Question bank — Phase equilibrium — Clausius-Clapeyron equation

False. Slope hai $\frac{L}{T\Delta v}$; $L>0$ hamesha, lekin $\Delta v$ negative ho sakta hai (ice→water mein volume ghatta hai), isliye ice-melting curve peeche ki taraf jhukti hai. Dekho Phase Diagrams and Triple Point.

False. Dono phases equal $T$, $P$, aur $\mu$ share karte hain — kabhi equal entropy nahi. Asliyat mein difference $\Delta s=L/T$ hi slope drive karta hai; equal entropy ka matlab hoga $L=0$.

True. ${\mu }_1={\mu }_2$ hi coexistence ki poori definition hai, aur $\mu$ hi molar Gibbs energy hai. Agar ye alag hoti, to matter lower-$\mu$ phase ki taraf flow karta jab tak ek phase khatam na ho jaaye.

False. Exact form $\frac{dP}{dT}=\frac{L}{T\Delta v}$ thermodynamically rigorous hai. Approximation vapour form hai (jo additionally ideal gas aur negligible liquid volume assume karta hai).

True (fixed $T$ aur $\Delta v$ par). $\frac{dP}{dT}\propto L$ directly, isliye bada heat-of-transition curve ko steeper banata hai — har degree ke liye zyada pressure change chahiye.

False. $\ln P$ ka $1/T$ ke against plot straight hota hai, slope $-L/R$ ke saath, kyunki $\ln P=-\frac{L}{R}\cdot \frac{1}{T}+C$. $T$ ke against plot karo to curve aata hai.

False. $L$ genuinely $T$ ke saath vary karta hai (critical point ke paas zero ki taraf girta hai). $L$ ko constant treat karna sirf modest temperature ranges par valid ek convenience hai.

Error: woh form assume karta hai ki ek phase ideal vapour hai ($\Delta v\approx v_{gas}=RT/P$). Solid aur liquid dono condensed hain, isliye exact $\frac{dP}{dT}=\frac{L}{T\Delta v}$ use karo real molar volumes ke saath.

Error: $\Delta v$ ko $L$ ki direction se match karna chahiye. Agar $L$ woh heat hai jo water→ice jaate waqt absorb hoti hai... lekin freezing heat release karti hai, isliye melting ka $L$, ice→water ke liye hai aur $\Delta v=v_{water}-v_{ice}<0$. Dono quantities ko ek hi direction ke liye rakho.

Error: relation $\Delta s=L/T$ ke liye heat reversibly aur isothermally transfer honi chahiye. Sirf transition temperature par reversible latent heat count karti hai; ek irreversible splash of heat $q/T$ se zyada entropy deta hai.

Error: $-SdT$ aur $+VdP$ survive karte hain kyunki first-law substitution $dU=TdS-PdV$ ne $TdS$ aur $-PdV$ terms ko cancel kar diya. Do surviving terms cancellation ke baad ke bache hue hain, naa ki first-law inputs. Dekho First Law of Thermodynamics.

Error (ek universal law ke roop mein). Normal liquids ke liye haan, lekin logic yeh hai ki $dT/dP=T\Delta v/L>0$ sirf isliye kyunki $\Delta v>0$ (gas, liquid se bada hai). Ye $\Delta v$ ka sign hai, naa ki koi magic rule, jo decide karta hai.

Error: derivation isse bahut pehle breakdown ho jaati hai. Critical point ke paas $v_{liquid}$ ab negligible nahi rehta, vapour ab ideal nahi rehta, aur $L\to 0$. Formula sirf critical temperature se kaafi neeche trustworthy hai.

Single equation ${\mu }_1(T,P)={\mu }_2(T,P)$ do variables par ek constraint hai, jisse ek free parameter bachta hai — ek one-dimensional curve. Ek area ke liye chahiye hoga ki ${\mu }_1={\mu }_2$ ek 2-D patch par ho, jo impossible hai jab tak phases identical na hon.

