= Maxwell relations
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{title2=$(\partial S/\partial V)_{T,N}=(\partial P/\partial T)_{V,N}$}
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= Maxwell relation
{c}
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Maxwell relations turn <pressure>, <volume (thermodynamics)> and <temperature> measurements into information about <entropy>. For example, the <pressure> rise when heating a rigid sealed vessel determines the <entropy> change when expanding the same sample isothermally:
$$
\left(\frac{\partial S}{\partial V}\right)_{T,N}
=\left(\frac{\partial P}{\partial T}\right)_{V,N}.
$$
The quantities on the two sides describe different experiments, yet both are <mixed partial derivatives> of one <Helmholtz free energy>. This is the useful surprise: a mechanical measurement can reveal a thermal property. Similarly, the <coefficient of thermal expansion> gives $(\partial S/\partial P)_{T,N}=-\alpha V$. https://ocw.mit.edu/courses/5-60-thermodynamics-kinetics-spring-2008/7bc33304ab1db561a64b4a6b12e25476_5_60_lecture13.pdf[MIT 5.60, Lecture 13: entropy derivatives from equations of state]{ref}
For a <simple compressible system>, hold particle number $N$ and composition fixed, and allow only <pressure>-<volume (thermodynamics)> <work>. Write $U$ for <internal energy>, $H=U+PV$ for <enthalpy>, $F=U-TS$ for <Helmholtz free energy>, and $G=H-TS$ for <Gibbs free energy>. Their natural-variable differentials are
$$
\begin{aligned}
dU(S,V,N)&=T\,dS-P\,dV, &dH(S,P,N)&=T\,dS+V\,dP,\\
dF(T,V,N)&=-S\,dT-P\,dV, &dG(T,P,N)&=-S\,dT+V\,dP.
\end{aligned}
$$
Thus the first <partial derivatives>, with the subscript recording the variables kept fixed, are
$$
\begin{aligned}
U_S\big|_{V,N}=T,&\quad U_V\big|_{S,N}=-P, &H_S\big|_{P,N}=T,&\quad H_P\big|_{S,N}=V,\\
F_T\big|_{V,N}=-S,&\quad F_V\big|_{T,N}=-P, &G_T\big|_{P,N}=-S,&\quad G_P\big|_{T,N}=V.
\end{aligned}
$$
The <symmetry of second derivatives> produces all four standard <Maxwell relations>, including their signs:
$$
\begin{aligned}
U:\quad\left(\frac{\partial T}{\partial V}\right)_{S,N}
&=-\left(\frac{\partial P}{\partial S}\right)_{V,N},\\
H:\quad\left(\frac{\partial T}{\partial P}\right)_{S,N}
&=\left(\frac{\partial V}{\partial S}\right)_{P,N},\\
F:\quad\left(\frac{\partial S}{\partial V}\right)_{T,N}
&=\left(\frac{\partial P}{\partial T}\right)_{V,N},\\
G:\quad\left(\frac{\partial S}{\partial P}\right)_{T,N}
&=-\left(\frac{\partial V}{\partial T}\right)_{P,N}.
\end{aligned}
$$
For a brief derivation of the $F$ relation, differentiate $F_T=-S$ with respect to $V$ and $F_V=-P$ with respect to $T$: $-S_V=F_{TV}=F_{VT}=-P_T$. Each other row follows in the same way. Assume continuous second <partial derivatives> of the relevant <thermodynamic potential> in the region considered. At a <phase transition>, the equilibrium potential may lose this smoothness; apply the identities within each smooth <thermodynamic phase>, and handle <latent heat> and <phase coexistence> separately. If particle numbers vary, the potential differentials acquire <chemical potential> terms, so the fixed-$N$ constraints must remain explicit. https://davidtong.org/teaching/statistical-physics/statmechhtml/S4#S4.SS4.SSS2[David Tong, Statistical Physics, section 4.4.2: the four Maxwell relations]{ref}
For an <ideal gas> with fixed $N$, the <ideal gas law> gives $P=Nk_BT/V$. At fixed <volume (thermodynamics)>, heating changes the <pressure> at rate $Nk_B/V$. Consequently an <isothermal expansion> at positive <temperature> has
$$
\Delta S=\int_{V_1}^{V_2}\left(\frac{\partial P}{\partial T}\right)_{V,N}\,dV
=Nk_B\ln\frac{V_2}{V_1}.
$$
For one <mole> with $R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}$ and $V_2=2V_1$, this is $\Delta S=R\ln2\simeq5.763\,\mathrm{J\,K^{-1}}$. At $300\,\mathrm K$, a reversible <isothermal expansion> absorbs $Q=T\Delta S\simeq1728.85\,\mathrm J$ and performs equal <work>, since its <internal energy> is unchanged. The <entropy> difference is a <state function> difference: the same equilibrium endpoints give the same $\Delta S$ in an insulated <free expansion>, although there $Q=0$ and the process is irreversible.
