# Presentation: 834878

## 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)*
- V/V0 = -1/K * P
- K = -V0* P/V (bulk modulus)
*Integrate*:  P = -K*ln(V/V0) =>
- 	  	 **V = V****0****exp(-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 = V****0**** (K****0**** +K’*P)****1/K’**

## Condensed Matter EOS (cont’d)

- 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= 1⁄2 [(V__0__/V)__2/3__ – 1]_

## B-M EOS

- _Birch-Murnaghan EOS_:
*expand strain energy in Taylor series*:
- F = a + bf + cf2 + df3 + ...
*look at 2**nd** 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 ](slide15.xml)__P = -dF/dV_
- P = -(dF/df)(df/dV)
*evaulate both parts*:
*(first part)* dF/df = b + 2cf;
*(second part)* df/dV = d(1⁄2 [(V0/V)2/3 – 1])/dV _[=](slide16.xml)__ -(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/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

*apply boundary conditions*:
*when f = 0 P = 0 (P(0) = 0)*
*this means b = 0*
- and _P = 2__c__f*(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/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 = 3K__0__f(1+2f)__5/2_

- 2nd order B-M EOS (cont’d)

## P = 3K0f(1+2f)5/2
substitute f= 1⁄2 [(V0/V)2/3 – 1]
P = 3K0/2 * [(V0/V)2/3 – 1] * (1 + 2* 1⁄2 [(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

*substitute* f= 1⁄2 [(V0/V)2/3 – 1]
- P = 3K0/2 * [(V0/V)2/3 – 1] * (1 + 2* 1⁄2 [(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 = 3K__0__/2 * [(V__0__/V)__7/3__ – (V__0__/V)__5/3__]_          *(**This is 2**nd** 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 2**nd** 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 3**rd** 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= 1⁄2 [(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 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
- (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

- 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)

## df/dV = d(1⁄2 [(V0/V)2/3 – 1])/dV 
	= d(1/2 V02/3* V-2/3 – 1/2 )
	=  1⁄2 * -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

- 	= d(1/2 V02/3* V-2/3 – 1/2 )
- 	=  1⁄2 * -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 = 1⁄2 [(V0/V)2/3 – 1]
- 2f + 1 = (V0/V)2/3
- (2f + 1)3/2 = V0/V