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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]

A frictionless piston of area $A$ and mass $m$ supported by gas has $p_0=p_{\mathrm{ext}}+mg/A$. Small displacement $x$ changes its volume by $Ax$. The local isentropic compressibility gives $\delta p=-\gamma p_{0}Ax/V_0$, so Newton's second law becomes $m\ddot{x}=-\gamma p_0{A}^{2}x/V_0$. The period follows. Gas inertia and heat leakage are neglected, and internal equilibration must be faster than the oscillation.

Compressibility measures how readily a sample's volume decreases when its pressure increases. For a simple compressible system with fixed particle number $N$ and composition, specify the constraint $X$:Here $P$ denotes pressure, also written $p$ elsewhere. The minus sign makes compressibility positive on a mechanically stable equilibrium branch. Its units are inverse pressure, $\mathrm{P}{\mathrm{a}}^{-1}$; its reciprocal $K_X=1/{\kappa }_X$ is the constrained bulk modulus, measured in pascal units. Holding temperature fixed gives isothermal compressibility; holding entropy fixed gives isentropic compressibility ${\kappa }_S$ along an isentropic process. The latter describes a reversible adiabatic compression, since an irreversible adiabatic process can increase entropy. ^[ref]

For an ideal gas, the ideal gas law $PV=N{k}_{B}T$ and the reversible ideal-gas adiabat give, at the same equilibrium state,Here $C_P$ and $C_V$ are the whole sample's heat capacity at constant pressure and heat capacity at constant volume, and $\gamma$ is the heat capacity ratio. Constant $\gamma$ is required for the finite-path formula $P{V}^{\gamma }=\mathrm{constant}$. Reversible adiabatic compression raises the temperature, making the pressure rise more steeply than in an isothermal compression. Thus keeping entropy fixed makes the gas harder to compress locally. For a classical monatomic gas at $100\mathrm{kPa}$, $\gamma =5/3$, so ${\kappa }_T=10^{-5}\mathrm{P}{\mathrm{a}}^{-1}$ and ${\kappa }_S=6\times 10^{-6}\mathrm{P}{\mathrm{a}}^{-1}$; the corresponding values of the bulk modulus are $100\mathrm{kPa}$ and $166.67\mathrm{kPa}$.

For a smooth, stable single thermodynamic phase at positive temperature, Maxwell relations also implywhere $\alpha$ is the coefficient of thermal expansion. The ideal gas result is one example of this broader link between mechanical and thermal responses. The equality ${\kappa }_T={\kappa }_S$ is possible when $\alpha =0$ and the other responses remain finite. These formulas compare equilibrium partial derivatives, rather than arbitrary fast or dissipative compression histories. ^[ref]

In a smooth fixed-composition thermodynamic region, write $a=(\partial S/\partial T{)}_V$, $b=(\partial S/\partial V{)}_T$, $c=(\partial p/\partial T{)}_V$, and $d=(\partial p/\partial V{)}_T$, with $a,d\ne 0$. The chain rule gives $(\partial S/\partial T{)}_p=a-bc/d$ and $(\partial p/\partial V{)}_S=d-bc/a$. Multiplying these expressions by $d$ and $a$, respectively, gives the same Jacobian determinant $ad-bc$.

Using heat capacity at constant pressure $C_p=T(\partial S/\partial T{)}_p$, heat capacity at constant volume $C_V=Ta$, isothermal compressibility ${\kappa }_T=-1/(Vd)$ and isentropic compressibility ${\kappa }_S=-1/[V(\partial p/\partial V{)}_S]$ proves the displayed ratio. The derivation does not require division by thermal expansion, so it still applies when that derivative vanishes. For a stable single phase, the Maxwell relation $b=c$ also explains why $C_p\ge C_V$ and ${\kappa }_T\ge {\kappa }_S$.

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 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.