Thermodynamic potentials — U, H, F, G (preview): Question bank

1.7.26 · D5 · Physics › Thermodynamics › Thermodynamic potentials — U, H, F, G (preview)

Shuru karne se pehle, teen words jo hum baar baar use karenge, bilkul scratch se:

Definition Symbol

$δ$ aur phrase "inexact differential" Jab koi quantity genuinely state ki property hoti hai (jaise $U$ , ya tank mein kitna paani hai), tab ek chhoti si change ko $d U$ likhte hain — ek exact differential: do states ke beech ka total change sirf endpoints par depend karta hai. Heat aur work stored properties nahi hain — ye path ke saath transfer ki gayi amounts hain. Heat ki thodi si matra ko $δ Q$ likhte hain (special $δ$ se, $d$ se nahi) taaki ye warn ho sake: ye inexact differential kaise pahunche us par depend karta hai, sirf endpoints par nahi. Toh $δ Q$ = "boundary cross karne wali heat ki chhoti si matra," $δ W$ = "kiye gaye work ki chhoti si matra." Hum label $δ Q_{rev}$ lagate hain jab path reversible ho (infinitely gently kiya gaya), kyunki tabhi $δ Q_{rev} = T d S$ hota hai. Dekho First Law of Thermodynamics aur Entropy and the Second Law.

Chaar differentials ke reminders jo hum quote karte rahenge (parent se): $d U = T d S - p d V, d H = T d S + V d p, d F = - S d T - p d V, d G = - S d T + V d p$

Do pictures jo tumhare dimaag mein rahein jab tum traps solve karo:

Figure — Thermodynamic potentials — U, H, F, G (preview)

True or false — justify

d U = T d S - p d V

kisi bhi process ke liye hold karta hai, reversible ho ya nahi.

False — state-function changes ki equality ke roop mein ye hamesha hold karta hai (saare terms exact differentials hain), lekin identifications

δ Q_{rev} = T d S

aur

δ W_{rev} = p d V

sirf reversible path ke liye hold karti hain. Toh equation universal hai; iska term-by-term interpretation nahi.

U

ki natural variables

T

aur

V

hain.

False — ye

S

aur

V

hain, kyunki

d U

d S

aur

d V

ke terms mein express kiya jaata hai. Tum

T

ko output ke roop mein paate ho (

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

), input knob ke roop mein nahi.

G

hamesha woh potential hai jo equilibrium par minimize hoti hai.

False —

G

sirf constant

T

aur

p

par minimize hoti hai. Constraints badlo (maan lo constant

S, V

) aur

U

boss ban jaati hai. Sahi potential wahi hoti hai jिसकी natural variables tum fixed rakh rahe ho.

"Free energy" ka matlab hai energy jo bina kisi cost ke extract kar sako.

False — ye fixed

T

par (aur

G

ke liye

p

par) extractable maximum useful work hai.

T S

term woh energy hai jo second law ke hisaab se surroundings ko "tax" ke roop mein lock ho jaati hai; "free" ka matlab work ke liye available hai, muft nahi.

Constant

T, p

par ek spontaneous reaction ke liye, system ka

G

decrease hona chahiye.

True — constant

T, p

par universe ki total entropy ka badhna exactly equivalent hai

Δ G_{system} < 0

se. Yahi translation hai kyun chemists

G

dekhte hain.

Enthalpy

H

generally ek body ka heat content hota hai.

False — sirf constant pressure par (aur ek reversible path ke saath)

d H = δ Q_{rev}

hota hai.

H

ko "heat content" sirf us setting mein kaho; constant volume par

U

ki change heat ke barabar hoti hai.

F = U - T S

U

se bada ho sakta hai.

Normal conditions mein False — usual positive absolute temperature (

T > 0

) aur non-negative entropy (

S \geq 0

) ke saath, product

T S \geq 0

hota hai, isliye

F = U - T S

hamesha

U

se kam ya barabar hota hai. Entropy tax sirf subtract karti hai; kabhi add nahi karti.

Maxwell relation nature ka ek extra experimental law hai.

False — ye pure mathematics hai: ek state function ke mixed second partial derivatives ki equality (differentiation ka order matter nahi karta). Koi naya physics nahi, bas yeh fact ki

F, G, U, H

exact differentials hain. Dekho Maxwell Relations.

Spot the error

d H = T d S - V d p

d U = T d S - p d V

ke analogy se."

V d p

term ka sign galat hai.

p V

add karne se

d (p V) = p d V + V d p

milta hai;

+ p d V

piece

d U

mein

- p d V

ko exactly cancel karta hai aur jo bachta hai woh

+ V d p

hai. Sahi:

d H = T d S + V d p

d F = - S d T - p d V

, isliye

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

Minus sign miss ho gayi.

d T

coefficient match karne se

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

milta hai, toh

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

. Yeh extra negative hi entropy ko positive banata hai.

"Kyunki

G = H - T S

, constant

T, p

par

G

minimize karna

H

minimize karne jaisa hai."

Nahi —

- T S

term matter karta hai. Ek reaction endothermic ho sakti hai (

Δ H > 0

) lekin phir bhi spontaneous, kyunki bada positive

Δ S

- T Δ S

ko dominate karwata hai aur

Δ G < 0

ho jaata hai. Sirf heat nahi, entropy bhi decide karti hai.

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

, isliye temperature volume ka derivative hai."

Kaunsa variable fixed hai, yahan confusion hai.

T

U

ka derivative hai

S

ke respect mein $V$ ko constant rakh kar; volume frozen variable hai, wo cheez nahi jise differentiate kar rahe hain.

"Legendre transform system ki physics badal deta hai."

