Phase equilibrium — Clausius-Clapeyron equation: Exercises

2.4.6 · D4 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Phase equilibrium — Clausius-Clapeyron equation

Constants jo poore use hote hain: gas constant $R = 8.314 J mo l^{- 1} K^{- 1}$ .

Do equations jinpar hum rely karte hain:

Level 1 — Recognition

Goal: sahi equation choose karo aur signs read karo. Heavy algebra nahi.

L1.1

Ek liquid aur uska vapour coexist kar rahe hain. Equilibrium par dono phases ki kaunsi quantities equal hoti hain?

Recall Solution

==Temperature $T$ , pressure $P$ , aur chemical potential $μ$ == (molar Gibbs free energy, dekho Gibbs Free Energy). KYU: agar $μ$ alag hota, toh particles higher- $μ$ phase se lower- $μ$ phase mein flow karte total $G$ ko kam karne ke liye — toh true equilibrium nahi hoga. Equal $T$ aur $P$ thermal aur mechanical equilibrium hain; equal $μ$ chemical equilibrium hai.

L1.2

Water→vapour transition (boiling) ke liye, coexistence curve ka slope $d P / d T$ positive hai ya negative? $Δ v$ aur $L$ ke sign se explain karo.

Recall Solution

Dono $L > 0$ (boil karne ke liye heat absorb hoti hai) aur $Δ v = v_{g a s} - v_{l i q} > 0$ (vapour kaafi bada hota hai). Toh $\frac{d P}{d T} = \frac{L}{T Δ v} > 0.$ Coexistence curve upar jaati hai — zyada $T$ ko dono phases ko balance mein rakhne ke liye zyada $P$ chahiye. Ye rozmarra ki zindagi se match karta hai: pressure cooker (high $P$ ) mein paani zyada temperature par boil hota hai.

Level 2 — Application

Goal: ek equation mein saaf numbers daalo.

L2.1

Ammonia ka $L_{v a p} = 23.3 kJ/mol$ hai aur ye $P_{1} = 101.3 kPa$ , $T_{1} = 239.8 K$ par boil karta hai. $T_{2} = 250 K$ par uska vapour pressure estimate karo.

Recall Solution

KYU vapour form: ek phase gas hai, $Δ v \approx v_{g a s} = R T / P$ (dekho Ideal Gas Law), toh use karte hain $ln \frac{P _{2}}{P _{1}} = - \frac{L}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$ Bracket compute karo: $\frac{1}{250} - \frac{1}{239.8} = 4.000 \times 1 0^{- 3} - 4.170 \times 1 0^{- 3} = - 1.702 \times 1 0^{- 4} K^{- 1} .$ Phir $ln \frac{P _{2}}{P _{1}} = - \frac{23300}{8.314} \times (- 1.702 \times 1 0^{- 4}) = + 0.4771.$ Toh $P_{2} = 101.3 e^{0.4771} = 101.3 \times 1.611 = 163.2 kPa$ . Sanity: zyada $T$ → zyada vapour pressure. ✓

L2.2

Ek metal ki melting curve ka measured slope $d P / d T = 3.0 \times 1 0^{7} Pa/K$ hai uske melting point $T = 600 K$ par, aur melting par molar volume change $Δ v = 6.0 \times 1 0^{- 7} m^{3} /mol$ hai. $L_{f u s}$ nikalo.

Recall Solution

KYU exact form: koi bhi phase gas nahi hai, toh ideal-gas shortcut illegal hai — use karo $\frac{d P}{d T} = \frac{L}{T Δ v}$ . Rearrange karo: $L = \frac{d P}{d T} \cdot T \cdot Δ v = (3.0 \times 1 0^{7}) (600) (6.0 \times 1 0^{- 7}) = 1.08 \times 1 0^{4} J/mol .$ Toh $L_{f u s} = 10.8 kJ/mol$ . Positive $Δ v$ aur positive slope consistent hain — ye metal bhi zyaadatar solids ki tarah freeze hone par contract karta hai (water ke unlike).

Level 3 — Analysis

Goal: equation ki structure ke saath reason karo, sirf plug in mat karo.

L3.1

Ek student kisi liquid ke vapour pressures measure karta hai aur $ln P$ ko $1/ T$ ke against plot karta hai, $- 4200 K$ slope aur $+ 22.5$ intercept wali straight line milti hai. (a) $L_{v a p}$ nikalo. (b) $T = 350 K$ par $P$ predict karo ( $P$ Pa mein).

Recall Solution

KYU straight line: vapour form integrate karne par milta hai $ln P = - \frac{L}{R} \cdot \frac{1}{T} + C$ . $y = m x + c$ se compare karo jahan $x = 1/ T$ : slope hai $m = - L / R$ aur $c$ intercept hai. Ye linearisation hi experiment ka poora point hai.

