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

2.4.5 · D3 · Physics › Thermodynamics & Statistical Mechanics (Advanced) › Chemical potential μ = (∂G - ∂N)_{T,P}

The scenario matrix

Kuch bhi solve karne se pehle, chalte hain har case class enumerate karte hain jo yeh topic tumpe throw kar sakta hai. Har worked example neeche tagged hai us cell se jo woh fill karta hai.

Cell	Case class	Kya tricky hai	Example
A	$μ_{A} - μ_{B} > 0$ (positive sign)	matter kis direction mein flow karta hai?	Ex 1
B	$μ_{A} - μ_{B} < 0$ (negative sign)	sign flow ko flip kar deta hai	Ex 1
C	$μ_{A} = μ_{B}$ (zero difference)	degenerate: equilibrium, koi flow nahi	Ex 2
D	Pressure change hua, $T$ fixed	$ln$ dependence, $Δ μ$ ka sign	Ex 3
E	Limit $P \to 0$ (dilute)	$μ \to - \infty$ : iska matlab kya hai?	Ex 4
F	Limit $P \to \infty$ (dense)	$μ$ bina bound ke badhta hai	Ex 4
G	Do phases coexist kar rahe hain	equal- $μ$ boundary line pick karta hai	Ex 5
H	Real-world word problem	osmosis / altitude	Ex 6
I	Multi-species mixture	partial molar, held-fixed subscripts	Ex 7
J	Exam twist (sign trap)	entropy-driven "uphill" flow	Ex 8

Har cell A–J neeche cover ki gayi hai. Chalte hain.

Cells A & B — $μ_{A} - μ_{B}$ ka sign direction decide karta hai

Worked example Example 1 — Matter kis taraf flow karta hai (dono signs)?

Do chambers ek porous wall share karte hain common $T, P$ pe. Dono cases ke liye flow direction padho:

Case A: $μ_{A} = - 0.30 eV$ , $μ_{B} = - 0.55 eV$ .
Case B: $μ_{A} = - 0.70 eV$ , $μ_{B} = - 0.40 eV$ .

Forecast: Har case ke liye guess karo: kya matter A→B jaata hai ya B→A? Aage padhne se pehle likh lo.

Step 1 — $d N$ particles ko A se B move karne ke liye $G$ mein change likhte hain. Parent se yaad karo ki $μ = (\partial G / \partial N)_{T, P}$ by definition woh change hai $G$ mein per particle jo fixed $T, P$ pe add kiya jaaye. Toh ek chamber mein $d N$ add karne se uska $G$ $μ d N$ se badalta hai; $d N$ hatane se $- μ d N$ se badalta hai. Jab $d N$ particles A se nikalte hain aur B mein jaate hain, $N_{A}$ $d N$ se girta hai aur $N_{B}$ $d N$ se badhta hai: $d G = μ_{A} (- d N) + μ_{B} (+ d N) = (μ_{B} - μ_{A}) d N .$ Yeh step kyun? $G$ equilibrium pe minimize hota hai (parent note). Koi move spontaneously tabhi hota hai jab woh $d G < 0$ banaye. Toh hume sirf $(μ_{B} - μ_{A})$ ka sign chahiye.

Step 2 — Case A. $μ_{B} - μ_{A} = - 0.55 - (- 0.30) = - 0.25 eV$ . $d N > 0$ ke liye, $d G = (- 0.25) d N < 0$ . Spontaneous: particles A→B flow karte hain. Yeh step kyun? Negative $d G$ = free energy mein downhill. Kyunki $μ_{A} > μ_{B}$ , yeh "high $\to$ low" rule action mein hai.

Step 3 — Case B. $μ_{B} - μ_{A} = - 0.40 - (- 0.70) = + 0.30 eV$ . $d N > 0$ ke liye, $d G = (+ 0.30) d N > 0$ — yeh $G$ ko raise karega, toh A→B forbidden hai. Reverse, B→A, $d G < 0$ deta hai. Particles B→A flow karte hain. Yeh step kyun? Yahan $μ_{B} > μ_{A}$ , toh B ab "mountain" hai aur A "mud" hai. Flow rule wahin ka wahin hai; sirf labels swap ho gaye.

