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Boyle's law states that a fixed amount of substance of an ideal gas at constant absolute thermodynamic temperature has constant product of absolute pressure and total volume:Here $p$ is measured relative to a vacuum, not relative to the surrounding atmospheric pressure. Halving the volume doubles the absolute pressure; doubling the volume halves it. A real gas approximately follows Boyle's law when it is sufficiently dilute and far from a phase transition.^[ref]

For an original calculated example, suppose a sealed gas sample has $p_1=100\mathrm{kPa}$ and $V_1=100\mathrm{c}{\mathrm{m}}^3$. After isothermal compression to $V_2=50\mathrm{c}{\mathrm{m}}^3$,Both states have $C=10000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10\mathrm{J}$, since $1\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10^{-3}\mathrm{J}$. The volume ratio is dimensionless, so using the same volume units in numerator and denominator leaves the pressure in kilopascals. Although $pV$ has units of energy, its value is not the compression work: isothermal ideal-gas work involves an integral along the path.

The blue curves show calculated ideal gas states, not measurements. The left panel is a branch of a hyperbola; the right panel plots exactly the same absolute pressures against $1/V$. The dashed line uses gauge pressure with a stipulated constant atmospheric pressure of $100\mathrm{kPa}$. Its displaced intercept explains why absolute pressure in Boyle's law matters.

A short derivation starts from the ideal gas law, $pV=nRT$. Fixing the amount of substance $n$ and absolute thermodynamic temperature $T$ fixes $C=nRT$, where $R$ is the molar gas constant. The microscopic picture in ideal gas pressure from wall collisions is equally appealing: at fixed temperature, the typical particle kinetic energy stays fixed, while compression increases the number density and wall momentum transfer per unit surface area and time. A larger absolute pressure follows without requiring hotter particles.

The constraints are essential. A leak changes the amount of substance. Rapid compression can raise the temperature before heat transfer restores thermal equilibrium, as in the isothermal gas syringe experiment. Molecular size and interactions in a dense real gas invalidate the ideal gas approximation; the Van der Waals equation describes one such correction. Equal endpoint temperatures allow an endpoint test of Boyle's law, but do not establish that every intervening state lies on an isothermal process.

For fixed mass $m$, mass density satisfies $\rho =m/V=mp/C$, so it is proportional to absolute pressure. In a dilute bubbly flow, Boyle's law allows bubble volume and its contribution to buoyancy to increase as hydrostatic pressure decreases, provided gas mass is retained, dissolution is negligible, and the temperature stays approximately fixed. Its usefulness is that a simple ratio predicts a substantial size change without tracking individual molecules.

For the insulated gas-only free expansion of an ideal gas, the equilibrium endpoints are $(T,V_1)$ and $(T,V_2)$ with $V_2>V_1$. Because entropy is a state function, calculate their difference along a hypothetical reversible thermodynamic process: an isothermal expansion using a frictionless piston and a thermal reservoir at $T$. This comparison process receives heat and delivers work, even though the actual free expansion does neither. Its reversible heat increment is $\delta Q_{\mathrm{rev}}=pdV=nRTdV/V$, soHere $R$ is the molar gas constant, and the logarithm acts on a dimensionless ratio. The reversible reference gives the correct endpoint entropy difference because entropy is a state function; putting the actual $Q=0$ into $\Delta S=Q/T$ would misuse a relation requiring reversible heat transfer. For the actual isolated gas, all this increase is entropy production. Thus an adiabatic process can increase entropy and need not be an isentropic process. ^[ref]

In the figure's volume doubling, the ideal gas law gives $p_1=99.768\mathrm{kPa}$ and $p_2=49.884\mathrm{kPa}$, while $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. The reference process receives and delivers $nRT\ln 2\simeq 1728.85\mathrm{J}$ as heat and work; its thermal reservoir loses the same entropy that the gas gains. The actual free expansion has $Q_{\mathrm{in}}=W_{\mathrm{out}}=0$ and produces that entropy internally. Reversibly compressing the gas back at $300\mathrm{K}$ would require $1728.85\mathrm{J}$ of work input and rejection of equal heat. Restoring the gas alone therefore changes the surroundings: the actual release is irreversible.

