Worked examples — Phase equilibrium — Clausius-Clapeyron equation

2.4.6 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Phase equilibrium — Clausius-Clapeyron equation

Yeh Phase equilibrium — Clausius-Clapeyron equation ka child page hai aur iska ek hi kaam hai: equation jo bhi situation de sake, unhe har tarah se drill karna. Kuch bhi compute karne se pehle, poora landscape samajhte hain taaki koi bhi case aisa na ho jo humne dikhaya nahi.

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Scenario matrix

Hum poore page mein do master forms use karenge:

Agar inme se kuch bhi unfamiliar lage toh pehle parent note, Latent Heat aur Ideal Gas Law revisit karo.

#	Case class	Kaunsa phase pair	Kaunsa form	$\Delta v$ ka sign	$dP/dT$ ka sign	Covered by
A	Normal vaporisation	liquid → vapour	vapour (log)	$+$ (huge)	$+$	Ex 1
B	Sublimation	solid → vapour	vapour (log)	$+$ (huge)	$+$	Ex 2
C	Normal melting	solid → liquid, solid denser	exact	$+$	$+$	Ex 3
D	Anomalous melting	ice → water, solid less dense	exact	$-$	$-$	Ex 4
E	Graph se $L$ measure karna	koi bhi vapour line	log form ka slope	—	—	Ex 5
F	Degenerate: $\Delta v\to 0$	critical point ke paas	exact (limiting)	$\to 0$	$\to \infty$	Ex 6
G	Real-world word problem	altitude par cooking	vapour (log)	$+$	$+$	Ex 7
H	Exam twist: mixed units / $L$ nikalna	vaporisation, $L$ solve karo	vapour (log)	$+$	$+$	Ex 8

Matrix ke do "axes" hain: (i) kya vapour involved hai? (form choose karta hai) aur (ii) $\Delta v$ ka sign kya hai? (slope ka sign choose karta hai). Baki sab details bas numbers plug karne ki hain.

Figure dekho: amber curves steeply rise karti hain (vapour cases A, B, G — $P$ mein chhota slope lekin relative change bahut bada), cyan solid–liquid lines almost vertical hain (cases C, D) kyunki $\Delta v$ tiny hai, aur case D left ki taraf jhukta hai (negative slope). Yeh picture dimag mein rakho — har example inhi lines mein se kisi ek par hai.

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Example 1 — Case A: normal vaporisation

1. Form choose karo. Ek phase vapour hai aur $v_{gas}\gg v_{liquid}$, isliye hum log form use karenge. Yeh step kyun? Matrix ka cell A kehta hai vapour → integrated form; exact form ke liye $v_{gas}(T)$ explicitly chahiye hoga.
2. Equation mein numbers likho. $\ln \frac{60}{101.3}=-\frac{40700}{8.314}\left(\frac{1}{T_2}-\frac{1}{373}\right)$ Yeh step kyun? Substituting hi poora computation hai; $L$ ko J/mol mein rakho taaki $R$ se match kare.
3. Left side evaluate karo. $\ln (0.5923)=-0.5238$. Yeh step kyun? $\ln$ pressure ratio ko us linear quantity mein convert karta hai jo formula predict karta hai.
4. $1/T_2$ ke liye solve karo. $L/R=4895.4\text{K}$ ke saath: $\frac{1}{T_2}-\frac{1}{373}=\frac{0.5238}{4895.4}=1.0700\times 10^{-4}$ $\frac{1}{T_2}=2.6810\times 10^{-3}+1.0700\times 10^{-4}=2.7880\times 10^{-3}$ Yeh step kyun? Hum $1/T_2$ isolate karte hain kyunki formula $1/T$ mein linear hai, $T$ mein nahi.
5. Invert karo. $T_2=358.7\text{K}$.