Verify: Dono cases mein matter us chamber se nikla jisme bada $μ$ tha. Case A: $- 0.30 > - 0.55$ , source hai A ✓. Case B: $- 0.40 > - 0.70$ , source hai B ✓. Units: eV energy hai, $d N$ dimensionless hai, toh $d G$ eV mein hai ✓.

Cell C — degenerate case: equal $μ$ , kuch nahi hilta

Cell D — fixed $T$ pe pressure change karo: $ln$ law

Worked example Example 3 — Ideal gas compress karo,

$μ$ ko badhte dekho $T = 300 K$ pe ek ideal gas ko $P^{\circ} = 1.00 atm$ se $P = 3.00 atm$ tak compress kiya jaata hai. $μ$ kitna change hota hai? $k_{B} = 8.617 \times 1 0^{- 5} eV/K$ use karo.

Forecast: Kya $μ$ upar jaata hai ya neeche jab tum gas squeeze karte ho?

Step 1 — Ideal-gas $μ$ yaad karo. Parent se, Example 2: $μ (T, P) = μ^{\circ} (T) + k_{B} T ln \frac{P}{P ^{\circ}} .$ Yeh step kyun? Yeh woh ek formula hai jo sari pressure dependence carry karta hai. Reference term $μ^{\circ} (T)$ cancel ho jaata hai jab hum difference lete hain.

Step 2 — $Δ μ$ paane ke liye subtract karo. $Δ μ = μ (T, P) - μ (T, P^{\circ}) = k_{B} T ln \frac{P}{P ^{\circ}} .$ Yeh step kyun? Hume sirf change ki parwah hai, toh (unknown) reference cleanly nikal jaata hai.

Step 3 — Numbers plug karo. $Δ μ = (8.617 \times 1 0^{- 5}) (300) ln (3.00) = 0.025851 \times 1.098612 \approx 0.0284 eV .$ Yeh step kyun? $ln 3 > 0$ , toh $Δ μ > 0$ : compress karne se $μ$ badhta hai.

Verify: $ln (3) \approx 1.0986$ , product $\approx 0.0284 eV$ ✓. Physical sense: ek squeezed gas "zyada eager hai escape karne ke liye" lower pressure ki taraf — higher $μ$ — exactly yahi wajah hai ki gas ek punctured tire se bahar bhagta hai. Units: $(eV/K) (K) = eV$ ✓.

Cells E & F — do limits, $P \to 0$ aur $P \to \infty$

Example 3 ka $ln$ do dramatic limits chupaata hai. Inhe seedha face karte hain.

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

Worked example Example 4 — Extreme dilution aur extreme density pe kya hota hai?

$μ (T, P) = μ^{\circ} + k_{B} T ln (P / P^{\circ})$ use karte hue, $P \to 0$ (Cell E) aur $P \to \infty$ (Cell F) mein $μ$ describe karo.

Forecast: Kya $μ$ kisi bhi limit mein finite rehta hai? Har ek ke liye ek sign guess karo.

Step 1 — Dilute limit, $P \to 0^{+}$ (Cell E). $ln \frac{P}{P ^{\circ}} \to ln (0^{+}) = - \infty \Rightarrow μ \to - \infty.$ Yeh step kyun? Figure mein pale-yellow curve ki taraf dekho left side par ( $P$ chota, vertical axis ke paas): woh horizontal dashed line se neeche ghus jaata hai aur bottom se nikal jaata hai. Infinitely dilute gas ka $μ$ infinitely negative hota hai — woh infinitely "bhukha" hota hai zyada particles receive karne ke liye. Isi liye ink ki ek boondi puri bathtub mein ghus jaati hai: khaali paani $μ \to - \infty$ pe hai.

