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Maxwell relations turn pressure, volume 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: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 $(\partial S/\partial P{)}_{T,N}=-\alpha V$. ^[ref]

For a simple compressible system, hold particle number $N$ and composition fixed, and allow only pressure-volume 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 areThus the first partial derivatives, with the subscript recording the variables kept fixed, areThe symmetry of second derivatives produces all four standard Maxwell relations, including their signs: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. ^[ref]

For an ideal gas with fixed $N$, the ideal gas law gives $P=N{k}_{B}T/V$. At fixed volume, heating changes the pressure at rate $N{k}_B/V$. Consequently an isothermal expansion at positive temperature hasFor one mole with $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $V_2=2V_1$, this is $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. At $300\mathrm{K}$, a reversible isothermal expansion absorbs $Q=T\Delta S\simeq 1728.85\mathrm{J}$ 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.

The same Maxwell relations let thermal measurements determine the difference between constant-pressure and constant-volume heat capacities. In a smooth single thermodynamic phase,using heat capacity at constant volume, heat capacity at constant pressure, coefficient of thermal expansion and isothermal compressibility. Along fixed pressure, $dV=\alpha VdT$, soThus 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 $C_P-C_V=PV/T=N{k}_B$, which is Mayer's relation.

When the necessary higher partial derivatives exist and commute, differentiating the Maxwell relations once more also givesIntegrating these slopes determines how a heat capacity changes with volume 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 or fixed pressure. Consequently $C_V$ and $C_P$ depend only on temperature at fixed $N$, although they need not be constant in temperature. ^[ref]

Historically, James Clerk Maxwell's 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 $\delta Q_{\mathrm{rev}}=TdS=-T\alpha VdP$. This explains why a substance with positive coefficient of thermal expansion emits heat when compressed isothermally, whereas one with negative coefficient absorbs it. ^{[ref] [ref]}

In a steady axisymmetric magnetohydrodynamic wind, axisymmetric pressure and Newtonian gravity exert no azimuthal torque. The azimuthal momentum equation is $\rho {\mathbf{u}}_p\cdot \nabla (s{u}_{\varphi })={\mu }_0^{-1}{\mathbf{B}}_p\cdot \nabla (s{B}_{\varphi })$. Mass conservation and the solenoidal vector field condition imply $\nabla \cdot {\mathbf{F}}_J=0$. With magnetohydrodynamic mass loading $\rho {\mathbf{u}}_p=\kappa {\mathbf{B}}_p$, the transported angular momentum per unit poloidal magnetic flux is $\ell =\kappa s{u}_{\varphi }-s{B}_{\varphi }/{\mu }_0$. It is constant on connected regular poloidal flux surfaces. The usual specific magnetohydrodynamic angular-momentum invariant is $L=\ell /\kappa$, provided $\kappa \ne 0$; confusing these two normalizations loses a factor of mass loading in the wind torque.

Mayer's relation states that the difference between heat capacity at constant pressure and heat capacity at constant volume for a fixed-composition ideal gas isHere both capacities are total heat capacities in $\mathrm{J}{\mathrm{K}}^{-1}$, $n$ is amount of substance, $N$ is the number of particles, $R$ is the molar gas constant and $k_B$ is the Boltzmann constant. In other normalizations the same statement isThe first is a difference of molar heat capacities, in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$; the second uses specific heat capacities, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Writing $C_p-C_V=R$ with total capacities would omit the sample's amount of substance. The relation holds pointwise even when the capacities depend on temperature. ^[ref]

The proof outline is short: enthalpy is $H=U+pV=U+nRT$, and internal energy of an ideal gas is independent of volume at fixed temperature. Differentiating at fixed composition and amount of substance therefore gives $C_p=dH/dT=dU/dT+nR=C_V+nR$. It uses the ideal gas law, rather than assuming any particular number of molecular degrees of freedom. At fixed pressure, the additional heat supplies expansion work, $pdV=nRdT$. The two processes can have the same temperature endpoints and equal $\Delta U$ while requiring different heat inputs.

For one mole of the rigid-rotation diatomic ideal gas model, $C_V=5R/2$ and $C_p=7R/2$. Heating by $10\mathrm{K}$ with $R\simeq 8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ gives $Q_V\simeq 207.86\mathrm{J}$ in a rigid sealed vessel and $Q_p\simeq 291.01\mathrm{J}$ under a frictionless piston at fixed pressure. Both gain $207.86\mathrm{J}$ of internal energy, while the piston delivers approximately $83.14\mathrm{J}$ of work. This is a calculated model budget. If the capacities vary with temperature, the corresponding finite statement is stillunder the fixed-pressure and fixed-volume heating assumptions in heat capacity. Their final volume values generally differ. Combining Mayer's relation with heat capacity ratio also gives $C_V=nR/(\gamma -1)$ and $C_p=\gamma nR/(\gamma -1)$ locally; constant values cannot be extrapolated across intervals in which $\gamma (T)$ changes.

