Maxwell relations — derivation from each potential: Question bank

2.4.3 · D5 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Maxwell relations — derivation from each potential

True or false — justify

True or false: Maxwell relation ek naya physics ka law hai jo pehle do laws se aage jaata hai.

False — yeh pure calculus hai (equality of mixed partials, dekho Equality of mixed partial derivatives (Schwarz theorem)) jo ek state function ke exact differential pe apply hota hai, jo khud pehle do laws se aaya tha.

True or false:

(\frac{\partial S}{\partial V})_{T} = (\frac{\partial P}{\partial T})_{V}

mein koi minus sign nahi hai, toh sab Maxwell relations sign-free hote hain.

False — yeh wala (

F

se) positive hota hai kyunki

- S d T

aur

- P d V

dono minus carry karte hain jo cancel ho jaate hain;

U

aur

G

se aane wale relations genuinely minus carry karte hain.

True or false: Maxwell relations ke liye substance ka ideal gas hona zaroori hai.

False — yeh kisi bhi simple compressible closed system ke liye hold karte hain kyunki yeh sirf potentials ke state functions hone se follow hota hai, kisi equation of state se nahi.

True or false:

d U = T d S - P d V

sirf reversible process ke liye likh sakte hain, toh Maxwell relations sirf reversibly apply hote hain.

False — yeh differential state variables ke beech relate karta hai, aur kyunki

U, S, V

sab state functions hain, yeh equation equilibrium states ke beech hold karta hai chahe path koi bhi ho.

True or false: Enthalpy relation

(\frac{\partial T}{\partial P})_{S} = (\frac{\partial V}{\partial S})_{P}

batata hai ki temperature pressure ke saath constant entropy pe kaise change hoti hai.

True — constant

S

exactly adiabatic-reversible condition hai, toh yeh quantity squeeze karne par reversible-adiabatic temperature response hai.

True or false: Kyunki

U

aur

F

dono mein

- P d V

hai, unke Maxwell relations identical lagte hain.

False —

U

natural variables

(S, V)

use karta hai jisse

(\partial T / \partial V)_{S} = - (\partial P / \partial S)_{V}

milta hai, jabki

F

(T, V)

use karta hai jisse

(\partial S / \partial V)_{T} = (\partial P / \partial T)_{V}

milta hai; alag independent variables, alag relations.

True or false: Ideal gas ke liye

(\partial U / \partial V)_{T} = 0

ek assumption hai jo hum haath se plug in karte hain.

False — yeh energy equation aur ideal-gas law se derived hota hai, jo dikhata hai ki

T (\partial P / \partial T)_{V} = P

exactly hold karta hai (dekho Joule expansion and internal energy).

Spot the error

"Kyunki

U = U (T, V)

bilkul valid hai, main

d U = T d S - P d V

(T, V)

as variables use karke read off kar sakta hoon." — error dhundho.

Clean coefficients

T

aur

- P

tabhi appear hote hain jab

U

apne natural variables

(S, V)

mein express ho;

(T, V)

mein differential

d U = C_{V} d T + [T (\partial P / \partial T)_{V} - P] d V

ban jaata hai, jo Maxwell-generating form nahi hai.

d G = - S d T + V d P

se mujhe

(\frac{\partial S}{\partial P})_{T} = (\frac{\partial V}{\partial T})_{P}

milta hai." — galti pakdo.

Minus missing hai:

M = - S

hai, toh

(\partial M / \partial P)_{T} = - (\partial S / \partial P)_{T}

hoga, jisse

(\partial S / \partial P)_{T} = - (\partial V / \partial T)_{P}

milta hai.

(\frac{\partial T}{\partial V})_{S} = (\frac{\partial T}{\partial V})_{P}

kyunki yeh same partial hai." — kya galat hai?

Subscript quantity ka hissa hai;

S

fixed rakhna versus

P

fixed rakhna genuinely alag numbers deta hai, toh yeh do alag derivatives hain.

H = U + P V

, toh

d H = d U + P d V = T d S - P d V + P d V = T d S

." — dropped term dhundho.

Product rule se

d (P V) = P d V + V d P

hota hai, toh correct result

d H = T d S + V d P

hai;

V d P

bhoolne se enthalpy ki natural pressure dependence kho jaati hai.

F

se Maxwell relation kehta hai

(\partial S / \partial V)_{T} = (\partial P / \partial T)_{V}

, toh entropy hamesha badhti hai jab main volume badhata hoon." — flaw kahan hai?

(\partial S / \partial V)_{T}

ka sign

(\partial P / \partial T)_{V}

ke sign ke barabar hai, jo normal substances ke liye positive hota hai lekin negative bhi ho sakta hai (jaise

4^{\circ}

C ke paas water mein), toh yeh claim universal nahi hai.

"Kyunki mixed partials equal hain,

(\frac{\partial P}{\partial S})_{V} = (\frac{\partial T}{\partial V})_{S}

plus sign ke saath." — check karo.

U

-relation minus carry karta hai:

(\partial T / \partial V)_{S} = - (\partial P / \partial S)_{V}

, kyunki

N = - P

minus inherit karta hai

- P d V

term se.

"Main charon signs

-, +, +, -

yaad kar lunga aur kaam khatam." — yeh risky kyun hai?

