Phase equilibrium — Clausius-Clapeyron equation

1. What is phase equilibrium?

WHY chemical potential? $\mu$ is the Gibbs free energy per particle (or per mole). If $\mu_1<\mu_2$ , particles lower the total Gibbs energy $G$ by flowing from phase 2 into phase 1 — so phase 2 disappears. Equilibrium (no net flow at fixed $T,P$ ) demands $\mu_1=\mu_2$ . This single equation $\mu_1(T,P)=\mu_2(T,P)$ defines a curve in the $(T,P)$ plane, not an area: that's the coexistence line.

2. Deriving the equation from scratch

Step — Why use $d\mu$ ? Because both phases must remain equal as we walk along the curve; equal values + equal at the next step $\Rightarrow$ equal infinitesimal changes.

Now we need $d\mu$ . From the Gibbs free energy per mole $g=\mu$ , the fundamental relation (per mole) is: $dg = -s\,dT + v\,dP$ where $s$ = molar entropy, $v$ = molar volume.

Apply $d\mu_1=d\mu_2$ : $-s_1\,dT + v_1\,dP = -s_2\,dT + v_2\,dP$

Group terms: $(v_1-v_2)\,dP = (s_1-s_2)\,dT$

$\boxed{\dfrac{dP}{dT}=\dfrac{s_2-s_1}{v_2-v_1}=\dfrac{\Delta s}{\Delta v}}$

Step — Why this is already the answer: the slope of the coexistence curve equals the ratio of the entropy jump to the volume jump between phases.

WHY $\Delta s=L/T$ ? Latent heat is absorbed isothermally and reversibly during the transition, so $\Delta S=\int dq_{rev}/T = L/T$ since $T$ is constant. This is the exact Clausius-Clapeyron equation — no approximation yet.

3. The approximate (vapour) form

Step — Why $\ln P$ vs $1/T$ is linear: rewriting, $\ln P = -\frac{L}{R}\cdot\frac1T + C$ . A plot of $\ln P$ against $1/T$ is a straight line of slope $-L/R$ — the standard experimental way to measure latent heat.

Figure — Phase equilibrium — Clausius-Clapeyron equation

4. Worked examples

Worked example Example 1 — Boiling point of water at altitude

At sea level water boils at $373\,\mathrm{K}$ , $P_1=101.3\,$ kPa. $L_{vap}=40.7\,$ kJ/mol. Find boiling point where $P_2=70\,$ kPa.

Use $\ln\frac{P_2}{P_1}=-\frac{L}{R}\left(\frac1{T_2}-\frac1{T_1}\right)$ .

Why this form? Vapour phase, large $\Delta v$ , so the integrated vapour formula applies. $\ln\frac{70}{101.3}=-\frac{40700}{8.314}\left(\frac1{T_2}-\frac1{373}\right)$ LHS $=\ln(0.691)=-0.369$ . So $-0.369=-4895\left(\frac1{T_2}-2.681\times10^{-3}\right)$ . $\frac1{T_2}-2.681\times10^{-3}=7.54\times10^{-5}$ ⇒ $\frac1{T_2}=2.757\times10^{-3}$ ⇒ $T_2\approx363\,$ K. Sanity: lower pressure → lower boiling point. ✓ (water boils ~90 °C up a mountain).

Worked example Example 2 — Melting curve slope of ice (the anomaly)

For ice→water at $0°\mathrm{C}$ ( $T=273\,$ K): $L_{fus}=6.01\,$ kJ/mol, molar volumes $v_{water}=18.0\times10^{-6},\ v_{ice}=19.66\times10^{-6}\,\mathrm{m^3/mol}$ . $\Delta v = v_{water}-v_{ice}=-1.66\times10^{-6}\ \mathrm{m^3/mol}\ (\text{negative!})$

Why negative? Ice is less dense than water — it expands on freezing. This is the famous water anomaly. $\frac{dP}{dT}=\frac{L}{T\,\Delta v}=\frac{6010}{273\times(-1.66\times10^{-6})}\approx -1.33\times10^{7}\ \mathrm{Pa/K}$ Interpretation: the melting curve has a negative slope — raising pressure lowers the melting point of ice. That's why pressure can help skating (a small effect) and why deep glaciers melt at base.

5. Common mistakes

Recall Feynman: explain to a 12-year-old

Imagine a tug-of-war between ice and water. They tie exactly at $0°$ C. If you change the temperature a tiny bit, the tie breaks — unless you also squeeze them (change pressure) by just the right amount to keep the tie. The Clausius-Clapeyron equation is the rulebook saying "to keep the tie when temperature goes up by 1, change the pressure by this much." The "this much" depends on how much heat the change soaks up (latent heat) and how much the stuff swells or shrinks (volume change). Water is weird: ice takes up more room than water, so squeezing it actually helps it melt.

