Thermodynamic potentials — U (internal), H (enthalpy), F (Helmholtz), G (Gibbs): Question bank

2.4.1 · D5 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Thermodynamic potentials — U (internal), H (enthalpy), F (He

True or false — justify

Har line: decide karo true hai ya false, phir kyun batao. Commit karne ke baad hi reveal karo.

d U = T d S - p d V

sirf reversible processes ke liye hold karta hai.

False.

U

ek state function hai, isliye

d U

sirf endpoints par depend karta hai; hum isse reversibly derive karte hain lekin yeh kisi bhi process ke liye hold karta hai unhi do states ke beech. Sirf individual pieces

T d S = δ Q

aur

p d V = δ W

reversible-path statements hain.

F

aur

G

bas "free energy" ke do naam hain aur interchangeable hain.

False. Ye

p V

se differ karte hain (yaad karo

G = F + p V

) aur inke different natural variables hain —

F = F (T, V)

G = G (T, p)

. Equilibrium par kaun minimize hoga yeh depend karta hai ki aap kya clamp kar rahe ho: volume (

F

) ya pressure (

G

Equilibrium par har thermodynamic potential minimize hota hai.

False. Har ek sirf tab minimize hota hai jab uske apne natural variables fixed hon:

U

fixed

(S, V)

par,

H

fixed

(S, p)

par,

F

fixed

(T, V)

par,

G

fixed

(T, p)

par. Apne constraints ke liye galat potential minimize karo toh criterion simply false ho jaata hai.

H = U + p V

, isliye enthalpy hamesha internal energy se badi hoti hai.

Generally false.

p V

gas ke liye positive hai isliye wahan

H > U

hota hai, lekin definition sirf ek Legendre transform hai convenience ke liye; physics mein kuch bhi require nahi karta ki

p V

kisi bhi comparison mein dominate kare, aur dusre systems ke liye (jaise negative pressures) yeh flip ho sakta hai.

H

paane ke liye Legendre transform

p V

add karta hai jabki

F

paane ke liye

T S

subtract karta hai, toh rule inconsistent hai.

False — yeh ek consistent rule hai. Aap hamesha conjugate-pair product subtract karte ho variable ko uske slope se swap karne ke liye; sign

d U = T d S - p d V

se inherited hota hai. Kyunki

V

minus carry karta hai (

- p d V

), isse swap karne par

p V

add hota hai; kyunki

S

plus carry karta hai, isse swap karne par

T S

subtract hota hai.

Constant

T

par

- Δ F

woh maximum work ke barabar hai jo system kar sakta hai.

True.

d F = - S d T - p d V

se, constant

T

par

d F = - p d V = - δ W

milta hai, isliye

- Δ F

woh total (maximum, reversible) work hai jo extract ki ja sakti hai jab temperature fixed ho.

(Δ H)_{p}

constant pressure par exchange ki gayi heat ke barabar hota hai.

True. Constant

p

par,

d H = T d S + V d p = T d S = δ Q

, isliye enthalpy change hi constant-pressure heat hai — yahi wajah hai ki chemists reaction heats ko

Δ H

ke roop mein tabulate karte hain.

Ideal gas ke liye,

G

pressure par depend karta hai lekin

H

nahi karta.

True. Ideal gas ke liye

(\partial H / \partial p)_{T} = 0

hai isliye

H = H (T)

only; lekin

(\partial G / \partial p)_{T} = V = n R T / p \neq = 0

, jisse famous

Δ G = n R T ln (p / p_{0})

milta hai.

Ek Maxwell relation ek unmeasurable derivative jaise

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

ko ek measurable se replace karne deta hai.

True.

d F

ki exactness force karti hai

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

; right side equation of state se padhi jaati hai, isliye aapko kabhi entropy ki volume-dependence directly measure nahi karni padti.

Why questions

Ek ya do sentences mein mechanistic reason do, restatement nahi.

Hume sirf

U

use karne ki bajay chaar potentials ki zaroorat kyun hai?

Kyunki aap lab mein wahi control karte ho jo aap clamp kar sako — usually

T

(thermostat) aur

p

(atmosphere),

S

aur

V

nahi. Har potential usi physics ko un variables mein re-express karta hai jo aap actually fixed rakh sakte ho, isliye uska minimum directly equilibrium signal karta hai.

Har

d Φ

ek exact differential kyun hona chahiye, aur iska kya matlab hai?

Kyunki har

Φ

ek state function hai — uski value sirf state par depend karti hai, isliye uska differential kisi bhi loop par close hota hai. Exactness mixed second partials ki equality force karti hai, jo hai hi har Maxwell relation ka source.

Chemists ki workhorse reactions ke liye

F

nahi balki

G

kyun hai?

Beaker mein reactions fixed

T

aur fixed

p

(room ke liye open) par hoti hain, aur yahi exactly

G

ke natural variables hain — isliye

G

ka minimize hona us setup ke liye correct equilibrium criterion hai.

