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

The scenario matrix

Before solving anything, let's enumerate every case class this topic can throw at you. Each worked example below is tagged with the cell it fills.

Cell	Case class	What makes it tricky	Example
A	$μ_{A} - μ_{B} > 0$ (positive sign)	which direction does matter flow?	Ex 1
B	$μ_{A} - μ_{B} < 0$ (negative sign)	the sign flips the flow	Ex 1
C	$μ_{A} = μ_{B}$ (zero difference)	degenerate: equilibrium, no flow	Ex 2
D	Pressure changed, $T$ fixed	$ln$ dependence, sign of $Δ μ$	Ex 3
E	Limit $P \to 0$ (dilute)	$μ \to - \infty$ : what does that mean?	Ex 4
F	Limit $P \to \infty$ (dense)	$μ$ grows without bound	Ex 4
G	Two phases coexisting	equal- $μ$ picks the boundary line	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

Every cell A–J is covered below. Let's go.

Cells A & B — the sign of $μ_{A} - μ_{B}$ decides direction

Worked example Example 1 — Which way does matter flow (both signs)?

Two chambers share a porous wall at common $T, P$ . Read off the flow direction for two cases:

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: Guess for each case: does matter go A→B or B→A? Write it down before reading on.

Step 1 — Write the change in $G$ for moving $d N$ particles from A to B. Recall from the parent that $μ = (\partial G / \partial N)_{T, P}$ is by definition the change in $G$ per particle added at fixed $T, P$ . So adding $d N$ to a chamber changes its $G$ by $μ d N$ ; removing $d N$ changes it by $- μ d N$ . When $d N$ particles leave A and enter B, $N_{A}$ drops by $d N$ and $N_{B}$ rises by $d N$ : $d G = μ_{A} (- d N) + μ_{B} (+ d N) = (μ_{B} - μ_{A}) d N .$ Why this step? $G$ is minimized at equilibrium (parent note). A move happens spontaneously only if it makes $d G < 0$ . So we just need the sign of $(μ_{B} - μ_{A})$ .

Step 2 — Case A. $μ_{B} - μ_{A} = - 0.55 - (- 0.30) = - 0.25 eV$ . For $d N > 0$ , $d G = (- 0.25) d N < 0$ . Spontaneous: particles flow A→B. Why this step? Negative $d G$ = downhill in free energy. Since $μ_{A} > μ_{B}$ , this is the "high $\to$ low" rule in action.

Step 3 — Case B. $μ_{B} - μ_{A} = - 0.40 - (- 0.70) = + 0.30 eV$ . For $d N > 0$ , $d G = (+ 0.30) d N > 0$ — that would raise $G$ , so A→B is forbidden. The reverse, B→A, gives $d G < 0$ . Particles flow B→A. Why this step? Here $μ_{B} > μ_{A}$ , so B is now the "mountain" and A is the "mud." The flow rule is unchanged; only the labels swapped.

Verify: In both cases matter left the chamber with the larger $μ$ . Case A: $- 0.30 > - 0.55$ , source is A ✓. Case B: $- 0.40 > - 0.70$ , source is B ✓. Units: eV is energy, $d N$ dimensionless, so $d G$ is in eV ✓.

Cell C — the degenerate case: equal $μ$ , nothing moves

Cell D — change pressure at fixed $T$ : the $ln$ law

Worked example Example 3 — Compress an ideal gas, watch

$μ$ rise An ideal gas at $T = 300 K$ is compressed from $P^{\circ} = 1.00 atm$ to $P = 3.00 atm$ . By how much does $μ$ change? Use $k_{B} = 8.617 \times 1 0^{- 5} eV/K$ .

Forecast: Does $μ$ go up or down when you squeeze the gas?

Step 1 — Recall the ideal-gas $μ$ . From the parent, Example 2: $μ (T, P) = μ^{\circ} (T) + k_{B} T ln \frac{P}{P ^{\circ}} .$ Why this step? This is the one formula that carries all pressure dependence. The reference term $μ^{\circ} (T)$ cancels when we take a difference.

Step 2 — Subtract to get $Δ μ$ . $Δ μ = μ (T, P) - μ (T, P^{\circ}) = k_{B} T ln \frac{P}{P ^{\circ}} .$ Why this step? We only care about the change, so the (unknown) reference drops out cleanly.

Step 3 — Plug numbers. $Δ μ = (8.617 \times 1 0^{- 5}) (300) ln (3.00) = 0.025851 \times 1.098612 \approx 0.0284 eV .$ Why this step? $ln 3 > 0$ , so $Δ μ > 0$ : compressing raises $μ$ .

Verify: $ln (3) \approx 1.0986$ , product $\approx 0.0284 eV$ ✓. Physical sense: a squeezed gas is "more eager to escape" to lower pressure — higher $μ$ — exactly why gas rushes out of a punctured tire. Units: $(eV/K) (K) = eV$ ✓.