For a real gas the final temperature may differ, so the ideal-gas logarithm formula cannot simply be copied. Evaluate $S(T_2,V_2)-S(T_1,V_1)$ using that gas's equilibrium properties or a suitable reversible comparison path. If the walls and water exchange heat with it, the Second law of thermodynamics constrains the total entropy of the isolated apparatus; the gas alone is then not an isolated system.

Free expansion occurs when a gas spreads into an initially evacuated region without driving an external load. The standard idealization uses a rigid, insulated vessel: initially a partition confines the gas to volume $V_1$, while the remaining compartment contains vacuum. Open a passage and wait for thermodynamic equilibrium; the same fixed amount of substance $n$ now occupies the larger accessible volume $V_2$. The outer vessel never changes size. This simple release separates three questions: what energy is conserved, whether temperature changes, and why entropy increases. ^[ref]

Choose the gas as the closed system, neglect heat exchange with the walls, and assume negligible changes in bulk kinetic energy and potential energy between the equilibrium endpoints. Insulation gives $Q_{\mathrm{in}}=0$; no moving outer boundary or external load receives work, so $W_{\mathrm{out}}=0$. The first law of thermodynamics therefore givesThe gas can develop internal motion during the release; this does not supply external work. The intermediate state need not have one equilibrium pressure or temperature, so there is no actual quasistatic process curve $p(V)$ to integrate. In particular, substituting the equilibrium ideal gas law into $\int pdV$ would describe a different process. ^[ref]

For an ideal gas, internal energy of an ideal gas depends only on temperature. With positive heat capacity at constant volume, $U(T_2)=U(T_1)$ implies $T_2=T_1$, and the ideal gas law gives $p_2/p_1=V_1/V_2$. Doubling the accessible volume thus halves the final pressure without changing the final temperature. This is an endpoint statement; it does not assert a spatially uniform, constant temperature during the rush. Insulation alone does not imply constant temperature: a reversible adiabatic expansion cools an ideal gas because it supplies work. ^[ref]

For a real gas, $U(T,V)$ can depend on volume as well as temperature, so unchanged internal energy need not mean unchanged temperature. For a fixed-composition simple compressible system in a single thermodynamic phase, equilibrium identities using a Maxwell relation giveHere $C_V$ is the whole sample's heat capacity at constant volume. The sign depends on the equation of state; for an ideal gas the numerator vanishes. For a finite free expansion, solve $U(T_2,V_2)=U(T_1,V_1)$, or integrate this slope along equilibrium states of equal internal energy; that mathematical curve does not describe the actual release. ^[ref]

Free expansion is useful because it tests how internal energy depends on number density while eliminating external expansion work. Its beautiful surprise is that an ideal gas can finish with the same temperature and internal energy but greater entropy: a thermometer alone cannot diagnose reversibility. See entropy change in free expansion for the calculation and temperature sensitivity of a free-expansion experiment for the experimental qualification. The continuous-flow throttling associated with the Joule-Thomson coefficient has a different balance: under its usual assumptions it preserves enthalpy, whereas the gas-only idealization here preserves internal energy. ^[ref]

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.

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]}

The ideal gas law is the equation of state relating the absolute pressure $P$, volume $V$, particle count $N$, and absolute thermodynamic temperature $T$ of a classical ideal gas:Here $k_B$ is the Boltzmann constant, $N_{\mathrm{A}}$ is the Avogadro constant, $R$ is the molar gas constant, and $n_{\mathrm{mol}}$ is the amount of substance in moles. The distinct symbol $n_{\mathrm{mol}}$ avoids confusing amount of substance with the number density $n=N/V$, measured in ${\mathrm{m}}^{-3}$: the same law can be written $P=n{k}_{B}T$. Use absolute pressure, relative to a vacuum, rather than an unconverted gauge pressure. ^[ref]

The exact SI definitions of the Avogadro constant and Boltzmann constant giveThis numerical value is exact; the applicability of the ideal gas model to a real gas is an approximation. ^{[ref] [ref]}

