1 / 16

Pressure-Volume Equations of State

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) = ?

milica
Download Presentation

Pressure-Volume Equations of State

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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/v2 (P+a/v2)*(v-b) = RT (VdW EOS)

  3. 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’

  4. 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]

  5. 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

  6. 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

  7. 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

  8. 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 + …]

  9. 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

  10. P-V vs F-f plots Ringwoodite Spinel (Mg0.75,Fe0.25) 2SiO4 in 4:1 ME

  11. P-V vs F-f plots (cont’d) Ringwoodite Spinel (Mg0.75,Fe0.25) 2SiO4 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 (Mg0.75,Fe0.25) 2SiO4 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. 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)

  16. 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

More Related