For a smooth stable single thermodynamic phase of a general simple compressible system, the broader Maxwell relation identity iswhere $\alpha$ is the coefficient of thermal expansion and ${\kappa }_T$ the isothermal compressibility, with composition fixed throughout. Substituting the ideal gas law gives $nR$; a real gas need not do so. This distinction matters near a phase transition or when interactions appreciably change the equation of state. It explains why Mayer's relation is useful: thermal measurements under different constraints become linked to a mechanical response, while the difference retains a particularly simple value for an ideal gas.^[ref]

Mean curvature measures the average signed bending of a smooth surface. Unlike Gaussian curvature, it adds the two principal curvatures rather than multiplying them. A circular cylinder therefore has nonzero mean curvature despite zero Gaussian curvature, whereas a catenoid has zero mean curvature despite negative Gaussian curvature. The cancellation on a catenoid explains how a curved minimal surface can be stationary for surface area.

Choose a unit normal $N$ and the shape operator $S=-dN$, as in the second fundamental form convention $II(v,w)=⟨Sv,w⟩$. For a smooth surface in Euclidean space of dimension three, define the signed scalar mean curvature byHere $E,F,G$ are the coefficients of the first fundamental form, and $e,f,g$ those of the second fundamental form. The factor $1/2$ makes $H$ an average; the scalar trace convention instead calls $k_1+k_2=2H$ the mean curvature. Authors who use $S=+dN$ reverse both signs. A formula involving mean curvature is therefore incomplete unless its shape operator, unit normal, and averaging convention are specified.

In orientation reversal of surface curvature, both principal curvatures and $H$ change sign, while Gaussian curvature and the mean curvature vector do not. The units of mean curvature are inverse length, compared with inverse squared length for Gaussian curvature. The distinction matters in soap film pressure balance: pressure couples to the sum of the signed principal curvatures.

Mechanical equilibrium requires force balance for the mechanical displacements that are allowed. Two uniform fluids separated by a freely moving, unloaded, frictionless piston with equal face areas require equal pressures. A supported piston can sustain a pressure difference, and a fluid in gravity instead obeys hydrostatic equilibrium. For an upward height coordinate $z$, that balance is $dp/dz=-\rho g$, where $\rho$ is mass density and $g$ is the downward gravitational acceleration. Mechanical equilibrium alone does not establish thermal equilibrium or chemical equilibrium.

A membrane osmometer connects osmosis to a visible liquid level or a measured balancing pressure. In a teaching version, a solute-containing pouch is joined to a narrow vertical tube and submerged in external water. Entry increases the enclosed volume and raises the meniscus. Both the column and any stretched pouch can contribute an opposing pressure.

The University of Toronto's Osmosis (Brine) by University of Toronto Department of Physics demonstration (activity Me-012b, Description, steps 1–14) specifies soaking a dialysis membrane tube, tying one end closed and fastening the other around a thistle tube's narrow end. A stand holds the tube. Colored brine is cooled to room temperature and injected using a syringe to reduce bubbles; the pouch is completely immersed in a $1$ or $2\mathrm{L}$ beaker of distilled water. A taped marker records the initial level; subsequent meniscus height is the readout. The protocol proposes another pouch containing distilled water as a comparison, and its older instructions explicitly check seals for leaks.^[ref]

A rise is consistent with net water entry. To infer osmotic pressure quantitatively, use a retained solute and establish a stable equilibrium after accounting for dilution. If both free surfaces are at the same atmospheric pressure, the liquids have approximately equal mass densities $\rho$, and the pouch adds negligible mechanical load, the hydrostatic pressure difference at membrane height is approximately $\Delta P=\rho gh$, where $h$ is the meniscus height above the external surface and $g$ is gravitational acceleration. This is the liquid-column inference, not a pressure calibration supplied by the Toronto demonstration. For unequal mass densities, compare the two column pressures separately.

Useful additional controls are matched initial levels and geometry, constant temperature, replicate runs, and monitoring external solute to test retention. These are proposed checks, not reported Toronto results. Leaks, trapped bubbles, surface tension in the narrow tube, evaporation and membrane stretching can alter the level. In particular, a dialysis membrane may pass dissolved salt: the brine setup illustrates transient movement but cannot be assumed to retain every dissolved ion. Color improves visibility without proving salt retention. A stationary level can reflect weakening concentration differences or leakage rather than osmotic thermodynamic equilibrium; a rising level alone supplies neither $n$ nor a verified value of $nRT/V$.

Membrane transport is the passage of substances across a cell membrane. For the small solutes considered here, distinguish the available route from its driving force: a lipid bilayer, ion channel, or carrier protein determines how passage can occur, while differences in chemical potential and membrane potential determine which direction is energetically downhill. A large driving force cannot make an unavailable route conduct. See Alberts et al.: principles of membrane transport.

