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

1.7.26 · D2 · Physics › Thermodynamics › Thermodynamic potentials — U, H, F, G (preview)

Prerequisite ideas jinka hum sahara lete hain: First Law of Thermodynamics, Entropy and the Second Law, Legendre Transform, aur baad mein Maxwell Relations. Yeh parent topic ka visual companion hai.

Step 1 — Wo ek cheez jisse hum shuru karte hain: ek tiny energy budget

KYA HAI. Ek system (maan lo ek cylinder mein gas) mein kuch hidden energy hoti hai jise hum ==internal energy $U$ == kehte hain — yeh uske saare molecules ki jiggling aur bonding energy ka sum hai. Hum kabhi uski total track nahi karte; hum track karte hain ki jab hum system ko thoda nudge karte hain toh yeh kaise badlti hai. Ek tiny change ko $d U$ likha jaata hai (" $U$ mein ek chhota sa change"). Woh chhota $d$ sirf iska matlab hai "ek infinitesimal step," jaise ek pixel hilana.

KYUN. Jo bhi potential hum banate hain woh $U$ plus ya minus kuch hota hai. Agar hum samajh lein ki $U$ kaise badalta hai, toh hum sab samajh lenge. Isliye humein $d U$ ka rule chahiye.

First Law of Thermodynamics kehta hai: energy change = jo heat daali, minus jo work system ne ki.

$d U = heat in δ Q - work out δ W$

$δ Q$ — heat ka ek chhota blob jo gas mein andar flow kar raha hai.
$δ W$ — work ka ek chhota blob jo gas bahar ke upar karti hai (piston ko push karke).

PICTURE. Gas ek bucket hai. Heat upar se andar girti hai ( $+ δ Q$ ), work side se bahar nikalti hai ( $- δ W$ ). Jo bhi bachta hai woh level $U$ ko badalta hai.

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

Step 2 — Dono blobs ko naam dena: $T d S$ aur $p d V$

KYA HAI. Ab hum kehte hain ki woh dono blobs kis cheez se bane hain, ek gentle (reversible) process ke liye.

Heat blob: $δ Q = T d S$ .
- $T$ — temperature (kitna garam, kelvin mein).
- $d S$ — ==entropy $S$ == mein ek tiny change, ek number jo measure karta hai ki energy kitni "spread out" ya disordered hai. Heat daalne se energy spread hoti hai, isliye heat aur $d S$ saath chalte hain; $T$ exchange rate hai.
Work blob: $δ W = p d V$ .
- $p$ — pressure jo gas push karta hai.
- $d V$ — ==volume $V$ == mein ek tiny change. Agar gas expand karta hai ( $d V > 0$ ) toh woh piston ko bahar push karta hai aur work karta hai.

YEH TOOLS KYUN? Hum $T d S$ use karte hain sirf " $δ Q$ " ki jagah kyunki $δ Q$ akela state ki property nahi hai — yeh liye gaye path pe depend karta hai. Lekin $T$ aur $S$ sach mein state properties hain. Heat ko $T d S$ ke roop mein likhna ek path-dependent blob ko state variables mein badal deta hai, jo exactly baad mein clean potentials banane deta hai. (Yeh Entropy and the Second Law ka payoff hai.)

Dono blobs substitute karne par:

$d U = hotness T disorder step d S - push p expansion step d V$

Yeh mother equation hai. Iska shape notice karo: (kuch)·(kisi cheez mein change) + (kuch)·(kisi cheez mein change).

PICTURE. Bucket ko feed karne wale do labelled arrows: ek laal $T d S$ arrow andar, ek neela $p d V$ arrow bahar, aur resulting $d U$ net change ke roop mein dikhaya gaya.

Step 3 — Equation se slopes padhna

KYA HAI. Do variables ka koi bhi smooth function chain rule maanta hai: ek chhota step hota hai (slope in $S$ )· $d S$ + (slope in $V$ )· $d V$ :

$d U = (\frac{\partial U}{\partial S})_{V} d S + (\frac{\partial U}{\partial V})_{S} d V$

$(\frac{\partial U}{\partial S})_{V}$ — $U$ kitni tezi se badhti hai agar tum $S$ ko nudge karo $V$ ko fixed rakhte hue (woh chhota subscript $V$ matlab "keep $V$ frozen").

