K0f(1+2f)5/2 substitutef= ½ [(V0/V)2/3 – 1] P = 3K0/2 * [(V0/V)2/3 – 1] * (1 + 2* ½ [(V0/V)2/3 – 1])5/2 = 3K0/2 * [(V0/V)2/3 – 1] * (1 +[(V0/V)2/3 – 1])5/2 = 3K0/2 * [(V0/V)2/3 – 1] * ((V0/V)2/3)5/2 = 3K0/2 * [(V0/V)2/3 – 1] * (V0/V)5/3 P = 3K0/2 * [(V0/V)7/3 – (V0/V)5/3](This is 2nd order BM EOS!) K = -V(dP/dV) = K0(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) K0’ = K’(f=0) = 4**3rd order B-M EOS**F = a + bf + cf2 + df3 apply boundary conditions and use derivative relations P = -dF/dV & K = -V(dP/dV) to solve for coefficients (just like in 2nd order B-M EOS) get another term, a lot more algebra & K’ not constrained to 4 P = 3K0/2 * [(V0/V)7/3 – (V0/V)5/3]*[1 + 3/4*(K0'-4) *((V/V0) -2/3 - 1)] = 3K0f(1+2f)5/2 * [1 + 3/4*(K0'-4) *((V/V0) -2/3 - 1)] (this is the 3rd order B-M EOS) and, in general, P = 3K0f(1+2f)5/2 * [1 + x1f + x2f2 + …]**F vs f**define a Normalized Pressure: F = P/{3/2 * [(V0/V)7/3 – (V0/V)5/3]}(yes, it’s confusing that there is another variable named F) remember(f= ½ [(V0/V)2/3 – 1]) = P/ 3f(1+2f)5/2 F = K0(1 + f(3/2*K0'-6)) (3rd order B-M EOS) *if you plot F vs f you get and equation of a line with a y-intercept of K0 and a slope of K0 *(3/2*K0'-6) *if K0’ 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 (Mg0.75,Fe0.25) 2SiO4 in 4:1 ME**P-V vs F-f plots (cont’d)**Ringwoodite Spinel (Mg0.75,Fe0.25) 2SiO4 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 (Mg0.75,Fe0.25) 2SiO4 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, 1257 - 1268 (1978). • T. Duffy, Lecture Notes, Geology 501, Princeton Univ. • W.A. Caldwell, Ph.D. thesis, UC Berkeley 2000**Thermodynamics refresher**(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)**nitty gritty**df/dV = d(½ [(V0/V)2/3 – 1])/dV = d(1/2 V02/3* V-2/3 – 1/2 ) = ½ * -2/3* V02/3*V-5/3 = -1/3 * V02/3*V-5/3 = -1/3 * (1/V0) * (V0/V)5/3 = -1/3 * (1/V0) * ((1 + 2f)3/2) 5/3 = -(1+2f)5/2/3V0 f = ½ [(V0/V)2/3 – 1] 2f + 1 = (V0/V)2/3 (2f + 1)3/2 = V0/V

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