Geochemistry. William M. White

Чтение книги онлайн.

Читать онлайн книгу Geochemistry - William M. White страница 57

Geochemistry - William M. White

Скачать книгу

2.8.) For liquids and particularly gases, the effects of pressure and temperature on ΔV are significant and cannot be ignored. The reference pressure is generally 0.1 MPa. For solids, however, we can often ignore the effects of temperature and pressure on ΔV so the first integral reduces to: images (see Example 2.9). On the other hand, we cannot ignore the temperature dependence of entropy. Hence we need to express ΔSr as a function of temperature. The temperature dependence of entropy is given by eqn. 2.105. Writing this in integral form, we have:

equation

      This is the change in entropy due to increasing the temperature from the reference state to T. The full change in entropy of reaction is then this plus the entropy change at the reference temperature:

      (2.131)equation

      Substituting this into 2.130, the second integral becomes:

      images, as we have defined it here, is the change in free energy of reaction as a result of increasing temperature from the reference state to T′.

      Using the thermodynamic reaction and data as in Example 2.7:

equation equation

      determine the pressure at which these two assemblages will be in equilibrium at 1000°C. Assume that the volume change of the reaction is independent of pressure and temperature (i.e., α and β = 0).

      Answer: These two assemblages will be in equilibrium if and only if the Gibbs free energy of reaction is 0. Mathematically, our problem is to solve eqn. 2.130 for P such that images.

      Our first step is to find ΔGr for this reaction at 1000°C (1273 K) using eqn. 2.130. Heat capacity data in Table 2.2 is in the form: images. Substituting for images, we have:

      (2.133)equation

      Performing the double integral and collecting terms, and letting images, this becomes:

      equation 2.134 is a general solution to eqn. 2.130 when the Maier-Kelley heat capacity is used.

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

      (2.135)equation

      Equation 2.130 may now be written as:

equation

      Let images. images (calculated from values in Table 2.2), so images. images 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

      With images, 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.

      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

Скачать книгу