Step 2 — Dense limit, $P \to \infty$ (Cell F). $ln \frac{P}{P ^{\circ}} \to + \infty \Rightarrow μ \to + \infty.$ Yeh step kyun? Figure ke right side pe (bada $P$ ) yellow curve bina bound ke chadhta hai. Hugely compressed gas ka $μ$ bahut bada hota hai — particles expel karne ke liye desperate. (Ek real gas ke liye, interactions eventually ideal formula tod deti hain, lekin qualitative " $μ$ climbs" bachi rehti hai.)

Step 3 — Crossing point. $P = P^{\circ}$ pe (yaani horizontal axis pe $P = 1$ ), $ln (1) = 0$ , toh $μ = μ^{\circ}$ . Woh chalk-blue dot hai jahan do dashed lines cross karti hain. Yeh step kyun? Har log curve apne reference se guzarta hai; ise name karne se poori curve pin ho jaati hai.

Verify: $ln (P / P^{\circ})$ monotonic increasing hai, dono taraf unbounded, $P = P^{\circ}$ pe zero ✓. Toh $μ$ ka koi maximum nahi aur koi minimum nahi pressure mein — yeh sare real values sweep karta hai. Yeh confirm karta hai ki Cells E aur F genuine limiting behaviours hain, artifacts nahi.

Cell G — do phases coexist: equal $μ$ line draw karta hai

Worked example Example 5 — Liquid–vapour coexistence

Ek pure liquid (L) aur uska vapour (V) ek container share karte hain. Kaunsi single condition unka coexistence fix karti hai, aur yeh Phase Equilibrium and Clausius-Clapeyron se kaise relate karti hai?

Forecast: Coexistence pe, kaun bada hota hai — $μ_{L}$ ya $μ_{V}$ ?

Step 1 — Equilibrium condition likhte hain. Particles interface cross kar sakte hain (evaporate/condense), toh yeh ek diffusive equilibrium hai: $μ_{L} (T, P) = μ_{V} (T, P) .$ Yeh step kyun? Wahi rule jaisa Example 2 mein, do boxes ki jagah do phases pe apply kiya gaya. Na koi bada hai — woh equal hain (figure mein yellow dot, jahan pink aur blue lines milti hain).

Recall

$d μ = - s d T + v d P$ kahan se aata hai? Parent ke master differential $d G = - S d T + V d P + μ d N$ ko $N$ se divide karo ek pure substance ke liye, $μ = G / N$ use karte hue. Per-particle entropy $s = S / N$ hai aur per-particle volume $v = V / N$ hai; ek short manipulation (Gibbs–Duhem relation, Euler Relation and Gibbs-Duhem mein derive ki gayi) $d μ = - s d T + v d P$ deta hai. Ise aise padho: " $μ$ girta hai jab $T$ badhta hai (rate $s$ ) aur badhta hai jab $P$ badhta hai (rate $v$ )."

Step 2 — $T$ slide karo aur maango ki equality phir bhi hold kare. Coexistence line ke saath $μ_{L} = μ_{V}$ differentiate karo abhi yaad ki gayi per-particle relation use karte hue, jahan $s, v$ per-particle entropy aur volume hain: $- s_{L} d T + v_{L} d P = - s_{V} d T + v_{V} d P .$ Yeh step kyun? Coexistence ek point nahi hai — yeh ek curve $P (T)$ hai. $μ_{L} = μ_{V}$ ko persist karne ki demand jab hum move karte hain woh curve ka slope deta hai.

Step 3 — Slope ke liye solve karo. $\frac{d P}{d T} = \frac{s _{V} - s _{L}}{v _{V} - v _{L}} = \frac{Δ s}{Δ v} = \frac{L}{T Δ v},$ jahan $L = T Δ s$ latent heat hai per particle. Yeh step kyun? Yeh Clausius–Clapeyron equation hai — yeh seedha " $μ$ boundary ke saath equal" se nikalti hai. Chemical potential ne saara kaam kiya.

