Maxwell relations — derivation from each potential: Worked examples

2.4.3 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Maxwell relations — derivation from each potential

Yeh page drill floor hai. Parent note ne chaar relations build kiye; yahaan hum har tarah ke situation unpe throw karte hain jab tak koi bhi tumhe surprise na kar sake. Pehle scenario matrix padho, phir dekho har cell kaise solve hoti hai.

Neeche sab kuch sirf wahi tools use karta hai jo parent mein already earn ho chuke hain: ek exact differential $d z = M d x + N d y$ , equality of mixed partials $(\partial M / \partial y)_{x} = (\partial N / \partial x)_{y}$ (yaani Schwarz theorem), aur chaar boxed relations. Agar koi naya symbol aaye, toh woh usi waqt define hoga jab woh aayega.

Definition Is page par use hone waale symbols (sab yahaan pehle define kiye gaye hain)

$R$ = universal gas constant, $R = 8.314 J K^{- 1} mol^{- 1}$ — ideal-gas law $P V = n R T$ mein fixed proportionality, jahan $n$ moles ki sankhya hai.
$α \equiv \frac{1}{V} (\frac{\partial V}{\partial T})_{P}$ = thermal expansion coefficient (fixed pressure par har kelvin pe fractional volume badhna), units $K^{- 1}$ .
$κ_{T} \equiv - \frac{1}{V} (\frac{\partial V}{\partial P})_{T}$ = isothermal compressibility (fixed $T$ par har pascal pe fractional volume ghatna), units $Pa^{- 1}$ . Ex 8 mein use hota hai.
$C_{V} \equiv (\frac{\partial U}{\partial T})_{V}$ = heat capacity at constant volume (system ko ek kelvin warm karne ke liye zaruri energy jab volume fixed ho), units $J K^{- 1}$ . Monatomic ideal gas ke liye $C_{V} = \frac{3}{2} n R$ .
Van der Waals equation mein, $a$ molecular attraction measure karta hai ( $Pa\cdotm^{6} mol^{- 2}$ ) aur $b$ excluded volume per mole hai ( $m^{3} mol^{- 1}$ ), yaani molecules ka finite size.

Formula Do workhorse identities jo hum ek baar re-derive karte hain, phir baar baar use

(i) Energy equation. Master differential $d U = T d S - P d V$ se shuru karo (parent se). $d V$ se divide karo $T$ fixed rakh ke, toh $d S \to (\partial S / \partial V)_{T} d V$ : $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial S}{\partial V})_{T} - P .$ Ab Helmholtz Maxwell relation $(\partial S / \partial V)_{T} = (\partial P / \partial T)_{V}$ daalo: $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial P}{\partial T})_{V} - P .$ (ii) Triple-product (cyclic) identity. Kisi bhi teen quantities ke liye jo ek equation of state se linked hain, $x, y, z$ mein se koi ek baar mein fixed rakhte hain, $(\frac{\partial x}{\partial y})_{z} (\frac{\partial y}{\partial z})_{x} (\frac{\partial z}{\partial x})_{y} = - 1 \Rightarrow (\frac{\partial x}{\partial y})_{z} = - \frac{( \partial x / \partial z ) _{y}}{( \partial y / \partial z ) _{x}} .$ $- 1$ kyun? $d z = (\partial z / \partial x)_{y} d x + (\partial z / \partial y)_{x} d y$ likhne par aur phir $z$ fixed rakhne par ( $d z = 0$ ) $d x$ aur $d y$ ek fixed ratio mein aa jaate hain; woh ratio solve karne par aur teeno variables ka loop close karne par exactly ek minus sign bachta hai. Yeh tab use karte hain jab humein swap karna ho ki kaun sa variable constant rakkha gaya hai.

Scenario matrix

Is topic se har problem inhi cells mein se ek hai. Sabse daayaan column us worked example ka naam deta hai jo usme aata hai.