Verify: Units — $\ln$ ka argument ratio $P_2/P_1$ hai, jo already dimensionless hai; separately exponent factor $\frac{L}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$ bhi dimensionless hona chahiye. Check karo: $\frac{L}{R}$ hai $\frac{\text{J/mol}}{{\text{J mol}}^{-1}{\text{K}}^{-1}}=\text{K}$, times $\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$ in ${\text{K}}^{-1}$, deta hai $\text{K}\cdot {\text{K}}^{-1}=$ dimensionless. ✓ Sanity — $358.7<373$: kam pressure se boiling point kam hota hai, forecast se match karta hai. ✓

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Example 2 — Case B: sublimation

1. Form choose karo. Solid → vapour mein bhi ek gas phase hai dominant volume ke saath, isliye log form bilkul waise hi apply hota hai jaise boiling mein. Yeh step kyun? Matrix vaporisation aur sublimation ko identically treat karta hai — sirf itna fark hai ki hum kis condensed phase se shuru kar rahe hain.
2. $P_2$ nikalne ke liye rearrange karo. $\ln \frac{P_2}{P_1}=-\frac{L}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$ Yeh step kyun? Yahan unknown pressure hai, temperature nahi, isliye $P_2$ ke liye solve karte hain.
3. Bracket compute karo. $\frac{1}{205}-\frac{1}{195}=4.8780\times 10^{-3}-5.1282\times 10^{-3}=-2.5019\times 10^{-4}{\text{K}}^{-1}$. Yeh step kyun? Note karo yeh negative hai ($T$ badhaya, toh $1/T$ ghata) — do minus signs milkar positive exponent denge.
4. Multiply karo. $-\frac{26100}{8.314}\times (-2.5019\times 10^{-4})=-3139.3\times (-2.5019\times 10^{-4})=0.7855$. Yeh step kyun? Yeh $\ln (P_2/P_1)$ hai; positive value matlab $P_2>P_1$, jaise forecast tha.
5. Exponentiate karo. $P_2=100\times e^{0.7855}=100\times 2.1935=219.4\text{kPa}$.

Verify: $P_2>P_1$, gas phase ki taraf heating ke saath consistent hai. ✓ Agar hum galti se sign flip kar dete toh $P_2\approx 45\text{kPa}$ milta (heating par pressure gir raha hai) jo physically absurd hai — ek achha sign-check.

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Example 3 — Case C: normal melting (positive $\Delta v$)

1. Exact form choose karo. Koi gas nahi hai, isliye log form forbidden hai. Yeh step kyun? Cell C: solid–liquid matlab hume $dP/dT=L/(T\Delta v)$ real volumes ke saath use karna hoga.
2. $\Delta v$ compute karo final minus initial (solid → liquid) ke roop mein. $\Delta v=v_{liquid}-v_{solid}=(8.20-8.00)\times 10^{-6}=+2.0\times 10^{-7}{\text{m}}^3/\text{mol}$ Yeh step kyun? $L$ ki direction (heat melting mein) $\Delta v$ ki direction (solid→liquid) se match karni chahiye, warna sign galat hoga.
3. Plug in karo. $\frac{dP}{dT}=\frac{9000}{600\times 2.0\times 10^{-7}}=\frac{9000}{1.2\times 10^{-4}}=7.5\times 10^7\text{Pa/K}$ Yeh step kyun? Yeh exact formula ka raw evaluation hai.

Verify: Units — J/mol ÷ (K · m³/mol) = J/(K·m³) = Pa/K. ✓ Sign — positive, ek "normal" denser solid se match karta hai. ✓ Magnitude — tens of MPa per kelvin: melting curves steep hain (almost vertical) kyunki $\Delta v$ tiny hai, bilkul Figure s01 mein cyan lines jaise.

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Example 4 — Case D: the anomaly (negative $\Delta v$)

1. $\Delta v$ melting ke liye (ice → water). $\Delta v=v_{water}-v_{ice}=(18.00-19.66)\times 10^{-6}=-1.66\times 10^{-6}{\text{m}}^3/\text{mol}$ Yeh step kyun? $\Delta v$ ka sign is cell ka poora point hai — direction solid→liquid rakho.
2. Slope. $\frac{dP}{dT}=\frac{6010}{273\times (-1.66\times 10^{-6})}=\frac{6010}{-4.5318\times 10^{-4}}=-1.326\times 10^7\text{Pa/K}$ Yeh step kyun? Negative, jaise forecast tha — melting curve left ki taraf jhukti hai (Figure s01 mein cyan D-line).
3. $dT/dP$ nikalne ke liye invert karo aur scale karo. $dT/dP=1/(dP/dT)=-7.54\times 10^{-8}\text{K/Pa}$. $T$ ko $1\text{K}$ drop karne ke liye: $\Delta P=\frac{-1}{-7.54\times 10^{-8}}=1.326\times 10^7\text{Pa}\approx 131\text{atm}$ Yeh step kyun? Yeh abstract slope ko ek tangible "kitna squeeze karna padega?" number mein convert karta hai.

