Pressure and Chemical Potential

Let two gases share not only energy but volume, through a movable partition, with \(U_A + U_B\) and \(V_A + V_B\) fixed. At equilibrium the total entropy \(S_{\text{total}} = S_A + S_B\) is at its maximum in both directions, \(\nabla S_{\text{total}} = 0\) in \((U_A, V_A)\). The energy derivative gives \(T_A = T_B\) as before; with \(dV_A = -dV_B\), the volume derivative gives

with the energies and particle numbers held fixed. Experience says their pressures must also be equal, so pressure must be some function of \(\partial S/\partial V\). Its units are \((\mathrm{J/K})/\mathrm{m^3} = \mathrm{Pa/K}\), so multiplying by \(T\) gives units of pressure. A system that gains a lot of entropy by expanding just a little “wants” to expand, which is a large pressure:

For a monatomic ideal gas, \(\Omega = f(N)V^NU^{3N/2}\), so \(S = Nk_{\mathrm{B}}\ln V + \tfrac32 Nk_{\mathrm{B}}\ln U + k_{\mathrm{B}}\ln f(N)\) and \(P = TNk_{\mathrm{B}}/V\): we have just derived the ideal gas law, \(PV = Nk_{\mathrm{B}}T\). Release the partition below and watch the system climb the entropy surface to its peak.

Change \(U\) and \(V\) together in two small steps and the entropy changes by \(dS = (\partial S/\partial U)_V\,dU + (\partial S/\partial V)_U\,dV = dU/T + P\,dV/T\), or

the thermodynamic identity, true for any infinitesimal change. Everything above follows from it: at fixed \(V\), \(1/T = (\partial S/\partial U)_V\); at fixed \(U\), \(P = T(\partial S/\partial V)_U\). It also gives \(C_V = T(\partial S/\partial T)_V\) and, from \(dH = T\,dS + V\,dP\), \(C_P = T(\partial S/\partial T)_P\) (Problem 3.33). It is tempting to match it with \(dU = Q + W\), with \(Q = T\,dS\) and \(W = -P\,dV\), but that holds only if the volume changes quasistatically (so the pressure is uniform throughout), no other work is done, and nothing else changes. Then

so \(\Delta S = Q/T\) even when work is being done, and a quasistatic adiabatic process (\(Q = 0\)) has no entropy change at all: it is isentropic.

Push a piston in very fast and molecules build up in front of it, pushing back hard: the work done exceeds \(-P\,dV\), so \(Q < T\,dS\), and \(dS > Q/T\). Extra entropy is created. Pull it out faster than the molecules can follow and you get free expansion: no work, no heat, \(\Delta U = 0\), yet \(T\,dS = P\,dV > 0\). In both cases a mechanical process creates entropy over and above any that flows in with heat, and once created it can never be made to disappear. Below, a one-dimensional gas (molecules bouncing back and forth, \(\Omega \propto L^NU^{N/2}\)) is squeezed to half its length, or pulled to twice it, at any speed.

Now let two systems of the same species exchange both energy and particles (with volumes fixed, for simplicity). Maximizing \(S_{\text{total}}\) with \(dN_B = -dN_A\) gives \(T_A = T_B\) and \(\partial S_A/\partial N_A = \partial S_B/\partial N_B\). Multiplying by \(-T\) gives a quantity with units of energy, the chemical potential:

with \(\mu_A = \mu_B\) in diffusive equilibrium. Out of equilibrium, the system with the larger \(\partial S/\partial N\), the smaller \(\mu\), gains particles, since it gains more entropy than the other loses. Particles flow from high \(\mu\) to low, like something running downhill, which is why the minus sign is the convention. Including \(N\) in the thermodynamic identity,

where \(-P\,dV\) is mechanical work and \(\mu\,dN\) chemical work. At fixed \(S\) and \(V\), \(\mu = (\partial U/\partial N)_{S,V}\): the energy change when one particle is added with the entropy and volume held fixed. Usually some energy must be removed to keep the multiplicity fixed, so \(\mu\) is negative; any potential energy the particle needs to enter the system adds to it.

From the Sackur–Tetrode equation, with \(U = \tfrac32 Nk_{\mathrm{B}}T\),

For helium at room temperature and atmospheric pressure, \(\mu \approx -0.32\) eV. Raising the concentration \(N/V\) at fixed temperature makes \(\mu\) less negative: the gas becomes more willing to give up particles to its neighbors.

Two chambers of the same gas at the same temperature, joined by a small opening. At fixed \(T\), \(\mu = k_{\mathrm{B}}T\ln(N/V) + \text{constant}\), so the more crowded side has the higher \(\mu\) and loses particles until the concentrations, and the chemical potentials, are equal: \(\mu_A - \mu_B = k_{\mathrm{B}}T\ln[(N_A/V_A)/(N_B/V_B)]\).

Problem 3.37. A gas at height \(z\) has potential energy \(mgz\) per molecule, which adds to its chemical potential: \(\mu(z) = -k_{\mathrm{B}}T\ln[(V/N)(2\pi mk_{\mathrm{B}}T/h^2)^{3/2}] + mgz\). For two chunks at the same temperature and volume in diffusive equilibrium, \(\mu(z) = \mu(0)\) gives \(\ln[N(z)/N(0)] = -mgz/k_{\mathrm{B}}T\):

the exponential atmosphere of Problem 1.16, now from the chemical potential. Higher up, each molecule has more potential energy, so the gas must be thinner there to keep \(\mu\) the same.

A system with several species has a chemical potential for each, \(\mu_i = -T(\partial S/\partial N_i)_{U,V,N_{j\ne i}}\), so \(dU = T\,dS - P\,dV + \sum_i\mu_i\,dN_i\), and in diffusive equilibrium each species’ \(\mu\) is separately equal. The chemical potential is central to equilibrium in chemical reactions and phase transformations, the subject of Chapter 5, and it plays a central role in quantum statistics, in Chapter 7.

For three model systems, the two-state paramagnet, the Einstein solid, and the monatomic ideal gas, the laws of microscopic physics give explicit formulas for the multiplicity \(\Omega\) and entropy \(k_{\mathrm{B}}\ln\Omega\), and from them heat capacities and a variety of other measurable quantities. That program is statistical mechanics.