Maxwell relations

Maxwell relations are based on simple partial differentiation rules, in particular the total differential of a function and the symmetry of evaluating second order partial derivatives.

Derivation

Derivation of the Maxwell relation can be deduced from the differential forms of the thermodynamic potentials: The differential form of internal energy U is ${\displaystyle {\begin{aligned}dU&=TdS-PdV\\\end{aligned}}\,\!}$ This equation resembles total differentials of the form ${\displaystyle dz=\left({\frac {\partial z}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial z}{\partial y}}\right)_{x}\!dy}$ It can be shown, for any equation of the form, ${\displaystyle dz=Mdx+Ndy\,}$ that ${\displaystyle M=\left({\frac {\partial z}{\partial x}}\right)_{y},\quad N=\left({\frac {\partial z}{\partial y}}\right)_{x}}$ Consider, the equation ${\displaystyle dU=TdS-PdV\,}$. We can now immediately see that ${\displaystyle T=\left({\frac {\partial U}{\partial S}}\right)_{V},\quad -P=\left({\frac {\partial U}{\partial V}}\right)_{S}}$ Since we also know that for functions with continuous second derivatives, the mixed partial derivatives are identical (Symmetry of second derivatives), that is, that ${\displaystyle {\frac {\partial }{\partial y}}\left({\frac {\partial z}{\partial x}}\right)_{y}={\frac {\partial }{\partial x}}\left({\frac {\partial z}{\partial y}}\right)_{x}={\frac {\partial ^{2}z}{\partial y\partial x}}={\frac {\partial ^{2}z}{\partial x\partial y}}}$ we therefore can see that ${\displaystyle {\frac {\partial }{\partial V}}\left({\frac {\partial U}{\partial S}}\right)_{V}={\frac {\partial }{\partial S}}\left({\frac {\partial U}{\partial V}}\right)_{S}}$ and therefore that ${\displaystyle \left({\frac {\partial T}{\partial V}}\right)_{S}=-\left({\frac {\partial P}{\partial S}}\right)_{V}}$ Derivation of Maxwell Relation from Helmholtz Free energy The differential form of Helmholtz free energy is ${\displaystyle {\begin{aligned}dF&=-SdT-PdV\\\end{aligned}}\,\!}$ ${\displaystyle -S=\left({\frac {\partial F}{\partial T}}\right)_{V},\quad -P=\left({\frac {\partial F}{\partial V}}\right)_{T}}$ From symmetry of second derivatives ${\displaystyle {\frac {\partial }{\partial V}}\left({\frac {\partial F}{\partial T}}\right)_{V}={\frac {\partial }{\partial T}}\left({\frac {\partial F}{\partial V}}\right)_{T}}$ and therefore that ${\displaystyle \left({\frac {\partial S}{\partial V}}\right)_{T}=\left({\frac {\partial P}{\partial T}}\right)_{V}}$ The other two Maxwell relations can be derived from differential form of enthalpy ${\displaystyle {\begin{aligned}dH&=TdS+VdP\\\end{aligned}}\,\!}$ and the differential form of Gibbs free energy ${\displaystyle {\begin{aligned}dG&=VdP-SdT\\\end{aligned}}\,\!}$ in a similar way. So all Maxwell Relationships above follow from one of the Gibbs equations.