Memorized signs exam pressure mein easily ulajh jaate hain; differential se re-derive karna (

M

N

unke signs ke saath read off karo, cross-differentiate karo) foolproof hai aur seconds mein hota hai.

Why questions

Legendre transform

H = U + P V

kyun

V

P

se natural variable ke roop mein swap karta hai?

P V

add karne se

+ P d V + V d P

inject hota hai;

+ P d V

existing

- P d V

ko annihilate kar deta hai,

V d P

bacha rehta hai toh

d P

(na ki

d V

) independent differential ban jaata hai (dekho Thermodynamic potentials & Legendre transforms).

F

aur

G

se aane wale relations ko "measurable" kyun kehte hain?

Yeh

S

(directly measure karna mushkil) ko left pe rakhte hain jo right pe

P, V, T

ke derivatives ke barabar hota hai, jisse tum ordinary lab quantities jaise thermal expansion se entropy changes infer kar sako.

Maxwell relation hold karne ke liye potential ka state function hona zaroori kyun hai?

Sirf state functions ke exact differentials hote hain jinke path-independent, smooth second partials hote hain, jo precisely wahi condition hai jab equality of mixed partials apply hoti hai.

(\partial S / \partial P)_{T} = - (\partial V / \partial T)_{P}

physically satisfying kyun hai?

Right side (

- V

tak) thermal expansion coefficient

α = \frac{1}{V} (\partial V / \partial T)_{P}

hai, toh jo substance garam hone pe expand hota hai woh compress hone pe entropy kho deta hai — expansion aur entropy-pressure response ek saath locked hain.

d U = T d S - P d V

ko Maxwell relation ke saath combine karne se

(\partial U / \partial V)_{T} = T (\partial P / \partial T)_{V} - P

kyun milta hai?

Fixed

T

d V

se divide karne par

(\partial U / \partial V)_{T} = T (\partial S / \partial V)_{T} - P

milta hai, phir

F

-relation unmeasurable

(\partial S / \partial V)_{T}

ko measurable

(\partial P / \partial T)_{V}

se replace kar deta hai.

In variables se paanchwa independent Maxwell relation kyun nahi bana sakte?

Sirf chaar tarike hain do conjugate couples

(T, S)

aur

(P, V)

ko ek potential mein pair karne ke, toh simple system ke liye exactly chaar distinct relations exist karte hain.

Edge cases

Edge case: Ideal gas ke liye,

F

-relation

(\partial S / \partial V)_{T}

ke liye kya deta hai?

(\partial P / \partial T)_{V} = n R / V

hai, toh

(\partial S / \partial V)_{T} = n R / V > 0

— fixed

T

pe entropy hamesha volume ke saath badhti hai, jo free (Joule) expansion mein disorder badhne se consistent hai.

Edge case: Kisi substance mein zero thermal expansion hone par

G

-relation ka kya hota hai?

Agar wahan

(\partial V / \partial T)_{P} = 0

hai, toh

(\partial S / \partial P)_{T} = 0

bhi — us special point pe fixed

T

pe squeeze karne se entropy nahi badlti (jaise water ke density maximum ke paas).

Edge case: Kya phase transitions ke dauran Maxwell relations hold karte hain?

Single phase ke andar hold karte hain, lekin first-order transition pe potentials mein kinks (discontinuous first derivatives) hoti hain, toh mixed-partial identity exactly coexistence line pe fail ho jaati hai jahan derivatives smooth nahi hote.

Edge case: Absolute zero pe, third law

(\partial S / \partial P)_{T}

aur

(\partial S / \partial V)_{T}

ke baare mein kya imply karta hai?

Kyunki

T \to 0

ke saath

S \to 0

P

aur

V

se independent hai, dono

(\partial S / \partial P)_{T}

aur

(\partial S / \partial V)_{T}

vanish ho jaate hain, jo Maxwell ke through force karta hai

(\partial V / \partial T)_{P} \to 0

aur

(\partial P / \partial T)_{V} \to 0

— thermal expansion absolute zero pe khatam ho jaati hai.

Edge case: Agar system closed nahi hai (matter enter kar sakta hai)?

Har differential mein

μ d N

term appear hota hai, toh simple four-corner Maxwell relations

μ

aur

N

wale naye relations se supplement hote hain; yeh chaar diye gaye fixed

N

assume karte hain.

Edge case: Kya

(\partial T / \partial V)_{S} = - (\partial P / \partial S)_{V}

kabhi dono sides pe trivially zero hota hai?

Aise system ke liye jahan entropy aur volume effectively independent hain (jaise ideal incompressible/no-work idealization), dono sides vanish ho sakti hain, lekin kisi bhi real gas ke liye reversible-adiabatic temperature-volume coupling nonzero hoti hai.

Connections

Maxwell relations — derivation from each potential — parent derivation
Equality of mixed partial derivatives (Schwarz theorem) — har relation ke peeche ka calculus engine
Thermodynamic potentials & Legendre transforms — kyun chaar potentials, kyun natural variables
Joule expansion and internal energy — $(\partial U / \partial V)_{T}$ application
Heat capacities $C_P - C_V$ aur Thermal expansion coefficient and isothermal compressibility — jahan yeh relations kaam aate hain