Visual walkthrough — Thermodynamic potentials — U (internal), H (enthalpy), F (Helmholtz), G (Gibbs)

2.4.1 · D2 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Thermodynamic potentials — U (internal), H (enthalpy), F (He

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Step 1 — Energy bookkeeping kya hai? (piggy bank)

KYA. Ek container mein band gas ek certain amount of internal energy rakhti hai. Use $U$ kaho — bas joules ki ek sankhya, jaise piggy bank ke andar ka paisa. Do cheezein us number ko badal sakti hain: hum gas ko heat kar sakte hain (walls ke through energy andar daalein) ya gas apni container wall ko baahir dhakkel ke work kar sakti hai (energy kharch karna).

YAHAN se kyun shuru karein. Jo bhi potential hum banate hain, woh $U$ ka hi rewrite hai. Agar tum $U$ ko change hota picture nahin kar sakte, toh baaki sab kuch samajh nahin aayega.

PICTURE. Figure dekho: orange arrow box ke andar hai — yeh added heat hai; teal arrow bahar hai — yeh gas ke piston ko push karne se kiya gaya work hai. Bank ka level heat ke saath badhta hai, work ke saath girta hai.

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Step 2 — "Slivers" ko naam dena: $TdS$ kya hai aur $pdV$ kya hai?

KYA. Ab hum $\delta Q$ aur $\delta W$ ko faces dete hain.

Work sliver dekhna aasaan hai: agar area $A$ ka piston thodi si door baahir khuske, gas us area par pressure $p$ se push karta hai, aur enclosed volume $dV$ se badh jaata hai. Force $\times$ distance $=(pA)\times (\text{distance})=pdV$. Toh $\delta W=pdV.$

• $p$ = pressure (gas kitni zyada push karta hai, per unit area).
• $dV$ = piston bahar khisakne par volume mein tiny increase.

Heat sliver ke liye ek naye character ki zaroorat hai. Jab hum gas ko reversibly heat karte hain (dheere, taki koi jhatka na lage), tab added heat hai $\delta Q=TdS.$

• $T$ = absolute temperature (kitna garam).
• $S$ = entropy — gas ki internal "spread-outness" ya disorder ka measure (dekho Entropy and the Second Law). $dS$ us disorder mein ek tiny change hai.

YEH do products kyun? Pattern dekho: har energy sliver ek intensive quantity hai (koi cheez jo ek probe se measure ho: $T$, $p$) times ek extensive quantity mein change (koi cheez jo system size ke saath scale hoti hai: $S$, $V$). Yeh "intensive × d(extensive)" shape hi neeche sab kuch ka core hai.

Entropy kyun? Kyunki "heat ki miqdar" gas ki property nahin hai — lekin $TdS$ ek exact bookkeeping entry hai. Entropy woh accountant hai jo nature ne ijaad ki taaki heat ko saaf tarike se track kiya ja sake.

PICTURE. Figure mein piston push dikhata hai (teal, $pdV$ deta hai) aur ek thermometer + dots ka ek scatter jo zyada spread hota ja raha hai (orange, $TdS$ deta hai).

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Step 3 — Yeh kisi bhi path par kyun hold karta hai (state function)

KYA. Humne $dU=TdS-pdV$ ek slow, reversible process assume karke derive kiya. Ab hum claim karte hain ki yeh hamesha hold karta hai, chahe change violent aur irreversible ho.

YEH allowed kyun hai. $U$, $S$, aur $V$ teeno state functions hain: inki values sirf is par depend karti hain ki gas abhi kahan hai (uska current $T$, $p$, etc.), na ki woh wahan kaise pahunchi. Ek map par do points imagine karo — unke beech ka altitude difference fix hai, chahe tum kaunsa bhi trail hike karo. Kyunki $dU$, $dS$, $dV$ state functions ke differences hain, unke beech ka relation landscape ka ek fixed feature hai, path-independent. Hum ise easy (reversible) trail ke along compute karne aur har jagah trust karne ke allowed hain.

PICTURE. Ek $(S,V)$ map par same start aur end dots ke beech do paths: ek wiggly irreversible path (plum, dashed) aur ek smooth reversible (teal). Altitude $U$ mein net change identical hai.