For $n_{\mathrm{mol}}=1\mathrm{mol}$, $T=300\mathrm{K}$, and $V=0.025{\mathrm{m}}^3$, direct substitution givesAt fixed temperature and particle count, doubling the volume to $0.050{\mathrm{m}}^3$ halves the pressure to approximately $49.887\mathrm{kPa}$: this is Boyle's law. At fixed volume and particle count, doubling the temperature from $300\mathrm{K}$ to $600\mathrm{K}$ doubles the pressure to approximately $199.547\mathrm{kPa}$. These ratios use kelvins, not readings in the degree Celsius scale: $300\mathrm{K}$ is $26.85{}^{\circ }\mathrm{C}$, and $600\mathrm{K}$ is $326.85{}^{\circ }\mathrm{C}$.

The curved slices show Boyle's law at three temperatures; the straight slice shows proportionality to absolute temperature at one fixed volume. They are relations among equilibrium states, rather than a prescribed process history. Compression could hold the temperature fixed through heat transfer, or raise it during adiabatic compression. To calculate a process's heat or work, supplement the ideal gas law with the process constraint, the first law of thermodynamics, and the relevant internal energy or heat capacity information. The graph's extrapolation toward $T=0$ is mathematical; it does not assert that a real gas remains classical or gaseous there.

For a stellar mixture with fixed mean molecular weight, the ideal gas law can be written $P=\mathcal{R}\rho T/\mu$, where $\rho$ is mass density and $\mathcal{R}$ is the stellar gas constant. It assumes negligible interaction corrections and a nondegenerate gas; excitation or ionization can change the particle count and thermodynamic response.

In his 1834 memoir on the motive power of heat, Émile Clapeyron combined gas laws into a relation between pressure, volume, and temperature while analysing heat engines (printed page 164). The equation above uses modern SI notation and absolute thermodynamic temperature; it is not a transcription of his historical temperature convention. ^[ref]

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.

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.

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]

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]

The Joule-Thomson coefficient is the local temperature response to changing pressure at fixed enthalpy and fixed amount and composition:It describes the equilibrium endpoints of a Joule-Thomson effect experiment meeting the steady-flow energy balance assumptions. With $\Delta p=p_2-p_1<0$, a sufficiently small drop gives $\Delta T\simeq {\mu }_{\mathrm{JT}}\Delta p$. Thus ${\mu }_{\mathrm{JT}}>0$ means cooling, ${\mu }_{\mathrm{JT}}<0$ warming, and ${\mu }_{\mathrm{JT}}=0$ no first-order change. At zero coefficient, a finite drop can still change temperature. For a chosen illustrative local value ${\mu }_{\mathrm{JT}}=0.20\mathrm{K}\mathrm{ba}{\mathrm{r}}^{-1}$ and $\Delta p=-0.50\mathrm{bar}$, the linear prediction is $\Delta T=-0.10\mathrm{K}$; these are stipulated numbers, not measured gas properties.

For a fixed-composition simple compressible system, the measurable form isHere $C_p$ is the whole sample's heat capacity at constant pressure and $\alpha$ its volumetric coefficient of thermal expansion; alternatively use molar volume and molar heat capacity consistently. An ideal gas has $\alpha =1/T$, so ${\mu }_{\mathrm{JT}}=0$. Its enthalpy depends only on temperature; positive heat capacity at constant pressure then makes any such finite throttle preserve temperature. A real gas can have either sign. For $C_p>0$, compare $T\alpha$ with one: the thermal expansion response decides the sign, not insulation alone.^[ref]

A short derivation uses $dH=TdS+Vdp$ and the Maxwell relation $(\partial S/\partial p{)}_T=-(\partial V/\partial T{)}_p$:Setting $dH=0$ gives the coefficient. These are equilibrium state function identities, not a claim of a reversible thermodynamic process inside the throttle. For a finite drop, solve the endpoint enthalpy equality or integratealong the constant-enthalpy equilibrium curve. The coefficient can change sign during that integration, as the Van der Waals throttling model illustrates.

