Categories: Physics, Thermodynamics.

Thermodynamic potential

Thermodynamic potentials are state functions whose minima or maxima represent equilibrium states of a system. Such functions are either energies (hence potential) or entropies.

Of the many options, which potential decides the equilibrium state for a given system? It depends on which variables are assumed to be in automatic equilibrium. Such variables are called the natural variables of that potential. For example, if a system can exchange heat with its environment, and is consequently at the same temperature T=TenvT = T_{\mathrm{env}}, then TT must be a natural variable.

The link from natural variables to potentials is established by thermodynamic ensembles.

Once the natural variables have been determined, the appropriate potential can be selected from the list below. All non-natural variables can then be calculated by taking partial derivatives of the potential with respect to the natural variables.

Mathematically, the potentials are related to each other by Legendre transformation.

Internal energy

The internal energy UU represents the capacity to do both mechanical and non-mechanical work, and to release heat. It is simply the integral of the fundamental thermodynamic relation:

U(S,V,N)≡TS−PV+μN\begin{aligned} \boxed{ U(S, V, N) \equiv T S - P V + \mu N } \end{aligned}

It is a function of the entropy SS, volume VV, and particle count NN: these are its natural variables, so an infinitesimal change dU\dd{U} is as follows:

dU=TdS−PdV+μdN\begin{aligned} \boxed{ \dd{U} = T \dd{S} - P \dd{V} + \mu \dd{N} } \end{aligned}

The non-natural variables are temperature TT, pressure PP, and chemical potential μ\mu. These can be recovered by differentiating UU with respect to the natural variables SS, VV, and NN:

T=(∂U∂S)V,NP=−(∂U∂V)S,Nμ=(∂U∂N)S,V\begin{aligned} \boxed{ T = \Big( \pdv{U}{S} \Big)_{V,N} \qquad P = - \Big( \pdv{U}{V} \Big)_{S,N} \qquad \mu = \Big( \pdv{U}{N} \Big)_{S,V} } \end{aligned}

It is convention to write those subscripts, to help keep track of which function depends on which variables. They are meaningless; these are normal partial derivatives.

Enthalpy

The enthalpy HH of a system, in units of energy, represents its capacity to do non-mechanical work, plus its capacity to release heat. It is given by:

H(S,P,N)≡U+PV\begin{aligned} \boxed{ H(S, P, N) \equiv U + P V } \end{aligned}

It is a function of the entropy SS, pressure PP, and particle count NN: these are its natural variables, so an infinitesimal change dH\dd{H} is as follows:

dH=TdS+VdP+μdN\begin{aligned} \boxed{ \dd{H} = T \dd{S} + V \dd{P} + \mu \dd{N} } \end{aligned}

The non-natural variables are temperature TT, volume VV, and chemical potential μ\mu. These can be recovered by differentiating HH with respect to the natural variables SS, PP, and NN:

T=(∂H∂S)P,NV=(∂H∂P)S,Nμ=(∂H∂N)S,P\begin{aligned} \boxed{ T = \Big( \pdv{H}{S} \Big)_{P,N} \qquad V = \Big( \pdv{H}{P} \Big)_{S,N} \qquad \mu = \Big( \pdv{H}{N} \Big)_{S,P} } \end{aligned}

Helmholtz free energy

The Helmholtz free energy FF represents the capacity of a system to do both mechanical and non-mechanical work, and is given by:

F(T,V,N)≡U−TS\begin{aligned} \boxed{ F(T, V, N) \equiv U - T S } \end{aligned}

It depends on the temperature TT, volume VV, and particle count NN: these are its natural variables, so an infinitesimal change dH\dd{H} is as follows:

dF=−PdV−SdT+μdN\begin{aligned} \boxed{ \dd{F} = - P \dd{V} - S \dd{T} + \mu \dd{N} } \end{aligned}

The non-natural variables are the entropy SS, pressure PP, and chemical potential μ\mu. These can be recovered by differentiating FF with respect to the natural variables TT, VV, and NN:

S=−(∂F∂T)V,NP=−(∂F∂V)T,Nμ=(∂F∂N)T,V\begin{aligned} \boxed{ S = - \Big( \pdv{F}{T} \Big)_{V,N} \qquad P = - \Big( \pdv{F}{V} \Big)_{T,N} \qquad \mu = \Big( \pdv{F}{N} \Big)_{T,V} } \end{aligned}

Gibbs free energy

The Gibbs free energy GG represents the capacity of a system to do non-mechanical work:

G(T,P,N)≡U+PV−TS\begin{aligned} \boxed{ G(T, P, N) \equiv U + P V - T S } \end{aligned}

It depends on the temperature TT, pressure PP, and particle count NN: they are its natural variables, so an infinitesimal change dG\dd{G} is as follows:

dG=VdP−SdT+μdN\begin{aligned} \boxed{ \dd{G} = V \dd{P} - S \dd{T} + \mu \dd{N} } \end{aligned}

The non-natural variables are the entropy SS, volume VV, and chemical potential μ\mu. These can be recovered by differentiating GG with respect to the natural variables TT, PP, and NN:

S=−(∂G∂T)P,NV=(∂G∂P)T,Nμ=(∂G∂N)T,P\begin{aligned} \boxed{ S = - \Big( \pdv{G}{T} \Big)_{P,N} \qquad V = \Big( \pdv{G}{P} \Big)_{T,N} \qquad \mu = \Big( \pdv{G}{N} \Big)_{T,P} } \end{aligned}

Landau potential

The Landau potential or grand potential Ω\Omega, in units of energy, represents the capacity of a system to do mechanical work, and is given by:

Ω(T,V,μ)≡U−TS−μN\begin{aligned} \boxed{ \Omega(T, V, \mu) \equiv U - T S - \mu N } \end{aligned}

It depends on temperature TT, volume VV, and chemical potential μ\mu: these are its natural variables, so an infinitesimal change dΩ\dd{\Omega} is as follows:

dΩ=−PdV−SdT−Ndμ\begin{aligned} \boxed{ \dd{\Omega} = - P \dd{V} - S \dd{T} - N \dd{\mu} } \end{aligned}

The non-natural variables are entropy SS, pressure PP, and particle count NN. These can be recovered by differentiating Ω\Omega with respect to the natural variables TT, VV, and μ\mu:

S=−(∂Ω∂T)V,μP=−(∂Ω∂V)T,μN=−(∂Ω∂μ)T,V\begin{aligned} \boxed{ S = - \Big( \pdv{\Omega}{T} \Big)_{V,\mu} \qquad P = - \Big( \pdv{\Omega}{V} \Big)_{T,\mu} \qquad N = - \Big( \pdv{\Omega}{\mu} \Big)_{T,V} } \end{aligned}

Entropy

The entropy SS, in units of energy over temperature, is an odd duck, but nevertheless used as a thermodynamic potential, to be maximized instead of minimized. It is given by:

S(U,V,N)≡1TU+PTV−μTN\begin{aligned} \boxed{ S(U, V, N) \equiv \frac{1}{T} U + \frac{P}{T} V - \frac{\mu}{T} N } \end{aligned}

It depends on the internal energy UU, volume VV, and particle count NN: they are its natural variables, so an infinitesimal change dS\dd{S} is as follows:

dS=1TdU+PTdV−μTdN\begin{aligned} \boxed{ \dd{S} = \frac{1}{T} \dd{U} + \frac{P}{T} \dd{V} - \frac{\mu}{T} \dd{N} } \end{aligned}

The non-natural variables are 1/T1/T, P/TP/T, and μ/T\mu/T. These can be recovered by differentiating SS with respect to the natural variables UU, VV, and NN:

1T=(∂S∂U)V,NPT=(∂S∂V)U,NμT=−(∂S∂N)U,V\begin{aligned} \boxed{ \frac{1}{T} = \Big( \pdv{S}{U} \Big)_{V,N} \qquad \frac{P}{T} = \Big( \pdv{S}{V} \Big)_{U,N} \qquad \frac{\mu}{T} = - \Big( \pdv{S}{N} \Big)_{U,V} } \end{aligned}

References

  1. H. Gould, J. Tobochnik, Statistical and thermal physics, 2nd edition, Princeton.