Phase transitions fixed $T$ aur fixed $P$ par hote hain, aur $G$ hi woh free energy hai jo exactly un dono constraints ke under minimise hoti hai (iske natural variables $T$ aur $P$ hain, jo $dg=-sdT+vdP$ se dikhta hai).

Hotter phase (liquid ke upar gas, solid ke upar liquid) ka usually $v$ zyada hota hai, isliye $\Delta v>0$; $L>0$ aur $T>0$ ke saath ye $\frac{dP}{dT}>0$ banata hai. Paani ki solid–liquid line exception hai kyunki ice, water se less dense hai.

Kyunki ${\mu }_1={\mu }_2$ curve par har point par hold karta hai. Agar do functions $(T,P)$ par equal hain aur $(T+dT,P+dP)$ par bhi equal hain, to un points ke beech unke infinitesimal changes bhi equal hone chahiye.

Vapour form integrate hokar straight line $\ln P=-\frac{L}{R}\cdot \frac{1}{T}+C$ deta hai jiska slope $-L/R$ hai. Do $(P,T)$ points slope fix kar dete hain, isliye $L=-R\times (\text{slope})$ — ek linear fit se poora thermodynamic quantity. Dekho Latent Heat.

Dono follow karte hain $\frac{dT}{dP}=\frac{T\Delta v}{L}$. Melting mein $\Delta v<0$ hai (ice→water mein volume ghatta hai) → negative slope → zyada $P$ matlab kam melt $T$. Boiling mein $\Delta v>0$ → positive slope → zyada $P$ matlab zyada boil $T$.

Slope ek ratio $\Delta s/\Delta v$ hai; ek ki direction flip karo doosre ki nahi to iska sign flip ho jaata hai aur ek falsely negated slope milta hai. Dono "final minus initial" hain ek consistent choice ke liye ki kaunsa phase "final" hai. Linked idea: Entropy and the Second Law.

Slope $\frac{L}{T\Delta v}$ vertical (infinite) tak blow up karta hai — coexistence curve momentarily $P$–$T$ plane mein seedha upar chali jaati hai. Physically wahan dono phases ka molar volume identical hai.

Phases indistinguishable ho jaate hain: $\Delta v\to 0$ aur $L\to 0$ saath mein, curve simply critical point par khatam ho jaati hai, aur vapour approximation collapse ho jaati hai (liquid volume ab negligible nahi rehta).

Teen coexistence curves (solid–liquid, liquid–vapour, solid–vapour) milte hain, har ek ka apna $\frac{L}{T\Delta v}$ slope hai. Wahan unka sublimation heat fusion plus vaporisation heat ke equal hota hai. Dekho Phase Diagrams and Triple Point.

$L_{\text{sub}}\approx L_{\text{fus}}+L_{\text{vap}}$, kyunki energetically sublimation = pehle melt phir boil. Ye sublimation curve ko triple point ke theek neeche vaporisation curve se steeper banata hai.

$\frac{dP}{dT}=\frac{0}{T\Delta v}=0$ — flat slope — lekin $\Delta v\to 0$ bhi ke saath formula genuinely $0/0$ ban jaata hai aur inapplicable ho jaata hai; aisi transitions ke liye Maxwell Relations se built Ehrenfest relations chahiye.

Better. Kam $P$ par vapour zyada dilute hota hai, isliye ye Ideal Gas Law zyada closely follow karta hai, aur $v_{gas}$ solid ke comparison mein itna enormous hota hai ki $\Delta v\approx v_{gas}$ extremely accurate hota hai.

Sirf tab agar $L=0$ ho jabki $\Delta v\ne 0$ — horizontal line matlab $T$ badhne par pressure unchanged hai aur phases coexisting hain. Ordinary first-order transitions mein $L\ne 0$, isliye truly horizontal segment occur nahi hoti.