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The Gibbs free energy iswhere $H$ is enthalpy. For a closed system of fixed composition, with only pressure-volume work, spontaneous relaxation at fixed temperature and pressure cannot increase $G$; stable thermodynamic equilibrium minimizes it over the allowed states. A loaded piston can move while maintaining the imposed pressure, so the enthalpy term includes the displacement cost. Along a smooth equilibrium branch of fixed composition,Thus its temperature slope at fixed pressure is $-S$, while its pressure slope at fixed temperature is $V$. See MIT 5.60, Lecture 13, Gibbs free energy and its derivatives.

Consider two illustrative thermodynamic phases, A and B, of the same substance at one fixed pressure. Write B minus A, using quantities per mole:Treat these differences as approximately constant over a limited temperature window. ThenAt $280\mathrm{K}$, $\Delta G=+0.8\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, so A is preferred. At $320\mathrm{K}$, $\Delta G=-0.8\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, so B is preferred. At $300\mathrm{K}$, both have equal molar Gibbs free energy, permitting phase coexistence in this bulk model. B's greater entropy outweighs its greater enthalpy above the crossing.

Each plotted point compares alternatives at the same temperature; moving along the curve changes the imposed temperature. The model predicts preference, not instantaneous conversion: nucleation barriers can sustain metastability. Real enthalpy and entropy differences generally vary with temperature through heat capacities, bending this line outside the approximation.

At fixed pressure, the first law of thermodynamics gives $Q=\Delta H$ under the stated work restriction, so the same thermal reservoir accounting yields $\Delta S_{\mathrm{total}}=-\Delta G/T\ge 0$. Variable composition restores ${\sum }_i{\mu }_{i}d{n}_i$ to $dG$; constrained Gibbs minimization for chemical equilibrium handles those permitted changes.

The historical foundation is Gibbs's ^[ref]. Its equations (87) and (91) introduce the combinations now called Helmholtz free energy and Gibbs free energy, connecting entropy criteria, chemical potentials and phase coexistence within one framework.

For a single-phase ideal gas mixture with molar amounts $N_i$, common temperature $T$, and pressure $P$, extensivity and ideal mixing giveHere $N={\sum }_i{N}_i$, $x_i$ is the mole fraction, $\mathcal{R}$ is the molar gas constant, and $P^{\circ }$ is the standard pressure. Integrating $\partial {\mu }_i/\partial P=\mathcal{R}T/P$ and including the ideal mixing entropy proves ${\mu }_i={\mu }_i^{\circ }+\mathcal{R}T\log (x_{i}P/P^{\circ })$. Reactions can change $N$, so fractions alone do not set the extensive Gibbs free energy. Its composition Hessian is $\mathcal{R}T({\delta }_{ij}/N_i-1/N)$; it is positive semidefinite by the weighted Cauchy-Schwarz inequality.

For a monatomic gas obeying the ideal gas law, the specific internal energy is $e=3P/(2\rho )$. With no stellar nuclear fusion heating, the first law of thermodynamics makes the effective stellar energy-generation rate $ϵ=-{\partial }_{t}e-P{\partial }_t(1/\rho )$, with the derivatives taken at fixed enclosed mass. Differentiating gives $ϵ=-3{\partial }_{t}P/(2\rho )+5P{\partial }_t\rho /(2{\rho }^2)$. In a fixed-mass stellar homology Kelvin-Helmholtz contraction, $\rho \propto R^{-3}$ and $P\propto R^{-4}$ at each mass label. Hence $ϵ=-3P\dot{R}/(2\rho R)>0$ when $\dot{R}<0$. This includes internal energy storage as well as pressure work; using the gravitational potential-energy decrease alone would overestimate the outgoing luminosity.