\Image[/maxwell-entropy-pressure-derivatives.png]
{title=<Ideal gas> <entropy> slopes equal measured <pressure> responses to heating}
{description=Left: <entropy> relative to one common reference state $(T_0,V_0)$ for a classical <monatomic gas>, $[S(T,V)-S(T_0,V_0)]/(Nk_B)=\frac32\ln(T/T_0)+\ln(V/V_0)$, with $T_0=300\,\mathrm K$. Each fixed-<temperature> curve gains $\ln2$ on doubling the <volume (thermodynamics)>. Right: both <partial derivatives>, multiplied by $V_0/(Nk_B)$, equal $V_0/V$; circles are independent finite differences of <entropy>, while the line is the <pressure> response.}
The same <Maxwell relations> let thermal measurements determine the <difference between constant-pressure and constant-volume heat capacities>. In a smooth single <thermodynamic phase>,
$$
dS=\frac{C_V}{T}\,dT+\frac{\alpha}{\kappa_T}\,dV
=\frac{C_P}{T}\,dT-\alpha V\,dP,
$$
using <heat capacity at constant volume>, <heat capacity at constant pressure>, <coefficient of thermal expansion> and <isothermal compressibility>. Along fixed <pressure>, $dV=\alpha V\,dT$, so
$$
C_P-C_V=\frac{TV\alpha^2}{\kappa_T}.
$$
Thus measured expansion and compression responses determine a <heat capacity> difference without measuring both <heat capacities> separately. They do not determine the absolute <heat capacities> or absolute <entropy> by themselves; an independent thermal measurement supplies that missing information. For an <ideal gas>, $\alpha=1/T$ and $\kappa_T=1/P$ reduce the result to $C_P-C_V=PV/T=Nk_B$, which is <Mayer's relation>.
When the necessary higher <partial derivatives> exist and commute, differentiating the <Maxwell relations> once more also gives
$$
\left(\frac{\partial C_V}{\partial V}\right)_{T,N}
=T\left(\frac{\partial^2P}{\partial T^2}\right)_{V,N},\qquad
\left(\frac{\partial C_P}{\partial P}\right)_{T,N}
=-T\left(\frac{\partial^2V}{\partial T^2}\right)_{P,N}.
$$
Integrating these slopes determines how a <heat capacity> changes with <volume (thermodynamics)> or <pressure> at fixed <temperature>. For an <ideal gas>, both right-hand sides vanish because the <ideal gas law> is linear in <temperature> at either fixed <volume (thermodynamics)> or fixed <pressure>. Consequently $C_V$ and $C_P$ depend only on <temperature> at fixed $N$, although they need not be constant in <temperature>. https://davidtong.org/teaching/statistical-physics/statmechhtml/S4#S4.SS4.SSS2[Tong, section 4.4.2: heat-capacity changes from thermal pressure and volume curves]{ref}
Historically, <James Clerk Maxwell>'s \i[Theory of Heat], first published in 1871, presents these connections using the geometry of <isothermal processes> and reversible <adiabatic processes>. His fourth edition, page 167, explicitly relates expansion on heating at fixed <pressure> to <heat> emitted during an <isothermal compression>: the modern $G$ relation gives $\delta Q_{\rm rev}=T\,dS=-T\alpha V\,dP$. This explains why a substance with positive <coefficient of thermal expansion> emits <heat> when compressed isothermally, whereas one with negative coefficient absorbs it. https://www.cambridge.org/core/books/theory-of-heat/7052E2DAA4CD862126D5105ABBFFAD0F/listing[Maxwell, Theory of Heat, publication record and chapter IX]{ref} https://en.wikisource.org/wiki/Page:Theory_of_Heat,_James_Clerk_Maxwell,_Fourth_Edition.djvu/183[Maxwell, Theory of Heat, fourth edition, page 167]{ref}
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