Verify: $131\text{atm}$ sirf $1\text{K}$ melting point shift karne ke liye — dikhata hai ki effect real hai par chhota — parent note ki is remark ke saath consistent hai ki skate blade ke neeche pressure-melting ek tiny effect hai. ✓ Sign negative ✓.

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Example 5 — Case E: straight-line fit se $L$ nikalna

1. Do axis values compute karo. $\ln P_1=\ln 40000=10.5966$, $\ln P_2=\ln 90000=11.4076$. Aur $1/T_1=2.8571\times 10^{-3}$, $1/T_2=2.6316\times 10^{-3}$. Yeh step kyun? Hume linearising coordinates $(1/T,\ln P)$ mein plot karna chahiye; raw $P$ vs $T$ curve hogi.
2. Line ka slope. $\text{slope}=\frac{\ln P_2-\ln P_1}{1/T_2-1/T_1}=\frac{11.4076-10.5966}{2.6316\times 10^{-3}-2.8571\times 10^{-3}}=\frac{0.8110}{-2.2556\times 10^{-4}}=-3595.4\text{K}$ Yeh step kyun? Formula $\ln P=-\frac{L}{R}\cdot \frac{1}{T}+C$ ka matlab hai yeh rise-over-run hi $-L/R$ hai.
3. $L$ recover karo. $L=-R\times \text{slope}=-8.314\times (-3595.4)=29.90\times 10^3\text{J/mol}\approx 29.9\text{kJ/mol}$ Yeh step kyun? $-R$ se multiply karna slope definition ko undo karta hai.

Verify: Units — slope K mein, $R$ J/(mol·K) mein, product J/mol mein. ✓ Magnitude $\sim 30\text{kJ/mol}$ ek believable vaporisation heat hai. ✓

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Example 6 — Case F: degenerate limit $\Delta v\to 0$

1. Diye gaye point par evaluate karo (sirf exact form). $\frac{dP}{dT}=\frac{2000}{600\times 5.0\times 10^{-6}}=\frac{2000}{3.0\times 10^{-3}}=6.667\times 10^5\text{Pa/K}$ Yeh step kyun? Log form yahan invalid hai — criticality ke paas vapour ideal nahi hai aur $v_{liquid}$ negligible nahi hai.
2. Limit lo. Jaise $\Delta v\to 0$ jabki $L$ uske relative stable rahe, $dP/dT=L/(T\Delta v)\to \infty$. Yeh step kyun? Yeh degenerate cell hai — formula "break" nahi hota, yeh ek diverging slope predict karta hai, jise experiment bhi confirm karta hai jab curve vertical ho jaati hai aur critical point par khatam hoti hai.
3. Validity ki boundary batao. Exactly critical point par dono $L=0$ aur $\Delta v=0$ hain, isliye $dP/dT$ indeterminate $0/0$ ban jaata hai — Clausius-Clapeyron akele slope decide nahi kar sakta, aur coexistence line simply terminate ho jaati hai.

Verify: Diye gaye numbers par, $6.667\times 10^5\text{Pa/K}$ finite aur positive hai. ✓ $\Delta v\to 0$ divergence aur $0/0$ endpoint milkar explain karte hain ki phase diagrams mein vapour curve ek point par kyun khatam hoti hai — dekho Phase Diagrams and Triple Point.