Flashcards

What three quantities must be equal for two phases to coexist?

Temperature

T

, pressure

P

, and chemical potential

\mu

(molar Gibbs free energy).

State the exact Clausius-Clapeyron equation.

\dfrac{dP}{dT}=\dfrac{L}{T\,\Delta v}

, where

L

is molar latent heat and

\Delta v

the molar volume change.

Derive

dg=-s\,dT+v\,dP

from.

From

g=u-Ts+Pv

dg=du-T\,ds-s\,dT+P\,dv+v\,dP

; substitute

du=T\,ds-P\,dv

⇒

dg=-s\,dT+v\,dP

Why does

\Delta s=L/T

at a phase transition?

The transition is isothermal and reversible, so

\Delta s=\int dq_{rev}/T = L/T

(T constant).

What two approximations give the vapour form

\dfrac{d\ln P}{dT}=\dfrac{L}{RT^2}

(1)

\Delta v\approx v_{gas}

(liquid volume negligible); (2) vapour is ideal,

v_{gas}=RT/P

What is the slope of a

\ln P

1/T

plot for vaporisation?

-L/R

, used to measure latent heat.

Why does ice have a negative

dP/dT

melting slope?

Because

v_{ice}>v_{water}

\Delta v<0

; with

L>0

dP/dT<0

— pressure lowers ice's melting point.

For water at higher altitude (lower pressure), does boiling point rise or fall? Why?

It falls; from

\ln(P_2/P_1)=-\frac LR(1/T_2-1/T_1)

, lower

P

⇒ lower

T_{boil}

Connections

Gibbs Free Energy — the $\mu$ equality is the foundation here.
First Law of Thermodynamics — used to simplify $dG$ .
Entropy and the Second Law — gives $\Delta s = L/T$ .
Phase Diagrams and Triple Point — CC gives slopes of every coexistence line.
Ideal Gas Law — supplies $v=RT/P$ for the vapour approximation.
Latent Heat — the $L$ in the equation.
Maxwell Relations — same Gibbs-potential machinery.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab do phases — jaise paani aur uski vapour — ek saath rehte hain (coexist karte hain), toh woh ek line pe baithte hain $(T,P)$ graph mein. Iss line ka slope hi Clausius-Clapeyron equation deti hai. Sabse important baat: equilibrium ka matlab hai dono phases ka chemical potential $\mu$ (yaani per-mole Gibbs free energy) barabar ho. Agar $\mu$ barabar nahi hua, toh particles us phase mein chale jaate hain jiska $\mu$ kam hai, jab tak ek phase khatam na ho jaaye.

Derivation simple hai: line pe har jagah $\mu_1=\mu_2$ , toh thoda aage badho toh $d\mu_1=d\mu_2$ . Aur hum jaante hain $dg=-s\,dT+v\,dP$ (ye first law se nikalta hai). Dono ko barabar karke seedha milta hai $\frac{dP}{dT}=\frac{\Delta s}{\Delta v}$ . Phase change isothermal hota hai, isliye $\Delta s = L/T$ , jisse final formula banta hai: $\frac{dP}{dT}=\frac{L}{T\,\Delta v}$ . Yahi pura khel hai — latent heat upar, temperature aur volume change neeche.

Vapour wale case mein ek shortcut hai: gas ka volume liquid se bahut bada hota hai aur gas ko ideal maan lo, toh $\frac{d\ln P}{dT}=\frac{L}{RT^2}$ , aur integrate karke $\ln\frac{P_2}{P_1}=-\frac{L}{R}(\frac1{T_2}-\frac1{T_1})$ . Iska sabse zabardast use: $\ln P$ vs $1/T$ ka graph ek straight line deta hai jiska slope $-L/R$ hota hai — bas do points se latent heat nikal lo!

Ek mazedaar baat: paani ke ice ki melting line ka slope negative hai, kyunki ice paani se halka (kam dense) hota hai, yaani $\Delta v<0$ . Isliye pressure badhane se ice ka melting point girta hai. Ye water ki anomaly hai, aur isi wajah se skating thoda easy ho jaati hai. Yaad rakho: $\Delta v$ ka sign hamesha check karo!

Phase equilibrium — Clausius-Clapeyron equation

1. What is phase equilibrium?

2. Deriving the equation from scratch

3. The approximate (vapour) form

4. Worked examples

5. Common mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Thermodynamics & Statistical Mechanics (Advanced)

Connections