Phase equilibrium — Clausius-Clapeyron equation: Visual walkthrough

2.4.6 · D2 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Phase equilibrium — Clausius-Clapeyron equation

Step 1 — Do phases, ek line

KYA. Pressure–temperature plane ki picture banao. Horizontal axis temperature $T$ hai (kitna hot). Vertical axis pressure $P$ hai (cheez kitni squeeze ho rahi hai). Paani jaisi koi substance liquid ya vapour ho sakti hai; woh "banana chahti hai" konsa — yeh depend karta hai ki tum is plane mein kahan ho.

WHY ek plane aur ek single point nahi. Agar liquid aur vapour sirf ek exact jagah coexist kar sakti, to boiling ek fluke hoti. Experiment kuch aur kehta hai: tum paani ko kaafi saari temperatures par boil kar sakte ho — bas matching pressure chahiye. To "dono phases ek saath khush" wali states ek poori curve banati hain, koi dot nahi.

PICTURE. Neeche ka board plane ko do parts mein baant ta hai: ek liquid region (left, cool+squeezed) aur ek vapour region (right, hot+loose). Unke beech chalk-blue curve coexistence curve $P (T)$ hai — woh jaghah jahan dono phases ek saath reh sakti hain.

Figure — Phase equilibrium — Clausius-Clapeyron equation

Step 2 — Referee: chemical potential $μ$

KYA. $μ$ (Greek "mu") introduce karo. Yeh ek phase ke ek mole ka Gibbs Free Energy hai. Ise padhho as "is $(T, P)$ par ek particle ke liye us phase mein rehna energy-wise kitna expensive hai."

WHY yeh quantity sab kuch decide karti hai. Nature lower Gibbs energy ki taraf slide karta hai. Agar liquid ka $μ$ vapour se chhota ho, to particles liquid mein pour ho jaate hain energy bachane ke liye, aur vapour gayab ho jaati hai. Tie — koi net flow nahi — tabhi hota hai jab dono costs equal hoon: $μ_{liquid} (T, P) = μ_{vapour} (T, P) .$ Term by term: $μ_{liquid}$ = phase 1 mein cost per mole; $μ_{vapour}$ = phase 2 mein cost per mole; arguments $(T, P)$ remind karte hain ki dono ek hi temperature aur pressure par measure ho rahe hain.

PICTURE. Do chalk buckets equal height par. Jab tak paani ki levels match karti hain, koi paani flow nahi karta. Woh level hi $μ$ hai. Step 1 ki coexistence curve exactly woh ridge hai jahan dono levels equal rehti hain.

Step 3 — Curve par ek tiny step lo

KYA. Curve par ek point choose karo, $(T, P)$ . Ek neighbour par nudge karo jo abhi bhi curve par ho, $(T + d T, P + d P)$ . Symbol $d T$ ka matlab hai " $T$ mein ek tiny change"; $d P$ ka matlab hai "matching tiny change in $P$ ". Unka ratio $\frac{d P}{d T}$ hi woh slope hai jo hum chahte hain.

WHY tie changes ko match karne par majboor karti hai. Pehle point par buckets level hain: $μ_{1} = μ_{2}$ . Doosre point par bhi abhi bhi level hain: $μ_{1} = μ_{2}$ . Agar do cheezein equal shuru hoon aur equal khatam hoon, to har ek kitni badli woh bhi equal hai: $d μ_{1} = d μ_{2} .$ Yahan $d μ_{1}$ = liquid ki cost mein tiny change, $d μ_{2}$ = vapour ki cost mein tiny change. Yeh ek line poori derivation ka engine hai.

PICTURE. Blue curve par ek doosre ke paas do points, chhote triangle ke saath jiske legs hain $d T$ (horizontal) aur $d P$ (vertical). Hypotenuse curve ke saath chalti hai; uski steepness hi slope hai.

Recall

$\frac{d P}{d T}$ slope kyun hai aur $\frac{d T}{d P}$ kyun nahi? Ek curve ka slope jo $P$ upar aur $T$ across draw hua ho woh hai "rise over run" = vertical change over horizontal change = $\frac{d P}{d T}$ .

Step 4 — Ek step $μ$ par kya karta hai: master relation

KYA. Hume formula chahiye ki $μ$ kaise change hota hai jab hum $T$ aur $P$ wiggle karein. Per mole kaam karte hue, result hai: $d μ = d g = - s d T + v d P .$ Term by term padhho: $s$ = molar entropy (Entropy and the Second Law dekho) — per mole spread-out-ness; $v$ = molar volume — per mole room; $d T, d P$ = hamare tiny nudges. To heating ( $T$ badhana) $μ$ ko rate $s$ par ghataata hai, aur squeezing ( $P$ badhana) $μ$ ko rate $v$ par badhaata hai.

