Phase Equilibria

Phase diagrams

A phase diagram is a representation of the states of matter, solid, liquid, or gas as a function of temperature and pressure. In the Figure shown below the regions of space indicate the three phases of carbon dioxide. The curved lines separating the regions of space indicate the pressures and temperatures where phases can coexist. At all points along the coexistence curves two phases are in equilibrium with one another. If we move off of the coexistence line then one phase or another is favored and will be the only phase present at equilibrium. Note that the three lines in the phase diagram intersect at a point where solid, liquid, and gas all coexist. This unique point is known as the triple point.

Within any one of the single-phase regions both temperature and pressure must be specified. Because two thermodynamic variables can be changed independently we say that the system has two degrees of freedom. Along any of the coexistence curves the pressure and temperature are coupled, i.e. any change in the temperature implies a change in pressure to remain on the line. Thus, along the curves there is only one degree of freedom. The triple point is a unique point in phase space and there is only one set of values of pressure and temperature consistent with the triple point. Thus, we say that at the triple point the system has zero degrees of freedom. If we follow the liquid-vapor coexistence curve towards higher temperature we find that it ends at the critical point. Above the critical point there is no distinction between liquid and vapor and we can think of the state of matter as a single fluid phase. In fact, as the critical point is approached the difference between liquid and vapor gradually becomes less distinct. A fluid near the critical point exhibits large fluctuations between the liquid and vapor states.

The various types of phase transitions are labeled on the Figure shown.

The chemical potential

In a system where two phases (e.g. liquid and gas) are in equilibrium the Gibbs energy is G = G^l + G^g, where G^l and G^g are the Gibbs energies of the liquid phase and the gas phase, respectively. If dn modes (a differential amount of n the number of moles) are transferred from one phase to another at constant temperature and pressure, the differential Gibbs energy for the process is

The number of moles transferred from the gas phase -dn^g is equal to the number transferred to the liquid phase dn^l. Since dn^l = -dn^g

The partial derivatives of the Gibbs energy with respect to number of moles are called chemical potential. The chemical potentials are

In terms of chemical potential, the Gibbs energy for the phase equilibrium is

Since the two phases are in equilibrium dG = 0 and since dn^g ¹ 0 we have m^g = m^l. If two phases of a single substance are in equilibrium their chemical potentials are equal.

If the two phases are not in equilibrium a spontaneous transfer of matter from one phase to the other will occur in the direction that dG < 0. Matter is transferred from a phase with higher chemical potential to a phase with lower chemical potential consistent with the negative sign of Gibb's free energy for a spontaneous process.

For a single substance G = nm. The Gibb's energy is an extensive quantity and the chemical potential, m is an intensive quantity.

The solid-liquid coexistence curve

To derive expressions for the coexistence curves on the phase diagram we use the fact that the chemical potential is equivalent in the two phases. We consider two phases a and b and write

m^a(T,P) = m^b(T,P)

Now we take the total derivative of both sides

The appearance of this equation is quite different from previous equations and yet you have seen this equation before. The reason for the apparent difference is the symbol m. Remember that m for a single substance is just the molar free energy.

Substituting these factors into the total derivative above we have

Solving for dP/dT gives

This equation is known as the Clapeyron equation. It gives the two-phase boundary curve in a phase diagram with D_trsH and D_trsV between them. The Clapeyron equation can be used to determine the solid-liquid curve by integration.

Starting with a known point along the curve (e.g. the triple point or the melting temperature at one bar) we can calculate the rest of the curve referenced to this point.

The liquid-vapor and solid-vapor coexistence curves

The Clapeyron equation cannot be applied to a phase transition to the gas phase since the molar volume in the gas phase is a function of the pressure. Making the assumption that Vg >> Vl we can use the ideal gas law to obtain a new expression for dP/dT.

The integrated form of this equation

yields the Clausius-Clapeyron equation.

If we use DH of evaporation this equation can be used to describe the liquid-vapor coexistence curve and if we use DH of sublimation this equation can be used to describe the solid-vapor curve.

