Chemical potential decides which phase wins, and phase diagrams map the winners

Chemical potential is the push that moves a substance to where it is happier

Intuition

Think of water running downhill. It does not care about the total amount of water, only about height. Chemical potential is the “height” for a substance: it flows from a place where its chemical potential is high to a place where it is low, and stops when the level is the same everywhere. Which phase is stable? The one sitting lowest.

Lecture 22 moves from fixed-composition systems to mixtures and phase equilibrium. Processes at constant pressure are best tracked with the Gibbs energy $G$. Its total differential is

$dG = \left(\frac{\partial G}{\partial T}\right)_{p,n_i} dT + \left(\frac{\partial G}{\partial p}\right)_{T,n_i} dp + \sum_i \left(\frac{\partial G}{\partial n_i}\right)_{T,p} dn_i$

The slide labels the three slopes: $-S$, $V$ and $\mu_i$. The last is the chemical potential. It is the change in Gibbs free energy per mole of substance $i$ added, in the limit of small changes. So it is the molar Gibbs energy:

$\mu_i = \Delta G_{m,i}$

At constant $T$ and $p$ the first two terms vanish and

$dG = \sum_i \mu_i\, dn_i$

This describes transport. Move an amount $dn_i$ from region I to region II. Region I loses it and region II gains it.

$dG = -\mu_i^{I}\,dn_i + \mu_i^{II}\,dn_i = \left(\mu_i^{II} - \mu_i^{I}\right) dn_i < 0$

For the move to happen by itself, $dG$ must be negative. That needs $\mu_i^{II} < \mu_i^{I}$, so species $i$ moves only from higher to lower chemical potential. At equilibrium the chemical potential of every species is the same throughout a mixture.

Every mu curve slopes down with T and up with p, and the crossings are melting and boiling

Start with $dG = V\,dp - S\,dT$. Per mole this is

$d\mu = V_m\,dp - S_m\,dT$

Hold $p$ fixed and the slope with temperature is

$\left(\frac{\partial \mu}{\partial T}\right)_p = -S_m < 0$

So the curves always slope down. Gas has the largest molar entropy, so its curve is the steepest. Hold $T$ fixed and the slope with pressure is

$\left(\frac{\partial \mu}{\partial p}\right)_T = V_m > 0$

Raising the pressure lifts the curves (the slide shows them dashed with primed temperatures). The slide’s diagram assumes $V_m^{\text{liquid}} > V_m^{\text{solid}}$.

The sketch is schematic and follows slide 8. The lowest curve at each temperature is the stable phase: solid left of $T_m$, liquid between $T_m$ and $T_b$, gas beyond $T_b$. Slide 10 shows two variants. In sublimation the solid and gas curves cross at $T_s$ before the liquid curve gets a chance. At the triple point all three curves cross at $T_{tp}$.

Compare

Rising temperature

$\mu$ falls, by $-S_m\,dT$. The gas curve falls fastest, so gas wins at high $T$.

Rising pressure

$\mu$ rises, by $V_m\,dp$. All curves lift, and the melting and boiling points shift to $T_m'$ and $T_b'$.

A phase boundary is where two potentials are equal, and its slope comes from that

On a coexistence curve the two phases $\alpha$ and $\beta$ have the same $\mu$:

$\mu_\alpha(p,T) = \mu_\beta(p,T)$

Try it: worked example

Derive the Clausius-Clapeyron equation from mu_alpha = mu_beta along a coexistence curve.

Show answer and steps

Answerdp/dT = dS_m,tr / dV_m,tr, where each is the change from phase alpha to phase beta.

1. Stay on the curve. If $\mu_\alpha = \mu_\beta$ at one point and at a neighbouring point, then the two potentials change by the same amount: $d\mu_\alpha = d\mu_\beta$.
2. Use $d\mu = V_m\,dp - S_m\,dT$ for each phase: $-S_{m,\alpha}\,dT + V_{m,\alpha}\,dp = -S_{m,\beta}\,dT + V_{m,\beta}\,dp$.
3. Collect the $dp$ terms on one side and the $dT$ terms on the other: $\left(V_{m,\beta} - V_{m,\alpha}\right)dp = \left(S_{m,\beta} - S_{m,\alpha}\right)dT$.
4. Name the differences. $\Delta V_{m,tr} = V_{m,\beta} - V_{m,\alpha}$ and $\Delta S_{m,tr} = S_{m,\beta} - S_{m,\alpha}$.
5. Divide both sides by $\Delta V_{m,tr}\,dT$: $\dfrac{dp}{dT} = \dfrac{\Delta S_{m,tr}}{\Delta V_{m,tr}}$. This is the Clausius-Clapeyron equation.

Read the result in plain words. The slope of a boundary is the entropy jump divided by the volume jump. The solid-liquid curve is usually steep because $\Delta V_{m,\text{melt}}$ is small, which makes the ratio large. The liquid-gas transition has a much bigger $\Delta V_{m,\text{vap}}$, is more sensitive to pressure, and is more curved.

Exam detail

Know the chain from $d\mu = V_m\,dp - S_m\,dT$ to $\mu_\alpha = \mu_\beta$ to $dp/dT = \Delta S_{m,tr}/\Delta V_{m,tr}$. Know the two slopes, $-S_m$ and $V_m$, and the sign of each.

Reading phase diagrams: boundaries, the triple point and the critical point

On a simple pT diagram (pressure vertical, temperature horizontal) the regions are solid, liquid and gas. The boundaries are melting and freezing, vaporisation and condensation, and sublimation and deposition. They meet at the triple point. The liquid-gas boundary stops at the critical point. The triple point of water is 0.01 °C and 612 Pa.

Critical behaviour. Beyond the critical point it is impossible to tell liquid from gas. There is only a supercritical fluid. Supercritical CO$_2$ is easy to reach and often used as a solvent, for example for decaffeinating coffee, because lowering the pressure removes it completely. The lecture’s ethane photos show a subcritical sample, the critical point with critical opalescence, and a supercritical fluid. The slide compares the structure to the Ising model: small sparse domains, large correlated clusters at the critical point, and fine disorder.

Solid phases. Tin has a $\beta$ tetragonal form (white tin) and an $\alpha$ cubic form (grey tin, diamond structure). Below 13 °C tin pest turns white tin to grey tin. In mediaeval times tin organ pipes in northern Europe would degrade. The water diagram on slide 16 has many ice phases (I, II, III, V, VI, VII, VIII, IX, X, XI) with a triple point and a critical point.

pV diagrams. For the van der Waals fluid the isotherms (243 K, 258 K, 274 K, 304.12 K critical, 334 K, 365 K, and a dashed ideal-gas curve at 334 K) enclose a shaded two-phase dome. A more general pV diagram adds the solid, a triple line and regions such as solid plus liquid and liquid plus gas. Slides 21 and 22 are labelled extra material: the 3D pVT surfaces for an ideal gas and for a real substance. They show isotherms, isochores, coexistence regions and projections onto the pT and pV planes, with no equations.

Pitfall

Do not mix up the slopes of $\mu$. With temperature the slope is $-S_m$ (negative, curves fall). With pressure it is $+V_m$ (positive, curves rise). The boundary slope $dp/dT$ is a different thing, the ratio of the two jumps.

Recall

Why does the gas line fall faster than the solid line on a mu against T plot?

The slope is $-S_m$ and gas has the largest molar entropy, so its slope is the most negative.