Phase rule — Gibbs phase rule: Worked examples — Hinglish deep dive

2.4.7 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Phase rule — Gibbs phase rule

Shuru karne se pehle, teen quantities yaad karo (from the parent note):

$C$ = independent components ki sankhya (species minus reactions minus extra constraints),
$P$ = coexisting phases ki sankhya (alag homogeneous regions),
$F$ = intensive knobs ki sankhya (jaise $T$ , pressure, composition) jo tum freely ghuma sakte ho.

Puri formula aur uska constant-pressure cousin (Condensed phase rule): $F = C - P + 2 and F_{condensed} = C - P + 1$

Scenario matrix

Har phase-rule problem in cells mein se kisi ek mein rehti hai. Neeche ke worked examples mein har ek ko us cell se tag kiya gaya hai jise woh cover karta hai, aur saath milkar yeh sabhi ko hit karte hain.

Cell	Isme kya mushkil hai	Example
A. Ek component, $P$ sweep karna	Dekho $F$ kaise 2→1→0 girti hai jab phases stack hote hain	Ex 1
B. Do components, single vs multi phase	Composition real freedom add karti hai	Ex 2
C. Reaction, $C$ ko reduce karti hai	Reactions subtract karne padenge	Ex 3
D. Fixed pressure (condensed rule)	$+ 2$ ban jaata hai $+ 1$	Ex 4
E. Degenerate $F = 0$ (invariant point)	Sab kuch locked; $P$ ya $T$ solve karo	Ex 5
F. Impossible / negative $F$	Formula scenario ko forbid karta hai	Ex 6
G. Extra constraint (fixed composition ratio)	Reactions ke aage bhi constraints subtract karo	Ex 7
H. Real-world word problem	Words ko $C, P$ mein translate karo	Ex 8
I. Exam twist (immiscible liquids / azeotrope)	Phases ko galat ginना**	Ex 9

Example 1 — Ek component, phases ke through sweeping (Cell A)

Worked example Pure carbon dioxide, teen situations

Pure $C O_{2}$ ( $C = 1$ ) ke liye, $F$ nikalо jab woh hai (a) sirf gas, (b) boiling line par liquid + gas, (c) triple point par solid + liquid + gas.

Forecast: padhne se pehle teen numbers guess karo. (Hint: har extra phase ek knob kha jaati hai.)

Step 1 — $C$ fix karo. Pure $C O_{2}$ ek chemical species hai, koi reaction nahi, isliye $C = 1$ . Yeh step kyun? $C$ sirf chemistry se set hoti hai, phases ginne se pehle.

Step 2 — Har phase count plug karo.

(a) $P = 1$ : $F = 1 - 1 + 2 = 2$ .
(b) $P = 2$ : $F = 1 - 2 + 2 = 1$ .
(c) $P = 3$ : $F = 1 - 3 + 2 = 0$ . Yeh step kyun? Teeno situations mein sirf $P$ badalta hai; $C$ aur $+ 2$ fixed rehte hain.

Step 3 — Geometry padho (figure dekho): $F = 2$ ek area hai, $F = 1$ ek line, $F = 0$ ek point.

Verify: numbers $2, 1, 0$ exactly 1 se ghatte hain har baar jab $P$ 1 se badhta hai — formula mein $- P$ se consistent. ✅

Yeh picture parent mein water diagram ke same shape ki hai — dekho Phase diagrams of pure substances. Lines ko Clausius–Clapeyron equation describe karti hai.

Example 2 — Do components: composition mein chhupi freedom (Cell B)

Worked example Salt water, teen phase counts

$NaCl$ aur $H_{2} O$ ka mixture ( $C = 2$ ). $F$ nikalо (a) ek single liquid solution ke liye, (b) solution + water vapour, (c) solution + solid salt + vapour.

Forecast: water se ek zyada component ke saath, kya tum Example 1 se zyada ya kam knobs expect karte ho?