Cell	Kya isko alag banata hai	Kahan cover hua
A. Pick-the-right-potential	Ek target derivative diya gaya hai; choose karo $U, H, F, G$ mein se kaun sa deta hai	Ex 1
B. Sign bookkeeping (+ case)	Ek relation bina minus sign ke ( $H$ )	Ex 2
C. Sign bookkeeping (− case)	Ek relation minus sign ke saath ( $G$ )	Ex 3
D. Numeric plug-in, ideal gas	Real numbers daalo, units ke saath number nikalo	Ex 4
E. Zero / degenerate input	Ek derivative jo exactly zero aata hai, aur kyun	Ex 5
F. Limiting behaviour	Kya hota hai jab $T \to 0$ ya koi coefficient $\to 0$	Ex 6
G. Real-world word problem	Rubber band / adiabatic cooling, physical story	Ex 7
H. Exam twist — chain the relations	Do Maxwell relations + ek triple product combine karo	Ex 8

Ex 1 — Cell A · potential choose karna

Worked example Statement

Tumhe $(\frac{\partial S}{\partial P})_{T}$ chahiye lekin tumhara lab sirf measure kar sakta hai ki volume temperature ke saath kaise change hota hai. Kaun sa Maxwell relation yeh deliver karta hai, aur tum kya measure karte ho?

Forecast: Pehle guess karo — kaun sa potential hai jiske natural variables $T$ aur $P$ hain?

Step 1. Natural-variable pairs list karo. $U : (S, V)$ , $H : (S, P)$ , $F : (T, V)$ , $G : (T, P)$ . Yeh step kyun? Target derivative $T$ constant rakhta hai aur $P$ se differentiate karta hai. Woh potential jiske natural variables exactly $(T, P)$ hain, wahi hai jiska Maxwell relation purely $T, P$ (aur unke partners $S, V$ ) mein expressed hota hai. Woh $G$ hai.

Step 2. Gibbs relation padho: $(\frac{\partial S}{\partial P})_{T} = - (\frac{\partial V}{\partial T})_{P} .$ Yeh step kyun? $G$ ka differential $d G = - S d T + V d P$ hai, toh $M = - S$ , $N = V$ , aur $(\partial M / \partial P)_{T} = (\partial N / \partial T)_{P}$ exactly yahi deta hai.

Step 3. Measurable ka naam lo. Upar define kiye gaye thermal expansion coefficient $α$ use karke, $(\frac{\partial S}{\partial P})_{T} = - V α .$

Verify: RHS units — $V$ hai $m^{3}$ , $α$ hai $K^{- 1}$ , product $m^{3} K^{- 1}$ . LHS hai entropy $/$ pressure $= (J/K) / (Pa) = J K^{- 1} Pa^{- 1} = m^{3} K^{- 1}$ (kyunki $J = Pa\cdotm^{3}$ ). ✓ Units match karte hain. Dekho Thermal expansion coefficient and isothermal compressibility.

Ex 2 — Cell B · plus-sign relation (enthalpy)

Worked example Statement

Directly dikhao ki $(\frac{\partial T}{\partial P})_{S} = + (\frac{\partial V}{\partial S})_{P}$ , aur confirm karo ki genuinely koi minus sign nahi hai.

Forecast: Kya minus aa jayega? $d P$ term ka sign dekho.

Step 1. $d H = T d S + V d P$ likho. $x = S$ , $y = P$ , $M = T$ , $N = + V$ identify karo. Yeh step kyun? Dono differential terms plus ke saath aate hain, toh na $M$ na $N$ koi minus carry karta hai.

Step 2. Schwarz apply karo: $(\frac{\partial M}{\partial P})_{S} = (\frac{\partial N}{\partial S})_{P} \Rightarrow (\frac{\partial T}{\partial P})_{S} = (\frac{\partial V}{\partial S})_{P} .$ Yeh step kyun? Yeh hai hi $\partial^{2} H / \partial S \partial P$ dono taraf se compute kiya; mixed partials ki equality dono sides ko equal force karti hai.