Figure — Phase equilibrium — Clausius-Clapeyron equation

(a)

L = - R \times (slope) = - 8.314 \times (- 4200) = 3.492 \times 1 0^{4} J/mol = 34.9 kJ/mol

. (b)

T = 350

par:

1/ T = 2.857 \times 1 0^{- 3}

ln P = - 4200 \times 2.857 \times 1 0^{- 3} + 22.5 = - 12.00 + 22.5 = 10.50.

P = e^{10.50} = 3.63 \times 1 0^{4} Pa \approx 36.3 kPa .

Graph par sanity check: steeper (zyada negative) slope matlab bada latent heat — vapour pressure temperature ke saath tez chadhta hai. Figure mein red line exactly ye fit hai.

L3.2

$\frac{d P}{d T} = \frac{L}{T Δ v}$ use karke explain karo ki sublimation curve (solid→vapour) triple point ke paas vapourisation curve (liquid→vapour) se steeper kyu hoti hai.

Recall Solution

Dono curves mein $Δ v \approx v_{g a s}$ hota hai (gas dono solid aur liquid se kaafi bada hota hai), aur dono triple point ke paas almost same $T$ share karte hain (dekho Phase Diagrams and Triple Point). Toh ek hi alag factor hai $L$ : $(\frac{d P}{d T})_{sub} / (\frac{d P}{d T})_{vap} \approx \frac{L _{sub}}{L _{vap}} .$ Kyunki $L_{sub} = L_{fus} + L_{vap}$ (solid→gas jaane ke liye energetically dono melt aur boil karna padta hai — dekho Latent Heat), hume milta hai $L_{sub} > L_{vap}$ . Isliye sublimation slope bada hai: sublimation curve steeper hoti hai. Ye har phase diagram ke triple point ki ek real, checkable feature hai.

Level 4 — Synthesis

Goal: CC ko entropy, Gibbs energy, ya two-step reasoning ke saath combine karo.

L4.1

Paani $373 K$ par $101.3 kPa$ mein boil karta hai aur $L_{v a p} = 40.7 kJ/mol$ hai. (a) $Δ s = L / T$ use karke molar entropy of vapourisation nikalo. (b) Exact form mein $Δ v \approx v_{g a s} = R T / P$ use karke boiling point par $d P / d T$ nikalo. (c) Us slope se $375 K$ par boiling pressure estimate karo (linear approximation) aur full log formula se compare karo.

Recall Solution

(a) Transition isothermal aur reversible hai, toh Entropy and the Second Law se, $Δ s = \frac{L}{T} = \frac{40700}{373} = 109.1 J mo l^{- 1} K^{- 1} .$ (Ye Trouton's rule $\approx 88$ ke kareeb hai; water usse upar hai kyunki hydrogen bonding hoti hai.) (b) Pehle gas molar volume: $v_{g a s} = \frac{R T}{P} = \frac{8.314 \times 373}{101300} = 0.03061 m^{3} /mol .$ Phir $\frac{d P}{d T} = \frac{L}{T Δ v} = \frac{40700}{373 \times 0.03061} = 3565 Pa/K .$ (c) $Δ T = 2 K$ par linear estimate: $P (375) \approx 101300 + 3565 \times 2 = 108430 Pa = 108.4 kPa .$ Full log formula comparison ke liye: $ln \frac{P _{2}}{101.3} = - \frac{40700}{8.314} (\frac{1}{375} - \frac{1}{373}) = - 4895 \times (- 1.430 \times 1 0^{- 5}) = 0.07001,$ $P_{2} = 101.3 e^{0.07001} = 108.6 kPa .$ Agreement: $108.4$ vs $108.6$ kPa. Linear tangent aur exact curve ek chhote 2 K step mein match karte hain — exactly wahi jo "slope $d P / d T$ hai" promise karta hai. Bade step par ye diverge hote kyunki $ln P$ curved hota hai.

L4.2

Dikhao ki exact Clausius-Clapeyron equation curve ke saath $d μ_{1} = d μ_{2}$ require karne se aati hai, phir ek sentence mein explain karo ki kaun si thermodynamic identity guarantee karti hai ki $d g = - s d T + v d P$ .

Recall Solution

Coexistence par $μ_{1} = μ_{2}$ har jagah hai, toh unke changes match karte hain: $d μ_{1} = d μ_{2}$ . Har phase ke liye $d μ = d g = - s d T + v d P$ ke saath (First Law of Thermodynamics ko $d G = d U - T d S - S d T + P d V + V d P$ mein plug karo, jahan $d U = T d S - P d V$ middle terms cancel karta hai): $- s_{1} d T + v_{1} d P = - s_{2} d T + v_{2} d P .$ Collect karo: $(v_{1} - v_{2}) d P = (s_{1} - s_{2}) d T$ , isliye $\frac{d P}{d T} = \frac{s _{2} - s _{1}}{v _{2} - v _{1}} = \frac{Δ s}{Δ v} = \frac{L}{T Δ v} .$ Ek-sentence identity: $d g = - s d T + v d P$ ye statement hai ki $g$ ke natural variables $T$ aur $P$ hain, jo exactly wahi setup hai jisse Maxwell Relations $(\frac{\partial s}{\partial P})_{T} = - (\frac{\partial v}{\partial T})_{P}$ $g$ ke mixed second derivatives ki equality se derive hoti hai.