We can use these equations to calculate a phase diagram. For example, we can begin with the CO₂ diagram shown above. The triple point for CO₂ is 5.11 atm and 216.15 K. The critical point for for CO₂ is 72.85 atm and 304.2 K. We also have the following data

Transition	DtrsH^o (kJ/mol)	Ttrs (K)
Fusion	8.33	217.0
Sublimation	25.23	194.6

Note that we can calculate the enthalpy of sublimation from

D_vapH^o = D_subH^o - D_fusH^o = 16.9 kJ/mol.

The densities of the solid and liquid are 1.53 g/cm³ and 0.78 g/cm³, respectively. The density r = m/V = nM/V so the molar volume is V_m = V/n = M/r where M is the molar mass. In units of L/mole we have

V^s_m = 44 g/mole/[1530 g/L] = 0.0287

V^l_m = 44 g/mole/[780 g/L] = 0.0564

D_fusV = V^l_m - V^s_m = 0.0564 - 0.0287 = 0.0277 L/mole

Starting with the triple point we use the Clausius-Clapeyron equation to calculate the liquid-vapor and solid-vapor coexistence curves.

Starting at the triple point

P = 5.11exp{D_vapH/R[T – 216.15]/216.15T}

P = 5.11exp{2,032[T – 216.15]/216.15T}

Notice that if we were to calculate the critical pressure using this formula we would obtain 77.3 atm which is about 5 atm larger than the experimental number. There are several sources of inaccuracy including mainly our neglect of the temperature dependence of the enthalpy. We can also begin a the critical point

P = 72.8 exp{D_vapH/R[T – 304.2]/304.2T}

P = 72.8 exp{2,032[T – 304.2]/304.2T}

Liquid-vapor

P (atm)	T (K)
5.11	216.15
6.03	220
9.0	230
13.0	240
24.9	260
41.0	280
63.7	300
72.8	304

Starting at the triple point

P = 5.11exp{D_vapH/R[T – 216.15]/216.15T}

P = 5.11exp{3,034[T – 216.15]/216.15T}

Solid-vapor

P (atm)	T (K)
0.111	170
0.298	180
0.725	190
1.64	200
3.38	210
5.11	216.15

Using the Clapeyron equation we calculate the solid-liquid coexistence curve

P = 5.11 + [D_fusH/D_fusV] ln{T/216.15}

P = 5.11 + [8,330 J/mole/0.0277L/mole/101.32 J/L-atm] ln{T/216.15}

P = 5.11 + 2,967 ln{T/216.15}

Solid-liquid

P (atm)	T (K)
5.11	216.15
5.24	216.16
5.39	216.17
8.57	216.4
11.34	216.6
14.1	216.8
16.9	217
30.6	218
44.3	219
58.0	220
124.2	225
189.3	225
124.2	230
253	235
319	240
559	260
780	280
987	300
1180	320

The phase diagram has the following appearance plotted on a log pressure scale.

The significance of vapor pressure

At each point along the liquid-vapor coexistence curve the pressure of phase transition is equal to the vapor pressure of the liquid. For example, the normal boiling point of water is 1 atm. This means that water boils at 373 K because its vapor pressure is equal to the atmospheric pressure. At temperatures below 373 K the vapor pressure of water is less than atmospheric pressure and it does not boil. However, water will boil if it is placed in container at reduced pressure. Conversely, in a pressure cooker, the pressure is increased requiring water to boil at a higher temperature.

The Clausius-Clapeyron equation can be used to calculate the boiling temperature of water at high elevation where the pressure is lower than atmospheric pressure. Recall that the barometric pressure formula allows one to estimate the pressure at a given elevation above sea level.

The Clapeyron and Clausius-Clapeyron equations were derived with the assumption that the enthalpies of phase transition are not a function of temperature. We have seen that this is not so. The indefinite integral of the C-C equation gives

Which implies that ln(P) can be plotted as a straight line vs. 1/T. In reality because of the temperature dependence of the enthalpy of transition this is not the case. We can use a function form for D_trsH prior to integration of the above equations to obtain more accurate results.