# 8. Phase Transitions

When chemical reactions occur, the system makes transition among multiple minima at the molecular level. The figure below illustrates the molecular connection between free energy $$G$$, its derivative $$\Delta G$$, and the free energy as a function of reaction coordinate, $$G(x)$$.

Figure 8.1: The free energy $$G(x)$$ as a function of coordinate (e.g. bond distance) has two local minima, A’ being lower in free energy.  When A converts to A’, $$\Delta G$$ kJ/mole are released per infinitesimal amount of reaction $$d\xi$$. Note that both x and $$\xi$$ can vary between 0 and 1, but the meaning is very different: when all of the substance is in state A, $$x=\xi=0$$, and when it is all in state A', $$x=\xi=1$$.  However, while the substance can have $$\xi=0.5$$ when the reaction is half completed, almost none of it will ever be at x=0.5.  Rather, half will be at x=1 and half at x=1.

So far, we have treated macroscopic pure substances and mixtures as though they had a single minimum in the free energy as a function of reaction coordinate. However, thermodynamics does not forbid multiple minima even at the macroscopic level, and can be used to make comparative statements about the minima.

Definition: Phase

A phase is a local minimum in the free energy surface.

Unlike ordinary chemical reactions, transition between phases can occur even when only one pure substance is present:

$A^{(1)} \rightarrow A^{(2)}$

For phase transitions, we call the reaction coordinate “order parameter.” The superscripts refer to the phases.

Definition: Order Parameter

An order parameter is a thermodynamic variable scaled to zero in one phase, nonzero in an(other) phase.

Example 8.1

Example:  a gas-liquid transition order parameter

$O = \rho - \rho_{gas}$

or

$O = \dfrac{ \rho - \rho_{gas}}{ \rho_{liq} - \rho_{gas}}$

in general

$O = X - X^{(2)}$

or

$O = \dfrac{ X - X^{(2)}}{ X^{(2)} - X^{(1)}}$

Thermodynamics cannot make statements about the details of the barrier (e.g. its height), or how fast a transition can occur. The transition itself is a rather delicate matter – it violates P2 since temporarily $$\Delta G >0$$ if the transition occurs at constant $$T$$ and $$P$$.

The solution to this dilemma: if climbing the barrier were required of the entire macroscopic system, phase transitions could indeed never occur. Rather, a small portion of phase (1), called a nucleus, fluctuates to look like phase (2). The nucleus is at the barrier top in fig. 8.1. From this nucleus, phase (2) grows downhill in chemical potential if it is at lower free energy (P2). Thus the transition itself relies on microscopic fluctuations, and microscopic information is required to determine the barrier height, which is rather small.  We need statistical mechanics to compute rates.

If we are interested only in equilibrium, not how we get there, we can treat the phase transition like any other chemical reaction:  $$A^{(1)}$$ and $$A^{(2)}$$ interconvert to yield mole numbers $$n^{(1)}_{eq}$$ & $$n^{(2)}_{eq}$$ or concentrations or pressures that minimize $$G$$:

$A{(1)} \rightarrow A^{(2)}$

$G(T,P,n^{(i)})=\mu^{(1)}n^{(1)} +\mu^{(2)}n^{(2)} = \mu^{(1)}n^{(1)} + \mu^{(2)}(n-n^{(1)})$

where $$n$$ is a constant. At equilibrium

$dG = 0$

$\mu^{(1)} n^{(1)} +\mu^{(2)} n^{(2)} = (\mu^{(1)}-\mu^{(2)}) dn^{(2)}$

$\mu^{(1)} = \mu^{(2)}$

Fig. 8.2: Chemical potential of two phases as a function of temperature and pressure (Gibbs ensemble). At high $$T$$, phase 1 is more stable, at (3)-(5) both phases coexist, at low $$T$$ phase 2 is more stable. In this diagram at high T and P, the two chemical potentials become degenerate and only one phase exists at (6).

Definition: First Order Phase Transition
A 1st order phase transition occurs when the chemical potential difference DG between two phases separated by a barrier vanishes.
Definition: Critical Phase Transition

A critical phase transition occurs when the chemical potential barrier between two phases just vanishes.

### Important Properties of 1st Order Phase transitions

The figure below show diagrams of the chemical potentials at the local minima, and of the resulting phase diagram:

Fig. 8.3 Phase diagram showing the nature of the chemical potential at various points.  It is a double well for first order phase transitions, and merges into a single quartic well at the critical point (6).

The properties below are illustrated by points in figure 6.

