Recall equation 4.39, the general equation for Gibbs energy change expressed as a function of T and p.

This equation includes the assumption that no variables other than temperature and pressure are varied.  In the case of phase changes, we do have moles of the two (or more) phases changing so a more general equation might be

The latter two terms in the equation would also be used in chemical reactions, one such term for every reactant or product.  We now define the chemical potential as the rate of change of Gibbs energy with a change in number of particles so,

In the case of one mole of a pure substance, we can substitute m directly for G_m.  Just like Gibbs energies, a spontaneous process will result in a lower chemical potential.

We are now ready to explore the phase diagrams.

We recall from our discussion leading to the Maxwell relation (equation 4.40) that the rate of change of Gibbs energy with temperature is simply the negative of the entropy.  For a pure substance, we can substitute chemical potential for Molar Gibbs energy so

From this equation we can see that as the temperature goes up, the chemical potential decreases (S is always positive).

This Figure shows the change in chemical potential for a system that is being warmed up from the solid state through the liquid state and finally into the gaseous state.  Note that the slope is steeper for the gas than for the liquid, which is steeper than for the solid.  This comes from the fact that the molar entropy is highest for the gas and lowest for the solid. 

Now follow the diagram from the solid phase into the liquid phase region.  If the substance doesn't melt at T_f, it's chemical potential would follow the light colored line.  However, as long as there is no kinetic barrier to the transition, there is a different state (liquid) available to it in this region and the solid will spontaneously change to the liquid, thus lowering its chemical potential. 

We can do the same logic for a phase change in the other direction a liquid is cooled below  T_f, without freezing.  It's chemical potential is clearly higher than that for the solid phase so the spontaneous phase change in the region below T_f, is from liquid to solid. 

We can use this logic for all phase change processes.  Consider a solid that is heated so much that it is now above T_b.  There are two phases (liquid and gas) available to it that have a lower chemical potential.  The liquid is only metastable at best so we see that the transition directly to the gas phase is the most likely transition. 

We can also explore the response of the system to pressure changes via the equation 4.40, making the same substitutions as we did to get to equation 5.5, we arrive at

	This shows that a plot of m versus p yields a slope equal to the molar volume.  Since molar volumes are always positive, the slope will be positive and steepest for gases. The inset phase diagram shows the path followed (constant T). If we trace the chemical potential as we increase p at this temperature we see that the system passes through gas, liquid and finally solid states at the chosen temperature
	This chosen temperature is lower than the triple point.  At this temperature, the chemical potential of the liquid is always higher than that of the solid or the gas so liquid is never thermodynamically stable.
	As the temperature is set at the triple point temperature, we see that all three phases have the same chemical potential when pressure is also at the triple point.
	In this graph, the temperature is set above the critical temperature.  This graph of a supercritical fluid shows that the slope shifts from gas-like slopes to liquid like slopes smoothly, there is no disconnect in the slope, i.e., no equilibrium point with two phases

We know that most materials experience an elevation in melting point when there is an increase in external pressure.  In other words, as we squeeze on a substance, it will tend to freeze at a higher temperature.  This works for most materials except water, where the reverse trend happens.  This is due to the molar volume difference between the liquid and solid phases.  Generally, the liquid has a larger molar volume.  In water, the solid has the larger molar volume (less dense) compared to the liquid.  Figure 4.10 (From Atkins) illustrates the effect this has on the chemical potential versus temperature .

As the system experiences an increase in pressure, the chemical potential will change, depending on the molar volume.  Thus, in (a), the molar volume of the liquid is greater than that of the solid and hence, the increase in molar volume is greater for the liquid than for the solid.  This results in an increased melting point.

In (b), the solid's chemical potential increases more, resulting in a lower melting point.

Example:  calculate the change in chemical potential for ice and for water at 0ºC as the pressure is changed from 1.00 bar to 2.00 bar.  the density of ice is 0.917 g/cm³ and of water is 0.999 g/cm³.

Using equation 5.6 we can quickly integrate to get Δm = V_m×Δp .  The molar volume is calculated from the density and the molar mass, thus V_m = M/r and thus, Δm = M/r ×Δp

Thus, if the system was at equilibrium at 1.00 bar then it is no longer at equilibrium at 2.00 bar and the spontaneous direction is to change the ice into water, so as to lower the overall chemical potential.

