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

The master differential (derived in the parent, Step 4) is $dG=-SdT+VdP+\mu dN.$ This equation says: if you nudge $T$, $P$, and $N$, the change in $G$ is the sum of three independent contributions. Each partial derivative just isolates one nudge while freezing the others.

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

Nature moves particles from high $\mu$ to low $\mu$ (like heat from hot to cold). Here ${\mu }_A=-0.10>{\mu }_B=-0.45$, so A is the "high" side. $\text{Particles flow }A\to B.$ Check by energy: moving $dN>0$ from A to B changes $G$ by $dG=({\mu }_B-{\mu }_A)dN=(-0.45-(-0.10))dN=-0.35dN<0$. Since spontaneous processes lower $G$, $A\to B$ is indeed spontaneous. ✓

For a pure substance, $\mu =G/N$ (parent's Euler shortcut — valid only because $G$ is extensive, i.e. homogeneous of degree 1 in $N$). $\mu =\frac{-2.40\times 10^5\text{J}}{3.0\text{mol}}=-8.0\times 10^4\text{J/mol}.$ ✓

Only the $\ln$ term changes when $T$ is fixed: $\Delta \mu =k_{B}T\ln \frac{P}{P^{\circ }}.$ Why $\ln$ and not linear? Because $d\mu =(k_{B}T/P)dP$ (from $dG=VdP$ with $V=N{k}_{B}T/P$), and integrating $1/P$ gives a logarithm. Numerically: $\Delta \mu =(1.381\times 10^{-23})(300)\ln 4=(4.143\times 10^{-21})(1.3863)=5.744\times 10^{-21}\text{J}.$ Convert to eV: $\Delta \mu =\frac{5.744\times 10^{-21}}{1.602\times 10^{-19}}=0.0359\text{eV}\approx +0.036\text{eV}.$ Sign check: compressing (raising $P$) raises $\mu$ — the gas becomes "more eager to escape." ✓

$\Delta \mu =k_{B}T\ln \frac{0.25}{1.00}=(4.143\times 10^{-21})\ln (0.25).$ $\ln (0.25)=-1.3863$, so $\Delta \mu =-5.744\times 10^{-21}\text{J}=-0.036\text{eV}.$ Dilution lowers $\mu$ — exactly the mirror image of L2.1 (same magnitude, since $\ln (1/4)=-\ln 4$). ✓

Equilibrium demands equal chemical potentials: ${\mu }_A={\mu }_B$. Since both are the same species at the same $T$, and $\mu ={\mu }^{\circ }+k_{B}T\ln (P/P^{\circ })$ is monotonically increasing in $P$, equal $\mu$ ⇔ equal $P$: $\boxed{P_A^{\text{final}}=P_B^{\text{final}}=P_f.}$ Initially ${\mu }_A>{\mu }_B$ because $P_A>P_B$, so particles flow A → B (high $\mu$ → low $\mu$) until the pressures equalise. ✓

Moving $dN$ from A to B: $G$ loses ${\mu }_{A}dN$ from A and gains ${\mu }_{B}dN$ in B, net $dG=({\mu }_B-{\mu }_A)dN=(-0.50-(-0.30))(1)=-0.20\text{eV}.$ $dG<0$ ⇒ spontaneous ⇒ particles flow A → B, and $G$ drops by $0.20\text{eV}$ per particle transferred, until ${\mu }_A={\mu }_B$. ✓

Ideal-gas molecules have no interaction potential energy, so the "raw energy per particle" is identical in A and B. The driver is entirely the entropic part of $\mu$: $\mu ={\mu }^{\circ }(T)+k_{B}T\ln \frac{P}{P^{\circ }}.$ ${\mu }_B$ has the larger $P$, hence larger $\ln P$, hence higher $\mu$. So particles flow B → A (from high $P$ to low $P$), i.e. from the crowded box to the sparse box. Why this is "uphill in energy but downhill in μ": spreading into the low-pressure box raises entropy $S$; since $\mu$ contains $-T(\partial S/\partial N)$, the entropy term makes the sparse side cheaper. Nothing about potential energy — pure entropy. ✓

Incorrect. The master differential $dG=-SdT+VdP+\mu dN$ shows that to isolate $\mu$ as $(\partial G/\partial N)$, you must kill the $dT$ and $dP$ terms — that is, hold $T$ and $P$ fixed: $\mu ={\left(\frac{\partial G}{\partial N}\right)}_{T,P}.$ Holding $S,V$ instead does not zero out $-SdT+VdP$ (since $T,P$ would then vary as $N$ changes), so $(\partial G/\partial N{)}_{S,V}$ is a different, less useful quantity. Each thermodynamic potential has its own natural variables; $\mu$ appears cleanly only when you differentiate along them. ✓

Step 1 — Isolate the per-particle change. For a pure substance $\mu =G/N$, so $d\mu =dG/N$ at fixed $N$. From the master differential at constant $T$ ($dT=0$, $dN=0$): $dG=VdP$. Hence $d\mu =\frac{V}{N}dP.$ Why: we want how $\mu$ (energy per particle) responds to pressure at fixed $T$; $dG=VdP$ is the only surviving term.

Step 2 — Insert the ideal-gas volume. From $PV=N{k}_{B}T$, $\frac{V}{N}=\frac{k_{B}T}{P}$. So $d\mu =\frac{k_{B}T}{P}dP.$ Why $1/P$? This is what forces a logarithm on integration — the response weakens as $P$ grows.

