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Internal energy $U$ is the energy stored in the microscopic degrees of freedom of a thermodynamic system: random molecular kinetic energy, internal molecular excitations, and interaction potential energy. Its value requires a specified system and an energy reference. In thermodynamic equilibrium, internal energy is a state function; it records the present state, without remembering whether a previous change came from stirring, compression or heat transfer. This makes internal energy the common accounting quantity connecting mechanical experiments to thermal ones.^[ref]

Heat and work describe energy crossing the system boundary during a process. They are not separately stored contents of the system. For a closed system whose bulk kinetic energy and external potential energy have unchanged endpoint values, the first law of thermodynamics givesHere $Q_{\mathrm{in}}$ is positive for incoming heat, and $W_{\mathrm{on}}=-W_{\mathrm{out}}$ is positive for net work done on the system. Two routes between the same equilibrium states have the same $\Delta U$, although their heat and work can differ. The infinitesimal notation $dU=\delta Q_{\mathrm{in}}+\delta W_{\mathrm{on}}$ distinguishes a state function change from path-dependent transfers.^[ref]

For a calculated example, take one mole of a classical monatomic gas in the ideal gas regime, at fixed volume, from $300$ to $310\mathrm{K}$. With stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, its internal energy change is $3nR\Delta T/2=124.71\mathrm{J}$. A purely thermal route supplies $Q_{\mathrm{in}}=124.71\mathrm{J}$ with $W_{\mathrm{on}}=0$. An ideal insulated stirring route supplies $W_{\mathrm{on}}=124.71\mathrm{J}$ with $Q_{\mathrm{in}}=0$, measured between equilibrium endpoints after the motion dies away. The same stored increment has two different transfer histories; neither history defines a separate kind of stored energy.

The familiar internal energy of an ideal gas is a useful special case. Neglect intermolecular interactions, hold the amount of substance $n$ and composition fixed, and suppose that $f$ independent quadratic terms per particle obey the classical equipartition theorem. Choosing the constant reference contribution as zero givesHere $T$ is absolute temperature, $R$ is the molar gas constant, and $C_V$ is the whole sample's heat capacity at constant volume. A monatomic gas has $f=3$ translational terms. A rigid diatomic molecule has $f=5$ when its two rotational terms are active and vibration is frozen; activating one harmonic oscillator vibration adds two terms, one kinetic energy and one potential energy, giving $f=7$. The counting is of quadratic contributions, so a vibrational coordinate is not just one contribution.^[ref]

This constant-$f$ formula applies only over a regime with those same active terms, classical translational statistics, nonrelativistic motion and unchanged composition. Quantum freezing, quantum degeneracy, chemical reactions or appreciable intermolecular interactions invalidate that simple counting. More generally, at fixed amount and composition an ideal gas can still have $U=U(T)$ with varying heat capacity at constant volume, so use $\Delta U={\int }_{T_1}^{T_2}{C}_V(T)dT$. A nonzero constant reference adds $U_{\ast }$ to $fnRT/2$ without altering $\Delta U$.

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.

Maxwell relations turn pressure, volume and temperature measurements into information about entropy. For example, the pressure rise when heating a rigid sealed vessel determines the entropy change when expanding the same sample isothermally:The quantities on the two sides describe different experiments, yet both are mixed partial derivatives of one Helmholtz free energy. This is the useful surprise: a mechanical measurement can reveal a thermal property. Similarly, the coefficient of thermal expansion gives $(\partial S/\partial P{)}_{T,N}=-\alpha V$. ^[ref]

For a simple compressible system, hold particle number $N$ and composition fixed, and allow only pressure-volume work. Write $U$ for internal energy, $H=U+PV$ for enthalpy, $F=U-TS$ for Helmholtz free energy, and $G=H-TS$ for Gibbs free energy. Their natural-variable differentials areThus the first partial derivatives, with the subscript recording the variables kept fixed, areThe symmetry of second derivatives produces all four standard Maxwell relations, including their signs:For a brief derivation of the $F$ relation, differentiate $F_T=-S$ with respect to $V$ and $F_V=-P$ with respect to $T$: $-S_V=F_{TV}=F_{VT}=-P_T$. Each other row follows in the same way. Assume continuous second partial derivatives of the relevant thermodynamic potential in the region considered. At a phase transition, the equilibrium potential may lose this smoothness; apply the identities within each smooth thermodynamic phase, and handle latent heat and phase coexistence separately. If particle numbers vary, the potential differentials acquire chemical potential terms, so the fixed-$N$ constraints must remain explicit. ^[ref]