#### i) Cooperativity: (2), (3), (4)

Although 1st order phase transitions can be thought of as ‘chemical reactions’ they are far more cooperative than chemical reactions. The reason is that the “order parameter” reaction coordinate corresponds to many degrees of freedom changing in concert, not just a few bonds. Once a nucleus has formed and the top of the barrier is reached, all the molecules in the less stable phase ‘react’ by adding to the nucleus. Although $$\Delta \mu^{(24)}$$ per degree of freedom may be very small, $$N \Delta \mu^{(24)}$$where $$N \sim O(10^{20})$$ may  be very large, so equilibrium switches rapidly.

#### ii) Coexistence curve: (3), (5), (6)

Everywhere along the coexistence curve, $$\mu^{(1)} = \mu^{(2)} \Rightarrow d\mu^{(1)} = d\mu^{(2)}$$ as shown in chapter 4 and again above. From the Gibbs-Duhem equation

$\Rightarrow -s^{(1)} dT + v^{(1)}dP = -s^{(2)} dT + v^{(2)}dP$

or

$\left( \dfrac{dP}{dT} \right)_{\mu} = \dfrac{s^{(1)}-s^{(2)}}{v^{(1)}-v^{(2)}} = \dfrac{\Delta s^{(12)}}{\Delta v^{(12)}} = \dfrac{\Delta h^{(12)}}{T \Delta v^{(12)}}$

#### iii) Discontinuity of thermodynamic variables in 1st order phase transitions: (2),(3),(4)

From the plot of µs, derivatives of the chemical potentials are clearly discontinuous across the coexistence curve, according to ii).

$\left( \dfrac{\partial \mu}{\partial T} \right)_P |_{\text{side 1}} - \left( \dfrac{\partial \mu}{\partial T} \right)_P |_{\text{side 2}} = \Delta \left( \dfrac{\partial \mu}{\partial T} \right)_P = \Delta s_{(12)}$

and

$\Delta \left( \dfrac{\partial \mu}{\partial P} \right)_T = \Delta v^{(12)}$

If $$\Delta s^{(12)} = \Delta h^{(12)}/T$$ is discontinuous, so is $$\Delta h^{(12)}$$. To account for the latent heat $$\Delta h^{(12)}$$ released during the phase transition, $$c_p$$ must have a $$\delta$$-function type discontinuity. In reality any sample of substance is always finite in size and not perfectly pure, so $$c_p$$ just has a very narrow sharp spike:

$\int _ {T_o}^T c_p \, dT = H-H_0$

Fig. 8.4: relationship between heat capacity and enthalpy.  At the transition temperature where $$\mu_1=\mu_2$$, heat is released while T remains constant. Thus the heat capacity has a sharp spike at the transition temperature.

#### iv) Number of coexisting phases: (1) vs. (3)

Usually, only two phases coexist. However, more phases can coexist (have the same chemical potential at the same T, P). There is however a limit. The maximum number of coexisting phases is given by Gibbs’ phase rule:

$f = r-m +2$

with

• $$f$$ is the number of independent variables,
• $$r$$ is the number of components=substances,
• $$m$$ is the number of phases

Proof

$$f$$ = total number of variables – number of constraints

$$T$$,$$P$$ and $$r$$ of the $$\chi_q^{(1)}$$ in each of m phases = 2 +$$rm$$

$$\mu_{i=1L}^{(1)} =\mu_i^{(2)} = ... = \mu_i^{(m)},$$ and $$\sum_{q=1}^r \chi_q^{(i)} =1$$

$-r(m-1) \;\;\;\; - \;\;\;\;m$

Example
For a one component system, the maximum value of $$m$$ is achieved when $$f = 0$$: $$0=1-m+2 \Rightarrow m=3$$; at most three phases can coexist, and it occurs at a single point in the phase diagram, since $$f=0$$. Hence the name ‘triple point’ point for point (1) in figure 8.3. Two phases ($$m = 2$$) leave $$f = 1$$ and coexist on a line, for example $$P(T)$$ in figure 8.3. One phase leaves $$f = 2$$ independent variables, so phases of single component systems generally occupy two-dimensional patches on the $$P(T)$$ phase diagram.
Example

For $$r = 2$$ substances and $$m = 2$$ phases, $$f=2-2+2=2$$. The phase transition reaction would be

$A_1^{(1)} + A_2^{(1)} A_1^{(2)} + A_2^{(2)}$

where subscripts refer to different substances (e.g. $$A_1$$ = methanol, $$A_2$$ = water) and superscripts refer to different phases (e.g. (1)=liquid, (2)=gas). There are now two degrees of freedom, so instead of a coexistence line $$P(T)$$, there is a coexistence surface. Picking $$T,P,\chi_1$$) as variable, $$P=P(T,\chi_1)$$. $$\chi_2 = 1-\chi_1$$ is not linearly independent of $$\chi_1$$). A sketch of the coexistence surface and a 2-D cut through it is shown below.

