Visual walkthrough — Phase equilibrium — Clausius-Clapeyron equation

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Step 1 — Two phases, one line

WHAT. Picture the pressure–temperature plane. The horizontal axis is temperature $T$ (how hot). The vertical axis is pressure $P$ (how hard the stuff is squeezed). A substance like water can be liquid or vapour; which one it "wants" to be depends on where you are in this plane.

WHY a plane and not a single point. If liquid and vapour could only coexist at one exact spot, boiling would be a fluke. Experiment says otherwise: you can boil water at many temperatures — you just need the matching pressure. So the "both phases happy together" states form a whole curve, not a dot.

PICTURE. The board below splits the plane into a liquid region (left, cool+squeezed) and a vapour region (right, hot+loose). The chalk-blue curve between them is the coexistence curve $P(T)$ — the only places where the two phases tie.

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Step 2 — The referee: chemical potential $\mu$

WHAT. Introduce $\mu$ (Greek "mu"). It is the Gibbs Free Energy carried by one mole of a phase. Read it as "how expensive it is, energy-wise, for a particle to live in this phase at this $(T,P)$."

WHY this quantity decides everything. Nature slides toward lower Gibbs energy. If liquid has smaller $\mu$ than vapour, particles pour into the liquid to save energy, and the vapour vanishes. A tie — no net flow — happens only when the two costs are equal: ${\mu }_{\text{liquid}}(T,P)={\mu }_{\text{vapour}}(T,P).$ Term by term: ${\mu }_{\text{liquid}}$ = cost per mole in phase 1; ${\mu }_{\text{vapour}}$ = cost per mole in phase 2; the arguments $(T,P)$ remind us both are measured at the same temperature and pressure.

PICTURE. Two chalk buckets at equal height. As long as the water levels match, no water flows. That level is $\mu$. The coexistence curve from Step 1 is precisely the ridge where the two levels stay equal.

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Step 3 — Take one tiny step along the curve

WHAT. Pick a point on the curve, $(T,P)$. Nudge to a neighbour still on the curve, $(T+dT,P+dP)$. The symbol $dT$ means "a tiny change in $T$"; $dP$ means "the matching tiny change in $P$". Their ratio $\frac{dP}{dT}$ is the slope we want.

WHY the tie forces the changes to match. At the first point the buckets are level: ${\mu }_1={\mu }_2$. At the second point they are still level: ${\mu }_1={\mu }_2$. If two things start equal and end equal, the amount each changed is also equal: $d{\mu }_1=d{\mu }_2.$ Here $d{\mu }_1$ = tiny change in liquid's cost, $d{\mu }_2$ = tiny change in vapour's cost. This single line is the engine of the whole derivation.

PICTURE. Two points a hair apart on the blue curve, with the little triangle whose legs are $dT$ (horizontal) and $dP$ (vertical). The hypotenuse rides along the curve; its steepness is the slope.

Recall Why is

$\frac{dP}{dT}$ the slope and not $\frac{dT}{dP}$? Slope of a curve drawn with $P$ up and $T$ across is "rise over run" = change in vertical over change in horizontal = $\frac{dP}{dT}$.

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Step 4 — What one step does to $\mu$: the master relation

WHAT. We need a formula for how $\mu$ changes when we wiggle $T$ and $P$. Working per mole, the result is: $d\mu =dg=-sdT+vdP.$ Read it term by term: $s$ = molar entropy (see Entropy and the Second Law) — spread-out-ness per mole; $v$ = molar volume — room per mole; $dT,dP$ = our tiny nudges. So heating (raising $T$) lowers $\mu$ at rate $s$, and squeezing (raising $P$) raises $\mu$ at rate $v$.

WHY these two rates. Do the algebra on the total $G$ first, then divide by moles. Start from $G=U-TS+PV$ (energy $U$, penalised for disorder $TS$, plus push-work $PV$). Differentiate: $dG=dU-TdS-SdT+PdV+VdP.$ Now feed in the First Law of Thermodynamics, $dU=TdS-PdV$. The $TdS$ cancels $-TdS$; the $-PdV$ cancels $+PdV$. What survives is $dG=-SdT+VdP.$ Divide by the number of moles to go per-mole ($G\to g$, $S\to s$, $V\to v$), then rename $g\to \mu$: $dg=-sdT+vdP⟺d\mu =-sdT+vdP.$ This clean cancellation is why $G$ (and hence $\mu$) is the natural variable of $T$ and $P$ (more in Maxwell Relations).

PICTURE. A tilted plane over the $(T,P)$ floor: its downhill slope along $T$ has size $s$, its uphill slope along $P$ has size $v$. Two chalk arrows label the two rates.

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Step 5 — Impose the tie, harvest the slope

WHAT. Put Step 4 into Step 3's condition $d{\mu }_1=d{\mu }_2$: $-s_{1}dT+v_{1}dP=-s_{2}dT+v_{2}dP.$ Left side = how phase 1's cost changed; right side = how phase 2's cost changed. They must be equal to keep the tie.

WHY collect terms. Gather the $dP$'s on one side, the $dT$'s on the other, keeping the fixed phase 1 → phase 2 ordering ($\text{2}-\text{1}$): $(v_2-v_1)dP=(s_2-s_1)dT.$ Divide to isolate the slope: $\boxed{\frac{dP}{dT}=\frac{s_2-s_1}{v_2-v_1}=\frac{\Delta s}{\Delta v}}$ Here $\Delta s=s_2-s_1$ is the entropy jump (how much more spread-out phase 2 is than phase 1), and $\Delta v=v_2-v_1$ is the volume jump (how much more room phase 2 takes than phase 1). The slope is their ratio — pure geometry of the two phases, with the direction already fixed.

