Question bank — Phase equilibrium — Clausius-Clapeyron equation

False. The slope is $\frac{L}{T\Delta v}$; $L>0$ always, but $\Delta v$ can be negative (ice→water shrinks in volume), so the ice-melting curve slopes backward. See Phase Diagrams and Triple Point.

False. They share equal $T$, $P$, and $\mu$ — never equal entropy. In fact the difference $\Delta s=L/T$ is exactly what drives the slope; equal entropy would mean $L=0$.

True. ${\mu }_1={\mu }_2$ is the whole definition of coexistence, and $\mu$ is the molar Gibbs energy. If they differed, matter would flow to the lower-$\mu$ phase until one phase vanished.

False. The exact form $\frac{dP}{dT}=\frac{L}{T\Delta v}$ is thermodynamically rigorous. The approximation is the vapour form (which additionally assumes an ideal gas and negligible liquid volume).

True (at fixed $T$ and $\Delta v$). $\frac{dP}{dT}\propto L$ directly, so a larger heat-of-transition makes the curve steeper — more pressure change is needed per degree.

False. It is $\ln P$ against $1/T$ that is straight, with slope $-L/R$, because $\ln P=-\frac{L}{R}\cdot \frac{1}{T}+C$. Plotted against $T$ it curves.

False. $L$ genuinely varies with $T$ (it falls toward zero at the critical point). Treating $L$ as constant is a convenience valid only over modest temperature ranges.

Error: that form assumes one phase is an ideal vapour ($\Delta v\approx v_{gas}=RT/P$). Solid and liquid are both condensed, so use the exact $\frac{dP}{dT}=\frac{L}{T\Delta v}$ with real molar volumes.

Error: $\Delta v$ must match the direction of $L$. If $L$ is heat absorbed going water→ice... but freezing releases heat, so $L$ for melting is ice→water and $\Delta v=v_{water}-v_{ice}<0$. Keep both quantities for the same direction.

Error: the relation $\Delta s=L/T$ needs the heat to be transferred reversibly and isothermally. Only the reversible latent heat at the transition temperature counts; an irreversible splash of heat gives more entropy than $q/T$.

Error: the $-SdT$ and $+VdP$ survive because the first-law substitution $dU=TdS-PdV$ cancels the $TdS$ and $-PdV$ terms. The two surviving terms are the leftovers after cancellation, not the first-law inputs. See First Law of Thermodynamics.

Error (as a universal law). For normal liquids yes, but the logic is $dT/dP=T\Delta v/L>0$ only because $\Delta v>0$ (gas bigger than liquid). It's the sign of $\Delta v$, not a magic rule, that decides.

Error: the derivation breaks down long before that. Near the critical point $v_{liquid}$ is no longer negligible, the vapour is no longer ideal, and $L\to 0$. The formula is only trustworthy well below the critical temperature.

The single equation ${\mu }_1(T,P)={\mu }_2(T,P)$ is one constraint on two variables, leaving one free parameter — a one-dimensional curve. An area would need ${\mu }_1={\mu }_2$ over a 2-D patch, impossible unless the phases were identical.

Phase transitions happen at fixed $T$ and fixed $P$, and $G$ is the free energy minimised under exactly those two constraints (its natural variables are $T$ and $P$, seen from $dg=-sdT+vdP$).

The hotter phase (gas above liquid, liquid above solid) usually has larger $v$, so $\Delta v>0$; with $L>0$ and $T>0$ that makes $\frac{dP}{dT}>0$. Water's solid–liquid line is the exception because ice is less dense than water.

Because ${\mu }_1={\mu }_2$ holds at every point on the curve. If two functions are equal at $(T,P)$ and equal again at $(T+dT,P+dP)$, their infinitesimal changes between those points must also be equal.

The vapour form integrates to a straight line $\ln P=-\frac{L}{R}\cdot \frac{1}{T}+C$ whose slope is $-L/R$. Two $(P,T)$ points fix the slope, so $L=-R\times (\text{slope})$ — a whole thermodynamic quantity from one linear fit. See Latent Heat.

Both follow $\frac{dT}{dP}=\frac{T\Delta v}{L}$. Melting has $\Delta v<0$ (ice→water shrinks) → negative slope → higher $P$ means lower melt $T$. Boiling has $\Delta v>0$ → positive slope → higher $P$ means higher boil $T$.

The slope is a ratio $\Delta s/\Delta v$; flipping the direction of one but not the other flips its sign and gives a spuriously negated slope. Both are "final minus initial" for one consistent choice of which phase is "final." Linked idea: Entropy and the Second Law.

The slope $\frac{L}{T\Delta v}$ blows up to vertical (infinite) — the coexistence curve momentarily rises straight up in the $P$–$T$ plane. Physically the two phases have identical molar volume there.

The phases become indistinguishable: $\Delta v\to 0$ and $L\to 0$ together, the curve simply ends at the critical point, and the vapour approximation collapses (liquid volume no longer negligible).

Three coexistence curves (solid–liquid, liquid–vapour, solid–vapour) meet, each with its own $\frac{L}{T\Delta v}$ slope. Their sublimation heat equals fusion plus vaporisation heat there. See Phase Diagrams and Triple Point.

$L_{\text{sub}}\approx L_{\text{fus}}+L_{\text{vap}}$, because energetically sublimation = melt then boil. This makes the sublimation curve steeper than the vaporisation curve just below the triple point.

$\frac{dP}{dT}=\frac{0}{T\Delta v}=0$ — a flat slope — but with $\Delta v\to 0$ too the formula becomes $0/0$ and is genuinely inapplicable; such transitions need the Ehrenfest relations built from Maxwell Relations instead.

Better. At low $P$ the vapour is more dilute, so it obeys the Ideal Gas Law more closely, and $v_{gas}$ is enormous compared to the solid, making $\Delta v\approx v_{gas}$ extremely accurate.

Only if $L=0$ while $\Delta v\ne 0$ — a horizontal line means pressure is unchanged as $T$ rises with the phases still coexisting. For ordinary first-order transitions $L\ne 0$, so a truly horizontal segment does not occur.