For an ideal gas with fixed $N$, the ideal gas law gives $P=N{k}_{B}T/V$. At fixed volume, heating changes the pressure at rate $N{k}_B/V$. Consequently an isothermal expansion at positive temperature hasFor one mole with $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $V_2=2V_1$, this is $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. At $300\mathrm{K}$, a reversible isothermal expansion absorbs $Q=T\Delta S\simeq 1728.85\mathrm{J}$ and performs equal work, since its internal energy is unchanged. The entropy difference is a state function difference: the same equilibrium endpoints give the same $\Delta S$ in an insulated free expansion, although there $Q=0$ and the process is irreversible.

The same Maxwell relations let thermal measurements determine the difference between constant-pressure and constant-volume heat capacities. In a smooth single thermodynamic phase,using heat capacity at constant volume, heat capacity at constant pressure, coefficient of thermal expansion and isothermal compressibility. Along fixed pressure, $dV=\alpha VdT$, soThus measured expansion and compression responses determine a heat capacity difference without measuring both heat capacities separately. They do not determine the absolute heat capacities or absolute entropy by themselves; an independent thermal measurement supplies that missing information. For an ideal gas, $\alpha =1/T$ and ${\kappa }_T=1/P$ reduce the result to $C_P-C_V=PV/T=N{k}_B$, which is Mayer's relation.

When the necessary higher partial derivatives exist and commute, differentiating the Maxwell relations once more also givesIntegrating these slopes determines how a heat capacity changes with volume or pressure at fixed temperature. For an ideal gas, both right-hand sides vanish because the ideal gas law is linear in temperature at either fixed volume or fixed pressure. Consequently $C_V$ and $C_P$ depend only on temperature at fixed $N$, although they need not be constant in temperature. ^[ref]

Historically, James Clerk Maxwell's Theory of Heat, first published in 1871, presents these connections using the geometry of isothermal processes and reversible adiabatic processes. His fourth edition, page 167, explicitly relates expansion on heating at fixed pressure to heat emitted during an isothermal compression: the modern $G$ relation gives $\delta Q_{\mathrm{rev}}=TdS=-T\alpha VdP$. This explains why a substance with positive coefficient of thermal expansion emits heat when compressed isothermally, whereas one with negative coefficient absorbs it. ^{[ref] [ref]}

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]

The Einstein model of a solid replaces the vibrations of $N$ identical atoms by $3N$ independent quantum harmonic oscillators with a common frequency ${\omega }_E$. Each atom has three displacement directions; the restoring force is harmonic, coupling between different atoms is replaced by a fixed local restoring potential, and anharmonicity and frequency dispersion are neglected. The crystal volume and ${\omega }_E$ are held fixed when calculating the vibrational heat capacity at constant volume.

The one-oscillator partition function isConsequently, with $T_E=\hslash {\omega }_E/k_B$,Differentiating at fixed volume givesThe zero-point energy does not contribute to this derivative. The result tends to the Dulong-Petit law $3N{k}_B$ at high temperature and vanishes exponentially at low temperature; the latter is a limitation of the single-frequency model for a real acoustic spectrum.

At fixed particle number, expand the entropy differential in $(T,p)$. The definition of heat capacity at constant pressure and the Maxwell relation $(\partial S/\partial p{)}_T=-(\partial V/\partial T{)}_p$ giveSimilarly expansion in $(T,V)$, using heat capacity at constant volume and $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$, givesAt constant pressure, $dV=V\alpha dT$. Also $dV/V=\alpha dT-\kappa dp$, so $(\partial p/\partial T{)}_V=\alpha /\kappa$. Equating the two entropy expressions at constant pressure yieldsFor a mechanically stable system with positive isothermal compressibility, this also gives $C_p\ge C_V$.