KYUN. Humare paas $d U$ do tarikoon se likha hai: ek baar physically ( $T d S - p d V$ ), ek baar mathematically (chain rule). Agar do expressions har $d S, d V$ ki choice ke liye equal hain, toh unke matching coefficients equal hone chahiye. Term by term match karo:

$(\frac{\partial U}{\partial S})_{V} = T, (\frac{\partial U}{\partial V})_{S} = - p$

Energy ka entropy ke saath slope hi temperature hai. Volume ke saath slope hi minus pressure hai. Jinhe symbols hum abstractly introduce kar chuke hain unka ab physical meaning hai.

PICTURE. Surface $U (S, V)$ ek hill ki tarah; east ki taraf chalte hue ( $S$ badhata hai) steepness $T$ read hoti hai; north ki taraf chalte hue ( $V$ badhata hai) downhill steepness $- p$ read hoti hai.

Step 4 — Problem: lab mein $S$ pakad nahi sakte

KYA HAI. $U (S, V)$ chahta hai ki hum entropy $S$ aur volume $V$ control karein. Volume theek hai (piston hilao). Lekin koi "entropy knob" nahi hota — tum directly disorder dial nahi kar sakte. Jo tum actually set kar sakte ho woh hai temperature $T$ (ek thermostat) aur pressure $p$ (ek open beaker atmosphere feel karta hai).

KYUN NAYA TOOL. Humein awkward variable $S$ ko uske friendly partner $T$ se swap karna hai, aur $V$ ko $p$ se, bina koi information khoye. Yeh kaam exactly karne wala tool hai Legendre Transform. Yeh ek precise sawaal ka jawab deta hai: "Main ek energy kaise rewrite karun taaki ek slope variable ban jaye, aur purana variable gaayab ho jaye?"

Recipe: product subtract karo un dono partners ka jo tum swap karna chahte ho.

$S \to T$ trade karne ke liye (partners, kyunki $T = \partial U / \partial S$ ): $T S$ subtract karo.
$V \to p$ trade karne ke liye (partners, kyunki $- p = \partial U / \partial V$ ): $p V$ add karo.

PICTURE. Ek knob panel: " $S$ " knob cross out hai (aisa koi knob exist nahi karta), " $T$ " aur " $p$ " knobs glow kar rahe hain — yeh woh hain jo ek real lab turn kar sakti hai.

Step 5 — Pehla swap: $p V$ add karo enthalpy $H$ pane ke liye

KYA HAI. Enthalpy define karo $H = U + p V$ . Product rule $d (p V) = p d V + V d p$ se iska small change lo:

$d H = d U + p d V + V d p$

Ab $d U = T d S - p d V$ plug in karo:

$d H = (T d S - p d V) + p d V + V d p$

Seed se $- p d V$ aur $d (p V)$ se $+ p d V$ equal aur opposite hain → woh cancel ho jaate hain.
Jo bachta hai: $T d S$ (untouched) aur ek brand-new $+ V d p$ .

$d H = T d S + V d p natural variables (S, p)$

KYUN. Cancellation hi puri baat hai: humne $d V$ term ko $d p$ term se trade kiya, isliye pressure ab ek natural variable hai. Sign flip dekho — $- p d V$ ban gaya $+ V d p$ . Yeh classic mistake trap hai; plus sign cancellation se forced hai, yeh optional nahi hai.

PICTURE. Do stacked bars: $d U$ ka $- p d V$ block aur $d (p V)$ ka $+ p d V$ block annihilate ho jaate hain; leftover $+ V d p$ block slide in karta hai.

Step 6 — Doosra swap: $T S$ subtract karo Helmholtz $F$ pane ke liye

KYA HAI. Helmholtz free energy define karo $F = U - T S$ . Iska change product rule $d (T S) = T d S + S d T$ use karta hai:

$d F = d U - T d S - S d T$

Seed plug in karo:

$d F = (T d S - p d V) - T d S - S d T$

Seed se $+ T d S$ aur $- d (T S)$ se $- T d S$ cancel ho jaate hain.
Bachnے wale: $- p d V$ (untouched) aur ek naya $- S d T$ .

$d F = - S d T - p d V natural variables (T, V)$

KYUN. $T d S$ ko khatam karna $T$ ko natural variable promote karta hai — thermostat + rigid box ke liye perfect. Sign pattern note karo: $S$ , $T$ ke saath $- S d T$ ke roop mein ride karta hai. (Compare Step 5 se jahan $V$ , $p$ ke saath $+ V d p$ ke roop mein ride karta tha.)

PICTURE. Step 5 ka mirror, lekin ab $T d S$ blocks annihilate ho jaate hain aur $- S d T$ bachta hai.