Verify: Figure mein chalk-blue vapour line aur chalk-pink liquid line ke saath, do $μ$ curves ek single temperature pe cross karti hain (fixed $P$ pe); $P$ slide karne se woh crossing move hoti hai, coexistence curve trace karti hai jiska slope $Δ s /Δ v$ hai ✓. Dimensional check: $[L] =$ energy/particle, $[T Δ v] =$ K·(volume/particle), aur $d P / d T$ ke units pressure/temperature hain — consistent hai jab $Δ s$ entropy per particle ho ✓.

Cell H — real-world word problem

Definition Is problem ke liye do building blocks

Pehle hume chahiye ki ek component ka $μ$ pressure aur concentration par kaise depend karta hai.

Pressure dependence. Per-particle Gibbs–Duhem relation $d μ = - s d T + v d P$ se (Example 5 mein yaad kiya gaya), fixed $T$ pe hume milta hai $(\partial μ / \partial P)_{T} = v$ , jahan $v$ partial molar volume per particle hai — woh volume jo ek water molecule liquid mein occupies karta hai (units: volume/particle, e.g. $m^{3}$ per molecule). Fixed $T$ pe integrate karo: $μ_{w} (P) = μ_{w} (P_{0}) + v_{w} (P - P_{0})$ . Toh ek pressure excess $Π$ water ke $μ$ mein ek term $v_{w} Π$ add karta hai.
Concentration dependence. Sugar dissolve karna water ko mole fraction $x_{w} < 1$ tak dilute karta hai, jo ek entropic term $k_{B} T ln x_{w}$ contribute karta hai. Saath mein rakhte hue, sugar side pe excess pressure $Π$ ke saath: $μ_{w} = μ_{w}^{\circ} + k_{B} T ln x_{w} + v_{w} Π.$ Yeh ek equation poora example drive karti hai — $v_{w} Π$ term exactly wahan hai jahan pressure knob enter karta hai.

Worked example Example 6 — Osmosis: sugar water vs pure water

Ek membrane water ko through jaane deti hai lekin sugar ko nahi. Left side: pure water. Right side: water mein dissolved sugar. Water kis taraf flow karta hai, aur kaunsa pressure $Π$ ise rokta hai?

Forecast: Water sugar mein jaata hai, ya usse bahar?

Step 1 — Dono sides pe $μ_{w}$ compare karo equal $T, P$ pe (abhi tak koi excess pressure nahi, $Π = 0$ ).

Pure side: $x_{w} = 1 \Rightarrow μ_{w} = μ_{w}^{\circ} + k_{B} T ln 1 = μ_{w}^{\circ}$ .
Sugar side: $x_{w} < 1 \Rightarrow ln x_{w} < 0 \Rightarrow μ_{w} < μ_{w}^{\circ}$ .

Yeh step kyun? 1 se chhote number ka $ln$ negative hota hai, toh sugar dissolve karna water ka $μ$ lower karta hai. Sugar side "mud" hai.

Step 2 — Flow rule apply karo. Water high $μ$ (pure side) se low $μ$ (sugar side) ki taraf flow karta hai. Water sugar solution mein cross karta hai — yeh osmosis hai. Yeh step kyun? Wahi high→low rule (Cell A); drive purely entropic $ln x_{w}$ term hai, koi energy nahi.

Step 3 — Ise pressure se roko (osmotic pressure). Sugar side ka pressure $Π$ se badhao. Boxed formula $μ_{w} = μ_{w}^{\circ} + k_{B} T ln x_{w} + v_{w} Π$ use karte hue, equilibrium demand karta hai ki water ka $μ$ pure side ke $μ_{w}^{\circ}$ se match kare: $μ_{w}^{\circ} + k_{B} T ln x_{w} + v_{w} Π = μ_{w}^{\circ} \Rightarrow v_{w} Π = - k_{B} T ln x_{w} \approx k_{B} T x_{s},$ chhote sugar fraction $x_{s} = 1 - x_{w}$ ke liye. $n_{s}$ = sugar particles per volume ke saath, yeh van 't Hoff's law $Π = n_{s} k_{B} T$ mein rearrange hota hai. Yeh step kyun? Lowered $μ_{w}$ ko wapas raise karne ke liye sirf ek free knob hai pressure, aur $(\partial μ / \partial P)_{T} = v_{w}$ batata hai exactly kitna har unit pressure ise lift karta hai. Hum $v_{w} Π = - k_{B} T ln x_{w}$ solve karte hain.

