Thermodynamic Relations and Maxwell Equations

Thermodynamic property relations and Maxwell relations

Thermodynamic property relations connect partial derivatives of state properties, so a quantity that is difficult to measure can be found from quantities that are easy to measure. They rest on three facts: the differential of a state function is exact, the mixed second derivatives of a state function are equal, and a property depends only on the state, not on the path. For a simple compressible substance, the basic Gibbs equations are du = T\,ds - P\,dv and dh = T\,ds + v\,dP. Holding pressure constant in the enthalpy equation gives \left(\partial h/\partial s\right)_P = T, which is the slope of an isobar on the T-s plane. The temperature is therefore the isobaric derivative of enthalpy with respect to entropy.

The Maxwell relations follow from the thermodynamic potentials u, h, the Helmholtz function \psi = u - Ts (also written a) and the Gibbs function g = h - Ts. Since dg = -s\,dT + v\,dP, cross-differentiation gives \left(\partial s/\partial P\right)_T = -\left(\partial v/\partial T\right)_P. The minus sign is essential and is the most frequently missed element of this relation. From d\psi = -s\,dT - P\,dv, the isothermal slope of \psi against v is \left(\partial \psi/\partial v\right)_T = -P, which is negative and finite for a gas at finite volume. The four first-order derivatives tied to the potentials are \left(\partial \psi/\partial v\right)_T = -P, \left(\partial \psi/\partial T\right)_v = -s, \left(\partial h/\partial P\right)_s = v and \left(\partial h/\partial s\right)_P = T. Matching a physical quantity to its correct partial derivative is a standard exam task, and the sign of each entry must be checked.

The definitions of the potentials are used directly in matching problems. The Helmholtz function is a = u - Ts, and the Gibbs function is g = h - Ts. The T-ds relation rearranges to du = T\,ds - P\,dv. Along the saturation curve, the Clapeyron-Clausius equation takes the form \frac{d(\ln P)}{dT}_{sat} = \frac{h_{fg}}{RT^2} when the liquid volume is neglected and the vapour is treated as an ideal gas.

For a reversible non-flow process, \delta q = du + P\,dv is an inexact differential, because \delta q depends on the path. Dividing by T gives \delta q / T = ds, and ds is exact because entropy is a state function. The integrating factor that makes the energy equation exact is therefore 1/T.

When a Maxwell relation is combined with a derivative of a thermodynamic property, the continuity of s and its derivatives allows the order of differentiation to be interchanged. Using c_v = T\left(\partial s/\partial T\right)_v together with \left(\partial s/\partial v\right)_T = \left(\partial P/\partial T\right)_v gives \left(\partial c_v/\partial v\right)_T = T\left(\partial^2 P/\partial T^2\right)_v. The prefactor is T, not 1/T.