Step 7 — Dono swaps ek saath: Gibbs $G$

KYA HAI. Dono tricks karo: $p V$ add karo aur $T S$ subtract karo. Gibbs free energy define karo $G = U + p V - T S$ . Tab

$d G = d U + p d V + V d p - T d S - S d T$

$d G = (T d S - p d V) + p d V + V d p - T d S - S d T$

$p d V$ pair aur $T d S$ pair dono cancel ho jaate hain. Bachnے wale:

$d G = - S d T + V d p natural variables (T, p)$

KYUN. $G$ ke natural variables $T$ aur $p$ hain — woh do cheezein jo ek chemist ek open flask mein bench pe actually control karta hai. Isliye $G$ chemistry mein raja hai (dekho Gibbs Free Energy and Chemical Equilibrium).

PICTURE. Char blocks feed in karte hain; dono cancelling pairs vanish ho jaate hain, $- S d T + V d p$ bachta hai.

Step 8 — Degenerate cases: kya hoga agar ek step zero ho?

KYA HAI. Differentials tab bhi sense dene chahiye jab koi nudge switch off ho. Har edge check karo:

Constant pressure ( $d p = 0$ ): $d H = T d S = δ Q$ . Enthalpy change hi absorbed heat hai — yahi reason hai ki $H$ open-flask chemistry ke liye "heat content" hai, aur kyun $C_{p} = (\partial H / \partial T)_{p}$ (dekho Heat Capacities Cp and Cv).
Constant volume ( $d V = 0$ ): $d U = T d S = δ Q$ , isliye $C_{v} = (\partial U / \partial T)_{V}$ .
Constant temperature ( $d T = 0$ ): $d F = - p d V$ . Helmholtz energy mein change equals minus the work — woh leftover hi woh useful work hai jo tum fixed $T$ pe extract kar sakte ho; isliye "free" energy.
Constant $T$ aur $p$ ( $d T = d p = 0$ ): $d G = 0$ . Bench conditions ke under equilibrium pe $G$ badalna band kar deta hai — woh apne minimum pe baith jaata hai.
Fully isolated & rigid ( $d S$ system se set, $d V = 0$ , koi heat nahi): $U$ boss hai; kuch bhi baahri kuch nahi karta.

KYUN. Yeh koi nayi physics nahi hai — yeh wahi chaar boxed equations hain jinmein se ek term delete ho gaya hai. Jo reader koi bhi real experiment se milta hai woh dekh sakta hai ki kaun sa term marta hai aur seedha meaning read kar sakta hai.

PICTURE. Chaar differentials ka 2×2 grid; har cell mein ek term grey out hai dikhane ke liye ki kaun sa constraint use kill karta hai.

Ek-picture summary

Sab kuch ek square pe rehta hai. Char potentials corners pe baithe hain; unke beech ke moves Legendre trades hain ( $+ p V$ across, $- T S$ down). Har potential ke do natural variables uske corner se latak rahe hain, aur uska differential seedha padha jaata hai.

Jab yeh visual ho jaate hain, toh Maxwell Relations free mein nikal aate hain har corner ke mixed second derivatives ko equate karke.

Recall Feynman: poori kahani seedhe shabdon mein bolo

Humne ek honest sentence se shuru kiya: energy change = heat in minus work out. Humne heat ko "temperature times a change in disorder" rakha aur work ko "pressure times a change in size," ek seed equation diya, $d U = T d S - p d V$ . Trouble: yeh humse disorder control karne ko bolta hai, jo koi lab nahi kar sakti. Toh hum ek trick khelते hain — ek awkward variable ko us friendly slope se swap karne ke liye jo woh hide karta hai, hum un dono ka product subtract (ya add) karte hain. $p V$ add karo: volume term cancel hoti hai aur enthalpy $H$ appear karta hai, heat aur pressure se ruled. $T S$ subtract karo: disorder term cancel hoti hai aur Helmholtz $F$ appear karta hai, temperature aur volume se ruled. Dono karo: Gibbs $G$ appear karta hai, temperature aur pressure se ruled — chemist ki duniya. Koi bhi ek term delete karo aur tum ek real experiment read kar lete ho: constant pressure heat deta hai $Δ H$ ke roop mein, constant temperature extractable work deta hai $- Δ F$ ke roop mein, aur fixed $T, p$ pe system $G$ ke bottom ki taraf slide karta hai. Char wallets, ek seed, aur ek square jo sab yaad rakhta hai.