Thermodynamic potentials — U, H, F, G (preview): Exercises

1.7.26 · D4 · Physics › Thermodynamics › Thermodynamic potentials — U, H, F, G (preview)

Definition Teen heat symbols jo tumhe baar baar milenge:

$δ Q$ , $δ Q_{rev}$ , $δ Q_{p}$ $d$ (jaise $d H$ , $d U$ mein) ek exact differential hai — ek state function mein ek chhoti si change, yaani ek aisi quantity jo sirf is baat par depend karti hai ki system kahan hai, na ki woh wahan kaise pahuncha. Poori change $\int d H = H_{final} - H_{initial}$ path-independent hoti hai. $δ Q$ (curly $δ$ ke saath, na ki $d Q$ ) ek inexact differential hai — boundary ke paar jaata hua thoda sa heat. Heat $Q$ ek state function nahi hai; kitna flow hoga yeh path par depend karta hai, isliye koi function $Q$ nahi hai jise difference kiya ja sake. $δ$ ek warning sign hai. $δ Q_{rev}$ wahi heat hai jo ek reversible path ke along measure ki gayi ho — yeh wahi ek case hai jiske liye $δ Q_{rev} = T d S$ bilkul sahi hota hai. $δ Q_{p}$ (L2.2 mein use kiya gaya) wahi inexact heat $δ Q$ hai, lekin extra tag " $p$ " ke saath jiska matlab hai pressure ko constant rakh ke measure kiya gaya. Yeh koi naya object nahi hai — bas $δ Q$ jo ek constant-pressure path ke liye specialized hai (aur reversible li gayi hai, isliye wahan $δ Q_{p} = T d S$ ). Subscript ko experimental condition ka naam samjho, bilkul waise jaise partial derivative par subscript hota hai.

Intuition Har exercise ke peeche ek tasveer (Level 1 se pehle padho)

Neeche ke do panels dekho. Left panel: Legendre transform koi algebra ka jaadu nahi hai — yeh ek tasveer hai. Orange curve $U (S)$ hai; har point par iska slope temperature $T$ hai. Variable $S$ ko uske slope $T$ se badalne ke liye, hum ek seedhi line jo slope $T$ ki ho use slide karte hain neeche tab tak jab tak woh curve ko just touch na kare (tangent na ho), aur jahan woh vertical axis ko cross kare woh record karte hain — woh intercept hai $F = U - T S$ . Magenta double-arrow wahi subtracted product $T S$ hai. Yeh poora matlab hai " $F = U - T S$ " ka: ek point ko ek tangent line se trade karo. Right panel: kyun "potential ka minimum = equilibrium." Violet bowl ek potential hai (maan lo $F$ ) jo kisi internal coordinate ke against plot ki gayi hai jise system adjust kar sakta hai; koi bhi ball kahin bhi rakh do woh neeche roll kar jaati hai. Neeche woh jagah hai jahan slope zero hota hai — equilibrium. Is page par sab kuch ya toh (a) ek differential se slope padhna hai, ya (b) aisa minimum dhundhna hai.

Level 1 — Recognition

Yahan hum tumse sirf sahi formula, naam, ya variable padhne ko bolte hain. Koi calculation nahi, koi trap nahi — bas yeh jaanna ki kaun sa wallet kaun sa hai.

L1.1 — Natural variables ka naam batao

Har potential ke liye, uske do natural variables batao (woh do jinke differentials $d (\cdot)$ mein aate hain): $U$ , $H$ , $F$ , $G$ .

Recall Solution L1.1

Chaaron differentials se seedha padho: $d U = T d S - p d V \Rightarrow (S, V)$ $d H = T d S + V d p \Rightarrow (S, p)$ $d F = - S d T - p d V \Rightarrow (T, V)$ $d G = - S d T + V d p \Rightarrow (T, p)$ Rule of thumb: natural variables exactly woh do cheezein hain jinke differentials ( $d S, d V, d p, d T$ ) right-hand side par baithte hain.