For a simple system of fixed composition, the first law of thermodynamics andgiveThusEquality of mixed derivatives gives the Gibbs free-energy Maxwell relation

The heat capacity at constant volume and heat capacity at constant pressure areRegarding $S$ as a function of $T,V$ and differentiating along a constant-pressure curve givesThe Maxwell relation from the Helmholtz free energy isTherefore the difference between constant-pressure and constant-volume heat capacities is

Along the liquid-gas phase coexistence curve, the molar Gibbs free energies agree. Differentiating $G_{\mathrm{l}}=G_{\mathrm{g}}$ along the curve givesHence the Clausius-Clapeyron relation iswhere $L=T(S_{\mathrm{g}}-S_{\mathrm{l}})$ is the molar latent heat.

If $V_{\mathrm{g}}\gg V_{\mathrm{l}}$ and the gas is ideal, then $V_{\mathrm{g}}=RT/P$, soFor constant $L$, integration yieldswhere $P_0$ is constant over the range in which the approximations apply.

For a fixed number $N$ of particles, the ideal gas equation of state isIts internal energy depends only on temperature, andWhen the heat capacity at constant volume is constant, a choice of the energy zero gives

For $\alpha =1/2$, the preceding density of states hasWhen $T\gg T_D$, the Bose-Einstein distribution has the classical approximationEach of the $2n$ modes therefore contributes $k_{B}T$ to the energy, so the Dulong-Petit law gives

When $T\ll T_D$, the upper integration limit tends to infinity. The Bose integralthen yieldsHence the heat capacity at constant volume isso $d=4$. The high-temperature constant is the classical equipartition theorem result, while the low-temperature power law makes $C_V\to 0$ as required by the third law of thermodynamics.

If $Pdx+Qdy=dF$ for a twice continuously differentiable potential $F$, then $P=F_x$ and $Q=F_y$. Equality of mixed partial derivatives gives the necessary condition for an exact differential:This condition alone does not ensure global exactness on a domain with holes.

For the Helmholtz free energy $A=U-TS$, use the stated identity for internal energy and the product rule:Thus $A_T{|}_V=-S$ and $A_V{|}_T=-p$. Equality of mixed partial derivatives yields the Maxwell relationThe temperature and volume are the independent variables here, with composition and amount of gas fixed.

Next express $S=S(T,V)$ and substitute its total differential into $dU=TdS-pdV$:Matching the coefficient of $dV$ givesThis is the thermodynamic internal pressure identity: an equation of state determines how internal energy changes under an isothermal change of volume.

For the Dieterici equation, set $c=an/(VR)$ while differentiating at fixed $V$. ThenSubstitute into the thermodynamic internal pressure identity. The term proportional to $T$ cancels $p$, leavingThe physical gas domain has $T>0$ and $V>nb$.

For the ideal gas instead, $p=nRT/V$ gives $T(\partial p/\partial T{)}_V-p=0$. Therefore $U$ is independent of volume at fixed amount of gas. Integrating the remaining condition $(\partial U/\partial T{)}_V=C_V$, with constant heat capacity at constant volume, giveswhere $U_0$ is independent of both $T$ and $V$ and fixes the reference for internal energy.

Differentiate $H=U+pV$ at fixed pressure and subtract the heat capacity at constant volume:Now use $U=U(T,V)$ and the multivariable chain rule along a constant-pressure path:Substitution cancels the fixed-volume derivative and gives the volume derivative form of the heat-capacity difference,For the ideal gas, the first term in brackets vanishes and $V=NkT/p$, so $(\partial V/\partial T{)}_p=Nk/p$. Consequently $\boxed{C_p-C_V=Nk}$ for the entire gas, with fixed particle number $N$. The extra heat capacity accounts for expansion work at constant pressure.

The required Legendre transform is the Helmholtz free energy, $\boxed{F=U-TS}$ up to an additive constant. Its differential is $dF=dU-TdS-SdT=-SdT-pdV$, so $F_T{|}_V=-S$ and $F_V{|}_T=-p$. Equality of mixed partial derivatives for a smooth thermodynamic state gives the Maxwell relationNext use the total differential $dS=S_T{|}_{V}dT+S_V{|}_{T}dV$. Along a fixed-pressure path, divide by $dT$ to get $S_T{|}_p=S_T{|}_V+S_V{|}_T{V}_T{|}_p$. Substitute the Maxwell relation to obtainMultiplying by $T$ gives the usual difference between the heat capacity at constant pressure and the heat capacity at constant volume.