WHY yeh do rates. Pehle total $G$ par algebra karo, phir moles se divide karo. $G = U - T S + P V$ se shuru karo (energy $U$ , disorder $T S$ ke liye penalised, plus push-work $P V$ ). Differentiate karo: $d G = d U - T d S - S d T + P d V + V d P .$ Ab First Law of Thermodynamics, $d U = T d S - P d V$ feed karo. $T d S$ , $- T d S$ ko cancel karta hai; $- P d V$ , $+ P d V$ ko cancel karta hai. Jo bachta hai woh hai $d G = - S d T + V d P .$ Per-mole jaane ke liye moles ki sankhya se divide karo ( $G \to g$ , $S \to s$ , $V \to v$ ), phir $g \to μ$ rename karo: $d g = - s d T + v d P ⟺ d μ = - s d T + v d P .$ Yeh clean cancellation hi woh reason hai ki $G$ (aur isliye $μ$ ) $T$ aur $P$ ka natural variable hai (zyada Maxwell Relations mein).

PICTURE. $(T, P)$ floor ke upar ek tilted plane: $T$ ke along uska downhill slope $s$ size ka hai, $P$ ke along uska uphill slope $v$ size ka hai. Do chalk arrows do rates label karti hain.

Step 5 — Tie impose karo, slope harvest karo

KYA. Step 4 ko Step 3 ki condition $d μ_{1} = d μ_{2}$ mein daalo: $- s_{1} d T + v_{1} d P = - s_{2} d T + v_{2} d P .$ Left side = phase 1 ki cost kaise badli; right side = phase 2 ki cost kaise badli. Tie rakhne ke liye yeh equal hone chahiye.

WHY terms collect karo. $d P$ wale ek side mein gather karo, $d T$ wale doosre side mein, fixed phase 1 → phase 2 ordering ( $2 - 1$ ) rakhte hue: $(v_{2} - v_{1}) d P = (s_{2} - s_{1}) d T .$ Slope isolate karne ke liye divide karo: $\frac{d P}{d T} = \frac{s _{2} - s _{1}}{v _{2} - v _{1}} = \frac{Δ s}{Δ v}$ Yahan $Δ s = s_{2} - s_{1}$ entropy jump hai (phase 2 kitna zyada spread-out hai phase 1 se), aur $Δ v = v_{2} - v_{1}$ volume jump hai (phase 2 kitna zyada room leta hai phase 1 se). Slope unka ratio hai — do phases ki pure geometry, direction already fixed.

PICTURE. Ek rectangle: bottom par width $Δ v$ , side par height $Δ s$ ; corner se corner tak diagonal ka slope $Δ s /Δ v$ hai — literally $d P / d T$ .

Step 6 — Entropy ko latent heat se swap karo

KYA. Ek phase jump ki entropy directly measure karna awkward hai, lekin Latent Heat $L$ (fixed $T$ par phase 1 → phase 2 convert karne par absorb hua heat — Step 5 jaisi hi direction) easy hai. Dono linked hain: $Δ s = \frac{L}{T} .$

WHY. Transition isothermally aur reversibly hoti hai — temperature kabhi nahi hilti jab substance phase change karti hai. Entropy change hai $\int \frac{d q _{rev}}{T}$ , aur kyunki $T$ constant hai woh bahar aa jaata hai: $Δ s = \frac{1}{T} \int d q_{rev} = \frac{L}{T}$ . Step 5 mein substitute karo: $\frac{d P}{d T} = \frac{L}{T Δ v}$ Term by term: $L$ = heat jo transition peeta hai (J/mol); $T$ = fixed transition temperature (K); $Δ v$ = room jo transition banata hai (m³/mol), abhi bhi $v_{2} - v_{1}$ . Yeh exact Clausius-Clapeyron equation hai — kahin bhi koi approximation nahi.

PICTURE. Ek heating graph par flat plateau: temperature pinned rehti hai jabki heat $L$ pour hoti rahti hai aur substance phase flip karti hai. Heat mein plateau ki length $L$ hai; uski height $T$ hai; ratio slope ko feed karta hai.

Step 7 — Edge case: $Δ v$ ka sign (ice backwards kyun jhukta hai)

KYA. Slope sign $Δ v$ se inherit karta hai, kyunki $L$ aur $T$ hamesha positive hote hain. Step 5 ki fixed direction rakho: phase 1 = ice (solid), phase 2 = water (liquid) kaho, to $L > 0$ woh heat hai jo ice melt karne ke liye chahiye, aur $Δ v = v_{water} - v_{ice}$ . Zyaatar substances mein liquid solid se zyada room leta hai, to $Δ v > 0$ aur $\frac{d P}{d T} > 0$ — curve right ki taraf jhukti hai. Lekin water rebel hai: ice liquid water se less dense hai, isliye ice actually zyada room leta hai. Tab $v_{ice} > v_{water}$ , giving $Δ v = v_{water} - v_{ice} < 0$ .