Step 3 — Integrate from a reference $P^{\circ }$ to $P$ (at fixed $T$, so $k_{B}T$ is a constant pulled out): ${\int }_{{\mu }^{\circ }}^{\mu }d\mu =k_{B}T{\int }_{P^{\circ }}^P\frac{dP}{P}=k_{B}T[\ln P{]}_{P^{\circ }}^P.$ $\mu (T,P)-{\mu }^{\circ }(T)=k_{B}T\ln \frac{P}{P^{\circ }}.$

Step 4 — Read the result: $\boxed{\mu (T,P)={\mu }^{\circ }(T)+k_{B}T\ln \frac{P}{P^{\circ }}.}$ Here ${\mu }^{\circ }(T)$ is the value at the reference pressure — the integration constant, carrying all the temperature dependence. ✓

Coexistence = phase equilibrium = diffusive equilibrium between phases, so ${\mu }_{\text{liquid}}(T,P)={\mu }_{\text{vapour}}(T,P).$ If they were unequal, molecules would stream from the higher-$\mu$ phase to the lower until they matched — that streaming is boiling or condensing. At coexistence the streaming stops. Why equal despite very different densities: $\mu$ is a free energy, not a crowding count. The dense liquid pays a low energy cost per particle (strong attractive bonds lower $U$) but a low entropy; the sparse vapour has high entropy ($k_{B}T\ln$ term) but weak bonds. At the boiling point these two effects exactly balance, giving equal $\mu$. This is precisely the machinery behind Phase Equilibrium and Clausius-Clapeyron. ✓

The extensive Euler relation (see Euler Relation and Gibbs-Duhem) generalises $G=N\mu$ to mixtures: $G=N_1{\mu }_1+N_2{\mu }_2=(2.0)(-5.0)+(3.0)(-8.0)=-10.0-24.0=-34.0\text{kJ}.$ Note you cannot write $G=N\mu$ with a single $\mu$ here — each species contributes its own partial molar term. ✓

Step 1 — Differentiate the Euler relation (product rule on each $N_i{\mu }_i$): $dG={\sum }_i{\mu }_{i}d{N}_i+{\sum }_i{N}_{i}d{\mu }_i.$ Why: Euler is an identity in the extensive variables, so its differential must also hold.

Step 2 — Set it equal to the master differential: ${\sum }_i{\mu }_{i}d{N}_i+{\sum }_i{N}_{i}d{\mu }_i=-SdT+VdP+{\sum }_i{\mu }_{i}d{N}_i.$

Step 3 — The ${\sum }_i{\mu }_{i}d{N}_i$ terms cancel on both sides, leaving ${\sum }_i{N}_{i}d{\mu }_i=-SdT+VdP,$ i.e. $\boxed{SdT-VdP+\sum _i{N}_{i}d{\mu }_i=0.}\text{(Gibbs–Duhem)}$ Why it matters: the chemical potentials are not independent — they are tied to $T,P$ and each other.

Step 4 — Single component at fixed $T,P$. With one species the sum has a single term, ${\sum }_i{N}_{i}d{\mu }_i=Nd\mu$, and setting $dT=0,dP=0$ kills the first two terms of Gibbs–Duhem: $S(0)-V(0)+Nd\mu =0⟹Nd\mu =0.$ Since $N\ne 0$, this forces $\boxed{d\mu =0.}$ Physical interpretation. For a single pure substance you cannot change $\mu$ at fixed $T$ and $P$ — $\mu$ is entirely determined by $T$ and $P$, so $\mu =\mu (T,P)$ is a genuine function of only those two variables (adding more of the substance changes $N$ but not $\mu$, consistent with $\mu =G/N$ being intensive). This is exactly why phase coexistence, where ${\mu }_{\text{liquid}}(T,P)={\mu }_{\text{vapour}}(T,P)$, is one equation in the two unknowns $T,P$ and therefore traces out a line in the $(T,P)$ plane — the coexistence curve of Phase Equilibrium and Clausius-Clapeyron. Gibbs–Duhem is what pins that degree of freedom down. ✓

(a) $\epsilon =\mu$: the exponent is $0$, so $e^0=1$ and $\bar{n}=\frac{1}{1+1}=\frac{1}{2}.$ The state at exactly the chemical potential is half-filled — this defines the Fermi level.

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

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

So at $T=0$ the occupancy is a sharp step: every state below $\mu$ is full, every state above is empty. $\mu$ is the energy at which the filling switches — the top of the filled sea. It is exactly the "energy cost of adding one more particle," now visible as the energy of the last-added electron. ✓

When $P<P^{\circ }$, the term $k_{B}T\ln (P/P^{\circ })<0$ (log of a number below 1 is negative). If this outweighs ${\mu }^{\circ }(T)$, then $\mu <0$. Physically a very dilute gas has enormous entropy per particle, and $\mu$ contains $-T(\partial S/\partial N)$ — a large positive entropy contribution makes $\mu$ negative, meaning the system is eager to accept more particles (adding one lowers $G$ via entropy). So "adds energy" is only the internal-energy part; the free-energy $\mu$ can sit below zero. This is the same lesson as the L3 trap: free energy, not raw energy, sets the sign. ✓

• Which way do particles flow between two reservoirs, and what does $dG$ have to do with it?
• Why does compressing an ideal gas raise $\mu$, and why is the dependence logarithmic?
• State and derive Gibbs–Duhem; what does it force for a single pure substance at fixed $T,P$?
• At the Fermi level $\epsilon =\mu$, what is the fermionic occupancy, and why?