Extended derivation

Combined form first and second law of thermodynamics, ${\displaystyle TdS=dU+PdV}$ (Eq.1) U, S, and V are state functions. Let, ${\displaystyle U=U(x,y)}$ ${\displaystyle S=S(x,y)}$ ${\displaystyle V=V(x,y)}$ ${\displaystyle dU=\left({\frac {\partial U}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial U}{\partial y}}\right)_{x}\!dy}$ ${\displaystyle dS=\left({\frac {\partial S}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial S}{\partial y}}\right)_{x}\!dy}$ ${\displaystyle dV=\left({\frac {\partial V}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial V}{\partial y}}\right)_{x}\!dy}$ Substitute them in Eq.1 and one gets, ${\displaystyle T\left({\frac {\partial S}{\partial x}}\right)_{y}\!dx+T\left({\frac {\partial S}{\partial y}}\right)_{x}\!dy=\left({\frac {\partial U}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial U}{\partial y}}\right)_{x}\!dy+P\left({\frac {\partial V}{\partial x}}\right)_{y}\!dx+P\left({\frac {\partial V}{\partial y}}\right)_{x}\!dy}$ And also written as, ${\displaystyle \left({\frac {\partial U}{\partial x}}\right)_{y}\!dx+\left({\frac {\partial U}{\partial y}}\right)_{x}\!dy=T\left({\frac {\partial S}{\partial x}}\right)_{y}\!dx+T\left({\frac {\partial S}{\partial y}}\right)_{x}\!dy-P\left({\frac {\partial V}{\partial x}}\right)_{y}\!dx-P\left({\frac {\partial V}{\partial y}}\right)_{x}\!dy}$ comparing the coefficient of dx and dy, one gets ${\displaystyle \left({\frac {\partial U}{\partial x}}\right)_{y}=T\left({\frac {\partial S}{\partial x}}\right)_{y}-P\left({\frac {\partial V}{\partial x}}\right)_{y}}$ ${\displaystyle \left({\frac {\partial U}{\partial y}}\right)_{x}=T\left({\frac {\partial S}{\partial y}}\right)_{x}-P\left({\frac {\partial V}{\partial y}}\right)_{x}}$ Differentiating above equations by y, x respectively ${\displaystyle \left({\frac {\partial ^{2}U}{\partial y\partial x}}\right)=\left({\frac {\partial T}{\partial y}}\right)_{x}\left({\frac {\partial S}{\partial x}}\right)_{y}+T\left({\frac {\partial ^{2}S}{\partial y\partial x}}\right)-\left({\frac {\partial P}{\partial y}}\right)_{x}\left({\frac {\partial V}{\partial x}}\right)_{y}-P\left({\frac {\partial ^{2}V}{\partial y\partial x}}\right)}$ (Eq.2) and ${\displaystyle \left({\frac {\partial ^{2}U}{\partial x\partial y}}\right)=\left({\frac {\partial T}{\partial x}}\right)_{y}\left({\frac {\partial S}{\partial y}}\right)_{x}+T\left({\frac {\partial ^{2}S}{\partial x\partial y}}\right)-\left({\frac {\partial P}{\partial x}}\right)_{y}\left({\frac {\partial V}{\partial y}}\right)_{x}-P\left({\frac {\partial ^{2}V}{\partial x\partial y}}\right)}$ (Eq.3) U, S, and V are exact differentials, therefore, ${\displaystyle \left({\frac {\partial ^{2}U}{\partial y\partial x}}\right)=\left({\frac {\partial ^{2}U}{\partial x\partial y}}\right)}$ ${\displaystyle \left({\frac {\partial ^{2}S}{\partial y\partial x}}\right)=\left({\frac {\partial ^{2}S}{\partial x\partial y}}\right)}$ ${\displaystyle \left({\frac {\partial ^{2}V}{\partial y\partial x}}\right)=\left({\frac {\partial ^{2}V}{\partial x\partial y}}\right)}$ Subtract eqn(2) and (3) and one gets ${\displaystyle \left({\frac {\partial T}{\partial y}}\right)_{x}\left({\frac {\partial S}{\partial x}}\right)_{y}-\left({\frac {\partial P}{\partial y}}\right)_{x}\left({\frac {\partial V}{\partial x}}\right)_{y}=\left({\frac {\partial T}{\partial x}}\right)_{y}\left({\frac {\partial S}{\partial y}}\right)_{x}-\left({\frac {\partial P}{\partial x}}\right)_{y}\left({\frac {\partial V}{\partial y}}\right)_{x}}$ Note: The above is called the general expression for Maxwell's thermodynamical relation. Maxwell's first relation Allow x = S and y = V and one gets ${\displaystyle \left({\frac {\partial T}{\partial V}}\right)_{S}=-\left({\frac {\partial P}{\partial S}}\right)_{V}}$ Maxwell's second relation Allow x = T and y = V and one gets ${\displaystyle \left({\frac {\partial S}{\partial V}}\right)_{T}=\left({\frac {\partial P}{\partial T}}\right)_{V}}$ Maxwell's third relation Allow x = S and y = P and one gets ${\displaystyle \left({\frac {\partial T}{\partial P}}\right)_{S}=\left({\frac {\partial V}{\partial S}}\right)_{P}}$ Maxwell's fourth relation Allow x = T and y = P and one gets ${\displaystyle \left({\frac {\partial S}{\partial P}}\right)_{T}=-\left({\frac {\partial V}{\partial T}}\right)_{P}}$ Maxwell's fifth relation Allow x = P and y = V ${\displaystyle \left({\frac {\partial T}{\partial P}}\right)_{V}\left({\frac {\partial S}{\partial V}}\right)_{P}}$${\displaystyle -\left({\frac {\partial T}{\partial V}}\right)_{P}\left({\frac {\partial S}{\partial P}}\right)_{V}}$ = 1 Maxwell's sixth relation Allow x = T and y = S and one gets ${\displaystyle \left({\frac {\partial P}{\partial T}}\right)_{S}\left({\frac {\partial V}{\partial S}}\right)_{T}-\left({\frac {\partial P}{\partial S}}\right)_{T}\left({\frac {\partial V}{\partial T}}\right)_{S}}$ = 1