Notice that as the pressure changes, the liquid's chemical potential changes.  This implies that the liquid's vapour pressure will change when external pressure is applied.  For example, if a closed system containing a liquid and vapour at equilibrium is pressurized by adding an inert gas, then the extra pressure exerted on the liquid will change it's chemical potential, as we calculated above but the gas (assuming ideal) pressure will not change.  Hence, a system that was formerly at equilibrium will no longer be and the vapour pressure will increase.  This goes against assumptions made in first year that the vapour pressure depended only on the temperature of the liquid.

Following the phase boundaries

The entire system at equilibrium must have uniform chemical potential throughout.  Otherwise, there would be a flow of particles from region of high potential to low potential until equilibrium is reached.  Thus, if two phases are in equilibrium as depicted here, along the phase transition line, then both phases have the same chemical potential.  If a pressure is applied, which shifts the system out of equilibrium then the temperature will change (as a result of some particles migrating from one phase to the other) until equilibrium is re-established.

The slope of the phase boundary is dp/dT.  We can change equation 4.39 by substituting chemical potential for molar Gibbs energy (along with molar other parameter too) to get

Since phase α and phase β must have equal chemical potentials at equilibrium, we can write

The Clapeyron equation is an exact equation describing the slope of any phase change boundary.

Solid-liquid boundary

If the two phases in question are the solid phase and the liquid phase then the transition in question will be fusion.  We can rewrite our Clapeyron equation specific to this equilibrium.  in this case, the molar entropy of fusion is the same as the molar enthalpy divided by the transition temperature.

since the enthalpy of melting is positive, we see that the slope of this transition will depend on the relative molar volumes of the solid and liquid  If the molar volume of the solid is less than that of the liquid then the slope of the line will be positive.  This is the most common case.  If the molar volume of the solid is greater than the molar volume of the liquid, as in water, then the slope of this phase boundary will be negative.

If we wish to determine a macroscopic change, we need to integrate:

Now, by assuming the enthalpy changes and volume changes are negligible, we simplify to

Finally, if we are taking relatively small steps in temperature we can assume and so

This is the equation of a straight line with a very steep slope.

Liquid-vapour boundary

If the two phases in question are the liquid phase and the vapour phase then the transition in question will be vaporization.  We can rewrite our Clapeyron equation specific to this equilibrium.  As in the solid/liquid equilibrium, the molar entropy of vaporization is the same as the molar enthalpy divided by the transition temperature.

This time, we have a situation where the molar volume of the liquid phase is very small compared to the molar volume of the gas.  So, we will assume that the change in volume is simply the molar volume of the gas   If the gas behaves ideally, then V_m = RT/p. to give us

This is the the Clausius-Clapeyron equation, which after integration looks a bit more familiar

Example:  What is the slope of the solid/liquid curve at the normal melting point for water?  Δ_fusH ⁰ = 6010 J/mol.

Use equation 5.8. Since this we're sitting on the equilibrium line, there is no change in internal energy so ΔU ⁰ = 0, hence,   Δ_fusS ⁰  = Δ_fusH ⁰/T

Note the very large value for the slope (very steep) and it's negative since the change in volume as water goes from solid to liquid is negative.

Order of phase transitions

First-order phase transitions

Consider two phases , α and β, in equilibrium.  Most phase transitions involve changes in enthalpy and in volume.  These changes figure into the differences in the slope of the chemical potential curve on either side of the transition point.

In words, the difference in the slope of chemical potential versus pressure is simply the difference in the molar volumes of the two phases.

The difference in the slopes of the chemical potential versus temperature is the difference in entropy or the difference in enthalpy over temperature for the transition in question.  The slope is different on either side of the transition point.  It is discontinuous at the transition.

This is a first-order phase transition.  Remember that at the transition point, we are changing the enthalpy of the system but not its temperature.  Thus, the heat capacity C_p at the transition point is infinite.  In other words, as we heat the system that is at the transition point, no temperature change happens because all the heat is going into the phase transition.

Fig. 4.16  (from Atkins) The changes in thermodynamic properties accompanying (a) first-order and (b) second-order phase transitions.

Second-order Phase transitions

In a second-order phase transition, the first derivative of the slope of m is not discontinuous but it's second derivative is.  In other words, there is an inflection point at the phase transition.   This type of transitions occurs between conducting and superconducting phases of metals at low temperatures.

l transitions

These transitions are not first order yet their heat capacity goes to infinity at the transition point.  We saw such a transition between the two different liquid phases of helium at very low temperature.  These transitions tend to include order/disorder transitions, paramagnet/ferromagnetic transitions and the fluid/superfluid transition of He.