L1.2 — Kaun sa potential minimize hota hai?

Ek reaction lab bench par ek open beaker mein chalti hai — atmosphere pressure fix karta hai, room temperature fix karta hai. Equilibrium par system kaun sa potential minimize karta hai?

Recall Solution L1.2

Constant $T$ aur constant $p$ → matching potential hai Gibbs free energy $G$ , jiske natural variables $(T, p)$ hain. Yeh chemistry wala case hai (dekho Gibbs Free Energy and Chemical Equilibrium). Upar ki figure ke right-hand bowl ki tasveer mein $F$ ki jagah $G$ rakh lo: system neeche ki taraf roll karta hai.

L1.3 — Definition dhundho

Har blank bharo: $H = U + \underline{p V}$ , $F = U - \underline{T S}$ , $G = U + \underline{p V} - \underline{T S}$ .

Recall Solution L1.3

$H = U + p V, F = U - T S, G = U + p V - T S .$ Parent note se mnemonic: $p V$ jodo → enthalpy; $T S$ ghataao → free; dono → Gibbs.

Level 2 — Application

Ab differentials mein plug karo taaki ek derivative, ek heat, ya ek heat capacity nikaal sako.

L2.1 — $F$ se Entropy

$F (T, V)$ diya hua hai, $S$ ko $F$ ke partial derivative ke roop mein likho.

Recall Solution L2.1

Step 1 (KYA): $d F = - S d T - p d V$ se shuru karo. Step 2 (KYUN): $V$ ko fixed rakho toh $d V = 0$ ; phir $d F = - S d T$ , aur $d T$ ke coefficient se match karo: $S = - (\frac{\partial F}{\partial T})_{V}$ Minus sign seedha $- S d T$ term se aata hai — "S follows T with a minus."

L2.2 — $H$ se Heat capacity $C_{p}$

Dikhao ki constant-pressure heat capacity $C_{p} = (\frac{\partial H}{\partial T})_{p}$ hai, aur compute karo $Δ H$ jab $2 mol$ ideal monatomic gas ko constant pressure par $300 K$ se $400 K$ tak garam kiya jaaye. ( $C_{p} = \frac{5}{2} n R$ , $R = 8.314 J mol^{- 1} K^{- 1}$ use karo.)

Recall Solution L2.2

Step 0 (notation): $δ Q$ , $δ Q_{rev}$ aur $δ Q_{p}$ sab is page ke top par box mein define hain. Yahan $δ Q_{p}$ ka matlab hai "inexact heat $δ Q$ jo ek reversible, constant-pressure path ke along measure ki gayi hai," isliye is path par $δ Q_{p} = δ Q_{rev} = T d S$ . Step 1: $d H = T d S + V d p$ . Ab $p$ ko fixed rakho, toh $d p = 0$ , bacha $d H = T d S$ . Reversible path ke along $T d S = δ Q_{rev}$ , isliye constant $p$ par: $d H = δ Q_{p} (constant p) .$ Saral bhasha mein: enthalpy mein exact change inexact heat ke barabar hai kyunki constant- $p$ constraint doosra term ( $V d p$ ) hata deta hai. Isliye $Δ H$ ko open-flask chemistry ke liye "heat content" kaha jaata hai. Step 2: Heat capacity ki definition se, $δ Q_{p} = C_{p} d T$ . $d H = δ Q_{p}$ rakh ke aur $d T$ se divide karo fixed $p$ par: $C_{p} = (\frac{\partial H}{\partial T})_{p} .$ Step 3 (numbers): $C_{p} = \frac{5}{2} n R = \frac{5}{2} (2) (8.314) = 41.57 J K^{- 1}$ . $Δ H = C_{p} Δ T = 41.57 \times (400 - 300) = 4157 J \approx 4.16 kJ .$ $C_{p} > C_{v}$ kyun hota hai yeh jaanne ke liye dekho Heat Capacities Cp and Cv.