Fig. 8.5: Liquid-gas coexistence surface (shaded) when there are two components and two phases. (a) is the boiling point of pure substance “$$A_2$$”, (b) is the boiling point of pure substance “$$A_1$$”, whose mole fraction is given by $$\chi_1$$. An important question we would like to answer is what the composition of the liquid and of the gas is when we are at the point  $$\bullet$$. This can be answered by the lever rules.

#### v) Unstable equations of state and lever rules

In a diagram like the above, it would be nice to have a rule for the composition $$X_A^{(l)}$$ and $$X_A^{(g)}$$ in each phase. Such ‘lever rules’ exist and are best derived by analyzing the equations of state. The prototype such equation is the van der Waals equation of sate, a modification of the ideal gas law:

$P =\dfrac{nRT}{V-nb}-a \left( \dfrac{n}{V} \right)^2$

We already saw in a homework set how additional terms to the ideal gas equation can arise. By solving the full $$W = N!/[(N-M)! M!]$$ without the assumption of small number of particles compared to volume, the lowest order correction increased the pressure and was of the type $$+a’(n/V)^2$$. Expansion to all orders actually yields the $$1/(V-nb)$$ term, whereas the net $$-a(n/V)^2$$ in the van der Waals equation comes from intramolecular attractions and has a (-) sign. We will later derive the van der Waals equation of state and related equations using statistical mechanics. In thermodynamics, we must take it as a given.

Fig. 8.6: Plot of the van der Waals equation phase diagram. It contains a region where k<0. which is unphysical by P2 and the proof in chapter 7. Instead, the sample begins to coexist in two phases, one corresponding to the dense branch of P(T) on the left, one to the low density branch. At high T, this region eventually disappears at the critical temperature $$T_c$$.

Consider the figure above.  In the region CD,

$\left( \dfrac{\partial P}{\partial V} \right) > 0 \Rightarrow \kappa < 0.$

This violates the stability criterion entirely and cannot be realized as a metastable state. Points C and D are known as spinodal points: it is where the barrier in the free energy just vanishes and two stable phases cannot be maintained separately.

Even regions BC and DE are not usually realized, as seen in figure 8.7, which is a 1-D projection of figure 8.2: $$\mu^{(1)} =\mu^{(2)}$$ at B and E, and the gas (1) is converted to a liquid (2) while the pressure is unchanged and the volume gradually decreases (line BE): \mu^{(gas)}\) first increases; at B and E $$\mu_1=\mu_2$$ and $$mu$$ remains unchanged; finally $$mu$$ increases with a smaller slope in the liquid state.

$\Rightarrow \left( \dfrac{\partial \mu^{(1)}}{\partial P} \right)_T > \left( \dfrac{\partial \mu^{(2)}}{\partial P} \right)_T$

or

$$v^{(1)} > v^{(2)}$$ (i.e., liquid is always denser than gas).

By Le Châtelier's principle, you must go from the larger slope $$\partial P/\partial V$$ to the smaller slope when pressure increases.

Fig. 8.7: The chemical potentials of gas and liquid intersect: as P is increased, both chemical potentials rise, but the one of the gas rises faster because its volume decreases more rapidly with pressure than the liquid volume: gases are more compressible than liquids. The surfaces $$\mu^(i)(P)$$ intersect at points B and E, shown here in the m-P diagram but also shown in the P-V diagram in fig. 8.6.  Inset: lever rule integral, like fig. 8.6 rotated by 90° angle.

Note that $$\kappa$$ from B to C: the gas is stable in the sense that $$d^2\mu^{(1)} > 0$$, even though the liquid chemical potential $$\mu ^{(2)}$$ minimum is lower. To get over the barrier, a nucleation site (liquid droplet) must first form by fluctuation. Then the whole sample can liquefy by condensing on the nucleus. Nucleation is the result of density fluctuations, which increase from B to C until $$|\Delta \rho| \sim \rho_{liq}$$ and nucleation occurs without barrier. The likelihood of density fluctuations cannot be computed in the framework of postulates P0 – P4; we will need statistical mechanics. Between B and C we call the state “metastable:” it is a local free energy minimum, but not the global free energy minimum. Only the lack of formation of a nucleus prevents part or all of the substance in going from B to E.

Now to the famous lever rule. The line BE has a simple geometrical interpretation:

$\mu_R = \mu_B \Rightarrow \int \nu(P)\,dp = 0$

at constant temperature. This is because m is a state function, and because the Gibbs-Duhem relation relates m to the integral over $$VdP$$. It follows that

$\int_E^D vP + \int_D^0 vdP + \int_0^C vdP + \int_C^B vdP = 0$

or area 1 = area 2 in the V-P plot inset in figure 8.7. Thus the onset of the phase transition assuming the barrier can be crossed the moment $$\mu_1 – \mu_2$$ is when the areas of the equation of state below and above the line segment EB in figure 8.6 are equal.

