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

Master differential (parent mein Step 4 mein derive kiya gaya) yeh hai: $dG=-SdT+VdP+\mu dN.$ Yeh equation yeh kehti hai: agar tum $T$, $P$, aur $N$ ko thoda nudge karo, toh $G$ mein change teen independent contributions ka sum hai. Har partial derivative sirf ek nudge isolate karta hai baaki ko freeze karte hue.

• $T$ aur $N$ freeze karo ($dT=0,dN=0$): $dG=VdP$, toh ${\left(\frac{\partial G}{\partial P}\right)}_{T,N}=V$.
• $T$ aur $P$ freeze karo ($dT=0,dP=0$): $dG=\mu dN$, toh ${\left(\frac{\partial G}{\partial N}\right)}_{T,P}=\mu$. ✓

Nature particles ko high $\mu$ se low $\mu$ ki taraf move karta hai (jaise heat hot se cold ki taraf). Yahaan ${\mu }_A=-0.10>{\mu }_B=-0.45$, toh A "high" side hai. $\text{Particles flow }A\to B.$ Energy se check karo: A se B mein $dN>0$ move karne par $G$ mein change: $dG=({\mu }_B-{\mu }_A)dN=(-0.45-(-0.10))dN=-0.35dN<0$. Kyunki spontaneous processes $G$ ko lower karti hain, $A\to B$ indeed spontaneous hai. ✓

Pure substance ke liye, $\mu =G/N$ (parent ka Euler shortcut — valid sirf isliye kyunki $G$ extensive hai, yani $N$ mein degree 1 mein homogeneous hai). $\mu =\frac{-2.40\times 10^5\text{J}}{3.0\text{mol}}=-8.0\times 10^4\text{J/mol}.$ ✓

Jab $T$ fixed hota hai toh sirf $\ln$ term badlta hai: $\Delta \mu =k_{B}T\ln \frac{P}{P^{\circ }}.$ $\ln$ kyun aur linear kyun nahi? Kyunki $d\mu =(k_{B}T/P)dP$ (from $dG=VdP$ with $V=N{k}_{B}T/P$), aur $1/P$ integrate karne par logarithm aata hai. Numerically: $\Delta \mu =(1.381\times 10^{-23})(300)\ln 4=(4.143\times 10^{-21})(1.3863)=5.744\times 10^{-21}\text{J}.$ eV mein convert karo: $\Delta \mu =\frac{5.744\times 10^{-21}}{1.602\times 10^{-19}}=0.0359\text{eV}\approx +0.036\text{eV}.$ Sign check: compress karna ($P$ raise karna) $\mu$ raise karta hai — gas "escape karne ko zyada eager" ho jaata hai. ✓

$\Delta \mu =k_{B}T\ln \frac{0.25}{1.00}=(4.143\times 10^{-21})\ln (0.25).$ $\ln (0.25)=-1.3863$, toh $\Delta \mu =-5.744\times 10^{-21}\text{J}=-0.036\text{eV}.$ Dilution $\mu$ lower karta hai — exactly L2.1 ka mirror image (same magnitude, kyunki $\ln (1/4)=-\ln 4$). ✓

Equilibrium ke liye equal chemical potentials chahiye: ${\mu }_A={\mu }_B$. Kyunki dono same species hain same $T$ pe, aur $\mu ={\mu }^{\circ }+k_{B}T\ln (P/P^{\circ })$ $P$ mein monotonically increasing hai, equal $\mu$ ⇔ equal $P$: $\boxed{P_A^{\text{final}}=P_B^{\text{final}}=P_f.}$ Initially ${\mu }_A>{\mu }_B$ kyunki $P_A>P_B$, toh particles A → B flow karte hain (high $\mu$ → low $\mu$) jab tak pressures equalize nahi ho jaate. ✓

A se B mein $dN$ move karne par: $G$ A se ${\mu }_{A}dN$ lose karta hai aur B mein ${\mu }_{B}dN$ gain karta hai, net $dG=({\mu }_B-{\mu }_A)dN=(-0.50-(-0.30))(1)=-0.20\text{eV}.$ $dG<0$ ⇒ spontaneous ⇒ particles A → B flow karte hain, aur $G$ har transfer particle pe $0.20\text{eV}$ drop karta hai, jab tak ${\mu }_A={\mu }_B$ nahi ho jaata. ✓