Cells E & F — the two limits, $P \to 0$ and $P \to \infty$

The $ln$ in Example 3 hides two dramatic limits. Let's face them head-on.

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

Worked example Example 4 — What happens at extreme dilution and extreme density?

Using $μ (T, P) = μ^{\circ} + k_{B} T ln (P / P^{\circ})$ , describe $μ$ as $P \to 0$ (Cell E) and as $P \to \infty$ (Cell F).

Forecast: Does $μ$ stay finite in either limit? Guess a sign for each.

Step 1 — Dilute limit, $P \to 0^{+}$ (Cell E). $ln \frac{P}{P ^{\circ}} \to ln (0^{+}) = - \infty \Rightarrow μ \to - \infty.$ Why this step? Look at the pale-yellow curve on the left of the figure ( $P$ small, near the vertical axis): it plunges through the horizontal dashed line and off the bottom. An infinitely dilute gas has infinitely negative $μ$ — it is infinitely "hungry" to receive more particles. That is why a drop of ink invades a whole bathtub: the empty water is at $μ \to - \infty$ .

Step 2 — Dense limit, $P \to \infty$ (Cell F). $ln \frac{P}{P ^{\circ}} \to + \infty \Rightarrow μ \to + \infty.$ Why this step? On the right of the figure (large $P$ ) the yellow curve climbs without bound. A hugely compressed gas has enormous $μ$ — desperate to expel particles. (For a real gas, interactions eventually break the ideal formula, but the qualitative " $μ$ climbs" survives.)

Step 3 — The crossing point. At $P = P^{\circ}$ (i.e. $P = 1$ on the horizontal axis), $ln (1) = 0$ , so $μ = μ^{\circ}$ . That is the chalk-blue dot where the two dashed lines cross. Why this step? Every log curve passes through its own reference; naming it pins the whole curve.

Verify: $ln (P / P^{\circ})$ is monotonic increasing, unbounded both ways, zero at $P = P^{\circ}$ ✓. So $μ$ has no maximum and no minimum in pressure — it sweeps all real values. This confirms Cells E and F are genuine limiting behaviours, not artifacts.

Cell G — two phases coexisting: equal $μ$ draws the line

Worked example Example 5 — Liquid–vapour coexistence

A pure liquid (L) and its vapour (V) share a container. What single condition fixes their coexistence, and how does it relate to Phase Equilibrium and Clausius-Clapeyron?

Forecast: At coexistence, which is bigger — $μ_{L}$ or $μ_{V}$ ?

Step 1 — Write the equilibrium condition. Particles can cross the interface (evaporate/condense), so it is a diffusive equilibrium: $μ_{L} (T, P) = μ_{V} (T, P) .$ Why this step? Same rule as Example 2, applied to two phases instead of two boxes. Neither is bigger — they're equal (the yellow dot in the figure, where the pink and blue lines meet).

Recall Where does

$d μ = - s d T + v d P$ come from? Divide the parent's master differential $d G = - S d T + V d P + μ d N$ by $N$ for a pure substance, using $μ = G / N$ . The per-particle entropy is $s = S / N$ and per-particle volume $v = V / N$ ; a short manipulation (the Gibbs–Duhem relation, derived in Euler Relation and Gibbs-Duhem) gives $d μ = - s d T + v d P$ . Read it as: " $μ$ drops when $T$ rises (rate $s$ ) and rises when $P$ rises (rate $v$ )."

Step 2 — Slide $T$ and demand equality still holds. Differentiate $μ_{L} = μ_{V}$ along the coexistence line using the per-particle relation just recalled, with $s, v$ the per-particle entropy and volume: $- s_{L} d T + v_{L} d P = - s_{V} d T + v_{V} d P .$ Why this step? Coexistence isn't one point — it's a curve $P (T)$ . Requiring $μ_{L} = μ_{V}$ to persist as we move gives that curve's slope.

Step 3 — Solve for the slope. $\frac{d P}{d T} = \frac{s _{V} - s _{L}}{v _{V} - v _{L}} = \frac{Δ s}{Δ v} = \frac{L}{T Δ v},$ where $L = T Δ s$ is the latent heat per particle. Why this step? This is the Clausius–Clapeyron equation — it falls straight out of " $μ$ equal along the boundary." The chemical potential did all the work.

Verify: Along the chalk-blue vapour line and chalk-pink liquid line in the figure, the two $μ$ curves cross at a single temperature (fixed $P$ ); sliding $P$ moves that crossing, tracing the coexistence curve whose slope is $Δ s /Δ v$ ✓. Dimensional check: $[L] =$ energy/particle, $[T Δ v] =$ K·(volume/particle), and $d P / d T$ has units pressure/temperature — consistent once $Δ s$ is entropy per particle ✓.

