\partial T)_{V, N}$ )

Maxwell relations turn pressure, volume and temperature measurements into information about entropy. For example, the pressure rise when heating a rigid sealed vessel determines the entropy change when expanding the same sample isothermally:

(\frac{\partial S}{\partial V})_{T, N} = (\frac{\partial P}{\partial T})_{V, N} .

(1)

The quantities on the two sides describe different experiments, yet both are mixed partial derivatives of one Helmholtz free energy. This is the useful surprise: a mechanical measurement can reveal a thermal property. Similarly, the coefficient of thermal expansion gives

(\partial S / \partial P)_{T, N} = - α V

. ^[ref]

For a simple compressible system, hold particle number

N

and composition fixed, and allow only pressure-volume work. Write

U

for internal energy,

H = U + P V

for enthalpy,

F = U - TS

for Helmholtz free energy, and

G = H - TS

for Gibbs free energy. Their natural-variable differentials are

d U (S, V, N) d F (T, V, N) = T d S - P d V, = - S d T - P d V, d H (S, P, N) d G (T, P, N) = T d S + V d P, = - S d T + V d P .

(2)

Thus the first partial derivatives, with the subscript recording the variables kept fixed, are

U_{S}_{V, N} = T, F_{T}_{V, N} = - S, U_{V}_{S, N} = - P, F_{V}_{T, N} = - P, H_{S}_{P, N} = T, G_{T}_{P, N} = - S, H_{P}_{S, N} = V, G_{P}_{T, N} = V .

(3)

The symmetry of second derivatives produces all four standard Maxwell relations, including their signs:

U : (\frac{\partial T}{\partial V})_{S, N} H : (\frac{\partial T}{\partial P})_{S, N} F : (\frac{\partial S}{\partial V})_{T, N} G : (\frac{\partial S}{\partial P})_{T, N} = - (\frac{\partial P}{\partial S})_{V, N}, = (\frac{\partial V}{\partial S})_{P, N}, = (\frac{\partial P}{\partial T})_{V, N}, = - (\frac{\partial V}{\partial T})_{P, N} .

(4)

For a brief derivation of the

F

relation, differentiate

F_{T} = - S

with respect to

V

and

F_{V} = - P

with respect to

T

- S_{V} = F_{T V} = F_{V T} = - P_{T}

. Each other row follows in the same way. Assume continuous second partial derivatives of the relevant thermodynamic potential in the region considered. At a phase transition, the equilibrium potential may lose this smoothness; apply the identities within each smooth thermodynamic phase, and handle latent heat and phase coexistence separately. If particle numbers vary, the potential differentials acquire chemical potential terms, so the fixed-

N

constraints must remain explicit. ^[ref]

For an ideal gas with fixed

N

, the ideal gas law gives

P = N k_{B} T / V

. At fixed volume, heating changes the pressure at rate

N k_{B} / V

. Consequently an isothermal expansion at positive temperature has

Δ S = \int_{V_{1}}^{V_{2}} (\frac{\partial P}{\partial T})_{V, N} d V = N k_{B} ln \frac{V _{2}}{V _{1}} .

(5)

For one mole with

R = 8.314 J mo l^{- 1} K^{- 1}

and

V_{2} = 2 V_{1}

, this is

Δ S = R ln 2 ≃ 5.763 J K^{- 1}

. At

300 K

, a reversible isothermal expansion absorbs

Q = T Δ S ≃ 1728.85 J

and performs equal work, since its internal energy is unchanged. The entropy difference is a state function difference: the same equilibrium endpoints give the same

Δ S

in an insulated free expansion, although there

Q = 0

and the process is irreversible.

Figure 1.
Ideal gas entropy slopes equal measured pressure responses to heating
. Left: entropy relative to one common reference state $(T_{0}, V_{0})$ for a classical monatomic gas, $[S (T, V) - S (T_{0}, V_{0})] / (N k_{B}) = \frac{3}{2} ln (T / T_{0}) + ln (V / V_{0})$ , with $T_{0} = 300 K$ . Each fixed-temperature curve gains $ln 2$ on doubling the volume. Right: both partial derivatives, multiplied by $V_{0} / (N k_{B})$ , equal $V_{0} / V$ ; circles are independent finite differences of entropy, while the line is the pressure response. **Details**.

The same Maxwell relations let thermal measurements determine the difference between constant-pressure and constant-volume heat capacities. In a smooth single thermodynamic phase,

d S = \frac{C _{V}}{T} d T + \frac{α}{κ _{T}} d V = \frac{C _{P}}{T} d T - α V d P,

(6)

using heat capacity at constant volume, heat capacity at constant pressure, coefficient of thermal expansion and isothermal compressibility. Along fixed pressure,

d V = α V d T

, so

C_{P} - C_{V} = \frac{T V α ^{2}}{κ _{T}} .

(7)

Thus measured expansion and compression responses determine a heat capacity difference without measuring both heat capacities separately. They do not determine the absolute heat capacities or absolute entropy by themselves; an independent thermal measurement supplies that missing information. For an ideal gas,

α = 1/ T

and

κ_{T} = 1/ P

reduce the result to

C_{P} - C_{V} = P V / T = N k_{B}

, which is Mayer's relation.

When the necessary higher partial derivatives exist and commute, differentiating the Maxwell relations once more also gives

(\frac{\partial C _{V}}{\partial V})_{T, N} = T (\frac{\partial ^{2} P}{\partial T ^{2}})_{V, N}, (\frac{\partial C _{P}}{\partial P})_{T, N} = - T (\frac{\partial ^{2} V}{\partial T ^{2}})_{P, N} .

(8)

Integrating these slopes determines how a heat capacity changes with volume or pressure at fixed temperature. For an ideal gas, both right-hand sides vanish because the ideal gas law is linear in temperature at either fixed volume or fixed pressure. Consequently

C_{V}

and

C_{P}

depend only on temperature at fixed

N

, although they need not be constant in temperature. ^[ref]

Historically, James Clerk Maxwell's Theory of Heat, first published in 1871, presents these connections using the geometry of isothermal processes and reversible adiabatic processes. His fourth edition, page 167, explicitly relates expansion on heating at fixed pressure to heat emitted during an isothermal compression: the modern

G

relation gives

δ Q_{rev} = T d S = - T α V d P

. This explains why a substance with positive coefficient of thermal expansion emits heat when compressed isothermally, whereas one with negative coefficient absorbs it. ^[ref] ^[ref]