2.7: Thermodynamic Potentials

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Thermodynamic systems may do work on their environments. Under certain constraints, the work done may be bounded from above by the change in an appropriately defined thermodynamic potential.

Energy $E$

Suppose we wish to create a thermodynamic system from scratch. Let’s imagine that we create it from scratch in a thermally insulated box of volume $V$. The work we must to to assemble the system is then $\CW=E$ . After we bring all the constituent particles together, pulling them in from infinity (say), the system will have total energy $E$. After we finish, the system may not be in thermal equilibrium. Spontaneous processes will then occur so as to maximize the system’s entropy, but the internal energy remains at $E$.

We have, from the First Law, $dE=\dbar Q-\dbar W$  and combining this with the Second Law in the form $\dbar Q \le T\,dS$ yields $$dE\le T\,dS - \dbar W\quad. \label{dEeqn}$$ Rearranging terms, we have $\dbar W\le T\,dS-dE$ . Hence, the work done by a thermodynamic system under conditions of constant entropy is bounded above by $-dE$, and the maximum $\dbar W$ is achieved for a reversible process. It is sometimes useful to define the quantity $$\dbar W\ns_{free}=\dbar W- p\,dV\ ,$$ which is the differential work done by the system other than that required to change its volume. Then we have $$\dbar W\ns_{free}\le T\,dS - p\,dV - dE \quad,$$ and we conclude for systems at fixed $(S,V)$ that $\dbar W_{free}\le - dE$.

In equilibrium, the equality in Equation [dEeqn] holds, and for single component systems where $\dbar W = p\,dV - \mu\,dN$ we have $E=E(S,V,N)$ with $$T=\pabc{E}{S}{V,N} \qquad,\qquad -p=\pabc{E}{V}{S,N} \qquad,\qquad \mu=\pabc{E}{N}{S,V}\quad .$$ These expressions are easily generalized to multicomponent systems, magnetic systems,

Now consider a single component system at fixed $(S,V,N)$. We conclude that $dE\le 0$ , which says that spontaneous processes in a system with $dS=dV=dN=0$ always lead to a reduction in the internal energy $E$. Therefore, spontaneous processes drive the internal energy $E$ to a minimum in systems at fixed $(S,V,N)$.

Helmholtz free energy $F$

Suppose that when we spontaneously create our system while it is in constant contact with a thermal reservoir at temperature $T$. Then as we create our system, it will absorb heat from the reservoir. Therefore, we don’t have to supply the full internal energy $E$, but rather only $E-Q$, since the system receives heat energy $Q$ from the reservoir. In other words, we must perform work $\CW=E-TS$ to create our system, if it is constantly in equilibrium at temperature $T$. The quantity $E-TS$ is known as the Helmholtz free energy, $F$, which is related to the energy $E$ by a Legendre transformation, $$F=E-TS\ .$$ The general properties of Legendre transformations are discussed in Appendix II, §16.

Again invoking the Second Law, we have $$dF\le -S\,dT-\dbar W\quad. \label{dFeqn}$$ Rearranging terms, we have $\dbar W\le -S\,dT - dF$ , which says that the work done by a thermodynamic system under conditions of constant temperature is bounded above by $-dF$, and the maximum $\dbar W$ is achieved for a reversible process. We also have the general result $$\dbar W\ns_{free}\le -S\,dT - p\,dV - dF \ ,$$ and we conclude, for systems at fixed $(T,V)$, that $\dbar W_{free}\le -dF$.

Under equilibrium conditions, the equality in Equation [dFeqn] holds, and for single component systems where $\dbar W = p\,dV-\mu\,dN$ we have $dF=-S\,dT - p\,dV + \mu\,dN$ . This says that $F=F(T,V,N)$ with $$-S=\pabc{F}{T}{V,N} \qquad,\qquad -p=\pabc{F}{V}{T,N} \qquad,\qquad \mu=\pabc{F}{N}{T,V}\quad .$$ For spontaneous processes, $dF \le -S\,dT - p\,dV + \mu\,dN$ says that spontaneous processes drive the Helmholtz free energy $F$ to a minimum in systems at fixed $(T,V,N)$.