Numeric check. $x_{s} = 0.010$ (1% sugar) ke liye $T = 300 K$ pe: $- ln (0.990) = 0.010050$ . Osmotic drive hai $k_{B} T (0.010050) = (8.617 \times 1 0^{- 5}) (300) (0.010050) \approx 2.60 \times 1 0^{- 4} eV$ per water molecule.

Verify: $- ln (0.99) \approx 0.01005$ , aur linear approximation $x_{s} = 0.010$ 0.5% tak agree karti hai ✓; osmotic drive $\approx 2.60 \times 1 0^{- 4} eV$ ✓. Sanity: zyada sugar add karna (chota $x_{w}$ ) $μ_{w}$ ko zyada negative banata hai, zyada water pull karta hai — rozmarra ke experience se match karta hai ki ek kishmish paani mein phool jaati hai ✓.

Cell I — mixtures: partial molar aur held-fixed subscript

Worked example Example 7 — Do species, subscripts sahi rakho

Ek container mein species 1 ke $N_{1}$ molecules aur species 2 ke $N_{2}$ molecules hain fixed $T, P$ pe. $G$ ka correct differential likho aur har $μ_{i}$ identify karo. Phir, agar sirf species 1 add ki jaaye, toh $d G$ kya hai?

Forecast: Kya $μ_{1}$ abhi bhi $\partial G / \partial N_{1}$ hai, ya species 2 rule change karta hai?

Step 1 — Master differential ko do species ke liye generalize karo. $d G = - S d T + V d P + μ_{1} d N_{1} + μ_{2} d N_{2} .$ Yeh step kyun? Har species apna khud ka particle-number knob carry karta hai. Har additional species ek $μ_{i} d N_{i}$ term add karta hai.

Step 2 — Har partial molar chemical potential define karo. $μ_{i} = (\frac{\partial G}{\partial N _{i}})_{T, P, N_{j \neq = i}} .$ Yeh step kyun? Subscript ab keh raha hai: $T$ , $P$ , aur har doosri species ki count fixed rakhो. Agar tum " $N_{j \neq = i}$ " bhool gaye toh tum ek alag quantity compute kar rahe ho — yeh parent ka "subscript is essential" mistake mixtures tak le jaaya gaya hai.

Step 3 — Sirf species 1 add karo fixed $T, P$ pe $d N_{2} = 0$ ke saath. $d G = μ_{1} d N_{1} .$ Yeh step kyun? Sare doosre differentials vanish ho jaate hain, exactly "cost per particle of species 1" chhodke.

Verify: Note karo ki $μ = G / N$ parent se sirf ek pure substance ke liye hai; yahan $G = N_{1} μ_{1} + N_{2} μ_{2}$ (Euler, Euler Relation and Gibbs-Duhem) instead. Consistency check: agar species 2 absent hai ( $N_{2} = 0$ ), yeh $G = N_{1} μ_{1}$ pe collapse karta hai, yaani $μ_{1} = G / N_{1}$ ✓.

Cell J — exam twist: entropy se "uphill" flow

Worked example Example 8 — Kya particles higher

energy ki taraf flow kar sakte hain? Do ideal-gas boxes same $T$ pe. Box A dense hai ( $P_{A} = 4.0 atm$ ), box B dilute hai ( $P_{B} = 1.0 atm$ ). Ek student claim karta hai: "Box A compressed hai, toh wahan har particle ki zyada energy hai; particles ko A→B flow karna chahiye energy lower karne ke liye — flow energy follow karta hai." Claim diagnose karo aur real answer do. $T = 300 K$ lo.

Forecast: Kaun se box mein higher $μ$ hai? Kya energy ya $μ$ flow decide karta hai?