Verify: $U$ ke relation $(\partial T / \partial V)_{S} = - (\partial P / \partial S)_{V}$ se cross-check karo, jisme minus hota hai — kyunki $U$ mein $- P d V$ aata hai. Kyunki $H$ mein uski jagah $+ V d P$ aata hai, minus gayab ho jaata hai. Consistent. ✓

Ex 3 — Cell C · minus-sign relation, numerically (Gibbs)

Worked example Statement

$25^{\circ} C$ ke paas water ke liye, $α = 2.57 \times 1 0^{- 4} K^{- 1}$ aur molar volume $V_{m} = 1.81 \times 1 0^{- 5} m^{3} / mol$ . $(\frac{\partial S}{\partial P})_{T}$ per mole nikalo. Kya entropy badhti hai ya ghaati hai jab tum water ko isothermally squeeze karte ho?

Forecast: Sign guess — squeezing usually molecules ko order karta hai, toh entropy drop honi chahiye. Dekhte hain.

Step 1. Gibbs relation use karo $α$ ke saath: $(\frac{\partial S}{\partial P})_{T} = - V_{m} α .$ Yeh step kyun? Ex 1 ne yeh exact link already establish kiya; ab hum numbers daaalte hain.

Step 2. Plug in karo: $(\frac{\partial S}{\partial P})_{T} = - (1.81 \times 1 0^{- 5}) (2.57 \times 1 0^{- 4}) = - 4.6517 \times 1 0^{- 9} J K^{- 1} Pa^{- 1} mol^{- 1} .$ Yeh step kyun? Direct multiply; units $m^{3} K^{- 1}$ rakho jo $J K^{- 1} Pa^{- 1}$ ke barabar hai.

Step 3. Sign interpret karo. Negative ⇒ fixed $T$ par $P$ badhane se entropy ghaati hai. Yeh step kyun? Maxwell relation ek mushkil entropy measurement ko ek aasaan expansion measurement mein convert karta hai; $α$ ka sign (positive for water at $25^{\circ} C$ ) answer ka sign fix karta hai.

Verify: Hamara forecast (compression par entropy girati hai) negative sign se match karta hai. Magnitude $\sim 5 \times 1 0^{- 9}$ tiny hai kyunki liquids barely change hote hain — physically sensible. ✓

Ex 4 — Cell D · ideal gas ke liye numeric plug-in (energy equation)

Worked example Statement

Is page ke top par re-derive ki gayi energy equation use karke, $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial P}{\partial T})_{V} - P$ , isko $n = 2 mol$ ideal gas ke liye $T = 300 K$ , $V = 0.05 m^{3}$ par evaluate karo. Phir van der Waals gas ke liye repeat karo taaki non-zero result dekho.

Forecast: Ideal gas ke liye parent ne prove kiya tha yeh $0$ hai. Van der Waals value ka sign guess karo.

Step 1 (ideal). Gas constant $R = 8.314 J K^{- 1} mol^{- 1}$ ke saath, $P = \frac{n R T}{V} \Rightarrow (\frac{\partial P}{\partial T})_{V} = \frac{n R}{V}$ . Phir energy equation deta hai $T \cdot \frac{n R}{V} - P = \frac{n R T}{V} - \frac{n R T}{V} = 0.$ Yeh step kyun? $T (\partial P / \partial T)_{V}$ exactly $P$ reproduce karta hai, toh woh cancel ho jaate hain — Joule's law nikal aata hai.

Step 2 (van der Waals). Equation of state $P = \frac{n R T}{V - nb} - \frac{a n ^{2}}{V ^{2}}$ , jahan (upar define kiya gaya) $a$ molecular-attraction parameter hai aur $b$ excluded volume per mole hai. Kyunki sirf pehla term $T$ contain karta hai, $(\frac{\partial P}{\partial T})_{V} = \frac{n R}{V - nb} .$ Yeh step kyun? $- a n^{2} / V^{2}$ term temperature-independent hai, toh $T$ se differentiate karne par woh drop ho jaata hai. $b$ term pehle fraction ke andar survive karta hai lekin similarly $T$ -independent hai ek constant offset ke roop mein $V$ mein.

