4.2 The thermodynamic potentials

All the potentials come from the fundamental relation \(dU=T\,dS-P\,dV\) by Legendre transformation. Subtracting the product of a pair of conjugate variables swaps which member of the pair is independent.

(a) Internal energy \(U(S,V)\).

\[dU=T\,dS-P\,dV\ \Rightarrow\ T=\Big(\frac{\partial U}{\partial S}\Big)_V,\quad P=-\Big(\frac{\partial U}{\partial V}\Big)_S\ \Rightarrow\ \boxed{\Big(\frac{\partial T}{\partial V}\Big)_S=-\Big(\frac{\partial P}{\partial S}\Big)_V} .\]

(b) Enthalpy \(H=U+PV\), natural variables \((S,P)\). Here \(dH=dU+P\,dV+V\,dP=T\,dS+V\,dP\), so \(T=(\partial H/\partial S)_P\) and \(V=(\partial H/\partial P)_S\), and the corresponding Maxwell relation is \(\boxed{\Big(\dfrac{\partial T}{\partial P}\Big)_S=\Big(\dfrac{\partial V}{\partial S}\Big)_P}\). Physically, \(\Delta H\) is the heat absorbed at constant pressure; it is positive for endothermic and negative for exothermic reactions.

(c) Helmholtz free energy \(F=U-TS\), natural variables \((T,V)\). Here \(dF=dU-T\,dS-S\,dT=-S\,dT-P\,dV\), so \(P=-(\partial F/\partial V)_T\) and \(S=-(\partial F/\partial T)_V\), and the Maxwell relation is \(\boxed{\Big(\dfrac{\partial P}{\partial T}\Big)_V=\Big(\dfrac{\partial S}{\partial V}\Big)_T}\). In an isothermal reversible process, \(-\Delta F\) is the maximum work obtainable.

(d) Gibbs free energy \(G=H-TS=U+PV-TS\), natural variables \((T,P)\). Here \(dG=V\,dP-S\,dT\), so \(V=(\partial G/\partial P)_T\) and \(S=-(\partial G/\partial T)_P\), and the Maxwell relation is \(\boxed{\Big(\dfrac{\partial V}{\partial T}\Big)_P=-\Big(\dfrac{\partial S}{\partial P}\Big)_T}\). The quantity \(-\Delta G\) is the maximum non-expansion work obtainable from a closed system at constant \(T\) and \(P\).

With particle exchange. \(dU=T\,dS-P\,dV+\mu\,dN\), and the chemical potential is

\[\mu=\Big(\frac{\partial U}{\partial N}\Big)_{S,V}=\Big(\frac{\partial F}{\partial N}\Big)_{T,V}=\Big(\frac{\partial G}{\partial N}\Big)_{T,P} .\]

Because \(G\) is extensive and \(T\), \(P\) are intensive, \(G=\mu N\).

(e) Grand potential \(F'=F-\mu N\), natural variables \((T,V,\mu)\). A further Legendre transformation exchanges \(N\) for \(\mu\): \(dF'=-S\,dT-P\,dV-N\,d\mu\), so \(N=-(\partial F'/\partial\mu)_{T,V}\), \(S=-(\partial F'/\partial T)_{V,\mu}\) and \(P=-(\partial F'/\partial V)_{T,\mu}\). Since \(F=G-PV=\mu N-PV\), we have \(F'=-PV\). This is the potential of an open system at fixed \(T\), \(V\) and \(\mu\), and it is the basis of the grand canonical ensemble (Chapter 9), where \(F'=-k_BT\ln Z_\mu\).

Gibbs–Duhem relation. Euler’s theorem for the extensive function \(U(S,V,N)\) gives \(U=TS-PV+\mu N\). Differentiating and subtracting \(dU\):

\[S\,dT-V\,dP+N\,d\mu=0 .\]

Equilibrium conditions. A system in contact with its surroundings settles into the state that minimises

An isolated system (\(U\), \(V\) fixed) maximises \(S\).

Mnemonic (thermodynamic square). Write the potentials clockwise on the sides of a square — \(F\) (top), \(G\) (right), \(H\) (bottom), \(U\) (left) — with the variables \(V\), \(T\), \(P\), \(S\) at the corners, starting top-left. (“Valid Facts and Theoretical Understanding Generate Solutions to Hard Problems.”) Each potential depends on its two neighbouring corners. Each Maxwell relation pairs a side’s two corners with the opposite corners; the sign is negative when the arrows point the same way.

Gibbs–Helmholtz relations

Example 4.1 Prove that (a) \(U=\left[\dfrac{\partial(F/T)}{\partial(1/T)}\right]_V\) and (b) \(H=\left[\dfrac{\partial(G/T)}{\partial(1/T)}\right]_P\).

Solution.

(a) Since \(U=F+TS\) and \(S=-(\partial F/\partial T)_V\), we have \(U=F-T\Big(\dfrac{\partial F}{\partial T}\Big)_V\). Now note that \(\Big(\dfrac{\partial(F/T)}{\partial T}\Big)_V=\dfrac1T\Big(\dfrac{\partial F}{\partial T}\Big)_V-\dfrac{F}{T^2}=-\dfrac{U}{T^2}\). Hence \(U=-T^2\Big[\dfrac{\partial(F/T)}{\partial T}\Big]_V\), and since \(d(1/T)=-dT/T^2\), this is \(U=\Big[\dfrac{\partial(F/T)}{\partial(1/T)}\Big]_V\).

(b) The same steps with \(G=H-TS\) and \(S=-(\partial G/\partial T)_P\) give \(H=-T^2\Big[\dfrac{\partial(G/T)}{\partial T}\Big]_P=\Big[\dfrac{\partial(G/T)}{\partial(1/T)}\Big]_P\).