A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

A hard-sphere gas is a classical model of identical impenetrable spheres, each of diameter $\sigma$ and radius $\sigma /2$. Its pair potential is $u(r)=+\infty$ for centre separation $r<\sigma$ and $u(r)=0$ for $r\ge \sigma$; the convention at exact contact has no effect on the integral. It isolates the effect of particle size without an attractive potential energy.

The upper-left panel is a cross-section of a three-dimensional forbidden region: the second centre cannot enter a sphere of radius $\sigma$ around the first centre. The particles themselves have radius $\sigma /2$. Consequently, with particle volume $v_{\mathrm{s}}$ and pair excluded volume $v_{\mathrm{ex}}$,The last equality follows directly from the Mayer f-function: $f=-1$ throughout the forbidden volume and zero outside it. Confusing particle radius with contact separation changes the answer by a factor of eight; omitting the unordered-pair factor changes it by a further factor of two.

Define the packing fraction $\eta =n{v}_{\mathrm{s}}$, where $n$ is number density. Then the virial expansion beginsFor example, at the toy value $\eta =0.01$, the leading correction predicts $Z\approx 1.04$; the omitted terms still matter if accuracy beyond this approximation is needed. The classical second virial coefficient is independent of temperature because the geometry of forbidden configurations does not change when the gas is warmed. The pressure itself still scales with temperature at fixed number density.

This model makes the positive correction particularly clear: eliminating overlapping configurations changes the dependence of the available configuration space on volume, even though the allowed configurations all have zero interaction energy. The figure's other panels add a toy attractive shell and show how its Boltzmann factor competes with that geometry. The hard-sphere gas is a useful reference model, while a low-order virial expansion remains a dilute-gas approximation rather than a dense sphere packing formula.

Heat is energy transferred across a thermodynamic system's boundary because of a temperature difference. It is measured in joules, like work and internal energy. A hot cup transfers heat to cooler surroundings; its temperature describes its state, while the transferred heat describes a process. The cup stores internal energy, rather than a substance called heat. ^[ref]

For a closed system with unchanged bulk kinetic energy and potential energy, choose $Q>0$ for heat entering and $W>0$ for work done by the system. The first law of thermodynamics givesHere $U$ is the stored internal energy. For example, a sealed piston cylinder receiving $20\mathrm{kJ}$ of heat and delivering $5\mathrm{kJ}$ of work gains $15\mathrm{kJ}$ of internal energy. A closed system retains its matter while exchanging energy; it need not be an isolated system. ^[ref]

Unlike internal energy, heat is not a state function: different paths between the same equilibrium states can transfer different $Q$, with compensating differences in work. The symbol $\delta Q$ records this dependence on the path; there is no stored quantity whose difference is always $Q$. The two paths in first law of thermodynamics illustrate this directly. Supplied heat can raise temperature as sensible heat, change a phase fraction through latent heat, or help supply expansion work. Measuring temperature alone therefore does not determine transferred heat.

For a reversible thermodynamic process, $\delta Q_{\mathrm{rev}}=TdS$, where $T$ is absolute temperature and $S$ is entropy. This identification requires reversibility: an irreversible adiabatic process has $Q=0$ and can nevertheless increase entropy. Likewise $Q=\Delta H$, with enthalpy $H$, requires the particular fixed-pressure conditions in enthalpy. These distinctions make heat useful as a precise boundary-transfer quantity in heat engines, refrigerators and heating calculations.

Historically, James Prescott Joule's paddle experiments showed that stirring water with input work could raise its temperature, helping establish conservation of energy. The lesson is still useful: a temperature rise can result from work even without incoming heat. ^[ref]

Heat capacity measures the heat supplied per small temperature increase along a specified process:The subscript $X$ records the constraint, such as fixed volume or fixed pressure. The units are $\mathrm{J}{\mathrm{K}}^{-1}$. The symbol $\delta Q$ emphasizes that heat is a transfer along a path, rather than a state function with a differential $dQ$. Consequently, an unspecified heat capacity does not determine the required heat.