Step 3. Energy equation ke through assemble karo: $(\frac{\partial U}{\partial V})_{T} = \frac{n R T}{V - nb} - (\frac{n R T}{V - nb} - \frac{a n ^{2}}{V ^{2}}) = \frac{a n ^{2}}{V ^{2}} .$ Notice karo ki poori $b$ -dependence cancel ho jaati hai. $a = 0.14 Pa\cdotm^{6} mol^{- 2}$ , $n = 2$ , $V = 0.05 m^{3}$ ke saath: $\frac{( 0.14 ) ( 2 ) ^{2}}{( 0.05 ) ^{2}} = \frac{0.56}{0.0025} = 224 Pa = 224 J/m^{3} .$ Yeh step kyun? Bachha hua $a n^{2} / V^{2}$ molecular attraction se internal pressure hai — real gases sach mein volume mein energy store karte hain.

Verify: Ideal answer $= 0$ parent se match karta hai. Van der Waals answer positive (attraction) expected tha; units $Pa = J/m^{3}$ $\partial U / \partial V$ ke liye correct hain. ✓

Ex 5 — Cell E · ek derivative jo exactly zero hai (degenerate)

Worked example Statement

Ek ideal gas vacuum mein expand karta hai (free / Joule expansion): na work, na heat, toh $U$ constant hai. $(\frac{\partial T}{\partial V})_{U}$ predict karo aur confirm karo ki yeh zero hai.

Forecast: Kya ideal gas free expansion par cool hota hai? Guess karo haan ya nahi.

Step 1. Triple-product identity (page ke top par re-derive ki gayi) $T (U, V)$ par apply karo: $(\frac{\partial T}{\partial V})_{U} = - \frac{( \partial U / \partial V ) _{T}}{( \partial U / \partial T ) _{V}} .$ Yeh step kyun? Hum constant $U$ par derivative chahte hain, lekin hum constant $T$ par derivatives jaante hain (energy equation). Triple product exactly woh tool hai jo swap karne ke kaam aata hai ki kaun sa variable fixed rakha gaya.

Step 2. Numerator: Ex 4 se, ideal gas ke liye $(\partial U / \partial V)_{T} = 0$ . Denominator: $(\partial U / \partial T)_{V} = C_{V} = \frac{3}{2} n R \neq = 0$ , heat capacity at constant volume jo Symbols box mein define kiya gaya. Yeh step kyun? Ek zero numerator over finite denominator zero deta hai.

Step 3. $(\frac{\partial T}{\partial V})_{U} = - \frac{0}{C _{V}} = 0.$ Yeh step kyun? Free expansion par koi temperature change nahi — degenerate/zero case.

Verify: Forecast check — ek ideal gas cool nahi hota (real gases hote hain, Ex 4 ke $a n^{2} / V^{2}$ term se). Zero genuine hai, undefined nahi, kyunki $C_{V} > 0$ . ✓

Ex 6 — Cell F · limiting behaviour jab $T \to 0$

Worked example Statement

Third law kehta hai entropy $S \to$ const jab $T \to 0$ . Ek Maxwell relation use karke dikhao ki thermal expansion coefficient $α \to 0$ jab $T \to 0$ .

Forecast: Guess karo ki kya materials absolute zero tak expand karte rehte hain.

Step 1. Gibbs relation: $(\frac{\partial V}{\partial T})_{P} = - (\frac{\partial S}{\partial P})_{T} .$ Yeh step kyun? Hum $V$ ki temperature dependence (yaani $α$ ) chahte hain; Gibbs relation ise tie karta hai ki $S$ $P$ par kaise depend karta hai, jo third law control karta hai.