For an ideal dilute chemical solution at uniform temperature and pressure, with the same reference molar chemical potential on both sides, the molar electrochemical potential change for transferring a solute from compartment A to B isHere $c_{\mathrm{A}},c_{\mathrm{B}}$ are concentrations, $T$ is absolute temperature, $R$ is the molar gas constant, $z$ is signed ion charge number, $F$ is the Faraday constant, and $V_{\mathrm{B}}-V_{\mathrm{A}}$ is the electric potential difference. The natural logarithm contains a dimensionless ratio. The electrical term vanishes for an uncharged solute. More generally, use thermodynamic activities in place of the ideal concentration ratio. This combines the ideal-solution chemical potential difference with electrical energy per mole; see IUPAC Gold Book: electrochemical potential. For a biological cell, taking B inside and A outside makes the voltage term use the existing membrane potential convention.

Mixing calorimetry determines heat capacities by bringing bodies at different temperatures into thermal contact. The striking feature is that the final temperature is a weighted arithmetic mean with heat capacities as weights, rather than weights given by mass. The vessel participates in the exchange even though it is not the sample being studied.

For a hot nonreacting sample of mass $m_x$, initial temperature $T_x$ and unknown specific heat capacity at constant pressure $c_x$, let receiving water have mass $m_w$, initial temperature $T_w$ and known specific heat capacity at constant pressure $c_w$. Let the vessel and immersed parts start at $T_w$ and have combined heat capacity $C_{\mathrm{cal}}$, excluding that water. Assume a common final temperature $T_f$, approximately constant pressure and heat capacities, no phase transition, no chemical reaction and negligible external nonexpansion work. For signed outward heat loss $Q_{\mathrm{loss},\mathrm{out}}$, the combined enthalpy balance isHenceThe loss is positive if the room receives net heat, negative if the room supplies it. If the vessel starts at a different temperature $T_{\mathrm{cal},i}$, replace its warming term by $C_{\mathrm{cal}}(T_f-T_{\mathrm{cal},i})$; do not assume the vessel shared the water's initial temperature. For variable material properties, replace each $mc\Delta T$ by the corresponding integral $m\int c_p(T)dT$ and each vessel term by $\int C_{\mathrm{cal}}(T)dT$. The inferred constant $c_x$ then represents an average over the sample's cooling interval.^[ref]

For a synthetic no-loss example, mix $0.100\mathrm{kg}$ of hot water at $80{}^{\circ }\mathrm{C}$ with $0.200\mathrm{kg}$ of cold water at $20{}^{\circ }\mathrm{C}$ in a vessel initially at $20{}^{\circ }\mathrm{C}$. Stipulate $c_w=4180\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $C_{\mathrm{cal}}=100\mathrm{J}{\mathrm{K}}^{-1}$. The enthalpy balance givesThe hot water releases $17.337\mathrm{kJ}$: the cold water receives $15.485\mathrm{kJ}$ and the vessel $1.852\mathrm{kJ}$. Omitting the vessel predicts $40{}^{\circ }\mathrm{C}$, despite perfect insulation. These are computed numbers from stipulated properties, not laboratory observations.

For an unknown-sample example using the same cold water and vessel, stipulate $m_x=0.100\mathrm{kg}$, $T_x=80{}^{\circ }\mathrm{C}$ and $c_x=450\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With no loss, $T_f\approx 22.752{}^{\circ }\mathrm{C}$. Inverting the full balance returns $450\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; ignoring vessel storage returns about $402\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With a positive outward loss, ignoring that loss also biases the inferred $c_x$ downward under these hot-sample conditions. A small water temperature rise makes sensor offsets and loss estimates relatively influential; propagation of uncertainty must include the measured temperatures, masses and calibrated vessel term.

Molar enthalpy is enthalpy per amount of substance $n$, measured in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. With $u_{\mathrm{m}}=U/n$ and molar volume $v_{\mathrm{m}}=V/n$, $h_{\mathrm{m}}=u_{\mathrm{m}}+p{v}_{\mathrm{m}}$. Its fixed-pressure temperature derivative is the corresponding molar heat capacity $C_{p,\mathrm{m}}$, at fixed composition.

For a local illustrative ice–water melting calculation, stipulate $T_m=273.15\mathrm{K}$, latent heat of fusion $L=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$, solid mass density ${\rho }_s=917\mathrm{kg}{\mathrm{m}}^{-3}$, and liquid mass density ${\rho }_l=1000\mathrm{kg}{\mathrm{m}}^{-3}$. These are explicitly stipulated rounded inputs, not measurements made here. The specific volume change on melting isThe common-pressure melting-point slope therefore givesA small $1\mathrm{MPa}$ increase in the common pressure thus lowers the melting temperature by about $0.074\mathrm{K}$ in this local model. Positive latent heat does not force a positive phase coexistence curve slope: ice contracts on conversion to liquid water. Increasing pressure favors the smaller specific volume. This calculation compares equilibrium states of the same two thermodynamic phases; extending its tangent to large pressures would require updated mass densities, latent heat of fusion, and the relevant ice phase diagram.