Ideal-gas molecules ki koi interaction potential energy nahi hoti, toh A aur B mein "raw energy per particle" identical hai. Driver entirely $\mu$ ka entropic part hai: $\mu ={\mu }^{\circ }(T)+k_{B}T\ln \frac{P}{P^{\circ }}.$ ${\mu }_B$ ka $P$ zyada bada hai, isliye $\ln P$ zyada bada hai, isliye higher $\mu$. Toh particles B → A flow karte hain (high $P$ se low $P$ ki taraf), yani crowded box se sparse box mein. Yeh "energy mein uphill lekin $\mu$ mein downhill" kyun hai: low-pressure box mein spreading entropy $S$ raise karta hai; kyunki $\mu$ mein $-T(\partial S/\partial N)$ hai, entropy term sparse side ko sasta banata hai. Potential energy ke baare mein kuch nahi — pure entropy. ✓

Galat. Master differential $dG=-SdT+VdP+\mu dN$ dikhata hai ki $\mu$ ko $(\partial G/\partial N)$ ke roop mein isolate karne ke liye, tumhe $dT$ aur $dP$ terms ko khatam karna hoga — yani $T$ aur $P$ ko hold karna hoga: $\mu ={\left(\frac{\partial G}{\partial N}\right)}_{T,P}.$ $S,V$ hold karna $-SdT+VdP$ ko zero nahi karta (kyunki $T,P$ tab vary honge jab $N$ change hoga), toh $(\partial G/\partial N{)}_{S,V}$ ek alag, kam useful quantity hai. Har thermodynamic potential ke apne natural variables hote hain; $\mu$ cleanly sirf tabhi appear karta hai jab tum unke saath differentiate karte ho. ✓

Step 1 — Per-particle change isolate karo. Pure substance ke liye $\mu =G/N$, toh fixed $N$ par $d\mu =dG/N$. Master differential se constant $T$ par ($dT=0$, $dN=0$): $dG=VdP$. Isliye $d\mu =\frac{V}{N}dP.$ Kyun: hum jaanna chahte hain ki $\mu$ (energy per particle) fixed $T$ par pressure ke response mein kaise change hoti hai; $dG=VdP$ ek hi surviving term hai.

Step 2 — Ideal-gas volume daalo. $PV=N{k}_{B}T$ se, $\frac{V}{N}=\frac{k_{B}T}{P}$. Toh $d\mu =\frac{k_{B}T}{P}dP.$ $1/P$ kyun? Yahi integration par logarithm force karta hai — response $P$ badhne par weak hota jaata hai.

Step 3 — Reference $P^{\circ }$ se $P$ tak integrate karo (fixed $T$ par, toh $k_{B}T$ ek constant hai jo bahar aa jaata hai): ${\int }_{{\mu }^{\circ }}^{\mu }d\mu =k_{B}T{\int }_{P^{\circ }}^P\frac{dP}{P}=k_{B}T[\ln P{]}_{P^{\circ }}^P.$ $\mu (T,P)-{\mu }^{\circ }(T)=k_{B}T\ln \frac{P}{P^{\circ }}.$

Step 4 — Result padho: $\boxed{\mu (T,P)={\mu }^{\circ }(T)+k_{B}T\ln \frac{P}{P^{\circ }}.}$ Yahaan ${\mu }^{\circ }(T)$ reference pressure pe value hai — integration constant, jisme saari temperature dependence hai. ✓

Coexistence = phase equilibrium = phases ke beech diffusive equilibrium, toh ${\mu }_{\text{liquid}}(T,P)={\mu }_{\text{vapour}}(T,P).$ Agar woh unequal hote, toh molecules higher-$\mu$ phase se lower wale mein stream karte jab tak match nahi ho jaate — woh streaming hi boiling ya condensing hai. Coexistence par streaming ruk jaati hai. Bahut alag densities ke bawajood equal kyun: $\mu$ ek free energy hai, crowding count nahi. Dense liquid per particle low energy cost deta hai (strong attractive bonds $U$ lower karte hain) lekin low entropy; sparse vapour mein high entropy hai ($k_{B}T\ln$ term) lekin weak bonds. Boiling point par yeh dono effects exactly balance karte hain, equal $\mu$ dete hue. Yahi machinery hai Phase Equilibrium and Clausius-Clapeyron ke peeche. ✓

Extensive Euler relation (dekho Euler Relation and Gibbs-Duhem) $G=N\mu$ ko mixtures ke liye generalize karta hai: $G=N_1{\mu }_1+N_2{\mu }_2=(2.0)(-5.0)+(3.0)(-8.0)=-10.0-24.0=-34.0\text{kJ}.$ Note karo tum yahaan ek single $\mu$ ke saath $G=N\mu$ nahi likh sakte — har species apna partial molar term contribute karta hai. ✓

Step 1 — Euler relation differentiate karo (har $N_i{\mu }_i$ par product rule): $dG={\sum }_i{\mu }_{i}d{N}_i+{\sum }_i{N}_{i}d{\mu }_i.$ Kyun: Euler extensive variables mein ek identity hai, toh uska differential bhi hold karna chahiye.

