Chapter 1
CrystalâAmbient Phase Equilibrium
1.1Equilibrium of Infinitely Large Phases
The equilibrium between two infinitely large phases Îą and β is determined by the equality of their chemical potentials Οι and Οβ. The latter represent the derivatives of the Gibbs free energies with respect to the number of particles in the system at constant pressure P and temperature T, Îź = (âG/ân)P,T, or, in other words, the work which has to be done in order to change the number of particles in the phase by unity. In the simplest case of a single-component system, we have
The above equation means that the pressures and the temperatures in both phases are equal. The requirement Pι = Pβ = P is equivalent to the condition that the boundary dividing both phases is flat or, in other words, that the phases are infinitely large. This question will be clarified in the next section where the equilibrium of phases with finite sizes will be considered.
Let us assume now that the pressure and the temperature are infinitesimally changed in such a way that the two phases remain in equilibrium, i.e.,
It follows from (1.1) and (1.2) that
Recalling the properties of the Gibbs free energy (dG = VdP â SdT), we can rewrite (1.3) in the form
where sι and sβ are the molar entropies, and vι and vβ are the molar volumes of the two phases in equilibrium with each other.
Rearranging (1.4) gives the well-known Equation of Clapeyron
where Îs = sÎą â sβ, Îv = vÎą â vβ, and Îh = hÎą â hβ is the enthalpy of the corresponding phase transition.
Let us consider first the case when the phase β is one of the condensed phases, say, the liquid phase and the phase Îą is the vapor phase. Then the enthalpy change Îh will be the enthalpy of evaporation Îhev = hv â hl and vl and vv will be the molar volumes of the liquid and the vapor phases, respectively. The enthalpy of evaporation is always positive and the molar volume of the vapor vv is usually much greater than that of liquid vl. In other words, the slope dP/dT will be positive. We can neglect the molar volume of the crystal with respect to that of the vapor and assume that the vapor behaves as an ideal gas, i.e., P = RT/vv. Then Eq. (1.5) attains the form
which is well known as the Equation of ClapeyronâClausius. Replacing Îhev with the enthalpy of sublimation Îhsub, we obtain the equation which describes the crystalâvapor equilibrium.
Assuming Îhev (or Îhsub) does not depend on the temperature, Eq. (1.6) can be easily integrated to
where P0 is the equilibrium pressure at some temperature T0.
In the case of the crystalâmelt equilibrium, the enthalpy Îh is equal to the enthalpy of melting Îhm which is always positive and the equilibrium temperature is the melting point Tm.
Figure 1.1: Phase diagram of a single-component system in PâT coordinates. O and OⲠdenote the triple point and the critical point, respectively. The vapor pressure becomes supersaturated or undercooled with respect to the crystalline phase if one moves along the line AAⲠor AAâł. The liquid phase becomes undercooled with respect to the crystalline phase if one moves along the line BBâł. ÎP and ÎT are the supersaturation and undercooling, respectively.
As a result of the above considerations, we can construct the phase diagram of our single-component system in coordinates P and T (Fig. 1.1). The enthalpy of sublimation of crystals Îhsub is greater than the enthalpy of evaporation Îhev of liquid, and hence the slope of the curve in the phase diagram giving the crystalâvapor equilibrium is greater than the slope of the curve of the liquidâvapor equilibrium. On the other hand, the molar volume vl of the liquid phase is usually greater than that of the crystal phase vc (with some very rare but important exceptions, for example, in the cases of water and bismuth), but the difference is small so that the slope dP/dT is great, in fact, much greater than that of the other two cases and is also positive with the exception of the cases mentioned above. Thus, the PâT space is divided into three parts. The crystal phase is thermodynamically favored at high pressures and low temperatures. The liquid phase is stable at high temperatures and high pressures, and the vapor phase is stable at high temperatures and low pressures. Two phases are in equilibrium along the lines and the three phases are simultaneously in equilibrium at the so-called triple point 0. The liquidâvapor line terminates at the so-called critical point 0Ⲡbeyond which the liquid phase does not exist any more because the surface energy of the liquid becomes equal to zero and the phase boundary between both phases disappears.
1.2Supersaturation
When moving along the dividing lines, the corresponding phases are in equilibrium, i.e., Eq. (1.1) is strictly fulfilled. If the pressure or the temperature is changed in such a way that we deviate from the lines of the phase equilibrium one or another phase becomes stable. This means that its chemical potential becomes smaller than the chemical potentials of the phases in the other regions. Any change of the temperature and/or pressure which lea...