Legendre transform conjugate product subtract kyun karta hai, koi formula substitute karne ki bajay?

x y

subtract karna jahan

y

Φ

x

mein slope hai,

d x

term ko cancel kar deta hai aur ek

d y

term appear karta hai, cleanly natural variable ko uske conjugate se swap karta hai — koi equation-of-state input nahi chahiye.

Ideal gas ke liye

(\partial H / \partial p)_{T} = 0

kyun special hai, universal kyun nahi?

Iske liye

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

chahiye, jo ideal-gas law exactly satisfy karta hai; intermolecular forces wale real gases isse violate karte hain, nonzero

(\partial H / \partial p)_{T}

dete hain (Joule–Thomson cooling ka basis).

Chaar Maxwell relations mein different signs kyun hote hain?

Har ek ka sign directly us potential ke differential mein conjugate term ke sign se copy hota hai — jaise

- S d T

ek minus inject karta hai. Inhe kabhi guess nahi kiya jaata; yeh

d Φ

se padhe jaate hain.

- Δ F

maximum work kyun hai, sirf the work kyun nahi?

Sirf reversible path

δ W = p d V

satisfy karta hai; irreversibility available energy ka kuch hissa dissipate kar deta hai, isliye real processes

- Δ F

se kam deliver karte hain, ise ceiling banaate hue.

Edge cases

Boundary aur degenerate scenarios — carefully reason karo.

G = U - T S + p V

T \to 0

absolute zero par kya hoga?

T S

term vanish ho jaata hai (aur third law se

S \to

const ya

0

), isliye

G \to U + p V = H

;

T = 0

par "messiness tax" gayab ho jaati hai aur Gibbs energy enthalpy ke barabar ho jaati hai.

Agar ek process dono

S

aur

T

fixed rakhe, toh equilibrium kaun sa potential govern karta hai?

Yeh ek simple one-component system ko over-constrain karta hai —

S

aur

T

conjugate hain, isliye dono fix karne se generally poori state pin ho jaati hai. Koi free minimization nahi hai; "kaun sa potential" ka sawaal us pair ke liye ill-posed hai.

Incompressible solid (

d V = 0

hamesha) ke liye,

F

aur

G

, aur

U

aur

H

kaise relate karte hain?

V

effectively constant hone par,

H = U + p V

aur

G = F + p V

U

aur

F

pressure per unit ek additive constant se differ karte hain; changes

Δ H \approx Δ U

aur

Δ G \approx Δ F

, isliye volume-work distinction lagbhag collapse ho jaati hai.

First-order phase transition par (jaise boiling), kaun sa potential boundary ke across continuous hota hai aur kyun?

G

(per mole) continuous hota hai — do phases coexist precisely isliye karte hain kyunki woh same

T

p

, aur molar

G

share karte hain. Iske derivatives (

S = - (\partial G / \partial T)_{p}

V = (\partial G / \partial p)_{T}

) jump karte hain, jo ise "first-order" banata hai.

Ideal gas ki vacuum mein free (irreversible) isothermal expansion ke liye

Δ F

kya hai?

Kyunki

F = F (T, V)

aur ideal-gas free expansion mein

U

(hence

T

) unchanged rehta hai,

Δ F = - T Δ S < 0

kyunki

S

increase hoti hai — even though koi work extract nahi ki gayi, yeh dikhata hai ki

- Δ F

sirf potential work hai, realized work nahi.

Agar entropy

S = 0

ho (ek perfectly ordered

T = 0

crystal), toh kya

F = U

exactly?

Haan:

F = U - T S

aur

T S = 0

jab koi bhi factor zero ho, isliye

F = U

. "Free" energy total energy ke barabar hai kyunki koi unusable, entropy-locked portion nahi hai.

Kya fixed

T, p

par kisi process ke liye

Δ H

negative ho sakta hai jabki

Δ G

positive ho?

Haan.

Δ G = Δ H - T Δ S

; strongly negative

Δ S

(badi ordering)

- T Δ S

ko negative

Δ H

se outweigh kara sakta hai, deta hai

Δ G > 0

— reaction exothermic hai lekin non-spontaneous.

Recall Ek-line takeaways

Ek potential minimize hota hai sirf apne natural variables ki fixed values par. ::: $U : (S, V)$ , $H : (S, p)$ , $F : (T, V)$ , $G : (T, p)$ .
Maxwell-relation signs aate hain ::: matching $d Φ$ mein signs se — kabhi guess nahi kiye jaate.
$- Δ F$ (const $T$ ) aur $- Δ G$ (const $T, p$ ) hain ::: maximum obtainable work, sirf reversibly saturate hota hai.

Question bank — Thermodynamic potentials — U (internal), H (enthalpy), F (Helmholtz), G (Gibbs)

True or false — justify

Spot the error

Why questions

Edge cases