Temperature describes thermal state and determines the direction of spontaneous heat transfer. A quantity of heat is energy transferred because of a temperature difference. It depends on the process and is measured in energy units; it is not another name for temperature. Equal heat inputs need not produce equal temperature rises in different masses or substances.

For a specified process, the specific heat capacity isOver a modest range with nearly constant $c$ and no phase change, $Q=mc\Delta T$. For a gas the constraint matters: heat capacity at constant pressure and heat capacity at constant volume generally differ because expansion can do work.

In method-of-mixtures calorimetry, a body of known mass $m_s$ at $T_s$ is transferred into water of mass $m_w$ at $T_w$ in a calibrated calorimeter initially also at $T_w$. If the final uniform temperature is $T_f$, the vessel and immersed fittings have heat capacity $C_{\mathrm{cal}}$, and losses are negligible,ThusStirring, quick transfer, a lid, a temperature-time record and allowance for surrounding heat exchange improve the inference. Ignoring the calorimeter's heat capacity systematically misattributes part of the energy uptake.

An alternative is electrical calorimetry: measure $E=\int VIdt$ supplied to a heater and compare the temperature rise with a blank or separately calibrated vessel. At nearly constant properties, $E=(mc+C_{\mathrm{cal}})\Delta T+Q_{\mathrm{loss}}$. An ice calorimeter instead measures the mass melted at a known phase-change temperature, using $Q=m_{\mathrm{melt}}{L}_f$ with appropriate corrections. Steam condensation similarly uses latent heat. These methods infer transferred energy from different measurable effects; a thermometer reading alone does not measure a quantity of heat.

Assume sodium supplies one conduction electron per atom. The supplied Fermi energy givesThe gas is strongly degenerate at room temperature. For one mole, $N=N_A$ and $N_A{k}_B=R=8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. Using the requested estimate,The supplied mass density determines the answer per unit volume: $N/V=(970/0.023)N_A=2.54\times 10^{28}{\mathrm{m}}^{-3}$. ThereforeThe density is not needed for the molar heat capacity once $E_F$ is supplied. Using the precise Sommerfeld expansion coefficient instead gives $0.327\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

Either electronic estimate is only about $1\%$ of the stated measured total. Most of the total heat capacity comes from lattice phonons, whose high-temperature Dulong-Petit law limit is $3R=24.94\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The remaining difference from $28\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ need not be electronic: anharmonic lattice effects and the difference between heat capacity at constant pressure and heat capacity at constant volume also contribute. The paper does not specify the measurement constraint, so the two totals should not be equated exactly.

The Debye model approximates a three-dimensional crystal as an elastic continuum supporting quantized normal modes, or phonons. At long wavelengths, the acoustic phonon branches have approximately linear dispersion relations ${\omega }_s(k)=v_{s}k$: one longitudinal branch and two transverse branches. Replace their complicated actual spectra by these linear forms, then introduce a cutoff chosen to preserve the total $3N$ vibrational degrees of freedom for $N$ atoms. This is an approximation to the spectrum, not a claim that a real crystal has no non-acoustic branches.

For periodic boundary conditions in volume $V$, each allowed wavevector occupies $(2\pi {)}^3/V$ in the space of wavevectors. Counting a spherical shell for each branch gives the density of states per unit angular frequencyAn effective speed of sound $v_D$ can be defined by $3/v_D^3=1/v_l^3+2/v_t^3$. In the single-speed Debye model, choose ${\omega }_D=v_D{k}_D$ so thatThus the Debye wavevector, Debye frequency and Debye temperature satisfyHere $g$ has units of seconds when angular frequency is measured in ${\mathrm{s}}^{-1}$; its integral is the dimensionless number of normal modes. The spherical cutoff replaces the real Brillouin zone by an equal-count continuum construction.