Step 2. Third law ⇒ jab $T \to 0$ , $S$ same constant approach karta hai sab pressures ke liye, toh $(\frac{\partial S}{\partial P})_{T} \to 0$ . Yeh step kyun? Agar $S (T = 0)$ $P$ se independent hai, toh $T = 0$ par uski $P$ -derivative zero ho jaati hai.

Step 3. Therefore $(\frac{\partial V}{\partial T})_{P} \to 0$ , hence $α = \frac{1}{V} (\frac{\partial V}{\partial T})_{P} \to 0 (T \to 0) .$ Yeh step kyun? Ek finite $V$ times ek vanishing derivative $α \to 0$ deta hai.

Verify: Experimentally, solids ka thermal expansion absolute zero ke paas zero pe flatten ho jaata hai — match karta hai. Maxwell relation ne ek entropy statement (third law) ko ek mechanical prediction (no expansion) mein convert kar diya. ✓

Ex 7 — Cell G · real-world word problem (rubber band)

Figure — Maxwell relations — derivation from each potential

Figure padhna. Horizontal axis band ki length $L$ (metres) hai — yeh mechanical variable hai, wahi role play karta hai jo gas ke liye volume $V$ play karta hai. Neeche wali black wavy line relaxed band hai; upar wali red wavy line (key object) wahi band hai jo zyada $L$ tak stretch hua hai. Dono ends par black arrows tension $f$ hain (inward pull, newtons mein). Diagonal black arrow stretching process dikhata hai. Red curve jo pedagogical point visible karti hai: stretching se tangled polymer chains seedhi ho jaati hain, toh band zyada ordered ho jaata hai — uski entropy $S$ girती hai jab $L$ badhta hai. Neeche ke steps mein woh red "ordered" state yaad rakho.

Worked example Statement

Ek stretched rubber band ek "elastic" system hai: uska work term $δ W = - f d L$ hai jahan $f$ tension hai (force, N) aur $L$ length hai (m) — toh $L$ $V$ ka role play karta hai aur $- f$ $P$ ka role play karta hai. Experimentally, ek rubber band adiabatically stretch karne par garam hota hai. Helmholtz relation ke analogue use karke, $(\partial f / \partial T)_{L}$ ka sign predict karo (kya tension fixed length par temperature ke saath badhti hai?).

Forecast: Figure se guess karo: fixed stretch par, kya ek garam rubber band zyada hard ya zyada soft pull karta hai?

Step 1. Elastic Helmholtz differential banao. $d F = - S d T - P d V$ mein $- P d V \to + f d L$ replace karo: $d F = - S d T + f d L .$ Yeh step kyun? Rubber band ke liye tum usse stretch karke kaam karte ho, toh mechanical term gas ke relative sign flip kar leta hai; $L$ natural mechanical variable hai (figure mein red-curve axis).

Step 2. $M = - S$ , $N = f$ padho, Schwarz apply karo: $- (\frac{\partial S}{\partial L})_{T} = (\frac{\partial f}{\partial T})_{L} \Rightarrow (\frac{\partial f}{\partial T})_{L} = - (\frac{\partial S}{\partial L})_{T} .$ Yeh step kyun? Same cross-derivative logic gas ki tarah, bas elastic variables ke saath.

Step 3. Sign karo. Figure ne dikhaya ki stretching polymer chains ko line up karta hai, disorder ghataata hai, toh $(\partial S / \partial L)_{T} < 0$ . Hence $(\partial f / \partial T)_{L} = - (negative) > 0$ . Yeh step kyun? Minus ek negative ko positive mein flip karta hai.

Verify: $(\partial f / \partial T)_{L} > 0$ matlab ek garam band fixed length par zyada hard pull karta hai — yeh famous rubber-band effect hai (ek stretched band garam hone par contract karta hai). Aur adiabatic route se, warming-on-stretch usi $S$ -decrease se follow karta hai. Prediction everyday experiment se match karti hai. ✓

Ex 8 — Cell H · exam twist, relations chain karna ( $C_{P} - C_{V}$ )

Worked example Statement

General relation $C_{P} - C_{V} = T V \frac{α ^{2}}{κ _{T}}$ derive karo, $α$ aur $κ_{T}$ top par define kiye gaye hain. Phir $1 mol$ ideal gas ke liye $T = 300 K$ par evaluate karo taaki confirm ho ki $R$ milta hai.