L2.3 — $G$ se Volume

$G (T, p)$ diya hua hai, $S$ aur $V$ dono ko derivatives ke roop mein nikalo.

Recall Solution L2.3

$d G = - S d T + V d p$ se: $S = - (\frac{\partial G}{\partial T})_{p}, V = + (\frac{\partial G}{\partial p})_{T} .$ Signs kyun? $- S$ multiply karta hai $d T$ ko, $+ V$ multiply karta hai $d p$ ko — seedha padho. " $V$ follows $p$ with a plus."

Level 3 — Analysis

Yahan tum do baatein combine karte ho, ya yeh socha karte ho ki ek step kyun legal hai.

L3.1 — $G$ se ek Maxwell relation

$G$ ke andar chhupa Maxwell relation derive karo: $(\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p} .$

Recall Solution L3.1

Step 1: $d G = - S d T + V d p$ se, first derivatives padho (har ek ke saath uska held-fixed variable naam ke saath): $- S = (\frac{\partial G}{\partial T})_{p}, V = (\frac{\partial G}{\partial p})_{T} .$ Step 2 (key idea — aur uski permission): $G$ ek state function hai $(T, p)$ ka jiske second partial derivatives continuous hain interest ke region mein. Us continuity (regularity condition) ke under, Clairaut's / Schwarz's theorem guarantee karta hai ki do mixed second derivatives equal hain — differentiation ka order matter nahi karta: $[\frac{\partial}{\partial p} (\frac{\partial G}{\partial T})_{p}]_{T} = [\frac{\partial}{\partial T} (\frac{\partial G}{\partial p})_{T}]_{p} .$ "Order doesn't matter" kaisa dikhta hai (neeche figure dekho): surface $G (T, p)$ par, ek neighbouring point tak pahunchna "pehle $T$ mein step, phir $p$ mein step" karke usi height par utarta hai jaise "pehle $p$ mein step, phir $T$ mein step" — chhota sa quadrilateral band ho jaata hai. Equal mixed partials exactly yeh statement hai ki yeh loop band hota hai. Agar $G$ mein koi kink ya discontinuous curvature hoti, toh loop band hone mein fail ho sakta tha aur Maxwell relation toot jaata; second derivatives ki continuity exactly yahi forbid karti hai.

Figure — Thermodynamic potentials — U, H, F, G (preview)

Step 3: Step-1 identities ko dono sides mein substitute karo: $(\frac{\partial ( - S )}{\partial p})_{T} = (\frac{\partial V}{\partial T})_{p} \Rightarrow (\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p} .$ Yeh ek unmeasurable entropy slope ko ordinary thermal-expansion slope se trade karta hai. Dekho Maxwell Relations.

L3.2 — Check karo ki ek candidate valid $G$ hai

Koi propose karta hai $G (T, p) = a T p^{2}$ (constants $a > 0$ ). $S = - (\partial G / \partial T)_{p}$ aur $V = (\partial G / \partial p)_{T}$ use karke $S$ aur $V$ nikalo, phir verify karo ki L3.1 ka Maxwell relation automatically hold karta hai.

Recall Solution L3.2

Compute (har derivative mein bataao kya held fixed hai): $S = - (\frac{\partial}{\partial T} (a T p^{2}))_{p} = - a p^{2}$ ( $p$ ko fixed rakho, isliye $p^{2}$ ek constant multiplier hai). $V = (\frac{\partial}{\partial p} (a T p^{2}))_{T} = 2 a T p$ ( $T$ ko fixed rakho). Maxwell check: $(\frac{\partial S}{\partial p})_{T} = - 2 a p$ ( $T$ fixed) aur $- (\frac{\partial V}{\partial T})_{p} = - (2 a p) = - 2 a p$ ( $p$ fixed). Equal ✓ — aur note karo $G = a T p^{2}$ ek polynomial hai, isliye uske second derivatives poori jagah continuous hain; L3.1 ki regularity condition poori hoti hai. (Physical aside: $S = - a p^{2} < 0$ is cheez ko ek toy banata hai, na ki koi real substance, lekin calculus consistent hai — bas yahi point hai: koi bhi smooth $G$ automatically apna Maxwell relation satisfy karta hai.)