Step 1 — $C$ fix karo. Do independent species, koi reaction nahi, isliye $C = 2$ . Yeh step kyun? $NaCl$ aur $H_{2} O$ independently dial kiye ja sakte hain — kuch bhi unki amounts ko link nahi karta.

Step 2 — Har $P$ plug karo.

(a) $P = 1$ : $F = 2 - 1 + 2 = 3$ .
(b) $P = 2$ : $F = 2 - 2 + 2 = 2$ .
(c) $P = 3$ : $F = 2 - 3 + 2 = 1$ . Yeh step kyun? Same machine, lekin $C = 2$ har answer ko Example 1 ke relative ek se upar shift kar deta hai.

Step 3 — (a) interpret karo. Teen free knobs: $T$ , pressure, aur salt concentration. Isliye tum ek solution ko heat, pressurise, aur dilute sab independently kar sakte ho.

Verify: Example 1 se cell-by-cell compare karo — har value exactly $1$ zyada hai, jo $F = C - P + 2$ mein extra component se match karta hai. ✅

Example 3 — Ek reaction component count ko shrink karti hai (Cell C)

Worked example Ammonium chloride decomposing

$N H_{4} Cl (s) ⇌ N H_{3} (g) + HCl (g)$ , pure $N H_{4} Cl$ se start kiya. $F$ nikalo.

Forecast: 3 species hain — lekin kya $C = 3$ hai?

Step 1 — Species aur reactions gino. Species $= 3$ ( $N H_{4} Cl, N H_{3}, HCl$ ); independent reactions $= 1$ . Yeh step kyun? Reaction ka equilibrium constant teeno amounts ko link karta hai — ek dependency.

Step 2 — Starting material se extra constraint. Kyunki hum pure $N H_{4} Cl$ se start karte hain, gas phase mein $N H_{3}$ aur $HCl$ 1:1 ratio mein hone chahiye. Yeh ek aur constraint hai. Yeh step kyun? Source ki stoichiometry vapour mein composition ratio fix karti hai — Components and independent reactions ise added constraint kehta hai.

Step 3 — Components. $C = 3 - 1 - 1 = 1$ .

Step 4 — Phases. Solid $N H_{4} Cl$ + ek gas phase $\Rightarrow P = 2$ .

Step 5 — Phase rule. $F = 1 - 2 + 2 = 1$ .

Verify: $F = 1$ matlab $T$ choose karne se decomposition pressure fix ho jaata hai — experimentally sach hai, ek temperature ke liye ek pressure. ✅ Parent ke $CaC O_{3}$ example ( $C = 2$ ) se compare karo jahan extra 1:1 ratio absent tha kyunki solid product $CaO$ gas composition ke baare mein kuch lock nahi karta.

Example 4 — Fixed pressure: condensed rule (Cell D)

Worked example Metallurgical alloy at 1 atm

Ek two-metal alloy (maan lo Cu–Ni, $C = 2$ ) ko fixed pressure 1 atm par study kiya jaata hai. Maximum kitne phases coexist kar sakte hain?

Forecast: kya answer 3 (full rule) hoga ya 4?

Step 1 — Sahi rule choose karo. Pressure constant rakha gaya hai, isliye ek global variable remove ho jaata hai. $F = C - P + 1$ use karo (dekho Condensed phase rule). Yeh step kyun? $+ 2$ ne $T$ aur $P$ dono count kiye; $P$ freeze karne se sirf $T$ bachta hai, isliye $+ 1$ .

Step 2 — Maximum phases matlab minimum freedom, $F = 0$ . $0 = 2 - P + 1 \Rightarrow P = 3$ Yeh step kyun? $F$ negative nahi ja sakta (neeche Cell F), isliye $F = 0$ sab se bada allowed $P$ deta hai.

