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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 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]

The heat capacity ratio compares heat capacity at constant pressure $C_p$ with heat capacity at constant volume $C_V$ at the same equilibrium state and fixed composition:Here $C_p,C_V$ refer to the whole sample, in $\mathrm{J}{\mathrm{K}}^{-1}$; molar heat capacities $C_{p,\mathrm{m}}=C_p/n$ and $C_{V,\mathrm{m}}=C_V/n$ are per mole; specific heat capacities $c_p=C_p/m$ and $c_V=C_V/m$ are per unit mass. The amount of substance $n$ or mass $m$ cancels only when both capacities use the same normalization. Thus heat capacity ratio is dimensionless and independent of sample size. For a fixed-composition ideal gas with positive finite $C_V$, Mayer's relation giveswhere $R$ is the molar gas constant. More ways to store internal energy make a given temperature increase require more heat, bringing heat capacity ratio closer to unity. ^[ref]

The equipartition theorem predicts $C_{V,\mathrm{m}}=fR/2$ when $f$ independent quadratic terms in molecular energy are thermally active and their number stays fixed; consequently $\gamma =1+2/f$. A classical monatomic gas with no appreciable electronic excitation has $f=3$, giving $C_{V,\mathrm{m}}=3R/2$ and $\gamma =5/3$. For an ideal gas of diatomic molecules with three translational and two rotational terms active, but its vibrational normal mode effectively frozen, $f=5$ gives $C_{V,\mathrm{m}}=5R/2$ and $\gamma =7/5$. A fully classical harmonic vibration adds both kinetic energy and potential energy, giving $f=7$ and $\gamma =9/7$. These are model regimes, not universal constants of every gas: rotational freezing, electronic excitation, and changing composition invalidate the simple count.^[ref]

The mechanical significance of heat capacity ratio follows from the first law of thermodynamics. For a fixed amount of ideal gas in a reversible adiabatic process, $C_V(T)dT=-pdV$. Combining this with the ideal gas law gives the local relationThe partial derivative holds at fixed entropy and composition even when heat capacity ratio varies with temperature. The finite-path formulas $p{V}^{\gamma }=\mathrm{constant}$ and $T{V}^{\gamma -1}=\mathrm{constant}$ require the additional approximation of constant heat capacity ratio. A rapid compression also needs internal equilibration to approximate this path; rapid motion alone does not ensure reversibility. The same local response gives isentropic compressibility $1/(\gamma p)$ and, for small sound disturbances in this ideal gas regime, speed of sound $a=\sqrt{\gamma p/\rho }$, where $\rho$ is mass density. Thus thermal storage controls mechanical stiffness, sound propagation and the measurements in Rüchardt experiment and Clément-Desormes experiment.

Helium liquefaction converts helium gas into a collectable liquid. It creates a bath for investigating material properties at very low temperatures, including superconductivity. The central question is how the incoming specific enthalpy compares with the specific enthalpies of the outgoing liquid and vapor. The liquefaction enthalpy budget makes that comparison explicit; a temperature decrease by itself does not specify a liquid yield.

In a Joule-Thomson effect cooling stage, a throttling pressure drop approximately preserves enthalpy under the assumptions of the steady-flow energy balance. For an ideal gas, specific enthalpy depends only on temperature: with positive specific heat capacity at constant pressure, equal specific enthalpy means equal temperature. Thus its Joule-Thomson coefficient is zero. An ideal gas also supplies no liquid branch. Cooling by throttling depends on real gas properties; expansion is not an automatic source of refrigeration.

For a local drop $dp<0$, $dT={\mu }_{\mathrm{JT}}dp$ cools only where the Joule-Thomson coefficient is positive. Precooling must bring the real gas into a useful cooling region, with enough cooling to overcome unwanted heat transfer. A finite expansion can cross a Joule-Thomson inversion curve; its final state requires the full enthalpy balance, including phase coexistence if liquid forms. Kamerlingh Onnes's 1908 helium liquefaction experiment combined external precooling with regeneration to reach that regime. The Joule-Thomson effect article develops the throttling identities and their assumptions.