WHY slope flip hota hai. $Δ v < 0$ aur $L, T > 0$ ke saath: $\frac{d P}{d T} = \frac{L}{T Δ v} < 0.$ Negative slope ka matlab: pressure badhaao, aur melting temperature giregi. Ice ko itna hard squeeze karo aur woh $0^{\circ}$ C se neeche bhi melt ho jaata hai. Dhyaan do ki humne koi convention flip nahi kiya — humne bas phases ko name kiya aur fixed $v_{2} - v_{1}$ ko negative nikalne diya.

PICTURE. Board par side by side do coexistence curves: ek normal substance right jhukti hui (positive slope) aur water ki melting line left jhukti hui (negative slope). Pale-yellow arrow dikhata hai "curve par neeche push karo → melt."

Step 8 — Degenerate case: ek phase gas hai (vapour shortcut)

KYA. Jab ek phase vapour ho, hum simplify kar sakte hain. Direction rakho phase 1 = liquid, phase 2 = vapour. Ek gas liquid se vastly zyada room leta hai, to $Δ v = v_{2} - v_{1} \approx v_{gas}$ , aur Ideal Gas Law se $v_{gas} = \frac{R T}{P}$ ( $R$ gas constant hai). Plug in: $\frac{d P}{d T} = \frac{L}{T \cdot \frac{R T}{P}} = \frac{L P}{R T ^{2}} .$

WHY yeh straight line ban jaati hai. $P$ se divide karo: $\frac{1}{P} \frac{d P}{d T} = \frac{d ( ln P )}{d T} = \frac{L}{R T ^{2}}$ . $L$ roughly constant maankar integrate karo: $ln P = - \frac{L}{R} \cdot \frac{1}{T} + C .$ To $ln P$ ko $\frac{1}{T}$ ke against plot karo aur straight line of slope $- L / R$ milti hai — data se latent heat measure karne ka standard tarika.

PICTURE. Chalk-pink straight line of $ln P$ versus $1/ T$ , uska downhill slope $- L / R$ label kiya hua.

Ek picture summary

Yahan ek board par poora safar hai: master relation tie condition ko feed karti hai, tie $Δ s /Δ v$ deta hai, latent heat $Δ s$ ko rewrite karta hai, aur $Δ v$ ka sign set karta hai ki curve kidhar jhukti hai.

Recall Feynman: poora walk plain words mein

Do phases do dukaan ki tarah hain jo ek hi cheez bechti hain. Customers (particles) jahan sasta ho wahan jaate hain, aur "price" $μ$ hai. Jab prices tie karein, koi nahi hilta — woh tie temperature–pressure map par poori ek line par hoti hai. Ab us line par ek baby step lo: kyunki prices pehle bhi equal thi aur baad mein bhi, unhe same amount change karna pada. Lekin hum jaante hain ki price exactly kya change karta hai: heating use rate $s$ par girata hai jo batata hai cheez kitni spread-out hai, aur squeezing use rate $v$ par badhata hai jo batata hai kitna room leta hai. Do price-changes ko equal set karo aur clean up karo, to tie-line ki steepness entropy jump divided by volume jump nikalta hai. Entropy jump ko easier-to-measure latent heat se swap karo (heat jo phase flip karte waqt peeta hai, temperature se divided), aur famous slope milta hai $\frac{d P}{d T} = \frac{L}{T Δ v}$ . Volume jump ka sign decide karta hai line kidhar jhukegi — aur kyunki ice paani se zyada room leta hai, uski line backwards jhukti hai, to ice squeeze karne se woh melt hoti hai.

Quick self-check

Recall Slope ka sign kis quantity ke sign se set hota hai?

$Δ v$ se (kyunki $L$ aur $T$ hamesha positive hote hain).

Recall Curve ke along

$d μ_{1} = d μ_{2}$ kyun hold karna chahiye? Do phases shuru aur khatam dono jagah equal hain ( $μ_{1} = μ_{2}$ tiny step se pehle aur baad mein), to unke changes bhi match karne chahiye.

Recall Kaun se do terms cancel hoke

$d G$ ko $- S d T + V d P$ banate hain? $T d S$ aur $- T d S$ , aur $- P d V$ aur $+ P d V$ , first law $d U = T d S - P d V$ use karke.

Recall Vapour form ke liye kaun se do assumptions chahiye?

$Δ v \approx v_{gas}$ (liquid volume negligible) aur $v_{gas} = R T / P$ (ideal gas).