Forecast: Ideal gas ke liye tum jaante ho $C_{P} - C_{V} = R$ . Kya messy formula uspe collapse hoti hai?

Step 1. $C_{P} - C_{V} = T (\frac{\partial P}{\partial T})_{V} (\frac{\partial V}{\partial T})_{P}$ se shuru karo (standard result from $d S$ ; dekho Heat capacities $C_P - C_V$). Yeh step kyun? Ismein already ek awkward factor hai, $(\partial P / \partial T)_{V}$ , jise hum ab page ke top se triple product se rewrite karenge.

Step 2. Triple product: $(\frac{\partial P}{\partial T})_{V} = - \frac{( \partial V / \partial T ) _{P}}{( \partial V / \partial P ) _{T}} = \frac{α}{κ _{T}} .$ Yeh step kyun? Yeh ek constant- $V$ derivative (mushkil) ko do measurables $α$ aur $κ_{T}$ mein convert karta hai. Triple product ka minus $κ_{T}$ ki definition mein minus ke saath cancel ho jaata hai.

Step 3. Substitute karo aur $(\partial V / \partial T)_{P} = V α$ use karo: $C_{P} - C_{V} = T \cdot \frac{α}{κ _{T}} \cdot V α = \frac{T V α ^{2}}{κ _{T}} .$ Yeh step kyun? Do $α$ 's collect karne se $α^{2}$ milta hai; ho gaya.

Step 4 (ideal-gas check). $P V = n R T$ ke liye hum do coefficients cleanly nikalte hain:

\kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T = -\frac{1}{V}\left(-\frac{nRT}{P^2}\right) = \frac{1}{P}.$$ Boxed formula mein substitute karo: $$C_P-C_V = \frac{TV\,(1/T)^2}{1/P} = \frac{TV\cdot P}{T^2} = \frac{PV}{T} = \frac{nRT}{T} = nR.$$ $n=1$ ke saath: $C_P-C_V = R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$. **Yeh step kyun?** Ideal-gas $\alpha=1/T$ aur $\kappa_T=1/P$ substitute karne se sab kuch $PV/T=nR$ mein collapse ho jaata hai. **Verify:** General formula ek mole ke liye textbook $C_P-C_V=R$ reproduce karta hai. Saath hi $C_P>C_V$ hamesha kyunki $\alpha^2\ge0$, $\kappa_T>0$, $T>0$ — ek bonus universal fact jo derivation hume free mein de deta hai. ✓

Recall Quick self-test

Kaun sa relation $(\partial S / \partial V)_{T}$ deta hai aur kya usme minus hota hai? ::: Helmholtz $F$ : $(\partial S / \partial V)_{T} = (\partial P / \partial T)_{V}$ , nahi minus (do minuses cancel ho jaate hain). Ideal gas ke liye $(\partial U / \partial V)_{T} = 0$ kyun hai? ::: Kyunki $T (\partial P / \partial T)_{V} = P$ exactly, toh energy equation $P - P = 0$ deta hai. Jab $T \to 0$ , $α \to 0$ kyun hota hai? ::: Third law ⇒ $(\partial S / \partial P)_{T} \to 0$ , aur Gibbs relation isko $(\partial V / \partial T)_{P} \to 0$ se tie karta hai.

Connections

Parent: Maxwell relations
Joule expansion and internal energy — Ex 4 & 5
Heat capacities $C_P - C_V$ — Ex 8
Thermal expansion coefficient and isothermal compressibility — $α, κ_{T}$
First and Second Laws of Thermodynamics — master differential ka source
Equality of mixed partial derivatives (Schwarz theorem) — har step ke peeche ka engine