Hess's law states that compatible chemical reaction enthalpies add when their equations add. Reversing a chemical reaction reverses its enthalpy change; multiplying the entire chemical reaction equation by a factor multiplies that change by the same factor. The reason is that enthalpy is a state function: intermediate states cancel from the differences. A cycle supplies one independent balance relation; it cannot determine two unknown inputs without further information.

The law is named after Germain Henri Hess.

For specified initial and final thermodynamic states, the change is $\Delta H=H_{\mathrm{final}}-H_{\mathrm{initial}}$. Because enthalpy is an extensive variable, a quoted reaction $\Delta H$ belongs to the amount of substance specified for each species in its chemical equation. Reversing the equation changes the sign; scaling every coefficient scales $\Delta H$; adding compatible equations adds their $\Delta H$ values. A per-mole value must say which chemical equation and normalization it uses. Match temperature, pressure, thermodynamic phases, chemical solution compositions and reference states before cancellation; the same formula in a gas and a liquid represents different states.^[ref]

The value of Hess's law is that an inaccessible reaction enthalpy can be obtained from measurable transformations. The algebra does not require those transformations to be steps of the actual reaction mechanism. Neither an activation energy nor a reaction rate follows from the endpoint difference: a catalyst may change the reaction rate while preserving the compatible initial and final enthalpies.

High harmonic generation converts an intense driving electric field into radiation at high integer multiples of its frequency. High harmonic generation in a gas is a process in nonlinear optics: an electron can leave a neutral target by tunnel ionization, gain kinetic energy in the driving electric field, and return to radiate upon recombination. This three-step picture is an approximate account of a probability amplitude for emission in quantum mechanics, not a measured path of one electron.^[ref]

Why odd multiples? In an ideal centrosymmetric gas driven by a single frequency, assume the radiating electric dipole moment satisfies $d(t+T/2)=-d(t)$, where $T$ is the driving period. Its Fourier series coefficient obeys $d_q=-(-1{)}^q{d}_q$, so every even coefficient vanishes. A finite envelope, an oriented target or an additional driving frequency can invalidate that symmetry argument. Producing many odd components also does not prove that their spectral phases form a short optical pulse.

The 1988 report involving Anne L'Huillier observed odd multiples of a $1064\mathrm{nm}$ driving wavelength in rare gases, reaching order 33 in argon. The unexpectedly slow fall of yield beyond the fifth order supplied a broad spectral plateau. A spectral observation alone did not establish the relative spectral phases needed for an attosecond pulse train.^[ref]

A disk-gas trajectory that turns from one side of a planet's orbital radius to the other in the corotating frame. Its angular-momentum exchange contributes to the corotation torque; repeated libration without restoring gradients can saturate that contribution.

Suppose an isothermal gas with ${\rho }_g\propto r^{-2}$ is supported in a spherical halo with $M(r)=Mr/R$. Then hydrostatic equilibrium and $P={\rho }_g{k}_{B}T/\bar{m}$ give $-2{\rho }_g{k}_{B}T/(\bar{m}r)=-{\rho }_{g}GM/(Rr)$, hence $k_{B}T=\bar{m}GM/(2R)$. Here $\bar{m}$ is the dimensionful mean particle mass. The factor two follows from the density slope and isothermal pressure response; it is not the uniform-sphere virial coefficient. The singular profile and finite boundary are idealizations.