The Joule-Thomson effect is the temperature response of a fluid flowing from higher to lower pressure through a valve or a plug of porous medium, when the throttle is insulated and the other steady-flow energy balance assumptions hold. Its equilibrium endpoints have equal specific enthalpy. The Joule-Thomson coefficient determines whether a small drop cools or warms; cooling is not universal.

A useful contrast is between three insulated expansions. In free expansion of a gas-only closed system into a vacuum, no external work occurs and $U_2=U_1$; an ideal gas returns to its initial equilibrium temperature. In a throttle, continuous flow carries pressure work across the boundary and preserves specific enthalpy under the stated assumptions; an ideal gas again has no endpoint temperature change. In a reversible adiabatic expansion of a loaded closed system, the gas supplies work, loses internal energy and cools, even if it is an ideal gas. That process preserves entropy; throttling generally produces entropy. For a small constant-enthalpy step, $dH=TdS+Vdp=0$ gives $dS=-Vdp/T>0$ when $dp<0$ and $T,V>0$. This equilibrium comparison computes endpoint entropy differences without supplying a reversible path through the plug.

The effect is useful in refrigerators because an appropriately prepared real gas can cool while passing through a simple valve. The inlet must lie in a suitable positive-Joule-Thomson coefficient region. The Joule-Thomson inversion curve makes that requirement precise; a pressure drop by itself is not a guarantee of refrigeration.

On 10 July 1908, Heike Kamerlingh Onnes's Leiden laboratory circulated purified helium through liquid air and evaporatively cooled liquid-hydrogen precooling, a regenerative heat exchanger, and an expansion valve. Returning helium cooled incoming helium; liquid collected in a transparent vacuum vessel. Reflected light revealed its liquid meniscus, crossed by thermocouple wires. Subsequent mass density and spectroscopy checks tested contamination.^[ref]

A constant-volume helium gas thermometer below the valve supplied the temperature indication. An oxygen boiling reference check disagreed with its expected reading; the real gas correction to thermodynamic temperature remained provisional. A temperature plateau suggested immersion before the liquid meniscus was recognized.^[ref]

The measurement principles separate two questions. A visible liquid meniscus establishes an interface; a gas thermometer estimates temperature through an equation of state. A stationary temperature trace can also result from a steady state without phase coexistence, so it cannot replace the interface observation. Conversely, seeing a liquid does not calibrate a thermometer. Thermometer calibration, the gas thermometer dead-space correction, departures from the ideal gas law and incomplete thermal equilibrium all matter when assigning an absolute temperature. A check far from the intended measurement range tests the instrument but does not by itself bound every low-temperature error.

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]

Observe each gas volume, temperature, pressure and moisture condition, and reduce both volumes to the same temperature, pressure and dry-gas basis. Equal scale readings alone do not establish equal amounts of gas.

Read the calibrated volume of each gas in a graduated tube or eudiometer, with the eye at the meniscus to avoid parallax. Allow the gas and the surrounding bath to reach thermal equilibrium, then read a thermometer close to the gas. Heating expands a fixed amount of gas, so a larger reading could reflect a higher temperature rather than more gas. The temperature used in a gas law must be an thermodynamic temperature.

Read the atmospheric pressure with a barometer. If a gas is confined over a liquid, also observe the difference between the inside and outside liquid levels. With $h$ the internal surface height above the external surface, the total internal pressure isHere ${\rho }_{\ell }$ is the confining density of the liquid and $g$ the gravitational acceleration; the sign reverses automatically when $h$ is negative. Equalizing the levels before taking the volume reading removes this hydrostatic pressure correction. A mercury barometer can itself require temperature and calibration corrections for accurate work.

If the gas is saturated with water vapor, subtract the vapor pressure of water at its measured temperature:The observed pressure includes both the test gas and water vapor; counting it all as test gas overestimates the amount. The saturation assumption needs time for equilibration. For a dry gas, no water vapor subtraction is made; another collecting liquid may require its own vapor pressure correction. A collecting liquid must also be chosen so that it neither appreciably dissolves nor reacts with the gas: water, for instance, is unsuitable for preserving the whole amount of freely soluble Ammonia.