L3.3 — Constant $T$ par $F$ kyun, $U$ kyun nahi?

Ek clean paragraph mein explain karo kyun ek system jo constant temperature aur volume par rakha gaya hai woh $F$ ke minimum par settle karta hai, $U$ ke nahi.

Recall Solution L3.3

Second law kehta hai ki universe (system + surroundings) ki entropy sirf badh sakti hai. Jab system temperature $T$ par ek bath ke saath heat $δ Q$ exchange karta hai, toh bath ki entropy $- δ Q / T$ se change hoti hai. System ki apni entropy ke saath combine karo aur, fixed $T, V$ par, total-entropy condition $d S_{univ} \geq 0$ algebraically $d F \leq 0$ ban jaati hai. Isliye system spontaneously $F$ mein neeche slide karta hai (opening figure mein violet bowl ke neeche roll karo) aur minimum par ruk jaata hai. $U$ akela woh entropy ignore karta hai jo system bath mein dump karta hai, isliye uska minimum yahan equilibrium condition nahi hai. (Poora statement: Entropy and the Second Law.)

Level 4 — Synthesis

Kai potentials ko jodne wala multi-step result banao.

L4.1 — Gibbs–Helmholtz equation

$G = H - T S$ aur $S = - (\partial G / \partial T)_{p}$ se shuru karke prove karo $H = G - T (\frac{\partial G}{\partial T})_{p} aur isliye (\frac{\partial ( G / T )}{\partial T})_{p} = - \frac{H}{T ^{2}} .$

Recall Solution L4.1

Step 1: $S = - (\partial G / \partial T)_{p}$ ko $G = H - T S$ mein substitute karo: $G = H - T (- \frac{\partial G}{\partial T})_{p} = H + T (\frac{\partial G}{\partial T})_{p} .$ Rearrange karo: $H = G - T (\frac{\partial G}{\partial T})_{p} .$ ✓ Step 2 ( $G / T$ form kyun): $G / T$ ko $T$ ke respect mein differentiate karo $p$ fixed rakh ke, quotient rule use karke (note karo $T$ variable hai, isliye $1/ T$ constant nahi hai): $(\frac{\partial ( G / T )}{\partial T})_{p} = \frac{1}{T} (\frac{\partial G}{\partial T})_{p} - \frac{G}{T ^{2}} = \frac{1}{T ^{2}} [T (\frac{\partial G}{\partial T})_{p} - G] .$ Step 3: bracket exactly $- H$ hai Step 1 se, isliye $(\frac{\partial ( G / T )}{\partial T})_{p} = - \frac{H}{T ^{2}} .$ Yeh woh workhorse hai jo chemists ko equilibrium ki temperature-dependence se reaction enthalpies nikaalne deta hai.

L4.2 — Maxwell relation se $C_{p}$ aur $C_{v}$ relate karo

Pehle identity $C_{p} - C_{v} = T (\frac{\partial p}{\partial T})_{V} (\frac{\partial V}{\partial T})_{p}$ ko motivate karo, phir ek mole ideal gas ( $p V = R T$ ) ke liye evaluate karo aur confirm karo ki $C_{p} - C_{v} = R$ .