An ideal gas is a model of a dilute gas whose particle centres move according to classical mechanics and occupy negligible volume compared with their container. Treating the particles as point masses removes the molecular-size correction; neglecting interactions between separated particles removes the intermolecular contribution to potential energy. The usual kinetic theory model retains brief elastic collisions with the walls and, optionally, with other particles: these transfer momentum while conserving the total kinetic energy of the colliding bodies. There is no appreciable interaction energy stored between collisions. These assumptions lead to the ideal gas law, $PV=N{k}_{B}T$. ^{[ref] [ref]}

The model's attraction is its universality: at the same temperature and number density, different classical ideal gases have the same pressure, even when their particle masses differ. In ideal gas pressure from wall collisions, heavier particles move more slowly at a given temperature, and the two effects cancel. For example, one mole of any ideal gas at $300\mathrm{K}$ in $0.025{\mathrm{m}}^3$ has pressure approximately $99.774\mathrm{kPa}$, as computed under the ideal gas law.

The ideal gas law describes the translational contribution to pressure; it does not specify all of the internal energy or the heat capacity. A monatomic gas has three classical translational degrees of freedom, whereas a gas of molecules can also store rotational kinetic energy and energy in vibrational normal modes. An equation of state alone therefore does not determine how much heat a chosen temperature increase requires. It also does not select a process: isothermal expansion and adiabatic expansion need different additional constraints. ^[ref]

A real gas departs from this approximation when particle size or interactions become important, especially at high number density or low temperature near a phase transition. The Van der Waals equation adds such corrections. Independently, at sufficiently low temperature or high number density, quantum statistics can become essential: even particles with negligible interactions need not obey the classical ideal gas law. The classical statistical assumption is the nondegenerate gas regime, which is distinct from neglecting interactions. ^{[ref] [ref]}

Consider a particle of mass $m$ in a rigid rectangular box of length $L$ along $x$ and wall surface area $A$, so $V=AL$ is the volume. Choose $+x$ toward the right wall. For an incoming normal velocity $v_x=v>0$, an elastic collision reverses the normal momentum:The impulse delivered to the wall points outward, toward $+x$, by Newton's third law. The particle's opposite momentum change is not the wall's impulse. A tangential velocity component is unchanged by the ideal smooth-wall collision.

Ignoring intervening particle-particle collisions in this counting model, the particle travels a normal distance $2L$ before hitting the same wall again. Its hit rate is $|v_x|/(2L)$; multiplying by the impulse per hit gives the time-averaged normal force and its contribution to pressure:For example, stipulate $m=4\times 10^{-26}\mathrm{kg}$, $|v_x|=500\mathrm{m}{\mathrm{s}}^{-1}$, $L=0.10\mathrm{m}$, and $A=0.010{\mathrm{m}}^2$. Then $V=0.0010{\mathrm{m}}^3$, the impulse per hit is $4\times 10^{-23}\mathrm{N}\mathrm{s}$, the hit rate is $2500{\mathrm{s}}^{-1}$, and $P_1=10^{-17}\mathrm{Pa}$. This is one particle's tiny time-averaged contribution to one wall's pressure, not an isotropic equilibrium gas by itself.

For $N$ particles of equal mass, add their force contributions and define $⟨v_x^2⟩=N^{-1}{\sum }_{i=1}^N{v}_{x,i}^2$. In the gas's rest frame, isotropy gives $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=⟨v^2⟩/3$, where $v^2=v_x^2+v_y^2+v_z^2$. HenceThe final equality uses the classical equipartition theorem. The kinetic pressure of an isotropic gas therefore recovers the ideal gas law from momentum transfers. Doubling the normal speed would double both the impulse per hit and the hit rate, giving four times the pressure contribution; pressure measures a mean squared speed. Interparticle elastic collisions redistribute velocities and help establish isotropy; the individual round-trip times need not remain $2L/|v_x|$, but the equilibrium momentum flux result survives. ^[ref]

An ideal solution is a chemical solution whose component chemical potentials depend on composition throughHere $n_i$ is an amount of substance, $x_i$ its mole fraction, $R$ the molar gas constant, and ${\mu }_i^{\ast }$ the pure-component liquid reference at the same temperature $T$ and pressure $P$. This model applies to every component across the composition range; it is stronger than an approximation valid only near a pure solvent.^[ref]