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Step 4 — Problem: lab mein $S$ aur $V$ ko fix nahin kar sakte

KYA. $U(S,V)$ beautiful hai lekin impractical. Real lab mein tum shayad hi kabhi entropy ya volume clamp karte ho. Tum temperature (thermostat / water bath) aur pressure (beaker ko atmosphere ke liye khula chhod do) clamp karte ho.

YEH kyun matter karta hai. Hum ek energy function chahte hain jinke natural variables wahi knobs hain jo hum actually ghoomte hain. Isliye humein ek aisi variable jo hum control nahin kar sakte ($S$ ya $V$) ko ek aisi variable se trade karna hoga jo hum kar sakte hain ($T$ ya $p$).

PICTURE. Ek lab bench: ek sealed rigid box ($V$ control karta hai), ek thermostat bath ($T$ control karta hai), atmosphere ke neeche ek open beaker ($p$ control karta hai). Un knobs ko circle kiya gaya hai jinhein ek insaan grab kar sakta hai.

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Step 5 — Legendre move: ek rectangle subtract karo

KYA. $V$ (jise hum fix nahin kar sakte) ko $p$ (jise hum kar sakte hain) se swap karne ke liye, hum $U$ mein $pV$ add karte hain. $S$ ko $T$ se swap karne ke liye, hum $TS$ subtract karte hain. "Yahan add, wahan subtract" kyun? Geometry decide karti hai, hum nahin.

YEH kaise kaam karta hai — rectangle picture. Maano $U$ ek curve hai jo $V$ ke saath badhti hai (fixed $S$ par). Iska slope $-p$ hai. Hum ek naya function chahte hain jiska independent variable slope ho, na ki $V$. Trick: har point par, tangent-line ka rectangle subtract karo taaki jo bachta hai woh $V$ ki jagah slope ke saath change kare. Concretely, product $pV$ ka differential hai $d(pV)=pdV+Vdp.$

• $pdV$ piece exactly woh annoying term hai $dU$ mein jo hum cancel karna chahte hain.
• $Vdp$ piece naya term hai, slope $p$ ko variable ki tarah carry karta hai.

PICTURE. Curve $U$ vs $V$, ek point par tangent, aur area $pV$ ka shaded rectangle. $V$ ki jagah slope ke along slide karna hi transform ka poora idea hai.

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Step 6 — Enthalpy $H$: $V\to p$ swap karo

KYA. $pV$ add karke $(S,p)$ mein natural potential banao: $H\equiv U+pV.$

$pV$ add kyun karein. $dU$ mein $-pdV$ woh term hai jo $V$ ko slope $p$ banने se rok raha hai. $pV$ add karne se $+pdV+Vdp$ inject hota hai; $+pdV$, $-pdV$ ko cancel karta hai, aur fresh $Vdp$ reh jaata hai.

Differential, term by term: $dH=dU+pdV+Vdp=(\underset{\text{rakha}}{\underbrace{TdS}}\underset{\text{cancel}}{\underbrace{-pdV}})+\underset{\text{cancel}}{\underbrace{pdV}}+\underset{\text{naya}}{\underbrace{Vdp}}=TdS+Vdp$

Toh $H=H(S,p)$ hai, aur $T=(\partial H/\partial S{)}_p$, $V=(\partial H/\partial p{)}_S$.

PICTURE. Ek ledger-arrow diagram: $dU$ se shuru karo, $-pdV$ aur $+pdV$ tokens annihilate ho jaate hain (crossed out), $Vdp$ appear hota hai. Natural-variable label $V$ se $p$ flip ho jaata hai.

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Step 7 — Helmholtz $F$ aur Gibbs $G$: $S\to T$ swap karo (aur dono)

KYA. $TS$ subtract karke $(T,V)$ mein natural potential banao: $F\equiv U-TS,$ aur dono swaps karke $(T,p)$ mein natural potential: $G\equiv U-TS+pV=H-TS=F+pV.$

$TS$ subtract kyun karein. $+TdS$ term $S$ ko slope $T$ banने se rok raha hai. $TS$ subtract karne se $-TdS-SdT$ inject hota hai; $-TdS$, $+TdS$ ko cancel karta hai, naya $-SdT$ reh jaata hai.