Enthalpy $\CH$

Suppose that when we spontaneously create our system while it is thermally insulated, but in constant mechanical contact with a ‘volume bath’ at pressure $p$. For example, we could create our system inside a thermally insulated chamber with one movable wall where the external pressure is fixed at $p$. Thus, when creating the system, in addition to the system’s internal energy $E$, we must also perform work $pV$ in order to make room for it. In other words, we must perform work $\CW=E+pV$. The quantity $E+pV$ is known as the enthalpy, $\CH$. (We use the calligraphic font for $\CH$ for enthalpy to avoid confusing it with magnetic field, $H$.) The enthalpy is obtained from the energy via a different Legendre transformation than that used to obtain the Helmholtz free energy $F$ , $$\CH=E+pV \quad.$$

Again invoking the Second Law, we have $$d\CH\le T\,dS -\dbar W + p\,dV + V dp\quad,$$ hence with $\dbar W\ns_{free}=\dbar W - p\,dV$, we have in general $$\dbar W\ns_{free} \le T\,dS + V dp -d\CH\ ,$$ and we conclude, for systems at fixed $(S,p)$, that $\dbar W_{free}\le -d\CH$.

In equilibrium, for single component systems, $$d\CH=T\,dS + V dp + \mu\,dN\ ,$$ which says $\CH=\CH(S,p,N)$, with $$T=\pabc{\CH}{S}{p,N} \qquad,\qquad V=\pabc{\CH}{p}{S,N} \qquad,\qquad \mu=\pabc{\CH}{N}{S,p}\quad.$$ For spontaneous processes, $d\CH\le T\,dS + V dp +\mu\,dN$, which says that spontaneous processes drive the enthalpy $\CH$ to a minimum in systems at fixed $(S,p,N)$.

Gibbs free energy $G$

If we create a thermodynamic system at conditions of constant temperature $T$ and constant pressure $p$, then it absorbs heat energy $Q=TS$ from the reservoir and we must expend work energy $pV$ in order to make room for it. Thus, the total amount of work we must do in assembling our system is $\CW=E-TS+pV$. This is the Gibbs free energy, $G$. The Gibbs free energy is obtained from $E$ after two Legendre transformations, $$G=E-TS+pV$$ Note that $G=F+pV=\CH-TS$. The Second Law says that $$dG\le -S\,dT + V dp + p\,dV -\dbar W\quad,$$ which we may rearrange as $\dbar W\ns_{free}\le -S\,dT + V dp - dG$ . Accordingly, we conclude, for systems at fixed $(T,p)$, that $\dbar W_{free}\le - dG$.

For equilibrium one-component systems, the differential of $G$ is $$dG=-S\,dT + V dp + \mu\,dN\ ,$$ therefore $G=G(T,p,N)$, with $$-S=\pabc{G}{T}{p,N} \qquad,\qquad V=\pabc{G}{p}{T,N} \qquad,\qquad \mu=\pabc{G}{N}{T,p}\quad.$$ Recall that Euler’s theorem for single component systems requires $E=TS-pV+\mu N$  which says $G=\mu N$, Thus, the chemical potential $\mu$ is the Gibbs free energy per particle. For spontaneous processes, $dG\le -S\,dT + Vdp + \mu\,dN$ , hence spontaneous processes drive the Gibbs free energy $G$ to a minimum in systems at fixed $(T,p,N)$.

Grand potential $\Omega$

The grand potential, sometimes called the Landau free energy, is defined by $$\Omega=E-TS-\mu N\ .$$ Under equilibrium conditions, its differential is $$d\Omega=-S\,dT -p\,dV - N\,d\mu\ ,$$ hence $$-S=\pabc{\Omega}{T}{V,\mu} \qquad,\qquad -p=\pabc{\Omega}{V}{T,\mu} \qquad,\qquad -N=\pabc{\Omega}{\mu}{T,V}\ .$$ Again invoking Equation [ETS], we find $\Omega = -pV$, which says that the pressure is the negative of the grand potential per unit volume.

The Second Law tells us $$d\Omega \le - \dbar W - S\,dT - \mu\,dN - N\,d\mu\ ,$$ hence $$\dbar {\widetilde W}\ns_{free}\equiv \dbar W\ns_{free}+\mu\,dN\le -S\,dT -p\,dV -N\,d\mu -d\Omega\ .$$ We conclude, for systems at fixed $(T,V,\mu)$, that $\dbar {\widetilde W}_{free}\le - d\Omega$.