Exercises — Phase rule — Gibbs phase rule

WHAT: plug numbers into $F=C-P+2$. $F=1-1+2=2$ WHY the answer means something: two free knobs means you may independently move both $T$ and $P$ over a filled two-dimensional patch. On the $P$–$T$ plane a two-parameter freedom is an area. ✅ $F=2\Rightarrow$ area.

$F=1-2+2=1$ One free knob. Pick $T$ and the pressure is forced to be the vapour pressure at that $T$ (see the Clausius–Clapeyron equation for the shape). A one-parameter family of $(T,P)$ points is a curve/line. ✅ $F=1\Rightarrow$ line.

$F=1-3+2=0$ Zero knobs — everything is locked. A zero-parameter object is a single point: the triple point. ✅ $F=0\Rightarrow$ point.

Count $C$: two chemical species that can vary independently — $\mathrm{NaCl}$ and ${\mathrm{H}}_2\mathrm{O}$. No reaction between them changes the count, so $C=2$. Count $P$: liquid solution (1) + solid salt (1) + vapour (1) $\Rightarrow P=3$. $F=2-3+2=1$ WHY one knob survives: with an extra component there is more room than the pure case. Fix $T$ and everything else (the saturation concentration and the vapour pressure) is forced. So the coexistence is a line in the $T$-direction.

Count species: $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$, $\mathrm{N}{\mathrm{H}}_3$, $\mathrm{HCl}$ $\Rightarrow$ 3 species. Subtract independent reactions: 1 reaction. $\Rightarrow 3-1=2$ so far. Subtract extra constraints: because we started from pure $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$, the gas phase must obey $n_{\mathrm{N}{\mathrm{H}}_3}=n_{\mathrm{HCl}}$ (stoichiometry makes them in a fixed 1:1 ratio). That is one more composition constraint. $C=3-1-1=1$ Count $P$: solid $\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$ (1) + gas mixture (1) $\Rightarrow P=2$. $F=1-2+2=1$ Pick $T$ and the decomposition pressure is fixed — exactly what is measured.

A curve is a one-parameter object $\Rightarrow F=1$. Solve $F=C-P+2$ for $P$ with $C=1$, $F=1$: $1=1-P+2\Rightarrow P=2.$ Two phases coexist along any such curve — melting, boiling, or sublimation lines. The Clausius–Clapeyron equation gives the slope of exactly these $F=1$ lines.

Pressure is held fixed, so one global dial is gone — use the Condensed phase rule: $F=C-P+1.$ Here $C=2$ (two metals), $P=3$ (liquid + solid A + solid B). $F=2-3+1=0.$ Zero degrees of freedom at fixed pressure means a fixed temperature and fixed compositions — this is the eutectic point. Every alloy of these two metals melts completely at this one temperature.

Species: $\mathrm{C}$, $\mathrm{C}{\mathrm{O}}_2$, $\mathrm{CO}$ $\Rightarrow 3$. Reactions: 1 independent reaction $\Rightarrow 3-1=2$. Extra constraints: here there is no fixed composition ratio in the gas phase (starting amounts of $\mathrm{C}$ and $\mathrm{C}{\mathrm{O}}_2$ are chosen freely and $\mathrm{C}$ is a separate solid), so no extra subtraction. $C=3-1-0=2.$ Phases: solid carbon (1) + gas mixture (1) $\Rightarrow P=2$. $F=2-2+2=2.$ Two knobs: you may set both $T$ and total $P$ independently; the equilibrium gas ratio then adjusts. Contrast this with L2.2 — the purity constraint there was the whole difference.

Species: ions and water — but for phase-rule counting use neutral independent species: $\mathrm{KCl}$, $\mathrm{NaCl}$, ${\mathrm{H}}_2\mathrm{O}$ $\Rightarrow 3$. No reaction converts one salt into another; electroneutrality is automatically satisfied by using neutral salts, so no extra subtraction. $C=3.$ Phases: liquid solution (1) + solid KCl (1) + solid NaCl (1) + vapour (1) $\Rightarrow P=4$. $F=3-4+2=1.$ One knob survives — fix $T$ and both saturation concentrations plus the vapour pressure lock in.

WHAT/WHY: the most phases you can force together happens when the system is maximally constrained, i.e. $F$ is at its minimum. $F$ cannot be negative (you cannot have fewer than zero free variables), so set $F=0$: $0=C-P_{\max }+2\Rightarrow P_{\max }=C+2.$

• $C=1$: $P_{\max }=3$ (triple point of a pure substance — ice/water/vapour).
• $C=2$: $P_{\max }=4$ (a four-phase invariant point, as in L4.2 read at its locked state). What it looks like: each is an invariant point — a single dot in the full intensive-variable space, no freedom at all.

Fixed pressure $\Rightarrow$ condensed rule $F=C-P+1$ with $C=2$. You need $T$ locked, i.e. zero remaining freedom: $F=0$. $0=2-P+1\Rightarrow P=3.$ Three coexisting phases (a eutectic: liquid + two solids) fixes $T$ absolutely at 1 atm — a physically realizable temperature standard.

Freezing pressure removes exactly one free intensive variable from the "total variables" count in the derivation. In the parent's Step 1, total variables drop from $P(C-1)+2$ to $P(C-1)+1$. The constraint count $C(P-1)$ is unchanged. Subtract: $F=[P(C-1)+1]-C(P-1)=C-P+1.$ Cross-check invariant points: full rule at $C=2$ gives $P_{\max }=C+2=4$; condensed rule at $C=2$ gives $P_{\max }=C+1=3$. They differ by one precisely because fixing $P$ removed one dial — consistent, not contradictory.

Recognition ::: plug into $F=C-P+2$. Application ::: count $C=$ species $-$ reactions $-$ constraints; count $P$. Analysis ::: read $F$ backwards from geometry; use $+1$ when $P$ is fixed. Synthesis ::: purity/stoichiometry constraints and electroneutrality change $C$. Mastery ::: $F\ge 0\Rightarrow P_{\max }=C+2$ (or $C+1$ at fixed $P$); invariant points are real and useful.