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)

constant compressibility (F=k*x)

V/V0 = -1/K * P

K = -V0* P/V (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’

Solid (condensed matter) EOS

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]

Condensed Matter EOS (cont’d)

B-M EOSBirch-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)

so P = b*(1+2f)5/2/3V0 + 2cf*(1+2f)5/2/3V0

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/9V0

2 * (1+2f)5/2 * [5f(1+2f)3/2 + (1+2f)5/2]evaluate for f = 0 => K = K0 = 2cV0/9V0

2 = 2c/9V0

and c = 9V0K0/2so P = 3K0f(1+2f)5/2

2nd order B-M EOS (cont’d)

P = 3K0f(1+2f)5/2

substitute f= 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

2nd order B-M EOS (cont’d)

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)

Gross PV 300K

1480

1500

1520

1540

1560

1580

1600

1620

1640

1660

1680

0 5 10 15 20 25Pressure (GPa)

V (

ang

stro

m^

3)

OlijynkJGR

Olijynk EOS

DWALSJan04

APSEOS

JiangEOS

APSPostHeat

APSPreHeat

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

(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 [(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

nitty gritty

f = ½ [(V0/V)2/3 – 1]2f + 1 = (V0/V)2/3

(2f + 1)3/2 = V0/V