Verify: fixed $P$ par binary phase diagram mein, sab se crowded feature ek horizontal invariant line (eutectic) hai jo teen phases ke saath hai — $P = 3$ se match karta hai. ✅

Example 5 — Invariant point, sab kuch locked (Cell E)

Worked example Triple point par state solve karna

Pure sulfur ( $C = 1$ ) mein rhombic solid, monoclinic solid, aur liquid sab coexist karte hain. Degrees of freedom kitne hain, aur physically iska kya matlab hai?

Forecast: EK substance ke teen solids-ya-liquids — kitne knobs?

Step 1 — Count karo. $C = 1$ , $P = 3$ (do solid phases + ek liquid — alag crystal structures alag count hote hain). Yeh step kyun? Rhombic aur monoclinic sulfur alag distinct homogeneous solids hain, isliye yeh do phases hain, ek nahi.

Step 2 — Phase rule. $F = 1 - 3 + 2 = 0$ .

Step 3 — Matlab. $F = 0$ ⇒ invariant point: dono $T$ aur pressure nature se fixed hain. Tum inhe choose nahi kar sakte.

Verify: experimentally sulfur ka rhombic–monoclinic–liquid triple point ek fixed $(T, P)$ ≈ $(15 1^{\circ} C, 1290 atm)$ par baitha hai — ek single dot, exactly jaisa $F = 0$ predict karta hai. ✅

Example 6 — Jab formula ek scenario forbid karta hai (Cell F)

Worked example Kya pure water ke char phases coexist kar sakte hain?

Claim: ice + liquid + vapour + ek chautha ice polymorph, sab ordinary $H_{2} O$ ke, normal conditions ke neeche, ek saath coexist karte hain aur $T$ freely choose kar sakte ho. Ise test karo.

Forecast: kya yeh possible hai?

Step 1 — $C = 1$ , $P = 4$ plug karo. $F = 1 - 4 + 2 = - 1$ Yeh step kyun? Hum bas machine chalate hain aur sign inspect karte hain.

Step 2 — Negative interpret karo. $F$ free knobs count karta hai; yeh negative nahi ho sakta. Negative $F$ matlab system over-constrained hai: equations unknowns se zyada hain, isliye aisa koi equilibrium exist nahi karta. Yeh step kyun? Negative freedom formula ka tarika hai yeh chillane ka ki "impossible."

Verify: $F \geq 0$ rearrange karne par milta hai $P \leq C + 2 = 3$ . Isliye ek one-component system mein maximum 3 coexisting phases ho sakte hain — char phases ka claim ruled out hai. ✅ (Exotic high-pressure ice polymorphs naye triple points bana sakte hain, lekin fixed $C = 1$ ke liye ek saath kabhi chaar phases nahi.)

Example 7 — Reaction ke aage ek extra constraint (Cell G)

Worked example Water gas equilibrium with imposed composition

$CO (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)$ ek gas phase mein, is tarah prepare kiya ki $CO$ aur $H_{2} O$ equal moles mein fed kiye gaye. $F$ nikalo.

Forecast: 4 species, 1 reaction — lekin ek aur subtraction hai. Use pahchano.

Step 1 — Species aur reactions. Species $= 4$ ; independent reactions $= 1$ .

Step 2 — Added constraint. $CO$ aur $H_{2} O$ equally feed karna mole numbers ke beech ek fixed relationship impose karta hai — ek extra constraint. Yeh step kyun? Koi bhi imposed composition ratio independence ki ek degree remove karta hai, exactly jaise Example 3 mein 1:1 ratio ne kiya.

Step 3 — Components. $C = 4 - 1 - 1 = 2$ .

Step 4 — Phases. Sab ek gas phase mein, $P = 1$ .

Step 5 — Phase rule. $F = 2 - 1 + 2 = 3$ .

Verify: teen free knobs = $T$ , pressure, aur ek composition variable — ek homogeneous reacting gas ko indeed teen tareekon se freely adjust kiya ja sakna chahiye. ✅

Example 8 — Real-world word problem (Cell H)

Worked example Espresso steam wand

Ek barista ka boiler liquid water aur steam ko equilibrium mein hold karta hai. Barista steam ka temperature set karna chahta hai. Thermodynamics kitne independent settings allow karta hai, aur jab woh ek choose karte hain toh kya hota hai?

