Phase rule — Gibbs phase rule: Exercises — Hinglish deep dive

2.4.7 · D4 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Phase rule — Gibbs phase rule

Level 1 — Recognition

Tumhe $C$ aur $P$ diye gaye hain; bas substitute karo.

Recall Solution L1.1

KYA KARNA HAI: numbers ko $F = C - P + 2$ mein plug karo. $F = 1 - 1 + 2 = 2$ WHY answer ka matlab kuch hai: do free knobs ka matlab hai ki tum $T$ aur $P$ dono ko independently ek filled two-dimensional patch par move kar sakte ho. $P$ – $T$ plane par do-parameter freedom ek area hai. ✅ $F = 2 \Rightarrow$ area.

Recall Solution L1.2

$F = 1 - 2 + 2 = 1$ Ek free knob. $T$ fix karo aur pressure force ho jaata hai ki us $T$ par vapour pressure hi hoga (shape ke liye Clausius–Clapeyron equation dekho). $(T, P)$ points ki ek one-parameter family ek curve/line hoti hai. ✅ $F = 1 \Rightarrow$ line.

Recall Solution L1.3

$F = 1 - 3 + 2 = 0$ Zero knobs — sab kuch locked hai. Zero-parameter object ek single point hota hai: triple point. ✅ $F = 0 \Rightarrow$ point.

Level 2 — Application

Ab tumhe khud $C$ count karna hai ya $P$ decide karna hai.

Recall Solution L2.1

$C$ count karo: do chemical species jo independently vary kar sakti hain — $NaCl$ aur $H_{2} O$ . Unke beech koi reaction count nahi badlata, toh $C = 2$ . $P$ count karo: liquid solution (1) + solid salt (1) + vapour (1) $\Rightarrow P = 3$ . $F = 2 - 3 + 2 = 1$ WHY ek knob bachta hai: ek extra component hone se pure case se zyada room hai. $T$ fix karo aur baaki sab (saturation concentration aur vapour pressure) force ho jaata hai. Toh yeh coexistence $T$ -direction mein ek line hai.

Recall Solution L2.2

Species count karo: $N H_{4} Cl$ , $N H_{3}$ , $HCl$ $\Rightarrow$ 3 species. Independent reactions ghataao: 1 reaction. $\Rightarrow 3 - 1 = 2$ abhi tak. Extra constraints ghataao: kyunki humne pure $N H_{4} Cl$ se shuru kiya, gas phase ko $n_{N H_{3}} = n_{HCl}$ satisfy karna hoga (stoichiometry unhe fixed 1:1 ratio mein rakhti hai). Yeh ek aur composition constraint hai. $C = 3 - 1 - 1 = 1$ $P$ count karo: solid $N H_{4} Cl$ (1) + gas mixture (1) $\Rightarrow P = 2$ . $F = 1 - 2 + 2 = 1$ $T$ choose karo aur decomposition pressure fix ho jaata hai — yahi measure kiya jaata hai.

Level 3 — Analysis

$F$ ya ek scenario diya gaya hai, koi count ya geometric consequence deduce karo.

Recall Solution L3.1

Curve ek one-parameter object hai $\Rightarrow F = 1$ . $C = 1$ , $F = 1$ ke saath $F = C - P + 2$ mein se $P$ solve karo: $1 = 1 - P + 2 \Rightarrow P = 2.$ Kisi bhi aisi curve par do phases coexist karti hain — melting, boiling, ya sublimation lines. Clausius–Clapeyron equation exactly inhi $F = 1$ lines ka slope deti hai.

Recall Solution L3.2

Pressure fixed hai, toh ek global dial chali gayi — Condensed phase rule use karo: $F = C - P + 1.$ Yahaan $C = 2$ (do metals), $P = 3$ (liquid + solid A + solid B). $F = 2 - 3 + 1 = 0.$ Fixed pressure par zero degrees of freedom ka matlab hai fixed temperature aur fixed compositions — yeh eutectic point hai. In do metals ka har alloy ek hi temperature par completely melt hota hai.

Level 4 — Synthesis

Ek messy system mein counting rules, reactions, aur constraints combine karo.

