22 Jul 2025

Thermodynamic Potentials and Maxwell Relations

Natural variables, equilibrium potentials, derivation of the four Maxwell relations, and useful differential identities.

bsc semester-iv mj-6 heat-and-thermodynamics thermodynamic-potentials maxwell-relations

Consider a closed, fixed-composition, simple compressible system with only $pV$ work. Combining the first and second laws for a reversible differential change gives the fundamental relation

\[\boxed{dU=T\,dS-p\,dV}.\]

Because $U$ is a state function, this relation remains a property identity between neighbouring equilibrium states even if the actual process joining them is irreversible.

Four thermodynamic potentials

Internal energy has natural variables $(S,V)$:

\[U=U(S,V),\qquad dU=T\,dS-p\,dV.\]

The enthalpy $H=U+pV$ has natural variables $(S,p)$:

\[dH=dU+p\,dV+V\,dp =\boxed{T\,dS+V\,dp}.\]

The Helmholtz free energy $F=U-TS$ has natural variables $(T,V)$:

\[dF=dU-T\,dS-S\,dT =\boxed{-S\,dT-p\,dV}.\]

The Gibbs free energy $G=H-TS=U+pV-TS$ has natural variables $(T,p)$:

\[dG=dH-T\,dS-S\,dT =\boxed{-S\,dT+V\,dp}.\]

For a stable closed system, equilibrium minimizes the appropriate potential under the imposed constraints: $U$ at fixed $(S,V)$, $H$ at fixed $(S,p)$, $F$ at fixed $(T,V)$, and $G$ at fixed $(T,p)$.

Derivation of Maxwell relations

For any state function $X(x,y)$ with continuous second derivatives,

\[\frac{\partial}{\partial y} \left(\frac{\partial X}{\partial x}\right)_y = \frac{\partial}{\partial x} \left(\frac{\partial X}{\partial y}\right)_x.\]

Apply this equality to each potential. From $U(S,V)$,

\[T=\left(\frac{\partial U}{\partial S}\right)_V, \qquad -p=\left(\frac{\partial U}{\partial V}\right)_S,\]

so

\[\boxed{\left(\frac{\partial T}{\partial V}\right)_S =-\left(\frac{\partial p}{\partial S}\right)_V}.\]

From $H(S,p)$,

\[\boxed{\left(\frac{\partial T}{\partial p}\right)_S =\left(\frac{\partial V}{\partial S}\right)_p}.\]

From $F(T,V)$,

\[\boxed{\left(\frac{\partial S}{\partial V}\right)_T =\left(\frac{\partial p}{\partial T}\right)_V}.\]

From $G(T,p)$,

\[\boxed{\left(\frac{\partial S}{\partial p}\right)_T =-\left(\frac{\partial V}{\partial T}\right)_p}.\]

These relations replace difficult entropy derivatives by measurable $p$-$V$-$T$ derivatives.

Applications to $U$ and $H$

Write entropy as $S(T,V)$. Since

\[\left(\frac{\partial S}{\partial T}\right)_V=\frac{C_V}{T}, \qquad \left(\frac{\partial S}{\partial V}\right)_T =\left(\frac{\partial p}{\partial T}\right)_V,\]

we have

\[dS=\frac{C_V}{T}\,dT +\left(\frac{\partial p}{\partial T}\right)_VdV.\]

Substitution into $dU=T\,dS-p\,dV$ yields

\[\boxed{dU=C_V\,dT+ \left[T\left(\frac{\partial p}{\partial T}\right)_V-p\right]dV}.\]

Thus

\[\boxed{\left(\frac{\partial U}{\partial V}\right)_T =T\left(\frac{\partial p}{\partial T}\right)_V-p}.\]

Similarly, with $S(T,p)$ and the fourth Maxwell relation,

\[\boxed{dH=C_P\,dT+ \left[V-T\left(\frac{\partial V}{\partial T}\right)_p\right]dp}.\]

For an ideal gas, $(\partial p/\partial T)_V=p/T$, so $(\partial U/\partial V)_T=0$ and $U=U(T)$. Also $(\partial V/\partial T)_p=V/T$, so $(\partial H/\partial p)_T=0$ and $H=H(T)$.