Recall Solution L4.2

Identity kahan se aati hai ("kyun"). Dono heat capacities "heat per degree" measure karti hain, lekin alag constraints ke under, isliye unka farq exactly woh extra energy hai jo constant-pressure path kharaab karta hai. Step A: Definition se $C_{v} = T (\frac{\partial S}{\partial T})_{V}$ aur $C_{p} = T (\frac{\partial S}{\partial T})_{p}$ (from $δ Q_{rev} = T d S$ , har constraint ke under $d T$ se divide karo). Step B: Entropy ko $S (T, V)$ maano aur uska total differential expand karo, phir chain rule se constant- $p$ temperature slope padho: $(\frac{\partial S}{\partial T})_{p} = (\frac{\partial S}{\partial T})_{V} + (\frac{\partial S}{\partial V})_{T} (\frac{\partial V}{\partial T})_{p} .$ Step C: $T$ se multiply karo aur $C_{v}$ ghataao: $C_{p} - C_{v} = T (\frac{\partial S}{\partial V})_{T} (\frac{\partial V}{\partial T})_{p} .$ Step D (Maxwell laao): $F$ se Maxwell relation (parent note) hai $(\frac{\partial S}{\partial V})_{T} = (\frac{\partial p}{\partial T})_{V}$ . Substitute karo aur advertised identity tak pahuncho: $C_{p} - C_{v} = T (\frac{\partial p}{\partial T})_{V} (\frac{\partial V}{\partial T})_{p} .$ Yeh ek entropy quantity (mushkil) ko do easy gas-law slopes mein badal deta hai. Ab ideal gas $p V = R T$ (one mole) ke liye evaluate karo: Step 1: $V$ fixed rakho aur likho $p = R T / V \Rightarrow (\frac{\partial p}{\partial T})_{V} = \frac{R}{V}$ . Step 2: $p$ fixed rakho aur likho $V = R T / p \Rightarrow (\frac{\partial V}{\partial T})_{p} = \frac{R}{p}$ . Step 3: Multiply karo aur $T$ lagao: $C_{p} - C_{v} = T \cdot \frac{R}{V} \cdot \frac{R}{p} = \frac{T R ^{2}}{p V} = \frac{T R ^{2}}{R T} = R,$ kyunki $p V = R T$ denominator cancel kar deta hai. Toh $C_{p} - C_{v} = R = 8.314 J mol^{- 1} K^{- 1}$ . ✓ (Mayer's relation.)

Level 5 — Mastery

Full-power problems: ek potential reconstruct karo, ya kisi limiting/degenerate case ke baare mein reason karo.

L5.1 — Ideal gas ke liye $F$ reconstruct karo

Ek mole ideal monatomic gas ke liye entropy hai (Sackur–Tetrode, schematically) $S (T, V) = C_{v} ln T + R ln V + s_{0},$ $C_{v} = \frac{3}{2} R$ aur constant $s_{0}$ ke saath. $F = U - T S$ use karo jahan $U = C_{v} T + u_{0}$ hai, aur phir $p = - (\partial F / \partial V)_{T}$ se ideal-gas law $p = R T / V$ recover karo.

Recall Solution L5.1

Step 1: $F$ assemble karo: $F = U - T S = (C_{v} T + u_{0}) - T (C_{v} ln T + R ln V + s_{0}) .$ Step 2: $- (\partial F / \partial V)_{T}$ lo, yaani $V$ ke respect mein differentiate karo $T$ fixed rakh ke. Sirf $- T R ln V$ term $V$ par depend karta hai: $(\frac{\partial F}{\partial V})_{T} = - T R \cdot \frac{1}{V} = - \frac{R T}{V} .$ Step 3: Isliye $p = - (\frac{\partial F}{\partial V})_{T} = \frac{R T}{V} .$ Ideal-gas law potential se nikal aata hai. ✓ (Note karo $U$ $V$ -derivative se nikal gaya kyunki yahan $U$ mein koi $V$ -dependence nahi hai — ideal gas ki internal energy sirf $T$ par depend karti hai.)

L5.2 — Degenerate case $T \to 0$

Third law ( $S \to 0$ as $T \to 0$ ) aur $F = U - T S$ use karke argue karo ki $T \to 0$ par $F$ aur $(\frac{\partial F}{\partial T})_{V}$ ka kya hota hai.

