Chemical potential μ = (∂G - ∂N)_{T,P}: Question bank

2.4.5 · D5 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Chemical potential μ = (∂G - ∂N)_{T,P}

Shuru karne se pehle, ek word jis par hum baar baar lean karenge: koi quantity extensive hoti hai agar woh system size ke saath scale karti hai (box double karo, quantity double ho — jaise $N$ , $V$ , $G$ ), aur intensive hoti hai agar nahi hoti (jaise $T$ , $P$ , aur $μ$ khud). Yeh split apne dimaag mein rakho; neeche ke aadhe traps sirf log dono ko confuse karte hain isi wajah se hain.

Do chhote schematics poore page ka visual kaam karte hain — unhe dekho jab cards "the peaks picture" aur "the coexistence line" reference karein.

Figure — Chemical potential μ = (∂G - ∂N)_{T,P}

True or false — justify

Chemical potential μ hamesha negative hota hai

False. Ideal gas ke liye

μ = μ^{\circ} (T) + k_{B} T ln (P / P^{\circ})

pressure aur reference ke hisaab se positive, negative, ya zero ho sakta hai; iska sign convention-plus-conditions ki baat hai, koi law nahi.

μ ek intensive quantity hai

True. Yeh Gibbs energy per particle hai (

μ = G / N

ek pure substance ke liye) — system double karne par yeh unchanged rehta hai, bilkul

T

aur

P

ki tarah.

Do connected reservoirs ke beech equilibrium par, particle numbers

N_{A}

aur

N_{B}

equal hote hain

False. Jo equalize hota hai woh

μ_{A} = μ_{B}

hai, na ki

N

. Ek bada box aur ek chhota box bahut alag numbers hold kar sakte hain lekin identical chemical potential par baith sakte hain.

Agar system A mein B se zyada particles hain, toh

μ_{A} > μ_{B}

False.

μ

concentration/pressure aur interactions track karta hai, raw count nahi. Ek bada dilute box ek chhote dense wale se kam

μ

rakh sakta hai.

μ = G / N

kisi bhi thermodynamic system ke liye hold karta hai

False. Yeh sirf ek single pure substance ke liye hold karta hai, kyunki yeh

G

ke ek particle number mein homogeneous degree 1 hone par depend karta hai. Mixtures ke liye har species ka apna partial molar

μ_{i} = (\partial G / \partial N_{i})_{T, P, N_{j \neq = i}}

hota hai.

Particles spontaneously lower particle energy wale region se higher particle energy wale region mein ja sakte hain

True. Flow free energy

μ

follow karta hai, jisme ek

- T \partial S / \partial N

entropy term hota hai; agar entropy gain dominate kare, toh matter bare energy mein upar chadh jaata hai.

(\frac{\partial G}{\partial N})_{S, V}

bhi μ ke barabar hai

False. Sirf

(\partial U / \partial N)_{S, V}

μ ke barabar hai. Subscripts potential aur held variables dono choose karte hain;

G

S, V

-held ke saath mix karna bilkul alag quantity deta hai.

Do phases coexistence par (liquid + uska vapor) ka chemical potential equal hota hai

True. Coexistence hi phase boundary ke across diffusive equilibrium hai, isliye

μ_{liquid} = μ_{vapor}

bhale hi densities bahut alag hon — yeh woh coexistence line hai jo doosre figure mein draw ki gayi hai, aur yeh Phase Equilibrium and Clausius-Clapeyron se connect hoti hai.

Fixed temperature par ideal gas compress karne se uska chemical potential kam ho jaata hai

False.

μ

k_{B} T ln P

ki tarah badhta hai, isliye zyada

P

matlab zyada

μ

— gas "zyada eager" ho jaati hai lower-pressure regions mein escape karne ke liye.

Spot the error

"Kyunki

d U = T d S - P d V

, particles add karne se

U

nahi badalta."

Error yeh hai ki fixed-N first law use ki ja rahi hai. Jab

N

vary kar sakta hai toh sahi law hai

d U = T d S - P d V + μ d N

;

μ d N

term precisely woh energy hai jo naye particles saath laate hain.

μ = (\partial G / \partial N)

, toh μ sirf

G

ka ek derivative hai — koi bhi derivative chalega."

Subscript

T, P

decoration nahi hai. Alag held variables alag physical quantities select karte hain; sirf

(\partial G / \partial N)_{T, P}

hi μ hai.

"μ ko potential kaha jaata hai aur iske energy units hain, isliye yeh ek particle ki potential energy hai."

μ ek free energy per particle hai, jo

- T (\partial S / \partial N)

ke through entropy bundle karta hai. Equal energy lekin alag concentration alag μ deta hai, isliye yeh bare potential energy nahi hai.

"Particles high concentration se low concentration mein flow karte hain — yeh fundamental law hai."

Concentration flow ek special case hai. Fundamental driver

μ

hai: high

μ

→ low

μ

. Concentration sirf tab

μ

track karta hai jab kuch aur (interactions, external fields) compete nahi kar raha.

"Kyunki

G = N μ

, agar main particles add karun toh

G

badhega lekin

μ

hamesha ke liye fixed rahega."

μ generally

N

par depend karta hai (pressure/concentration ke through).

G = N μ

ek instantaneous relation hai, yeh promise nahi ki

μ

constant rahega jaise tum particles add karte raho.

d G = 0

par system equilibrium mein hai, isliye

d G = 0

matlab

μ_{A} = μ_{B}

automatically kisi bhi process ke liye."

Sirf jab constraint

d N_{A} = - d N_{B}

d N_{A}

ko akela free variable banata hai tab

d G = (μ_{A} - μ_{B}) d N_{A} = 0

equality force karta hai. Is logic ko woh particle-conservation link chahiye, akela

d G = 0

nahi.