Recall Solution L4.1

Species: $C$ , $C O_{2}$ , $CO$ $\Rightarrow 3$ . Reactions: 1 independent reaction $\Rightarrow 3 - 1 = 2$ . Extra constraints: yahaan gas phase mein koi fixed composition ratio nahi hai ( $C$ aur $C O_{2}$ ki starting amounts freely choose ki gayi hain aur $C$ ek alag solid hai), toh koi extra subtraction nahi. $C = 3 - 1 - 0 = 2.$ Phases: solid carbon (1) + gas mixture (1) $\Rightarrow P = 2$ . $F = 2 - 2 + 2 = 2.$ Do knobs: tum $T$ aur total $P$ dono independently set kar sakte ho; equilibrium gas ratio phir adjust ho jaata hai. Ise L2.2 se compare karo — wahaan purity constraint hi poora fark tha.

Recall Solution L4.2

Species: ions aur water — lekin phase-rule counting ke liye neutral independent species use karo: $KCl$ , $NaCl$ , $H_{2} O$ $\Rightarrow 3$ . Koi reaction ek salt ko dusre mein convert nahi karta; neutral salts use karne se electroneutrality automatically satisfy hoti hai, toh koi extra subtraction nahi. $C = 3.$ Phases: liquid solution (1) + solid KCl (1) + solid NaCl (1) + vapour (1) $\Rightarrow P = 4$ . $F = 3 - 4 + 2 = 1.$ Ek knob bachta hai — $T$ fix karo aur dono saturation concentrations plus vapour pressure lock in ho jaate hain.

Level 5 — Mastery

Rule ko uski limits tak push karo aur ek general statement prove karo.

Recall Solution L5.1

KYA/KYU: sabse zyada phases tab force hoti hain jab system maximally constrained ho, yaani $F$ apni minimum par ho. $F$ negative nahi ho sakta (zero se kam free variables nahi ho sakte), toh $F = 0$ set karo: $0 = C - P_{m a x} + 2 \Rightarrow P_{m a x} = C + 2.$

$C = 1$ : $P_{m a x} = 3$ (pure substance ka triple point — ice/water/vapour).
$C = 2$ : $P_{m a x} = 4$ (ek four-phase invariant point, jaise L4.2 mein apni locked state par). Yeh kaisa dikhta hai: har ek ek invariant point hai — full intensive-variable space mein ek single dot, bilkul koi freedom nahi.

Recall Solution L5.2

Fixed pressure $\Rightarrow$ condensed rule $F = C - P + 1$ with $C = 2$ . Tumhe $T$ locked chahiye, yaani zero remaining freedom: $F = 0$ . $0 = 2 - P + 1 \Rightarrow P = 3.$ Teen coexisting phases (ek eutectic: liquid + do solids) $T$ ko 1 atm par bilkul fix kar deti hain — ek physically realizable temperature standard.

Recall Solution L5.3

Pressure freeze karna derivation mein "total variables" count se exactly ek free intensive variable hata deta hai. Parent ke Step 1 mein, total variables $P (C - 1) + 2$ se $P (C - 1) + 1$ ho jaate hain. Constraint count $C (P - 1)$ unchanged rehta hai. Subtract karo: $F = [P (C - 1) + 1] - C (P - 1) = C - P + 1.$ Invariant points cross-check: full rule par $C = 2$ deta hai $P_{m a x} = C + 2 = 4$ ; condensed rule par $C = 2$ deta hai $P_{m a x} = C + 1 = 3$ . Yeh exactly ek se differ karte hain kyunki $P$ fix karne se ek dial hata di — consistent hai, contradictory nahi.

Recall Har level ki one-line summary

Recognition ::: $F = C - P + 2$ mein plug karo. Application ::: $C =$ species $-$ reactions $-$ constraints count karo; $P$ count karo. Analysis ::: $F$ ko geometry se ulta padho; jab $P$ fixed ho toh $+ 1$ use karo. Synthesis ::: purity/stoichiometry constraints aur electroneutrality $C$ badal dete hain. Mastery ::: $F \geq 0 \Rightarrow P_{m a x} = C + 2$ (ya fixed $P$ par $C + 1$ ); invariant points real aur useful hain.

Connections

Gibbs Phase Rule — parent note jise yeh exercises drill karte hain.
Chemical potential — har constraint ke peeche ki equalities $μ_{i}^{(α)} = μ_{i}^{(β)}$ .
Components and independent reactions — L2.2, L4.1 mein $C$ kaise shrink hota hai.
Condensed phase rule — L3.2, L5.2 mein use hone wala $+ 1$ version.
Phase diagrams of pure substances · Clausius–Clapeyron equation · Gibbs free energy and equilibrium.