Once we know where the line of phase coexistence lies, it is easy to derive the lever rule. At some volume $$V$$ during condensation (x-axis in figure 8.6, y-axis of inset in fig. 8.7)

$v=\chi_1v_1+\chi_2v_2$

$\Rightarrow (\chi_1+\xi_2)v=\chi_2v_2+\chi_2v_2$

$\Rightarrow \dfrac{\chi^{(1)}}{\chi^{(2)}}=\dfrac{v-v^{(2)}}{v^{(1)}-v}= \dfrac{v-v^{(liq)}}{v^{(gas)}-v}$

In words: from A to B, the gas compresses as we decrease $$V$$. Finally, the volume of the gas reaches $$v_B=v^{(gas)}$$. At B, a liquid phase begins to appear, with smaller molar volume $$v_E=v^{(liq)}$$. As we reduce volume further, more liquid appears and gas disappears. Therefore the pressure remains constant because $$v_E < v_B$$ as we move along line segment EB. The mole fraction of liquid to gas increases as $$V$$ (or $$v = V/n_{tot}$$) decreases further. The ratio of mole fractions is given by the ratio of the lever arms from $$v$$ to B or $$v$$ to E, respectively. Finally, we reach point E, and all gas has been converted to liquid. The gas lever $$\xi^{(1)} = \xi^{(gas)} \sim v-v^{(liq)}-v^{(liq)}$$ reaches 0 at point E, so all gas has disappeared (or almost all, as an infinitesimal amount given by a Boltzmann factor remains.)

The mole fractions are in inverse ratio of the lever arms $$v^{(1)} - v$$ and $$v-v^{(2)}$$. Similar lever rules apply in the multi-component system we discussed, or if parameters other than $$P$$ are held constant:

Fig. 8.8: Lever rule for the phase diagram coexistence surface in fig. 8.5. The solid horizontal lever arms give the mole fraction of substance $$A_1$$ in the liquid and gas. The starting point for the distillation was “start.”

#### vi)  Critical phrase transitions

Last, we briefly discuss critical phase transitions. Here we quickly get into a realm where statistical mechanics is needed because microscopic phenomena are amplified straight to the macroscopic level.

So far, we avoided point (6), where the phases $$A^{(1)}$$ and $$A^{(2)}$$ fuse such that their local minima become one minimum.  At that point, the system switches from having three places where $$dG(x)=0$$, to having only one, and also 2 positive+1 negative $$d^2G$$ to having $$d^2G=0$$ and only $$d^4G≠0$$. The figure below shows what happens when one approaches (6) along an extension of the coexistence curve, and what the order parameter looks like.

Fig. 8.9: $$d^2G(x)=0$$ at the critical point; above $$T_c$$, the system has one quadratic minimum, below $$T_c$$, it has two quadratic minima and a quadratic barrier. Right: two values of the order parameter (e.g. 0 and 1) merge to a single value at the critical point.  Near the merger the order parameter (e.g. density) obeys a power law.

At (6):

• a single minimum in free energy splits into two minima:  only one local minimum $$\mu_i(T,P)$$ exists above $$T_c$$, two $$\mu^{(1)}$$ and $$\mu^{(2)}$$ below $$T_c$$.
• $$d^2G=0$$ (i.e. $$G$$ is a function of the type $$aO^4-b(T)O^2$$; $$b$$ switches from positive to negative at $$T_c$$, and exactly equals 0 at $$T_c$$.
• the order parameter is zero above $$T_c$$, and can take on one of two equally likely values below $$T_c$$ (as long as we stay on the coexistence curve).

Because $$d^2G=0$$, thermodynamic stability analysis fails at $$T_c$$. The macroscopicity assumption of thermodynamics fails: $$N >> \sqrt{N}$$ usually implies small fluctuations so $$\sqrt {\delta S^2} = S$$, $$\sqrt {\delta U^2} = U$$, $$\sqrt {\delta V^2} = V$$, etc.  Because $$G$$ is flat to 4th order, fluctuations become comparable to the difference in the order parameter between phases.

Example

$$\sqrt{\delta\rho^2} \sim \rho_{liq}$$ at the critical point: bubbles of fluid of any density from gas to liquid coexist. As a result, quantities such as $$c_p$$ diverge not just at a point, but over a broad range of $$T$$.

Thermodynamics is a theory of averages and breaks down when $$\sqrt{\delta ^2x} \sim \bar{x}$$. It cannot predict critical points or the values of exponents for diverging parameters or order parameters. It cannot predict certain constants in equations of sate. We need statistical mechanics.