Each phonon mode is a quantum harmonic oscillator. Since phonon number is not conserved at thermal equilibrium, the chemical potential is zero and its Bose-Einstein distribution gives mean thermal energy $\hslash \omega /[e^{\hslash \omega /(k_{B}T)}-1]$. The zero-point energy is independent of temperature in the harmonic, fixed-volume approximation and does not contribute to the heat capacity at constant volume. Excluding that constant term,Differentiating the first integral at fixed volume avoids losing the temperature-dependent upper limit in the second one. The resulting Debye heat capacity isFor $T\ll {\Theta }_D$, extend the upper limit to infinity. Expanding $1/(e^x-1)={\sum }_{r\ge 1}{e}^{-rx}$ and integrating term by term gives ${\int }_0^{\infty }{x}^3/(e^x-1)dx=6\zeta (4)={\pi }^4/15$. ThereforeThis $T^3$ law follows from the three-dimensional ${\omega }^2$ density of states and the availability only of low-frequency modes at low temperature. It is a central success of the Debye model.

For $T\gg {\Theta }_D$, the Bose-Einstein distribution reduces to the classical equipartition theorem: every mode contributes thermal energy $k_{B}T$. Consequentlythe Dulong-Petit law, or $3R$ per mole of atoms. The simple model omits detailed phonon dispersion relation, distinct optical phonon branches, and anharmonic effects such as thermal expansion. In a metal, the low-temperature total heat capacity additionally contains a term approximately proportional to $T$ from electrons, so the lattice $T^3$ term should not be mistaken for the whole measurement.

For a reversible adiabatic process of an ideal gas, $\delta Q=0$ and $dU=-pdV$. At fixed composition and constant sample heat capacity at constant volume $C_V$,Mayer's relation gives $n_{\mathrm{mol}}R=C_p-C_V$, so $T{V}^{\gamma -1}$ is constant andThe finite-path power law requires approximately constant heat capacities as well as reversibility. When rotational or vibrational populations change appreciably, use $\int C_V(T)dT/T+n_{\mathrm{mol}}R\ln V=\mathrm{constant}$ instead; the local isentropic pressure exponent remains $\gamma (T)$.

For an intact heteronuclear diatomic gas, quantum excitation progressively activates molecular motion. With rotational and vibrational modes frozen, the three translational degrees of freedom give $C_{V,\mathrm{m}}=3R/2$ and $\gamma =5/3$. Two active rotations add $R$, giving $C_{V,\mathrm{m}}=5R/2$ and $\gamma =7/5$. One active harmonic vibration contributes a further $R$, through kinetic and potential terms, giving $C_{V,\mathrm{m}}=7R/2$ and $\gamma =9/7$.

For molecular moment of inertia $I={\mu }_{\mathrm{red}}{r}_e^2$, the rigid rotor levels are $E_J={\hslash }^{2}J(J+1)/(2I)$, so the rotational excitation scale isThe first gap corresponds to $2{\Theta }_{\mathrm{rot}}$, so the crossover occurs at temperatures of that order. For bond force constant $k_s$ and reduced molecular mass ${\mu }_{\mathrm{red}}$, the vibrational scale isTypically ${\Theta }_{\mathrm{vib}}\gg {\Theta }_{\mathrm{rot}}$. Real gases may condense at the low-temperature end or dissociate and excite electronic states at the high-temperature end; the three plateau values describe the ideal intact-molecule model.

For the piston apparatus, mechanical equilibrium requires $p_{0}A=p_{\mathrm{air}}A+mg$, hence $p_0=p_{\mathrm{air}}+mg/A$. Let $x$ be a small upward displacement, so $V=V_0+Ax$. The local isentropic compressibility givesConsequently the adiabatic gas-spring piston period isThis neglects friction, gas inertia, leakage and dissipation, with internal equilibration much faster than the oscillation and external thermal equilibration much slower. The small oscillation probes $\gamma$ locally; it need not keep $\gamma$ constant over a large temperature interval.

With $p_{\mathrm{air}}=101325\mathrm{Pa}$, the supplied dimensions give $p_0=101351.4\mathrm{Pa}$ and $B=2.20566\times 10^{-5}{\mathrm{s}}^2$. ThusHelium is close to the monatomic ideal gas value $5/3$; NO is close to the diatomic gas value $7/5$, with rotations active and vibration largely frozen. The numbers are approximate experimental inferences, rather than exact rational values.

The specific heat capacity at constant volume is the temperature derivative of specific internal energy at fixed density and composition: $c_V=(\partial u/\partial T{)}_{\rho }$. It is the corresponding extensive heat capacity at constant volume divided by material mass.