(i.e., if we know the temperature at which these three phases are in equilibrium, the pressure is also fixed).
Rearranging eqn. 3.2, we also can determine the maximum number of phases that can coexist at equilibrium in any system. The degrees of freedom cannot be less than zero, so for an invariant, one-component system, a maximum of three phases can coexist at equilibrium. In a univariant one-component system, only two phases can coexist. Thus, sillimanite and kyanite can coexist over a range of temperatures, as can kyanite and andalusite, but the three phases of Al2SiO5 coexist only at one unique temperature and pressure.
Let's consider the example of the three-component system Al2O3–H2O–SiO2 in Figure 3.2. Although many phases are possible in this system, for any given composition of the system only three phases can coexist at equilibrium over a range of temperature and pressure. Four phases (e.g, a, k, s, and p) can coexist only along a one-dimensional line or curve in P–T space. Such points are called univariant lines (or curves). Five phases can coexist at invariant points at which both temperature and pressure are uniquely fixed. Turning this around, if we found a metamorphic rock whose composition fell within the Al2O3–H2O–SiO2 system, and if the rock contained five phases, it would be possible to determine uniquely the temperature and pressure at which the rock equilibrated.
3.2.3 The Clapeyron equation
A common problem in geochemistry is to know how a phase boundary varies in P–T space, for example, how a melting temperature will vary with pressure. At a phase boundary, two phases must be in equilibrium, so ΔG must be 0 for the reaction Phase 1 ⇌ Phase 2. The phase boundary therefore describes the condition:
Thus, the slope of a phase boundary on a temperature-pressure diagram is:
(3.3)
where ΔVr and ΔSr are the volume and entropy changes associated with the reaction. Equation 3.3 is known as the Clausius–Clapeyron equation, or simply the Clapeyron equation. Because ΔVr and ΔSr are functions of temperature and pressure, this is, of course, only an instantaneous slope. For many reactions, however, particularly those involving only solids, the temperature and pressure dependencies of ΔVr and ΔSr will be small and the Clapeyron slope will be relatively constant over a large T and P range (see Example 3.1).
Because ΔS = ΔH/T, the Clapeyron equation may be equivalently written as:
(3.4)
Slopes of phase boundaries in P–T space are generally positive, implying that the phases with the largest volumes also generally have the largest entropies (for reasons that become clear from a statistical mechanical treatment). This is particularly true of solid–liquid phase boundaries, although there is one very important exception: water. How do we determine the pressure and temperature dependence of ΔVr and why is ΔVr relatively T- and P-independent in solids?
We should emphasize that application of the Clapeyron equation is not limited to reactions between two phases in a one-component system but may be applied to any univariant reaction.
Example 3.1 The graphite–diamond transition
At 25°C the graphite–diamond transition occurs at 1600 MPa (megapascals, 1 MPa =10 b). Using the standard state (298 K, 0.1 MPa) data below, predict the pressure at which the transformation occurs when temperature is 1000°C.
| Graphite | Diamond | |
| α (K–1) | 1.05 × 10−05 | 7.50 × 10−06 |
| β (MPa−1) | 3.08 × 10−05 | 2.27 × 10−06 |
| S° (J/K-mol) | 5.74 | 2.38 |
| V (cm3/mol) | 5.2982 | 3.417 |
Answer: We can use the Clapeyron equation to determine the slope of the phase boundary. Then, assuming that ΔS and ΔV are independent of temperature, we can extrapolate this slope to 1000°C to find the pressure of the phase transition at that temperature.
First, we calculate the volumes of graphite and diamond at 1600 MPa as (eqn. 2.140):
(3.5)
where ΔP is the difference between the pressure of interest (1600 MPa in this case) and the reference pressure (0.1 MPa). Doing so, we find the molar volumes to be 5.037 for graphite and 3.405 for diamond, so ΔVr is −1.6325 cc/mol. The next step will be to calculate ΔS at 1600 MPa. The pressure dependence of entropy is given by equation 2.106: (∂S/∂P)T = −αV. Thus, to determine the effect of pressure we integrate:
(3.6)
(We use Sp to indicate the entropy at the pressure of interest and S° the entropy at the reference pressure.) We need to express V as a function of pressure, so we substitute eqn. 3.5 into 3.6:
(3.7)
The reference pressure, Pref, is negligible compared with P1 (0.1 MPa vs 1600 MPa), so that this simplifies to:
For graphite, Sp is 5.66 J/K-mol, for diamond it is 2.34 J/K-mol, so ΔSr at 1600 MPa is −3.32 J-K−1-mol−1.
The Clapeyron slope is therefore:
One distinct advantage of the SI units is that cm3 = J/MPa, so the above units are equivalent to K/MPa. From this, the pressure of the phase change at 1000°C can be calculated as:
The Clapeyron slope we calculated (solid line) is compared with the experimentally determined phase boundary in