Geochemistry. William M. White

we may assume the phases are incompressible, the solution to the pressure integral is:

      (2.135)

      Equation 2.130 may now be written as:

      Let . (calculated from values in Table 2.2), so . is positive, meaning that the left side of the reaction is favored at 1000°C and atmospheric pressure, consistent with our prediction based on ∂G/∂T.

      Solving for pressure, we have

(2.136)

      With , we obtain a value of 1.49 GPa (14.9 kbar). Thus, assemblages on the right and left will be in equilibrium at 1.49 GPa and 1000°C. Below that pressure, the left is stable, and above that pressure, the right side is the stable assemblage, according to our calculation.

      The transformation from “plagioclase peridotite” to “spinel peridotite” actually occurs around 1.0 GPa in the mantle. The difference between our result and the real world primarily reflects differences in mineral composition: mantle forsterite, enstatite and diopside are solid solutions containing Fe and other elements. The difference does not reflect our assumption that the volume change is independent of pressure. When available data for pressure and temperature dependence of the volume change are included in the solution, the pressure obtained is only marginally different: 1.54 GPa.

Example 2.9 Volume and free energy changes for finite compressibility

      The compressibility (β) of forsterite (Mg₂SiO₄) is 8.33 × 10^–6 MPa⁻¹. Using this and the data given in Table 2.2, what is the change in molar volume and Gibbs free energy of forsterite at 100 MPa and 298 K?

      Answer: Let's deal with volume first. We want to know how the molar volume (43.79 cc/mol) changes as the pressure increases from the reference value (0.1 MPa) to 1 GPa. The compressibility is defined as:

(2.12)

      So the change in volume for an incremental increase in pressure is given by:

      (2.137)

      To find the change in volume over a finite pressure interval, we rearrange and integrate:

Performing the integral, we have:

      (2.138)

      This may be rewritten as:

(2.139)

      However, the value of P−P^o is of the order of 10^–2, and in this case, the approximation holds, so that eqn. 2.139 may be written as:

(2.140)

      equation 2.140 is a general expression that expresses volume as a function of pressure when β is known, small, and is independent of temperature and pressure. Furthermore, in situations where P > P^o, this can be simplified to:

(2.141)

      Using equation 2.141, we calculate a molar volume of 43.54 cc/mol (identical to the value obtained using eqn. 2.139). The volume change, ΔV, is 0.04 cc/mol.

      The change in free energy with volume is given by:

      so that the free energy change as a consequence of a finite change is pressure can be obtained by integrating:

      Into this we may substitute eqn. 2.141:

(2.142)

      Using eqn. 2.142 we calculate a value of ΔG of 4.37 kJ/mol.

2.12 THE MAXWELL RELATIONS*

The reciprocity* relationship, which we discussed earlier, leads to a number of useful relationships. These relationships are known as the Maxwell relations. Consider the equation:

(2.58)

      If we write the partial differential of U in terms of S and V we have:

      (2.143)

      From a comparison of these two equations, we see that:

      (2.144a)

and

      (2.144b)

      And since the cross-differentials are equal, it follows that:

      (2.145)

      The other Maxwell relations can be derived in an exactly analogous way from other state functions. They are:

       From dH (eqn 2.65):(2.146)

       from dA (eqn. 2.121)(2.147)

       from dG (