Cell H — real-world word problem

Definition The two building blocks for this problem

First we need how $μ$ of a component depends on pressure and on concentration.

Pressure dependence. From the per-particle Gibbs–Duhem relation $d μ = - s d T + v d P$ (recalled in Example 5), at fixed $T$ we get $(\partial μ / \partial P)_{T} = v$ , where $v$ is the partial molar volume per particle — the volume one water molecule occupies in the liquid (units: volume/particle, e.g. $m^{3}$ per molecule). Integrating at fixed $T$ : $μ_{w} (P) = μ_{w} (P_{0}) + v_{w} (P - P_{0})$ . So a pressure excess $Π$ adds a term $v_{w} Π$ to the water's $μ$ .
Concentration dependence. Dissolving sugar dilutes the water to mole fraction $x_{w} < 1$ , contributing an entropic term $k_{B} T ln x_{w}$ . Put together, on the sugar side at excess pressure $Π$ : $μ_{w} = μ_{w}^{\circ} + k_{B} T ln x_{w} + v_{w} Π.$ This one equation drives the whole example — the $v_{w} Π$ term is exactly where the pressure knob enters.

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

A membrane lets water through but not sugar. Left side: pure water. Right side: water with dissolved sugar. Which way does water flow, and what pressure $Π$ stops it?

Forecast: Water into the sugar, or out of it?

Step 1 — Compare $μ_{w}$ on the two sides at equal $T, P$ (no excess pressure yet, $Π = 0$ ).

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

Why this step? $ln$ of a number below 1 is negative, so dissolving sugar lowers the water's $μ$ . The sugar side is the "mud."

Step 2 — Apply the flow rule. Water flows from high $μ$ (pure side) to low $μ$ (sugar side). Water crosses into the sugar solution — this is osmosis. Why this step? Same high→low rule (Cell A); the drive is purely the entropic $ln x_{w}$ term, not any energy.

Step 3 — Stop it with pressure (osmotic pressure). Raise the sugar side's pressure by $Π$ . Using the boxed formula $μ_{w} = μ_{w}^{\circ} + k_{B} T ln x_{w} + v_{w} Π$ , equilibrium demands the water's $μ$ match the pure side's $μ_{w}^{\circ}$ : $μ_{w}^{\circ} + k_{B} T ln x_{w} + v_{w} Π = μ_{w}^{\circ} \Rightarrow v_{w} Π = - k_{B} T ln x_{w} \approx k_{B} T x_{s},$ for small sugar fraction $x_{s} = 1 - x_{w}$ . With $n_{s}$ = sugar particles per volume, this rearranges to van 't Hoff's law $Π = n_{s} k_{B} T$ . Why this step? The only free knob to re-raise the lowered $μ_{w}$ is pressure, and $(\partial μ / \partial P)_{T} = v_{w}$ tells us exactly how much each unit of pressure lifts it. We solve $v_{w} Π = - k_{B} T ln x_{w}$ .

Numeric check. For $x_{s} = 0.010$ (1% sugar) at $T = 300 K$ : $- ln (0.990) = 0.010050$ . The osmotic drive is $k_{B} T (0.010050) = (8.617 \times 1 0^{- 5}) (300) (0.010050) \approx 2.60 \times 1 0^{- 4} eV$ per water molecule.

Verify: $- ln (0.99) \approx 0.01005$ , and the linear approximation $x_{s} = 0.010$ agrees to 0.5% ✓; osmotic drive $\approx 2.60 \times 1 0^{- 4} eV$ ✓. Sanity: adding more sugar (smaller $x_{w}$ ) makes $μ_{w}$ more negative, pulling more water in — matches everyday experience of a raisin swelling in water ✓.

Cell I — mixtures: partial molar and the held-fixed subscript

Worked example Example 7 — Two species, get the subscripts right

A container holds $N_{1}$ molecules of species 1 and $N_{2}$ of species 2 at fixed $T, P$ . Write the correct differential of $G$ and identify each $μ_{i}$ . Then, if only species 1 is added, what is $d G$ ?

Forecast: Is $μ_{1}$ still $\partial G / \partial N_{1}$ , or does species 2 change the rule?

Step 1 — Generalize the master differential to two species. $d G = - S d T + V d P + μ_{1} d N_{1} + μ_{2} d N_{2} .$ Why this step? Each species carries its own particle-number knob. Every additional species adds one $μ_{i} d N_{i}$ term.

Step 2 — Define each partial molar chemical potential. $μ_{i} = (\frac{\partial G}{\partial N _{i}})_{T, P, N_{j \neq = i}} .$ Why this step? The subscript now says: hold $T$ , $P$ , and every other species' count fixed. If you forget " $N_{j \neq = i}$ " you're computing a different quantity — this is the parent's "subscript is essential" mistake taken to mixtures.

