Open Systems and Coexistence: Chemical Potential and Clausius–Clapeyron

The chemical potential

Open the box to particle exchange and the fundamental relation grows a term, dU=T\,dS-P\,dV+\mu\,dN. The chemical potential \mu is the energy cost of adding one more particle — and, like T and P, it can be read as a slope of any of the potentials, provided you hold that potential's own natural variables fixed.

\mu = \left(\frac{\partial U}{\partial N}\right)_{S,V} = \left(\frac{\partial F}{\partial N}\right)_{T,V} = \left(\frac{\partial G}{\partial N}\right)_{T,P}

Three equivalent faces of the chemical potential — the same \mu, read off different potentials.

Intuition: \mu is to particle number what T is to entropy and P is to volume — the intensive 'force' conjugate to N. Just as heat flows from hot to cold, particles flow from high \mu to low \mu until the two are equal. That equality is diffusive (chemical) equilibrium.

Extensivity, Euler and Gibbs–Duhem

Energy is extensive: double S, V and N together and U doubles. Applying Euler's theorem on homogeneous functions to U(S,V,N) gives an integrated form, and hence a clean statement of G for a single component.

U = TS - PV + \mu N \quad\Longrightarrow\quad G = U - TS + PV = \mu N

For one component, the Gibbs energy is simply \mu per particle times N.

Differentiate U=TS-PV+\mu N and subtract the fundamental relation, and the extensive terms cancel to leave a constraint among the intensive variables — the Gibbs–Duhem relation. You cannot vary T, P and \mu all independently.

S\,dT - V\,dP + N\,d\mu = 0 \quad\Longleftrightarrow\quad d\mu = -s\,dT + v\,dP

The Gibbs–Duhem relation (right side per particle, with s=S/N, v=V/N).

The conditions for phase coexistence

When can two phases — say liquid and vapour — sit together in stable equilibrium? They can exchange energy, volume and particles, and stability demands no net flow of any of them. So each conjugate intensive variable must match across the interface.

T_1 = T_2, \qquad P_1 = P_2, \qquad \mu_1(T,P) = \mu_2(T,P)

Thermal, mechanical and chemical equilibrium between two coexisting phases.

The first two hold automatically once the phases touch. The third, \mu_1(T,P)=\mu_2(T,P), is one equation in two unknowns — so it carves out a curve in the P–T plane. That is exactly why phases meet along coexistence lines, and why three phases coexist only where two such curves cross: the triple point.

Clausius–Clapeyron: the slope of the line

Now find how steep a coexistence line is. Step along it: since \mu_1=\mu_2 everywhere on the line, moving a little must keep them equal, so d\mu_1=d\mu_2. Expand each with Gibbs–Duhem per particle.

On the line, equality is preserved: d\mu_1 = d\mu_2.
For each phase, d\mu = -s\,dT + v\,dP (per-particle entropy s and volume v).
Set them equal: -s_1\,dT + v_1\,dP = -s_2\,dT + v_2\,dP.
Collect: (v_1-v_2)\,dP = (s_1-s_2)\,dT, so dP/dT = \Delta s/\Delta v.
The phase change is isothermal, so the latent heat absorbed is L=T\,\Delta s; substitute to get the Clausius–Clapeyron equation.

\frac{dP}{dT} = \frac{\Delta s}{\Delta v} = \frac{L}{T\,\Delta v}

The Clausius–Clapeyron equation: a coexistence line's slope from measurable latent heat and volume change.

Counting freedoms: the Gibbs phase rule

The coexistence conditions can be counted in general. With C chemical components and \Phi phases present, matching every chemical potential across every phase leaves a definite number F of intensive variables still free to vary.

F = C - \Phi + 2

The Gibbs phase rule: the number of independent intensive degrees of freedom at coexistence.

For a single substance (C=1): one phase gives F=2 (an area on the diagram), two phases give F=1 (a line), and three phases give F=0 (an isolated point, the triple point). The counting matches the diagram exactly — thermodynamics dictates its very geometry. See the Gibbs phase rule.