With this pure-liquid standard state convention, the thermodynamic activity satisfies ${\mu }_i={\mu }_i^{\ast }+RT\ln a_i$, so an ideal solution has $a_i=x_i$, or activity coefficient ${\gamma }_i=a_i/x_i=1$.^[ref] This statement concerns the liquid. Assuming that a coexisting vapor is an ideal gas mixture is an additional approximation. Its mole fractions $y_i$ generally differ from the liquid mole fractions $x_i$.

For an original two-component example, stipulate $T=300\mathrm{K}$ and pure-component vapor pressures $p_A^{\ast }=80\mathrm{kPa}$ and $p_B^{\ast }=20\mathrm{kPa}$. A and B are invented distinguishable components; these numbers are chosen model inputs, not material data. Combine an ideal solution with an ideal gas vapor, neglect the pure-liquid reference's pressure correction, and exclude other gases. At liquid mole fractions $x_A=0.4$ and $x_B=0.6$, Raoult's law and addition of partial pressures giveThus the coexisting vapor contains about $72.73\%$ A by amount of substance, although the liquid contains only $40\%$. The total pressure is neither pure-component vapor pressure, and $x_A$ cannot be substituted for $y_A$. This enrichment is the basis of distillation.

The usefulness of the ideal solution is that one composition law connects chemical potentials, vapor-liquid equilibrium, entropy of mixing and Gibbs free energy of mixing. Its liquid components can interact strongly: the defining condition is their composition dependence, rather than the absence of all interactions as in an ideal gas model.

For liquid–gas phase coexistence, make two distinct approximations: the gas obeys the ideal gas law, $v_{m,g}=RT/p$, and its molar volume greatly exceeds the liquid molar volume, $v_{m,g}\gg v_{m,l}$. Here $R$ is the molar gas constant and $p$ is the saturation pressure of the pure substance. Substitution into the Clapeyron equation giveswhere $p_{\ast }>0$ is any fixed reference pressure, making the logarithm dimensionless. Constant latent heat is a further approximation, needed for the elementary Integrated Clausius-Clapeyron relation below.^[ref]

The neglected liquid molar volume has a transparent error estimate. If the gas is ideal but $ϵ=v_{m,l}/v_{m,g}$ is retained, thenThus ignoring the liquid molar volume underestimates this slope by a relative amount $ϵ$. This estimate does not bound errors from treating a real gas as ideal or from freezing $L_m(T)$. Near a thermodynamic critical point, the gas and liquid molar volumes approach each other and these dilute-vapor assumptions fail.

If the molar latent heat $L_m$ is approximately constant on the chosen temperature interval, the ideal-vapor Clausius-Clapeyron approximation integrates toFor $L_m>0$, higher absolute temperature gives higher saturation pressure. A plot of $\ln (p/p_1)$ against inverse temperature $1/T$ is a straight line of slope $-L_m/R$. One reference pair $(T_1,p_1)$ fixes its intercept. On a per-unit-mass basis replace $L_m/R$ by $L/R_s$, with $R_s=R/M$ and $M$ the molar mass; using $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ with the molar $R$ would give incompatible units.

If only the constant-latent heat approximation is relaxed, the same ideal-vapor model instead givesConsequently an apparent straight-line fit across a broad temperature range need not represent a single material constant. The simple exponential also approaches a finite value as $T\to \infty$; this is a property of the stipulated formula, not a prediction that liquid–gas phase coexistence extends beyond a thermodynamic critical point.