PICTURE. A rectangle: width $\Delta v$ across the bottom, height $\Delta s$ up the side; the diagonal from corner to corner has slope $\Delta s/\Delta v$ — literally $dP/dT$.

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Step 6 — Swap entropy for latent heat

WHAT. Entropy of a phase jump is awkward to measure directly, but the Latent Heat $L$ (heat absorbed converting phase 1 → phase 2 at fixed $T$ — same direction as Step 5) is easy. They are linked: $\Delta s=\frac{L}{T}.$

WHY. The transition happens isothermally and reversibly — temperature never budges while the substance changes phase. Entropy change is $\int \frac{d{q}_{\text{rev}}}{T}$, and since $T$ is constant it pulls out: $\Delta s=\frac{1}{T}\int d{q}_{\text{rev}}=\frac{L}{T}$. Substituting into Step 5: $\boxed{\frac{dP}{dT}=\frac{L}{T\Delta v}}$ Term by term: $L$ = heat the transition drinks (J/mol); $T$ = the fixed transition temperature (K); $\Delta v$ = the room the transition makes (m³/mol), still $v_2-v_1$. This is the exact Clausius-Clapeyron equation — no approximation anywhere.

PICTURE. A flat plateau on a heating graph: temperature stays pinned while heat $L$ pours in and the substance flips phase. The plateau length in heat is $L$; its height is $T$; the ratio feeds the slope.

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Step 7 — Edge case: the sign of $\Delta v$ (why ice tilts backwards)

WHAT. The slope inherits the sign of $\Delta v$, because $L$ and $T$ are always positive. Keep the fixed direction from Step 5: call phase 1 = ice (solid), phase 2 = water (liquid), so $L>0$ is the heat to melt ice, and $\Delta v=v_{\text{water}}-v_{\text{ice}}$. For most substances the liquid takes more room than the solid, so $\Delta v>0$ and $\frac{dP}{dT}>0$ — the curve leans right. But water is the rebel: ice is less dense than liquid water, so ice actually takes more room. Then $v_{\text{ice}}>v_{\text{water}}$, giving $\Delta v=v_{\text{water}}-v_{\text{ice}}<0$.

WHY it flips the slope. With $\Delta v<0$ and $L,T>0$: $\frac{dP}{dT}=\frac{L}{T\Delta v}<0.$ A negative slope means: raise the pressure, and the melting temperature drops. Squeeze ice hard enough and it melts even below $0^{\circ }$C. Notice we never flipped any convention — we just named the phases and let the fixed $v_2-v_1$ come out negative.

PICTURE. Two coexistence curves side by side on the board: a normal substance leaning right (positive slope) and water's melting line leaning left (negative slope). The pale-yellow arrow shows "push down on the curve → melt."

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Step 8 — Degenerate case: one phase is a gas (the vapour shortcut)

WHAT. When one phase is vapour, we can simplify. Keep the direction phase 1 = liquid, phase 2 = vapour. A gas takes vastly more room than a liquid, so $\Delta v=v_2-v_1\approx v_{\text{gas}}$, and by the Ideal Gas Law $v_{\text{gas}}=\frac{RT}{P}$ (with $R$ the gas constant). Plug in: $\frac{dP}{dT}=\frac{L}{T\cdot \frac{RT}{P}}=\frac{LP}{R{T}^2}.$

WHY it becomes a straight line. Divide by $P$: $\frac{1}{P}\frac{dP}{dT}=\frac{d(\ln P)}{dT}=\frac{L}{R{T}^2}$. Integrating with $L$ roughly constant gives $\ln P=-\frac{L}{R}\cdot \frac{1}{T}+C.$ So plotting $\ln P$ against $\frac{1}{T}$ gives a straight line of slope $-L/R$ — the standard way to measure latent heat from data.

PICTURE. The chalk-pink straight line of $\ln P$ versus $1/T$, its downhill slope labelled $-L/R$.

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The one-picture summary

Here is the whole journey on one board: the master relation feeds the tie condition, the tie gives $\Delta s/\Delta v$, latent heat rewrites $\Delta s$, and the sign of $\Delta v$ sets which way the curve leans.

Recall Feynman: the whole walk in plain words

Two phases are like two shops selling the same thing. Customers (particles) go to whichever is cheaper, and "price" is $\mu$. When the prices tie, nobody moves — that tie happens along a whole line on the temperature–pressure map. Now take one baby step along that line: since the prices were equal before and after, they must have changed by the same amount. But we know exactly what changes a price: heating drops it by an amount set by how spread-out the stuff is ($s$), and squeezing raises it by how much room it takes ($v$). Setting the two price-changes equal and cleaning up, the steepness of the tie-line turns out to be the entropy jump divided by the volume jump. Swap the entropy jump for the easier-to-measure latent heat (heat drunk while flipping phase, divided by temperature), and you get the famous slope $\frac{dP}{dT}=\frac{L}{T\Delta v}$. The sign of the volume jump decides which way the line leans — and because ice takes up more room than water, its line leans backwards, so squeezing ice melts it.

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Quick self-check

Recall The slope's sign is set by the sign of which quantity?

$\Delta v$ (since $L$ and $T$ are always positive).

Recall Why must

$d{\mu }_1=d{\mu }_2$ hold along the curve? The two phases start and end equal (${\mu }_1={\mu }_2$ before and after the tiny step), so their changes must match too.

Recall Which two terms cancel to turn

$dG$ into $-SdT+VdP$? $TdS$ with $-TdS$, and $-PdV$ with $+PdV$, using the first law $dU=TdS-PdV$.

Recall What two assumptions give the vapour form?

$\Delta v\approx v_{\text{gas}}$ (liquid volume negligible) and $v_{\text{gas}}=RT/P$ (ideal gas).