Step 2 — Ise master differential ke equal set karo: ${\sum }_i{\mu }_{i}d{N}_i+{\sum }_i{N}_{i}d{\mu }_i=-SdT+VdP+{\sum }_i{\mu }_{i}d{N}_i.$

Step 3 — ${\sum }_i{\mu }_{i}d{N}_i$ terms dono sides cancel ho jaate hain, aur reh jaata hai ${\sum }_i{N}_{i}d{\mu }_i=-SdT+VdP,$ yani $\boxed{SdT-VdP+\sum _i{N}_{i}d{\mu }_i=0.}\text{(Gibbs–Duhem)}$ Kyun matter karta hai: chemical potentials independent nahi hain — woh $T,P$ aur ek dusre se tied hain.

Step 4 — Fixed $T,P$ par single component. Ek species ke saath sum mein ek hi term hai, ${\sum }_i{N}_{i}d{\mu }_i=Nd\mu$, aur $dT=0,dP=0$ set karne se Gibbs–Duhem ke pehle do terms kill ho jaate hain: $S(0)-V(0)+Nd\mu =0⟹Nd\mu =0.$ Kyunki $N\ne 0$, yeh force karta hai $\boxed{d\mu =0.}$ Physical interpretation. Single pure substance ke liye tum fixed $T$ aur $P$ par $\mu$ change nahi kar sakte — $\mu$ puri tarah se $T$ aur $P$ se determined hai, toh $\mu =\mu (T,P)$ genuinely sirf unhi do variables ka function hai (substance zyada add karne se $N$ change hota hai lekin $\mu$ nahi, consistent with $\mu =G/N$ intensive hone ke saath). Exactly yahi reason hai ki phase coexistence, jahaan ${\mu }_{\text{liquid}}(T,P)={\mu }_{\text{vapour}}(T,P)$, do unknowns $T,P$ mein ek equation hai aur isliye $(T,P)$ plane mein ek line trace karta hai — Phase Equilibrium and Clausius-Clapeyron ka coexistence curve. Gibbs–Duhem wahi degree of freedom pin down karta hai. ✓

(a) $\epsilon =\mu$: exponent $0$ hai, toh $e^0=1$ aur $\bar{n}=\frac{1}{1+1}=\frac{1}{2}.$ Chemical potential par exactly wali state half-filled hai — yeh Fermi level define karta hai.

(b) $T\to 0$ limit. Maano $x=(\epsilon -\mu )/k_{B}T$.

• Agar $\epsilon <\mu$: $x\to -\infty$, toh $e^x\to 0$ aur $\bar{n}\to \frac{1}{0+1}=1$. Fully occupied.
• Agar $\epsilon >\mu$: $x\to +\infty$, toh $e^x\to \infty$ aur $\bar{n}\to \frac{1}{\infty }=0$. Empty.

Toh $T=0$ par occupancy ek sharp step hai: $\mu$ se neeche har state full hai, upar har state empty hai. $\mu$ woh energy hai jis par filling switch hoti hai — filled sea ka top. Yeh exactly "ek aur particle add karne ki energy cost" hai, ab last-added electron ki energy ke roop mein visible. ✓

Jab $P<P^{\circ }$, term $k_{B}T\ln (P/P^{\circ })<0$ (1 se neeche ke number ka log negative hota hai). Agar yeh ${\mu }^{\circ }(T)$ ko outweigh kare, toh $\mu <0$. Physically ek bahut dilute gas mein per particle enormous entropy hoti hai, aur $\mu$ mein $-T(\partial S/\partial N)$ hai — ek bada positive entropy contribution $\mu$ ko negative banata hai, matlab system particles accept karne ko eager hai (ek add karna entropy ki wajah se $G$ lower karta hai). Toh "energy adds" sirf internal-energy part hai; free-energy $\mu$ zero se neeche ja sakta hai. Yeh wahi lesson hai jaise L3 trap: free energy, not raw energy, sign set karta hai. ✓

• Particles do reservoirs ke beech kis taraf flow karte hain, aur iska $dG$ se kya lena dena hai?
• Ideal gas compress karna $\mu$ kyun raise karta hai, aur dependence logarithmic kyun hai?
• Gibbs–Duhem state karo aur derive karo; yeh fixed $T,P$ par single pure substance ke liye kya force karta hai?
• Fermi level $\epsilon =\mu$ par, fermionic occupancy kya hai, aur kyun?