Step 1 — Ek ideal gas ke liye, per-particle energy $\frac{3}{2} k_{B} T$ hai — yeh sirf $T$ pe depend karta hai. Dono boxes same $T$ pe hain, toh unka energy per particle same hai. Student ka "compressed hone pe zyada energy" ideal gas ke liye galat hai. Yeh step kyun? Yeh trap expose karta hai: compression $μ$ change karta hai entropic $ln P$ term ke through, kinetic energy ke through nahi.

Step 2 — Ideal-gas law use karke $μ$ compare karo (dono $μ^{\circ}$ share karte hain). $μ_{A} - μ_{B} = k_{B} T ln \frac{P _{A}}{P _{B}} = k_{B} T ln \frac{4.0}{1.0} = k_{B} T ln 4.$ Numerically: $(8.617 \times 1 0^{- 5}) (300) ln 4 = 0.025851 \times 1.386294 \approx 0.0358 eV > 0$ . Yeh step kyun? $ln 4 > 0$ toh $μ_{A} > μ_{B}$ . Dense box "mountain" hai.

Step 3 — Flow rule apply karo. High $μ \to$ low $μ$ : particles A→B flow karte hain (dense se dilute tak). Yeh step kyun? Coincidence se student ne sahi direction guess ki — lekin galat wajah se. Yeh lower energy nahi hai (energies equal hain); yeh higher entropy hai jab gas spread out karta hai. Drive woh $- T \partial S / \partial N$ piece of $μ$ hai (parent's first steel-man mistake).

Step 4 — Dikhao ki yeh sach mein entropy-driven hai. Kyunki $Δ (energy) = 0$ lekin gas phir bhi flow karta hai, saara $Δ G$ $- T Δ S$ term se aaya: dilute box mein spread out hone se entropy per particle badhti hai, toh process $G$ lower karta hai even though koi kinetic energy release nahi hui. Lasting lesson: particles even higher potential energy ki taraf flow kar sakte hain agar entropy gain zyada ho — free energy, bare energy nahi, referee hai. Yeh step kyun? Yeh exam ka asli target hai: free energy ko bare energy se distinguish karna. Sahi diagnosis yeh hai ki student sahi direction (A→B) pe pahuncha ek galat argument (energy) se, aur sachcha driver woh entropic $k_{B} T ln (P_{A} / P_{B}) > 0$ hai jo $μ_{A} > μ_{B}$ banata hai.

Verify: $ln 4 \approx 1.3863$ , $μ_{A} - μ_{B} \approx 0.0358 eV$ , positive ✓. Direction A→B "high $μ \to$ low $μ$ " se match karta hai ✓. Aur energies equal hain ( $\frac{3}{2} k_{B} T$ dono mein) confirm karta hai ki flow purely entropic hai ✓.

Recall Har cell ke across self-test

Answers cover karo. Har ek ke liye, matrix cell aur result bolo.

$μ_{A} = - 0.2$ , $μ_{B} = - 0.9$ eV: kis taraf? ::: Cell A — A→B (A higher hai).
Do connected boxes $μ_{A} = μ_{B}$ ke saath: kya flows? ::: Cell C — kuch nahi; equilibrium.
$P \to 0$ pe, $μ \to$ ? ::: Cell E — $- \infty$ .
Liquid aur uske vapour ke liye coexistence condition? ::: Cell G — $μ_{L} = μ_{V}$ .
Osmosis: kya water sugar side mein jaata hai ya nikalta hai? ::: Cell H — jaata hai (sugar side mein $μ_{w}$ lower hai).
Ek mixture mein, $μ_{1}$ ke liye correct held-fixed set? ::: Cell I — $T, P, N_{j \neq = 1}$ .
Do equal- $T$ ideal gases, dense vs dilute — flow direction aur kyun? ::: Cell J — dense→dilute, entropy-driven.

Parent pe wapas jaao: Chemical potential note. Related depth: Grand Canonical Ensemble (jahan $μ$ control knob ban jaata hai), Fermi-Dirac and Bose-Einstein Statistics (quantum $μ$ ), Entropy and the Second Law (kyun $G$ minimize hota hai bilkul).