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Example 7 — Case G: real-world word problem

1. Form choose karo aur symbols ke saath likho, phir substitute karo. Gas phase present hai, isliye vapour log form use karo aur $P_2$ ke liye solve karo: $\ln \frac{P_2}{P_1}=-\frac{L}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)=-\frac{40700}{8.314}\left(\frac{1}{361}-\frac{1}{373}\right)$ Yeh step kyun? Matrix ka cell G Case A hai jo ek story ke roop mein dress hua hai — vapour involved hai ⇒ integrated log form; pehle symbolic equation likhne se substitution pattern Examples 1 aur 2 jaise hi rehta hai.
2. Bracket. $\frac{1}{361}-\frac{1}{373}=2.7701\times 10^{-3}-2.6810\times 10^{-3}=8.9086\times 10^{-5}{\text{K}}^{-1}$. Yeh step kyun? Positive kyunki humne temperature ghatayi ($1/T$ badhta hai).
3. Exponent. $-\frac{40700}{8.314}\times 8.9086\times 10^{-5}=-4895.4\times 8.9086\times 10^{-5}=-0.43613$. Yeh step kyun? Yeh $\ln (P_2/P_1)$ hai; negative matlab $P_2<P_1$.
4. Solve karo. $P_2=101.3\times e^{-0.43613}=101.3\times 0.6466=65.5\text{kPa}$.

Verify: $65.5\text{kPa}$ lagbhag $3500\text{m}$ altitude correspond karta hai — ek realistic high mountain. ✓ Kam pressure ↔ kam boiling point, forecast se match karta hai, aur explain karta hai ki upar chawal zyada time kyun lete hain (thanda boiling water).

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Example 8 — Case H: exam twist, $L$ solve karo mixed units ke saath

1. Recognize karo ki tumhe sirf ratio chahiye. $P_2/P_1=2$, isliye $\ln (P_2/P_1)=\ln 2=0.6931$. Yeh step kyun? Matrix ka cell H ek vaporisation problem hai jo $L$ ke liye backwards solve kiya gaya hai; log form sirf pressure ratio par depend karta hai, isliye absolute pressures kabhi bhi zaruri nahi hote — yahi trap hai.
2. Likho aur $L$ ke liye rearrange karo. $\ln 2=-\frac{L}{R}\left(\frac{1}{320}-\frac{1}{300}\right)\Rightarrow L=\frac{-R\ln 2}{\frac{1}{320}-\frac{1}{300}}$ Yeh step kyun? $L$ akela unknown hai; ise algebraically isolate karo.
3. Bracket compute karo. $\frac{1}{320}-\frac{1}{300}=3.1250\times 10^{-3}-3.3333\times 10^{-3}=-2.0833\times 10^{-4}{\text{K}}^{-1}$. Yeh step kyun? Negative kyunki $T$ badha — yeh negative leading minus ke saath milkar positive $L$ dega.
4. Finish karo. $L=\frac{-8.314\times 0.6931}{-2.0833\times 10^{-4}}=\frac{-5.7625}{-2.0833\times 10^{-4}}=2.766\times 10^4\text{J/mol}\approx 27.7\text{kJ/mol}$

Verify: Positive $L\approx 27.7\text{kJ/mol}$ — physically sensible hai. ✓ Sign logic cross-check karo: temperature upar + pressure upar ⇒ positive $L$, evaporation ke liye correct hai (Latent Heat). ✓

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Recall Kaunse cell ke liye kaunsa form?

Kya vapour involved hai? ::: Haan → integrated log form ($\ln P_2/P_1=-\frac{L}{R}(1/T_2-1/T_1)$); Nahi (solid–liquid) → exact form $dP/dT=L/(T\Delta v)$. Melting-curve slope ka SIGN kya decide karta hai? ::: $\Delta v=v_{liquid}-v_{solid}$ ka sign; water ke liye negative (ice kam dense hai) negative slope deta hai. $dP/dT$ ka kya hota hai jab $\Delta v\to 0$ critical point ke paas? ::: Yeh diverge karta hai (curve vertical ho jaati hai); exactly critical point par $L=0$ aur $\Delta v=0$ se $0/0$ aata hai aur line khatam ho jaati hai. "Doubling pressure" exam trick mein, tumhe actually kya info chahiye? ::: Sirf pressure ratio $P_2/P_1$ (yahan $=2$), absolute pressures nahi.

Yeh bhi dekho Gibbs Free Energy, Entropy and the Second Law, First Law of Thermodynamics aur Maxwell Relations ke liye ki exact form khud kahan se aata hai.