For dilute gases obeying the ideal gas law, the volume $V_0$ at a common chosen $T_0,p_0$ isConsequently the corrected comparison isThis is also the ratio of amounts of substance in the ideal gas approximation. Check for leaks, trapped air, changing composition and incomplete transfer. At high pressure or near condensation, real gas behaviour may require compressibility factors; with $pV=ZnRT$, the inferred amount ratio contains the further factor $Z_2/Z_1$.

This is the one-mole, or equivalently molar volume, form of the Van der Waals equation, an approximate equation of state for a real gas:Here $p$ is the gas pressure, $T$ its absolute temperature, $R$ the molar gas constant, and $V$ the volume occupied by one mole. If $V$ denotes molar volume explicitly, its units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $b$ represents excluded volume arising from the finite size and short-range repulsion of the molecules. The volume available for translational motion is reduced from $V$ to $V-b$. This is an effective excluded volume, not simply the sum of the molecules' geometric volumes. Its molar units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $a$ measures intermolecular attraction in a mean-field approximation. Attractions reduce the force exerted on the walls, lowering the measured pressure by $a/V^2$ relative to the repulsion-only expression. Thus $p+a/V^2$ is the pressure corrected for this attraction. The squared inverse-volume dependence reflects a contribution proportional to the square of the molecular number density. The molar units of $a$ are $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$.

For an arbitrary amount $n$ measured in moles, with total container volume $V_{\mathrm{tot}}$, the corresponding formula isAt low density, when $b\ll V$ and $a/V^2\ll RT/V$, it reduces to the ideal gas law. The model captures qualitative attraction, repulsion and liquid-gas behavior, but does not supply an exact equation of state for every material.

An equation of state relates equilibrium pressure, temperature and density. The ideal-gas form $p=n{k}_{B}T$ neglects intermolecular forces and excluded volume. A low-density real-gas expansion iswhere the virial coefficients encode interaction effects and generally depend on temperature.

A simple phenomenological example is the molar Van der Waals equationHere v is molar volume, b represents excluded volume and the attractive interaction lowers pressure through a. Its dilute expansion gives $pv/(RT)=1+[b-a/(RT)]/v+\cdots$. It recovers the ideal-gas limit but is not a quantitatively universal equation for all real gases.

Below its critical temperature, the raw isotherm contains a region with $(\partial p/\partial v{)}_T>0$, corresponding to negative isothermal compressibility and mechanical instability. Equilibrium phase separation replaces that segment by a coexistence pressure satisfying equal chemical potentials, equivalently the Maxwell equal-area construction. The critical point follows from $p_v=p_{vv}=0$:The distinction between an unstable homogeneous branch and a stable two-phase state is essential when interpreting a real-gas equation of state.

In the ideal solution and ideal gas vapor approximation, each component's partial pressure equals its liquid mole fraction times its pure-component vapor pressure at the same temperature:These pressure formulas neglect the change in the pure-liquid reference between its saturation pressure and $P$; appreciable real gas behavior also requires a more general treatment. The ideal solution definition in terms of chemical potentials does not require either approximation.^[ref]

A short derivation explains the reference choice. At vapor-liquid equilibrium, each component has the same chemical potential in the liquid and vapor. For an ideal gas mixture, chemical potential of an ideal gas giveswhere $p^{\circ }$ is a reference pressure. Approximate ${\mu }_i^{\ast }(T,P)$ by ${\mu }_i^{\ast }(T,p_i^{\ast })$ and subtract the pure-component equation from the mixture equation:Taking the exponential function gives Raoult's law. Summing partial pressures and normalizing gives $P$ and $y_i$; this last step uses the ideal gas vapor assumption separately from the liquid model.

The law is named after François-Marie Raoult, who measured vapor pressures and freezing behavior of chemical solutions.^[ref]

A real gas departs from the ideal gas law because interactions and molecular volume matter. Nonideal equations of state, including the Dieterici equation, model such departures.