Step 3 — Add only species 1 at fixed $T, P$ with $d N_{2} = 0$ . $d G = μ_{1} d N_{1} .$ Why this step? All other differentials vanish, leaving exactly the "cost per particle of species 1."

Verify: Note $μ = G / N$ from the parent is only for a pure substance; here $G = N_{1} μ_{1} + N_{2} μ_{2}$ (Euler, Euler Relation and Gibbs-Duhem) instead. Consistency check: if species 2 is absent ( $N_{2} = 0$ ), this collapses to $G = N_{1} μ_{1}$ , i.e. $μ_{1} = G / N_{1}$ ✓.

Cell J — the exam twist: "uphill" flow driven by entropy

Worked example Example 8 — Can particles flow toward higher

energy? Two ideal-gas boxes at the same $T$ . Box A is dense ( $P_{A} = 4.0 atm$ ), box B is dilute ( $P_{B} = 1.0 atm$ ). A student claims: "Box A is compressed, so each particle there has more energy; particles should flow A→B to lower energy — flow follows energy." Diagnose the claim and give the real answer. Take $T = 300 K$ .

Forecast: Which box has the higher $μ$ ? Does energy or $μ$ decide the flow?

Step 1 — For an ideal gas, per-particle energy is $\frac{3}{2} k_{B} T$ — it depends only on $T$ . Both boxes are at the same $T$ , so they have the same energy per particle. The student's "more energy when compressed" is wrong for an ideal gas. Why this step? This exposes the trap: compression changes $μ$ through the entropic $ln P$ term, not through kinetic energy.

Step 2 — Compare $μ$ using the ideal-gas law (both share $μ^{\circ}$ ). $μ_{A} - μ_{B} = k_{B} T ln \frac{P _{A}}{P _{B}} = k_{B} T ln \frac{4.0}{1.0} = k_{B} T ln 4.$ Numerically: $(8.617 \times 1 0^{- 5}) (300) ln 4 = 0.025851 \times 1.386294 \approx 0.0358 eV > 0$ . Why this step? $ln 4 > 0$ so $μ_{A} > μ_{B}$ . The dense box is the "mountain."

Step 3 — Apply the flow rule. High $μ \to$ low $μ$ : particles flow A→B (dense to dilute). Why this step? By coincidence the student guessed the right direction — but for the wrong reason. It's not lower energy (energies are equal); it's higher entropy when the gas spreads out. The drive is the $- T \partial S / \partial N$ piece of $μ$ (parent's first steel-man mistake).

Step 4 — Show it really is entropy-driven. Since $Δ (energy) = 0$ but the gas still flows, the entire $Δ G$ came from the $- T Δ S$ term: spreading out into the dilute box raises the entropy per particle, so the process lowers $G$ even though no kinetic energy was released. The lasting lesson: particles can even flow toward higher potential energy if the entropy gain is bigger — free energy, not bare energy, is the referee. Why this step? This is the exam's real target: distinguishing free energy from bare energy. The correct diagnosis is that the student reached the right direction (A→B) by a wrong argument (energy), and the true driver is the entropic $k_{B} T ln (P_{A} / P_{B}) > 0$ that makes $μ_{A} > μ_{B}$ .

Verify: $ln 4 \approx 1.3863$ , $μ_{A} - μ_{B} \approx 0.0358 eV$ , positive ✓. Direction A→B matches "high $μ \to$ low $μ$ " ✓. And energies equal ( $\frac{3}{2} k_{B} T$ both) confirms the flow is purely entropic ✓.

Recall Self-test across all cells

Cover the answers. For each, name the matrix cell and the result.

$μ_{A} = - 0.2$ , $μ_{B} = - 0.9$ eV: which way? ::: Cell A — A→B (A is higher).
Two connected boxes with $μ_{A} = μ_{B}$ : what flows? ::: Cell C — nothing; equilibrium.
As $P \to 0$ , $μ \to$ ? ::: Cell E — $- \infty$ .
Coexistence condition for a liquid and its vapour? ::: Cell G — $μ_{L} = μ_{V}$ .
Osmosis: does water enter or leave the sugar side? ::: Cell H — enters (sugar side has lower $μ_{w}$ ).
In a mixture, the correct held-fixed set for $μ_{1}$ ? ::: Cell I — $T, P, N_{j \neq = 1}$ .
Two equal- $T$ ideal gases, dense vs dilute — flow direction and why? ::: Cell J — dense→dilute, entropy-driven.

Back to the parent: Chemical potential note. Related depth: Grand Canonical Ensemble (where $μ$ becomes the control knob), Fermi-Dirac and Bose-Einstein Statistics (quantum $μ$ ), Entropy and the Second Law (why $G$ is minimized at all).