Ye sirf description badalta hai, swap karta hai ki kaunsa variable independent knob hai (jaise

S \to T

)

T S

product subtract karke. Same system, same physics, zyada convenient bookkeeping. Dekho Legendre Transform.

C_{p} = (\partial U / \partial T)_{p}

Potentials mix ho gaye. Constant pressure par heat capacity enthalpy use karta hai:

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

, kyunki constant

p

par

H

(nah ki

U

) ki change heat ke barabar hoti hai.

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

. Dekho Heat Capacities Cp and Cv.

Why questions

Hume sirf

U

ki jagah chaar potentials ki zaroorat kyun hai?

Kyunki alag alag experiments alag variables fix karte hain. Tum rarely entropy control karte ho, lekin tum

T

(thermostat) aur

p

(open beaker) control kar sakte ho. Har potential is tarah engineer ki gayi hai ki jo variables tum actually fix karte ho, woh uski natural variables ban jaayein.

T S

subtract karne se

T

"natural variable" kyun ban jaata hai?

Kyunki

d (T S) = T d S + S d T

, aur

T d S

piece exactly

d U

ke andar

T d S

ko cancel kar deta hai. Jo bachta hai woh ek

S d T

term hai, toh

d T

(hence

T

) ab differential ke roop mein appear karta hai — knob ban jaata hai.

G

"chemistry" potential kyun hai?

Open flasks mein reactions atmospheric pressure aur room temperature par hoti hain — constant

T

aur

p

— jo precisely

G

ki natural variables hain. Iska minimum usi exact lab setup ka equilibrium hai. Dekho Gibbs Free Energy and Chemical Equilibrium.

"Universe ki entropy maximize" kyun "ek system potential minimize" mein convert ho jaata hai?

Fixed

T, p

par surroundings ke saath heat/work exchange karne wale system ke liye, surroundings ki entropy change ko

- Δ H_{sys} / T

likha ja sakta hai. System ke apne

Δ S

ke saath combine karne par pata chalta hai: total entropy upar

\Leftrightarrow

(yaani if and only if)

Δ G_{sys}

neeche. Constraint sirf second law ko re-express karta hai.

d U = T d S - p d V

mein

p d V

par minus kyun hai lekin

T, p

ke potentials mein

V d p

par plus hai?

Minus encode karta hai ki system energy lose karta hai jab wo expansion work karta hai (

+ p d V

work bahar jaati hai). Jab hum

p

ko natural banane ke liye Legendre-transform karte hain, added

d (p V)

ise

+ V d p

mein flip kar deta hai: ek bookkeeping sign, physics change nahi.

Maxwell relation practically useful kyun ho sakta hai?

Ye ek measure karna mushkil quantity ko ek aasaan se swap karta hai. Jaise

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

ek entropy slope (directly unmeasurable) ko ek pressure–temperature slope se replace karta hai jo tum gauge se padh sakte ho. Dekho Maxwell Relations.

Edge cases

Constant volume par, kya

Δ H

abhi bhi absorbed heat ke barabar hoga?

Nahi — constant

V

par (aur ek reversible path ke saath)

Δ U

heat ke barabar hota hai (

d U = T d S = δ Q_{rev}

jab

d V = 0

ho).

Δ H = δ Q_{rev}

ke liye constant pressure chahiye. Inhe mix karna classic trap hai.

F

mein entropy tax

T S

ka kya hoga jab

T \to 0

Jab

T \to 0

, toh

T S \to 0

, isliye

F \to U

: absolute zero par free energy aur internal energy coincide karte hain. Saari internal energy "available" ho jaati hai, kyunki koi thermal disorder nahi hai jiske liye pay karna pade.

Agar ek process

S

aur

V

fixed rakhe, toh kaunsa potential equilibrium decide karta hai?

U

— kyunki

S, V

U

ki natural variables hain, aur fixed

S, V

par second law

U

ko minimum tak force karta hai. Ye isolated, rigid box hai (practice mein rare, lekin conceptual anchor).

Ek incompressible solid ke liye (

d V \approx 0

U

aur

H

, aur

F

aur

G

mein kya relation hai?

V

nearly constant aur

p

modest hone se

p V

changes tiny hote hain, isliye

H \approx U

aur

G \approx F

. "

+ p V

" pairs ke beech ka distinction collapse ho jaata hai jab volume change nahi ho sakta — isliye solid-state work mein aksar ise ignore kiya jaata hai.

Kya

Δ G > 0

kabhi ek real observed reaction describe kar sakta hai?

Haan, lekin sirf driven — ek aisi doosri process ke saath coupled jिसका

Δ G < 0

zyada bada ho (jaise biochemistry mein ATP hydrolysis ke saath coupling). Akele, constant

T, p

par, positive

Δ G

wali reaction forward direction mein spontaneously proceed nahi karti.

Kya

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

meaningful hai agar

S

har jagah fixed ho (adiabatic)?

Derivative ek local slope hai

U

-vs-

S

surface ka; ye defined hai chahe koi given process

S

change kare ya nahi. Yahan tak ki ek adiabatic step mein jahan

S

constant rehta hai, woh slope phir bhi system ka temperature equal karta hai.

Agar do potentials ki value same ho — toh kya state fix ho jaati hai?

Zarooori nahi. Ek akela number kabhi state pin nahi karta; tumhe natural-variable pair chahiye. Equilibrium is baare mein hai ki ek potential allowed states par apne minimum par ho, numerical coincidences ke baare mein nahi.

Recall Ek-line self-test

Woh constraint batao jo har potential ko boss banata hai. $U$ ::: constant $S, V$ $H$ ::: constant $S, p$ $F$ ::: constant $T, V$ $G$ ::: constant $T, p$