For a closed system of fixed composition undergoing a quasistatic process, assume that only pressure-volume work occurs, with no friction and no change in bulk kinetic energy or potential energy. The first law of thermodynamics and $H=U+pV$, where $U$ is internal energy and $H$ is enthalpy, then giveThese partial derivatives define heat capacity at constant volume and heat capacity at constant pressure, with the amount and composition also held fixed. Integrating along the selected constraint gives $Q={\int }_{T_1}^{T_2}{C}_X(T)dT$ within a single thermodynamic phase; $Q=C_X\Delta T$ requires constant $C_X$. ^[ref]

For a computed example, take a monatomic gas in the classical ideal gas regime, with amount of substance $n=1\mathrm{mol}$ and stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The equipartition theorem gives $U=3nRT/2$, and the ideal gas law gives $pV=nRT$. ThusFor the same temperature increase $\Delta T=10\mathrm{K}$, the two processes requireHere work done by the gas is positive. At fixed pressure, the extra heat supplies expansion work; the internal energy increase is the same because internal energy of an ideal gas depends only on temperature. Their final volume values differ. ^[ref]

The practical value of heat capacity is that a measured heat input and temperature response can determine it. Once its constraint and temperature dependence are known, it predicts heating requirements and supplies entropy changes through entropy from heat capacity. ^[ref]

For a fixed-composition closed system, the heat capacity at fixed pressure is $C_p=(\partial H/\partial T{)}_p$, with enthalpy $H$. Under the quasistatic process assumptions in heat capacity, only pressure-volume work is permitted, so $dp=0$ gives $\delta Q=dH=C_{p}dT$. Thus supplied heat is $Q=\Delta H$, or $Q=C_p\Delta T$ within a single thermodynamic phase when $C_p$ is constant. For an ideal gas, the expansion work explains why this exceeds the required heat at fixed volume for the same temperature increase. Its per-unit-mass counterpart is specific heat capacity at constant pressure.

For a fixed-composition closed system, the heat capacity at fixed volume is $C_V=(\partial U/\partial T{)}_V$, with internal energy $U$. Under the quasistatic process assumptions in heat capacity, only pressure-volume work is permitted, so $dV=0$ gives $\delta Q=dU=C_{V}dT$. A rigid vessel therefore needs $Q=\Delta U$, or $Q=C_V\Delta T$ when $C_V$ is constant. Its per-unit-mass counterpart is specific heat capacity at constant volume.

Heat-engine calorimetry measures heat transfers so that observed work can be compared with the correct energy input. Measuring motion alone cannot supply the denominator of thermal efficiency.

The institutional Laboratory 16: Stirling Engine describes a modern teaching arrangement: an ethanol lamp drives a Stirling engine coupled to an electrical output device, which supplies a resistor immersed in water in a calorimeter. The prescribed readouts include fuel mass consumed, the pressure-volume loop, electrical readings, and water temperature. The sequence separates fuel input, gas-cycle work, electrical output, and Joule heating of the water. Warming of the water is the final useful output; it is not the engine's rejected $Q_c$. The handout distinguishes warm-up from steady operation and asks students to compare the final warming with direct heating by the lamp. It gives a procedure, not a table of measured efficiencies. Its pp. 1–3 therefore establish an apparatus and measurement design without supplying a new experimental result.

For a proposed calibration, use a known input of electrical work to determine the effective heat capacity $C_{\mathrm{eff}}$ of the water, vessel and immersed parts; stir to make the measured temperature representative. A matched run with no input estimates exchange with the room. For constant $C_{\mathrm{eff}}$ and no phase transition, the resulting model iswhere $Q_{\mathrm{leak},\mathrm{out}}$ is the signed net heat lost to the room during the measurement. Here $E_{\mathrm{in}}$ is supplied electrical work when the resistor lies inside the calibration boundary; for a separate external heat input it is the received heat. These are proposed controls, not controls reported in that handout. Slow cooling, changing material properties and imperfectly matched room exchange limit this inference; the approach to thermal equilibrium and heat capacity matter as much as the thermometer reading.