Constant temperature does not generally mean constant internal energy. At fixed amount and composition, within one smooth thermodynamic phase of a simple compressible system, an equilibrium identity isThe bracket vanishes for the ideal gas law $pV=nRT$. It need not vanish for a real gas: changing the volume changes the average interaction potential energy, even when the random-motion contribution stays at the same temperature. The identity follows by writing $dU=TdS-pdV$ and using a Maxwell relation; it does not require the actual process to be reversible. ^[ref]

For a Van der Waals equation model with volume per mole $v=V/n$, $p=RT/(v-b)-a/v^2$ gives $(\partial U/\partial V{)}_{T,n}=a/v^2$. Thus an isothermal expansion from $v_1$ to $v_2>v_1$, at fixed amount of substance $n$, increases internal energy bySeparating attractive particles makes their interaction potential energy less negative. For stipulated $n=1\mathrm{mol}$, $a=0.10\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$, $b=10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, $T=300\mathrm{K}$, $v_1=0.010$ and $v_2=0.020{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, the calculated increase for this one-mole sample is $5\mathrm{J}$, despite $\Delta T=0$. This dilute stable-branch model example is not a measurement of a named gas. ^[ref]

A phase transition gives another reason. During equilibrium vaporization of a pure substance at fixed pressure, the temperature can stay at its phase coexistence value while the phase fraction changes. Molecular separation changes interaction potential energy.^[ref] If the closed system does only pressure-volume work, its latent heat input obeysHere $H=U+pV$ is enthalpy. A vaporization latent heat therefore is not automatically the internal energy increment: some input supplies expansion work. A fixed temperature reading alone cannot tell how much internal energy changed.

For a fixed amount of substance and composition of classical ideal gas, internal energy depends only on temperature: $U=U(T,n)$, with no independent volume dependence. This does not require constant heat capacity at constant volume. Between equilibrium endpoints,Here $C_V$ is total heat capacity at constant volume, $C_{V,\mathrm{m}}$ is molar heat capacity, and $U/m$ for sample mass $m$ is specific internal energy, whose change is ${\int }_{T_1}^{T_2}{c}_V(T)dT$. These endpoint relations apply even when the intervening process is not at fixed volume; they do not say that transferred heat always equals $\Delta U$. With constant $C_V$, the safe statement is $U(T)=U(T_{\ast })+C_V(T-T_{\ast })$. Writing $U=C_{V}T$ additionally chooses the reference energy; it does not prove that a physical gas remains classical and ideal down to zero temperature. ^[ref]

A thermodynamic proof outline uses the first law of thermodynamics in equilibrium form, $dU=TdS-pdV$, and the Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$:The last step uses the ideal gas law and the molar gas constant $R$, with fixed composition and amount of substance. Microscopically, separated ideal gas particles store no intermolecular interaction energy; their translational kinetic energy and any internal molecular energy are determined by temperature. Negligible intermolecular interactions still allow a molecule to have internal vibrational potential energy.

An original analytical model makes the distinction between an ideal gas and constant heat capacity concrete. Assume classical translation, two fully active rotational terms, and one independent quantum harmonic oscillator per diatomic molecule, with fixed angular frequency $\omega$. Define the vibrational temperature scale $\Theta =\hslash \omega /k_B$, using the reduced Planck constant $\hslash =h/(2\pi )$ and Boltzmann constant $k_B$. Summing its thermal occupations givesThe constant $U_{\ast }$ includes the chosen reference and zero-point energy, whose temperature derivative vanishes. When $T\ll \Theta$ the vibrational contribution to heat capacity is exponentially small; when $T\gg \Theta$ it tends to $R$ per mole. This model assumes rotations remain active throughout the interval and excludes dissociation, electronic excitation and real gas corrections. Its low-$T/\Theta$ edge is not a model of rotational freezing or of an actual substance at absolute zero.^[ref]

For this stipulated model, heating from $T_1=0.2\Theta$ to $T_2=\Theta$ gives $\Delta U/(nR\Theta )\simeq 2.57519$. Holding the frozen-vibration value $C_{V,\mathrm{m}}=5R/2$ throughout would give $2$, underestimating the required internal energy increase by about $22.3\%$ of the model's full value. The integral measures all thermal storage activated along the interval. At fixed volume with only heat input, this is also the supplied heat; at fixed pressure, expansion work must be added through Mayer's relation.