Do differentials, term by term: $dF=dU-TdS-SdT=(\underset{\text{cancel}}{\underbrace{TdS}}-pdV)\underset{\text{cancel}}{\underbrace{-TdS}}\underset{\text{naya}}{\underbrace{-SdT}}=-SdT-pdV$ $dG=dH-TdS-SdT=(\underset{\text{cancel}}{\underbrace{TdS}}+Vdp)\underset{\text{cancel}}{\underbrace{-TdS}}-SdT=-SdT+Vdp$

Toh $F=F(T,V)$ aur $G=G(T,p)$ hain.

PICTURE. Chaar potentials ka ek 2×2 map axes par rakha gaya hai: "pehle slot ke liye choose ki gayi variable" (S ya T) vs "doosra slot" (V ya p). $+pV$ aur $-TS$ label wale arrows tumhe corners ke beech move karte hain.

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Step 8 — Degenerate & edge cases (koi gap mat chhodo)

KYA. Woh corners check karo jahan koi term khatam ho jaata hai.

• Isochoric ($dV=0$, rigid box): $dU=TdS$, aur $dF=-SdT$. Heat ab seedha $U$ ko bump karti hai; koi work escape nahin karta.
• Isobaric ($dp=0$, open): $dH=TdS=\delta Q$ (Step 6). $H$ heat tracker ban jaata hai.
• Isothermal ($dT=0$, bath): $dF=-pdV$ (saari energy change work hai) aur $dG=Vdp$ (worked example (b) mein use hua: $\Delta G=nRT\ln (p/p_0)$, dekho Chemical potential and the grand potential).
• Adiabatic-reversible ($dS=0$, insulated & slow): $dU=-pdV$ aur $dH=Vdp$. Energy changes pure work hain.
• Zero-entropy limit ($T\to 0$): $TS$ correction vanish ho jaati hai, toh $F\to U$ aur $G\to H$ — "messiness tax" absolute zero ke paas gayab ho jaati hai (Third Law flavour).

SAAREY dikhana kyun. Har real experiment ek variable clamp karta hai; reader ko pehchanna chahiye ki kaunsa potential flat ho jaata hai aur kaunsa term har case mein survive karta hai.

PICTURE. Ek grid: rows = chaar constraints, columns = har $d\Phi$ ka kaunsa term survive karta hai vs khatam. Dead terms greyed, live terms glowing.

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Ek-picture summary

KYA. Ek diagram poori derivation compress karta hai: $U(S,V)$ se shuru karo, $+pV$ apply karo (right) aur $-TS$ (down), har corner mein differential padho. Yahi Legendre transforms square hai, aur iske exact-differential corners har Maxwell relation aur $C_p - C_V$ result generate karte hain.

Recall Feynman retelling (plain words, cover and recall)

Ek energy $U$ ki piggy bank se shuru karo. Do cheezein level ko move karti hain: use heat karna (woh hai $TdS$, temperature times messiness mein change) aur use apni wall ko push out karne do (woh hai $pdV$, pressure times size mein growth). Woh ek sentence hi $dU=TdS-pdV$ hai. Lekin real lab mein mujhe "messiness" ya "size" dial karne ka option nahin milta — main ek thermostat (temperature) dial karta hoon aur cheezein khuli hawa mein chhod deta hoon (pressure). Toh main same energy ko repackage karta hoon taaki iske handles wahi hon jo mere paas actually hain. "Size" ko "pressure" se swap karne ke liye, main $pV$ bolt on karta hoon; swap size-term ko cancel karta hai aur mujhe ek pressure-term deta hai — yahi enthalpy $H$ hai. "Messiness" ko "temperature" se swap karne ke liye, main $TS$ (woh tax jo main kharch nahin kar sakta) peel off karta hoon; yahi Helmholtz $F$ hai. Dono karo aur Gibbs $G$ milta hai, jinke handles $T$ aur $p$ exactly ek chemist ke knobs hain. Har "add $pV$" ya "subtract $TS$" pehle sentence mein already baitha plus/minus sign se forced hai — main kabhi guess nahin karta. Aur special cases mein (rigid box, open beaker, insulated, absolute zero ke paas), ek term simply switch off ho jaata hai, mujhe bata ta hai ki kaunsa bank balance dekhna hai.