Forecast: kya woh temperature aur pressure dono freely set kar sakte hain?

Step 1 — Symbols mein translate karo. Pure water: $C = 1$ . Liquid + vapour: $P = 2$ . Yeh step kyun? Word problems ko kisi bhi formula se pehle $C$ aur $P$ mein reduce karna padta hai.

Step 2 — Phase rule. $F = 1 - 2 + 2 = 1$ .

Step 3 — Interpret karo. Exactly ek free knob. Temperature choose karne se pressure force ho jaata hai (boiling-point relation), aur vice versa. Yeh step kyun? $F = 1$ $P$ – $T$ plane par ek curve hai — ek variable se us par ek point choose karo, doosra phir determined ho jaata hai.

Verify: real espresso boilers pressure YA temperature se rate kiye jaate hain lekin dono independently nahi — $F = 1$ se match karta hai. Dono ke beech link Clausius–Clapeyron equation hai. ✅

Example 9 — Exam twist: phases galat mat gino (Cell I)

Worked example Oil, water, aur unka shared vapour

Immiscible liquid oil aur liquid water ek sealed jar mein baithte hain jisme dono wala common vapour hai. Oil aur water ko do independent components maante hue, $F$ nikalo.

Forecast: trap $P$ mein hai. Phases kitne: 2 ya 3?

Step 1 — Phases dhyan se gino. Oil (liquid) aur water (liquid) mix nahi hote — yeh do alag liquid phases hain, plus ek vapour phase. Isliye $P = 3$ , 2 nahi. Yeh step kyun? Phases physically distinct homogeneous regions hote hain; do immiscible liquids do phases hain chahe dono "liquid" hon.

Step 2 — Components. Oil aur water: $C = 2$ .

Step 3 — Phase rule. $F = 2 - 3 + 2 = 1$ .

Step 4 — Interpret karo. Ek free knob. Practice mein mixture ek given pressure ke liye fixed temperature par boil karta hai (steam distillation ka basis) — pressure choose karo, temperature set ho jaata hai.

Verify: agar tumne galti se $P = 2$ use kiya hota toh $F = 2$ milta aur independent $T$ – $P$ control predict hota, jo observed fixed steam-distillation temperature ko contradict karta. Isliye $P = 3 \Rightarrow F = 1$ sahi hai. ✅ (Teeno phases mein Chemical potential ki equality underlying constraints supply karti hai; dekho bhi Gibbs free energy and equilibrium.)

Recall Quick self-test

Teen-component system do phases show karta hai fixed pressure par. $F = ?$ ::: Condensed rule: $F = 3 - 2 + 1 = 2$ . Pure substance, $F = 0$ . Kitne phases? ::: $P = C + 2 = 1 + 2 = 3$ . $F$ kabhi negative kyun nahi ho sakta? ::: Negative $F$ matlab constraints, variables se zyada hain — equilibrium impossible hai, isliye $P \leq C + 2$ . Do immiscible liquids + unka vapour, do components: $F = ?$ ::: $F = 2 - 3 + 2 = 1$ .

Connections

Phase rule — Gibbs phase rule (index 2.4.7) — parent note jo $F = C - P + 2$ derive karta hai.
Condensed phase rule — $+ 1$ version jo Examples 4 mein use hua.
Components and independent reactions — reactions aur constraints $C$ ko kaise kam karte hain (Examples 3, 7).
Chemical potential — har equilibrium constraint supply karta hai.
Phase diagrams of pure substances aur Clausius–Clapeyron equation — $F = 2, 1, 0$ ke peeche ki geometry.
Gibbs free energy and equilibrium — chemical potentials equalise kyun hote hain.
2.4.07 Phase rule — Gibbs phase rule (Hinglish) — Hinglish companion.