Joule's free-expansion experiment tests the independence from volume in a different way: an insulated rigid vessel allows gas into an evacuated compartment, so no external work is delivered and ideally no heat is exchanged. The first law of thermodynamics gives $\Delta U=0$; positive $C_V$ then predicts equal initial and final equilibrium temperatures for the ideal gas. The apparatus, thermal-capacity controls and limitations of this inference are described in Joule's free-expansion experiment and temperature sensitivity of a free-expansion experiment. This endpoint test differs from the reversible adiabatic expansion of a loaded piston, which delivers work and lowers internal energy.

Isothermal compressibility is the compressibility measured at fixed temperature, particle number and composition. A slowly compressed sealed piston in contact with a thermal reservoir can approach this constraint. For an ideal gas, differentiating $V=N{k}_{B}T/P$ at fixed temperature gives $(\partial V/\partial P{)}_{T,N}=-V/P$, hence ${\kappa }_T=1/P$. A small pressure increase $\Delta P$ therefore gives $\Delta V/V\simeq -\Delta P/P$. For a finite change, use the whole isothermal process curve $V_2/V_1=P_1/P_2$ instead.

Take one mole of classical monatomic gas, with stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, initially at $T_0=300\mathrm{K}$ and $P_0=100\mathrm{kPa}$. The initial volume is $V_0=nR{T}_0/P_0=0.024942{\mathrm{m}}^3$. Compressing to $V_0/2$ givesThe isothermal compression releases heat to the thermal reservoir; the reversible adiabatic compression retains the input work as internal energy. Reversing either curve gives the corresponding expansion, with adiabatic expansion cooling the gas.

The equilibrium curvature of Gibbs free energy provides another measurement of isothermal compressibility:This follows from $V=(\partial G/\partial P{)}_{T,N}$, with $G$ denoting Gibbs free energy. Within a stable single thermodynamic phase, increasing pressure reduces volume and makes this curvature negative. At phase coexistence, changing the phase fraction can change the volume at constant temperature and pressure; the usual finite single-phase isothermal compressibility then need not exist. Near a critical point, it can diverge. ^[ref]

An isothermal compression reduces the volume of a thermodynamic system at fixed temperature. For a fixed-composition ideal gas compressed reversibly from $V_i$ to $V_f<V_i$, isothermal ideal-gas work givesThe surroundings supply positive work $W_{\mathrm{on}}$, and the gas rejects the same amount of heat to the heat bath. Compressing $1\mathrm{mol}$ at $300\mathrm{K}$ from $20$ to $10\mathrm{L}$ requires approximately $1.729\mathrm{kJ}$ of work input and rejects $1.729\mathrm{kJ}$ of heat. If that heat transfer cannot keep up, the temperature rises: the process departs from isothermal compression toward adiabatic compression. The finite-rate isothermal piston model explains the required small temperature excess.

An isothermal expansion increases the volume of a thermodynamic system at fixed temperature. For a reversible ideal gas path, the gas delivers positive work and absorbs positive heat, with $Q=W_{\mathrm{by},\mathrm{rev}}=nRT\ln (V_f/V_i)$ and $\Delta U=0$. The heat bath supplies the expansion energy; without sufficient heat transfer, doing work instead lowers the temperature. This transfer is the hot step of a Carnot cycle, and it is why an expanding gas need not cool when its surroundings continuously replenish its internal energy.