G \equiv U - T S + P V

chemical potential ke liye invent kiya gaya tha."

Gibbs Free Energy G independently exist karta hai

T, P

mein natural potential ke roop mein; μ sirf

d G = - S d T + V d P + μ d N

mein

d N

ke coefficient ke roop mein emerge hota hai.

Why questions

N

ka conjugate kyun hai,

S

V

ka nahi?

d U = T d S - P d V + μ d N

mein har intensive quantity apni paired extensive variable ke change ke saath multiply hoti hai; μ

d N

ke saath baithta hai, isliye yeh woh "force" hai jo particle number changes drive karta hai — dekho First Law of Thermodynamics.

Hum aam taur par μ

U

se nahi balki

G

se kyun likhte hain?

G

ke natural variables

T, P, N

hain — woh quantities jo hum lab mein actually control karte hain — jabki

U

S, V

fixed rakhna padta hai, jo experimentally awkward hai.

Euler's theorem

G = N μ

sirf ek pure substance ke liye kyun deta hai?

Euler's homogeneous-function theorem ko

G

ka sab extensive arguments mein linearly scale karna zaroori hai; ek species ke saath woh sirf

N

hai, jo

G = N μ

deta hai, lekin multiple species har ek

N_{i} μ_{i}

term contribute karti hain — dekho Euler Relation and Gibbs-Duhem.

Nature particles ko high μ se low μ ki taraf kyun dhakelta hai?

Kyunki woh motion total

G

kam karta hai (

d G = (μ_{low} - μ_{high}) d N < 0

), aur fixed

T, P

par spontaneous processes

G

decrease karte hain — yeh Second Law ka consequence hai. Yeh pehle figure ka "downhill in μ" picture hai.

Ideal-gas μ mein

ln P

term aata hi kyun hai?

Fixed

T

par,

d μ = (V / N) d P

; ideal-gas law

P V = N k_{B} T

rearrange hoke

V / N = k_{B} T / P

deta hai, isliye

d μ = (k_{B} T / P) d P

, aur

1/ P

integrate karne par logarithm milta hai — yeh zyada volume mein particles phailne ki entropy encode karta hai.

Grand canonical ensemble mein μ natural variable kyun hai?

Wahan system particles ek reservoir ke saath exchange karta hai, isliye hum reservoir ka μ fix karte hain

N

ki jagah; μ average particle number ka control knob ban jaata hai — dekho Grand Canonical Ensemble.

Fermi-Dirac aur Bose-Einstein occupation numbers mein μ kyun aata hai?

Occupation energy cost relative to μ par depend karta hai,

(ε - μ)

, kyunki μ woh reference level set karta hai jis par ek particle add karna "free" hai — dekho Fermi-Dirac and Bose-Einstein Statistics.

Edge cases

Ek aisa system jahan particles bilkul exchange nahi ho sakte (sealed, no reactions), uska μ kya hai?

μ abhi bhi

(\partial G / \partial N)_{T, P}

ke roop mein well-defined hai, lekin woh kabhi act nahi karta —

d N = 0

ke saath

μ d N

term vanish ho jaata hai aur μ ka koi dynamical role nahi hai.

Ideal gas ke liye

P \to 0

par μ ka kya hoga?

μ = μ^{\circ} + k_{B} T ln (P / P^{\circ}) \to - \infty

, kyunki infinite dilution mein ek particle add karna almost "free" ho jaata hai — entropy gain diverge ho jaata hai.

Fermi gas ke liye absolute zero par μ kya hai?

Yeh ek finite positive value ki taraf approach karta hai jise Fermi energy kehte hain;

T = 0

par bhi ek fermion add karne par energy lagti hai kyunki low levels pehle se full hain (Pauli exclusion) — dekho Fermi-Dirac and Bose-Einstein Statistics.

Agar do ideal gases equal

P

aur

T

par lekin alag species ki milti hain, toh kya unka chemical potential equal hai?

Zaroori nahi — har species ka apna

μ_{i}

hota hai jo uski partial pressure se set hota hai. Equal total

P, T

individual partial-molar potentials ke baare mein kuch nahi kahta.

Kya μ exactly zero ho sakta hai, aur kya iska matlab "koi particles nahi hain"?

Haan, yeh specific

P, T

par zero ho sakta hai (photon gases ka

μ \equiv 0

hota hai), aur iska matlab hai ek particle add karne par koi free energy cost nahi — na ki particles absent hain.

Gravity ya electric field jaise external potentials μ ko kaise change karte hain?

Woh ek per-particle potential energy add karte hain, jo total (electrochemical) potential

\overset{μ}{ˉ} = μ + m g h + q V_{elec}

deta hai; equilibrium tab

\overset{μ}{ˉ}

equalize karta hai, bare

μ

nahi. Isliye gas pressure altitude ke saath girta hai aur battery unequal internal

μ

ke bawajood charge alag rakh sakti hai.

First-order phase transition par, μ boundary ke across continuous hai ya discontinuous?

Continuous — coexisting phases same μ share karte hain; yeh μ ke derivatives hain (jaise entropy aur volume per particle) jo jump karte hain — yahi woh hai jo Clausius-Clapeyron quantify karta hai, aur yeh doosre figure mein coexistence line par kink hai.

Recall Har trap ka ek-line summary

μ ek intensive free energy per particle hai, jo is baat se define hota hai ki kaunse variables fixed hain, flow ko high se low μ ki taraf drive karta hai (ya high se low electrochemical $\overset{μ}{ˉ}$ ki taraf jab fields present hon) chahe raw particle count ya bare energy kuch bhi ho.