1.7.26
Intuition The big pictureEk thermodynamic potential ek energy-jaisi quantity hoti hai jiske natural variables aap lab mein control kar sakte ho, aur jiska minimum aapko equilibrium state batata hai. Jaise ek ball valley ke neeche roll karti hai (potential energy ka minimum), waise hi ek thermodynamic system sahi potential ke minimum ki taraf "roll" karta hai — lekin kaun sa potential depend karta hai kya fix rakha hai (entropy? temperature? pressure? volume?) par.
WHY four of them? Kyunki alag-alag experiments mein alag-alag cheezein constant rakhi jaati hain. Aap entropy easily control nahi kar sakte, lekin aap temperature (thermostat se) aur pressure (khule beaker se) control kar sakte ho. Har potential iss tarah design kiya gaya hai ki ek khaas pair of variables uske liye "natural" ban jaaye.
Definition Internal energyU U combined 1st + 2nd law (closed system, koi aur work nahi) yeh hai:
d U = T d S − p d V
WHY this form? First law: d U = δ Q − δ W δ Q = T d S δ W = p d V d U = T d S − p d V .
Yeh equation sabhi potentials ki mother hai. Ise U = U ( S , V ) U natural variables hain == S and V == d U d S d V
Intuition What is a Legendre transform (Feynman style)?Maano tumhari energy ek aisi variable mein likhi hai jise tum control nahi kar sakte (jaise S partner T control kar sakte ho). Ek Legendre transform ek variable ko uske conjugate slope se swap karta hai, unka product subtract karke .
HOW it works. Hamare paas hai T = ( ∂ U / ∂ S ) V S T T S F = U − T S . d F = d U − T d S − S d T = ( T d S − p d V ) − T d S − S d T = − S d T − p d V . T d S cancel ho jaate hain, aur d F d T d V
Wohi trick − p d V → + V d p + p V
Derivation of d H H = U + p V ⇒ d H = d U + p d V + V d p = ( T d S − p d V ) + p d V + V d p = T d S + V d p . Yeh step kyun? d U − p d V d ( p V ) + p d V
Derivation of d F upar ho chuka hai. Kyun? T S T d S T
Derivation of d G G = U + p V − T S d G = d U + p d V + V d p − T d S − S d T = − S d T + V d p . Kyun? T d S p d V ( T , p )
Intuition Which potential is "the boss"?2nd law kehta hai universe ki total entropy kabhi kam nahi hoti . Us constraint ko sirf system par alag-alag lab conditions mein translate karne par "entropy maximized" ka matlab ban jaata hai "sahi potential minimized."
Held constant
Minimized potential
Lab situation
S , V U isolated, rigid (rare)
S , p H adiabatic, constant pressure
T , V F thermostat, rigid box
T , p G thermostat, open to atmosphere (chemistry! )
Intuition Why "free energy"?F = U − T S T free hai — tumhe T S Δ G < 0
Worked example Example 1 — Enthalpy = heat at constant pressureDikhao ki δ Q p = d H Step 1: d H = T d S + V d p Kyun? Hum us differential ka use karte hain jo humne derive kiya.
Step 2: Constant pressure par d p = 0 d H = T d S = δ Q rev Kyun? δ Q = T d S Result: Constant p Δ H C p = ( ∂ H / ∂ T ) p
Worked example Example 2 — Get an entropy fromG G ( T , p ) S Step 1: d G = − S d T + V d p Kyun? Natural-variable differential.
Step 2: Constant p ( ∂ T ∂ G ) p = − S Kyun? d T Result: S = − ( ∂ T ∂ G ) p V = ( ∂ p ∂ G ) T
Worked example Example 3 — A Maxwell relation fromF ( ∂ V ∂ S ) T = ( ∂ T ∂ p ) V Step 1: d F = − S d T − p d V S = − ( ∂ F / ∂ T ) V p = − ( ∂ F / ∂ V ) T Step 2: F ∂ V ∂ T ∂ 2 F = ∂ T ∂ V ∂ 2 F Kyun? F Step 3: Isse milta hai − ( ∂ V ∂ S ) T = − ( ∂ T ∂ p ) V Result: ( ∂ V ∂ S ) T = ( ∂ T ∂ p ) V
Common mistake "Free energy ka matlab hai energy jo free hai / kuch cost nahi karti."Kyun sahi lagta hai: "free" word ki wajah se. Sachchi baat: F G maximum useful work hai jo fixed T G p T S band ho jaati hai . Fix: "free" ko "kaam karne ke liye available" padho, na ki "muft."
Common mistake "Sabhii potentials har waqt equilibrium par minimize hote hain."Kyun sahi lagta hai: char definitions ki symmetry ki wajah se. Sachchi baat: minimization sirf matching constraints ke liye hota hai (jaise G T , p Fix: potential ko uske natural variables se match karo jinhein fix rakha gaya ho.
d H d H = T d S − V d p Kyun sahi lagta hai: d U Fix: joda gaya p V d ( p V ) + p d V − p d V + V d p
Recall Feynman: explain to a 12-year-oldEnergy bilkul wallet mein paise ki tarah hai. Lekin tum actually kitna kharch kar sakte ho, yeh depend karta hai tum kahan ho uske rules par. Ek garam kamre mein (fixed temperature) T S free energy F p V G U , H , F , G yeh account karne ke baad ki tumhare situation mein kaunse taxes lagte hain , aur system hamesha wahan settle karta hai jahan uska relevant wallet sabse chhota ho.
Mnemonic Remembering the four & their variables"Good Physicists Have Studied Under Very Fine Teachers" wala nahi — square use karo:
Corners clockwise: U H G F p V T S Sign rule: "S follows T, V follows p." Jab T − S d T p + V d p
What is the fundamental differential of U d U = T d S − p d V S , V
Enthalpy H d H H = U + p V d H = T d S + V d p S , p
Helmholtz free energy ko define karo aur d F F = U − T S d F = − S d T − p d V T , V
Gibbs free energy ko define karo aur d G G = U + p V − T S d G = − S d T + V d p T , p
Constant T , p Gibbs free energy G
Constant T , V Helmholtz free energy F
Constant-pressure heat ke liye enthalpy kyun matter karti hai? d p = 0 d H = T d S = δ Q Δ H
G ( T , p ) S V S = − ( ∂ G / ∂ T ) p V = ( ∂ G / ∂ p ) T
Yahan Legendre transform kya kar raha hai? Ek variable ko uske conjugate ke saath unka product subtract karke swap karna, taaki natural variables badal sakein
F ( ∂ S / ∂ V ) T = ( ∂ p / ∂ T ) V
Free energy mein "free" ka matlab kya hai? Fixed T G p
swaps variable for conjugate