Level 5 — Mastery

Goal: sign subtleties aur interpretation ke saath full multi-step problems.

L5.1 — Ice-skating myth, quantitatively

Ice $T = 273.15 K$ par $101.3 kPa$ mein melt hoti hai, $L_{f u s} = 6010 J/mol$ , $v_{w a t er} = 18.00 \times 1 0^{- 6}$ , $v_{i ce} = 19.66 \times 1 0^{- 6} m^{3} /mol$ ke saath. (a) $d T / d P$ compute karo (melting temperature har pascal par kitna shift karti hai). (b) Ek skater blade ke neeche roughly $Δ P = 5.0 \times 1 0^{6} Pa$ exert karta hai. Melting-point drop $Δ T$ estimate karo. (c) Kya pressure-melting skating ke liye convincing explanation hai? Comment karo.

Recall Solution

(a) $Δ v = v_{w a t er} - v_{i ce} = (18.00 - 19.66) \times 1 0^{- 6} = - 1.66 \times 1 0^{- 6} m^{3} /mol$ (negative — ice less dense hoti hai). $\frac{d P}{d T} = \frac{L}{T Δ v} = \frac{6010}{273.15 \times ( - 1.66 \times 1 0 ^{- 6} )} = - 1.326 \times 1 0^{7} Pa/K .$ Invert karo: $\frac{d T}{d P} = \frac{1}{- 1.326 \times 1 0 ^{7}} = - 7.54 \times 1 0^{- 8} K/Pa .$ (b) $Δ T \approx \frac{d T}{d P} Δ P = (- 7.54 \times 1 0^{- 8}) (5.0 \times 1 0^{6}) = - 0.377 K$ . (c) Sirf $\approx 0.38 K$ ki drop — melting temperature lagbhag $- 0.38 °C$ tak girta hai. Skating isse kaafi zyada thandi jagahon par hoti hai, toh pressure-melting skating explain karne ke liye bahut weak hai. Asli culprit frictional heating plus naturally pre-melted surface layer hai. Lekin sign hi physics ka gem hai: kyunki $Δ v < 0$ hai, pressure melting point ko neeche laata hai — ye water ki anomaly ka direct fingerprint hai.

L5.2 — Field data se two-point latent heat

Ek mountaineer paani ko $T_{1} = 373.15 K$ ( $P_{1} = 101.3 kPa$ ) par sea level par aur $T_{2} = 344.15 K$ ( $P_{2} = 31.0 kPa$ ) par summit par boil hote dekhta hai. Constant $L$ ke saath vapour form assume karo aur $L_{v a p}$ extract karo.

Recall Solution

KYU vapour form: liquid↔vapour, gas volume dominate karta hai. $ln \frac{P _{2}}{P _{1}} = - \frac{L}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ ko $L$ ke liye solve karo: $L = - R \frac{l n ( P _{2} / P _{1} )}{\frac{1}{T _{2}} - \frac{1}{T _{1}}} .$ Numerator: $ln \frac{31.0}{101.3} = ln (0.3060) = - 1.1841$ . Bracket: $\frac{1}{344.15} - \frac{1}{373.15} = 2.9057 \times 1 0^{- 3} - 2.6799 \times 1 0^{- 3} = 2.258 \times 1 0^{- 4} K^{- 1}$ . $L = - 8.314 \times \frac{- 1.1841}{2.258 \times 1 0 ^{- 4}} = 8.314 \times 5244 = 4.36 \times 1 0^{4} J/mol = 43.6 kJ/mol .$ Sanity: water ke liye accepted $L_{v a p}$ $\approx 40.7 kJ/mol$ hai; humara two-point estimate $43.6$ close hai — thoda excess $L$ ke 29 K span mein temperature ke saath drift karne se aata hai. Do pressures, do temperatures → ek latent heat, entirely CC se.

Recall Poore page ka one-line recap

Kisi bhi transition ke liye Exact form; log form tabhi jab ek phase gas ho; $ln P$ vs $1/ T$ ka slope $- L / R$ hota hai; aur signs matter karte hain — negative $Δ v$ (water) melting curve ko flip kar deta hai. ::: Agar tum kisi problem ko "gas phase present hai ya nahi" mein classify kar sako aur $Δ v$ ka sign track karo, toh har CC exercise solve ho jaata hai.