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Adiabatic compression decreases the volume of a sample without heat transfer across its boundary. For the fixed-composition ideal gas and endpoint assumptions in adiabatic process, compression work increases internal energy; positive heat capacity at constant volume then means a higher final temperature. The effect comes from supplied work, even though no heat enters. The amount of temperature increase depends on the work actually delivered to the gas, not on the adiabatic process condition alone.

The reversible ideal-gas adiabat supplies a particularly useful quantitative benchmark. Assume a fixed amount of substance $n$, a quasistatic process, negligible friction, only pressure-volume work, and constant heat capacity ratio $\gamma =C_p/C_V>1$. Here $C_p$ and $C_V$ are the whole sample's heat capacity at constant pressure and heat capacity at constant volume. With $R$ the molar gas constant, Mayer's relation gives $C_V=nR/(\gamma -1)$. The proof outline is $C_{V}dT=-pdV$ from the first law of thermodynamics, followed by $p=nRT/V$ from the ideal gas law and integration. The result isThese finite power laws belong to this reversible, constant-$\gamma$ ideal gas model. If heat capacity ratio varies, use the integral relation in heat capacity ratio; if the compression is irreversible, use the actual work balance, as in irreversible adiabatic piston compression. A short stroke by itself justifies neither power law. ^[ref]

Writing $r=V_1/V_2>1$ for the compression ratio, the reversible model predictsAll temperatures in these ratios are absolute. The larger pressure rise than in an isothermal compression, where $p_2/p_1=r$, explains the greater local stiffness described by isentropic compressibility.

For an original stipulated example, set $\gamma =7/5$, $p_1{V}_1=nR{T}_1=1\mathrm{kJ}$ and $r=2$. Then $T_2/T_1=2^{0.4}\simeq 1.31951$ and $p_2/p_1=2^{1.4}\simeq 2.63902$. The input work and internal energy increase are both $0.79877\mathrm{kJ}$, while $Q_{\mathrm{in}}=0$. No initial temperature or apparatus performance is implied by the chosen energy scale. Independent integration gives the signed outward workwhich is negative during compression and agrees with $-\Delta U$. This integral uses equilibrium gas pressure only because the path is reversible.

In the sensor experiment of adiabatic process, compare the pressure-volume integral with $C_V\Delta T$ as an energy check only when internal pressure is sufficiently uniform and piston dynamics permit that boundary-work estimate. Account for friction separately before comparing it with hand input work. PASCO's manual supplies a compression procedure and an illustrative logarithmic-slope result $1.39$, compared with the model value $1.40$ for a diatomic gas; that is its sample result, not new measurements here. ^[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.

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]