An insulated free expansion into an evacuated compartment is different. In the idealization with no heat exchange, no external work and no container energy change, $Q=W_{\mathrm{by}}=0$. Since internal energy of an ideal gas depends only on temperature, its final equilibrium temperature equals its initial value. The transient need not have a uniform temperature or pressure, so equal endpoint temperatures do not make it a reversible isotherm. Do not apply the logarithmic isothermal ideal-gas work formula to this irreversible process. Its entropy still increases by $\Delta S=nR\ln (V_f/V_i)$; entropy change in free expansion evaluates this state function difference using a hypothetical reversible path.^[ref]

For a real gas, even equal-temperature equilibrium endpoints need not have equal internal energy. An insulated free expansion can therefore change the temperature, while a controlled isothermal expansion must provide the heat needed for both its internal energy change and its delivered work. The internal energy at constant temperature article gives the Van der Waals equation example.^[ref]

For a reversible thermodynamic process that changes the volume of a fixed amount of ideal gas at absolute temperature $T$, the work done by the gas isHere $n$ is the amount of substance, $R$ the molar gas constant, $p$ the pressure, and $\ln$ the natural logarithm. The ideal gas law supplies $p=nRT/V$; the integral is the signed area under the pressure-volume curve. Isothermal expansion gives positive work; isothermal compression gives negative work. A convention counting work done on the gas instead has $W_{\mathrm{on}}=-W_{\mathrm{by}}$.^[ref]

For a calculated example, let $n=1\mathrm{mol}$, $T=300\mathrm{K}$ and $V_i=10\mathrm{L}$, $V_f=20\mathrm{L}$. The pressure falls from approximately $249.4$ to $124.7\mathrm{kPa}$, andTraversing the same isothermal process backward changes both transfer signs: $W_{\mathrm{by},\mathrm{rev}}=Q\approx -1.729\mathrm{kJ}$. Neither direction changes internal energy of an ideal gas. These stipulated values illustrate the energy balance; they are not experimental measurements.

The shaded geometric area is positive in both panels; the direction of the integral supplies the sign of work. Knowing only the two equilibrium endpoints does not justify shading this curve for an irreversible process. For an idealized piston moving against a specified external resisting pressure, boundary work is $W_{\mathrm{by}}=\int p_{\mathrm{ext}}dV$. Substitution of the equilibrium gas pressure requires the appropriate reversible limit. In free expansion into vacuum, $p_{\mathrm{ext}}=0$ and this boundary work is zero.

An isothermal process changes a thermodynamic system while keeping its temperature constant. It need not keep its pressure, volume or internal energy constant, and it does not generally have zero heat transfer. An adiabatic process specifies zero heat transfer instead. For example, a slowly expanding ideal gas can absorb heat from a heat bath and deliver the same amount of work while its temperature stays fixed.

For a fixed amount of substance and composition, internal energy of an ideal gas depends only on temperature. With $Q>0$ for heat entering and $W_{\mathrm{by}}>0$ for work delivered to the surroundings, the first law of thermodynamics therefore givesThis equality does not determine either transfer until the path is specified. Isothermal ideal-gas work gives the reversible value; an insulated free expansion has a different value even when its equilibrium endpoints have the same temperature. The closed system convention here excludes changes in bulk kinetic energy and external potential energy.

A conducting cylinder in a large, stirred heat bath approximates an isothermal process when the piston moves slowly enough for thermal conduction to keep up. A quasistatic process keeps the gas close to internal thermodynamic equilibrium; a reversible thermodynamic process additionally requires negligible friction and vanishing driving differences in pressure and temperature. Exact equality of the gas and heat bath temperatures gives no finite heat transfer rate through a finite thermal resistance. The reversible description is a limit of arbitrarily small differences and arbitrarily slow motion, illustrated by the finite-rate isothermal piston model.^[ref]

For a real gas, internal energy at constant temperature can change with volume because molecular interaction potential energy changes. In one smooth thermodynamic phase of a fixed-composition simple compressible system,Thus an isothermal process generally has $Q=\Delta U+W_{\mathrm{by}}$, with no reason to set $\Delta U=0$. The right-hand side vanishes for the ideal gas law, but not generally for another equation of state.^[ref]