To measure the actual engine-input $Q_h$, one must instrument its hot-side transfer separately, or close a calibrated energy balance around a specified heater boundary. A heater's electrical work includes changes in its stored internal energy and heat transfer that bypasses the engine. At periodic operation a cold-side flowing-water measurement can instead estimate ${\dot{Q}}_c=\dot{m}{c}_p(T_{\mathrm{out}}-T_{\mathrm{in}})$, if mass flow rate $\dot{m}$ and specific heat capacity at constant pressure $c_p$ are known, with negligible storage and other heat exchange in that measurement boundary. It measures only the rejection captured by that stream. Include all remaining transfers before using the residual to infer ${\dot{Q}}_h=P_{\mathrm{out}}+{\dot{Q}}_c$. This measurement discipline makes a comparison with Carnot's theorem meaningful.

The Helmholtz free energy iswhere $U$ is internal energy, also denoted $E$ elsewhere here, $T$ is absolute temperature, and $S$ is entropy. For the closed system of fixed composition and only pressure-volume work described above, spontaneous relaxation at fixed temperature and volume cannot increase $F$. Stable thermodynamic equilibrium is a global minimum over the permitted alternatives, including different thermodynamic phases. A rigid vessel fixes the total volume; it does not require each coexisting thermodynamic phase to retain its own volume.

For a constructed comparison at $300\mathrm{K}$, suppose one candidate has $1.5\mathrm{kJ}$ more internal energy but $6\mathrm{J}{\mathrm{K}}^{-1}$ more entropy than another at the same temperature and volume. Its difference isThus the candidate with higher internal energy has lower Helmholtz free energy. This is the practical reason for retaining the entropy term when selecting a state.

For equilibrium states within a smooth branch of fixed composition, the first law of thermodynamics gives $dU=TdS-pdV$, henceThese derivatives extract entropy and pressure; the differential is not an evolution equation for relaxation. If chemical species amounts $n_i$ vary, add ${\sum }_i{\mu }_{i}d{n}_i$, with molar chemical potentials ${\mu }_i$. A closed system can still change chemical species amounts through chemical reactions; the fixed-composition simplification then needs the conserved-inventory constraints of chemical equilibrium. See MIT 5.60, Lecture 14, fundamental equations and chemical potentials.

The minimum principle has a short explanation. At fixed volume, no boundary work occurs, so $Q=\Delta U$. A thermal reservoir at positive temperature $T$ loses entropy $Q/T$. Consequentlyby the Second law of thermodynamics. Here both endpoint states share the imposed temperature. See MIT 5.60, Lecture 12, criteria for spontaneous change.

The microscopic counterpart is $F=-k_{B}T\ln Z$ in the canonical ensemble, with Boltzmann constant $k_B$ and partition function $Z$. The partition function examples compute the balance between internal energy and entropy; the chemical equilibrium example already displays an ideal-mixture Helmholtz free energy minimum.

At an initially resting instant with negligible pressure, sum the horizontal components of the magnetized-fluid tensor virial theorem. The horizontal magnetic trace is $-B_z^2/{\mu }_0$ and the gravitational force-stress trace is $({\partial }_z\Phi {)}^2/(4\pi G)$. Their negatives give the displayed radial second-moment balance. Vacuum field stresses contribute too; being initially at rest does not imply zero acceleration.

Hydraulic resistance relates the pressure drop across a resistive fluid element to its volumetric flow rate. In a linear model, $\Delta p=RQ$ with $R>0$, so dissipated mechanical power is $\Delta pQ=R{Q}^2$. A peripheral circulatory bed can be represented by such a resistance when its detailed dynamics are not retained. This terminal pressure-to-flow resistance differs from the area-dependent friction rate multiplying velocity in a distributed tube momentum equation.