Recall Solution L5.2

Step 1: Jaise $T \to 0$ , product $T S \to 0$ (kyunki dono $T \to 0$ aur $S \to 0$ ), isliye $F = U - T S \to U$ . Free energy absolute zero par internal energy mein merge ho jaati hai — entropy ka koi "tax" nahi bachta. Step 2: Yaad karo $(\frac{\partial F}{\partial T})_{V} = - S$ . Kyunki $S \to 0$ , $F$ ka $T$ ke saath slope zero ho jaata hai: $F (T)$ curve $U$ ke paas horizontally approach karta hai.

Neeche ki figure padho. Horizontal axis temperature $T$ hai, vertical axis energy hai. Orange curve internal energy $U (T)$ hai (steadily $T$ ke saath badhti hai); magenta curve free energy $F (T) = U - T S$ hai. Vertical navy double-arrow ek temperature par unke beech ka gap mark karta hai — woh gap exactly entropy tax $T S$ hai, jo garam hone par badhta hai. Bilkul left mein, $T = 0$ par, magenta $F$ curve orange $U$ curve ko violet dot par touch karti hai, aur wahan tangentially (flat slope, kyunki $- S \to 0$ ) pahunchti hai. Toh dono curves absolute zero par "kiss" karti hain aur $T$ badhne ke saath alag ho jaati hain — Step 1 aur Step 2 ka poora matlab ek picture mein.

L5.3 — $Δ G$ ka sign aur spontaneity

$298 K$ par ek reaction ka $Δ H = - 50.0 kJ$ aur $Δ S = + 100 J K^{- 1}$ hai. $Δ G = Δ H - T Δ S$ compute karo aur batao ki yeh spontaneous hai ya nahi. Phir woh temperature dhundho jahan $Δ G = 0$ (equilibrium crossover), aur $T \to \infty$ case describe karo.

Recall Solution L5.3

Step 1: Units convert karo: $Δ H = - 50000 J$ , $T Δ S = 298 \times 100 = 29800 J$ . $Δ G = - 50000 - 29800 = - 79800 J = - 79.8 kJ .$ Kyunki $Δ G < 0$ , reaction $298 K$ par spontaneous hai. Step 2 (crossover): $Δ G = 0 = Δ H - T Δ S$ set karo: $T = \frac{Δ H}{Δ S} = \frac{- 50000}{100} = - 500 K .$ Negative temperature unphysical hai, matlab koi real crossover nahi hai — $Δ H < 0$ aur $Δ S > 0$ ke saath, dono terms $Δ G$ ko negative drive karte hain, isliye reaction har positive temperature par spontaneous hai. Step 3 (limit): $T \to \infty$ par, $- T Δ S$ term dominate karta hai aur $Δ G \to - \infty$ : abhi bhi spontaneous, aur bhi zyada. Yeh Ellingham-type sign analysis ka "dono favourable" quadrant hai.

Recall Jaane se pehle ek-line self-check

Page cover karo. Kya tum (a) har potential ke natural variables bata sakte ho, (b) $G$ se $S$ aur $V$ nikal sakte ho, (c) ek Maxwell relation derive kar sakte ho, aur (d) sahi units ke saath $Δ G$ compute kar sakte ho? Agar chaaron mein haan, toh yeh topic tumhara hai.

Constant

T, V

par kaun sa potential minimize hota hai?

Helmholtz free energy

F

Ek mole ideal gas ke liye,

C_{p} - C_{v} = ?

R = 8.314 J mol^{- 1} K^{- 1}

\partial (G / T) / \partial T

at fixed

p

ka Gibbs–Helmholtz form?

- H / T^{2}

T \to 0

par,

F

tend karta hai

U

ki taraf (kyunki

T S \to 0

), slope

- S \to 0

ke saath

δ

(jaise

δ Q

mein) ka

d

se kya matlab?

δ

= inexact differential (path-dependent, state function nahi);

d

= state function ka exact differential

Maxwell relation mein equal mixed partials ke liye kaun si condition permission deti hai?

second partial derivatives ki continuity (Clairaut/Schwarz theorem)