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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]

An adiabatic process has no heat transfer across the chosen boundary throughout the process: $\delta Q=0$, rather than merely zero net heat after incoming and outgoing transfers cancel. It is a condition on energy transfer, whereas an isothermal process holds temperature constant. A sealed cylinder can approach an adiabatic process through insulation or through motion short compared with the time needed for appreciable heat transfer. Its temperature can nevertheless change because work crosses the boundary.^[ref]

Take the trapped gas as a closed system, with fixed amount of substance and composition and no net change in bulk kinetic energy or potential energy between the endpoints. Define $Q_{\mathrm{in}}>0$ as heat entering and $W_{\mathrm{out}}>0$ as work delivered to the surroundings. The first law of thermodynamics givesHere $U$ is internal energy. An adiabatic compression normally supplies work, so $W_{\mathrm{out}}<0$; an adiabatic expansion against a resisting load delivers work, so $W_{\mathrm{out}}>0$. An adiabatic process need not preserve internal energy and need not make the closed system an isolated system. ^[ref]

For a reversible thermodynamic process, $dS=\delta Q_{\mathrm{rev}}/T$ at positive absolute temperature $T$. A reversible adiabatic process therefore has $dS=0$: it is an isentropic process. Irreversible adiabatic processes can instead generate entropy; free expansion is the existing counterexample, with its calculation in entropy change in free expansion. Conversely, constant entropy alone does not certify an adiabatic process or reversibility: outward heat can export the entropy generated internally. For a closed system the macroscopic entropy balance iswhere $T_b$ is the absolute boundary temperature where each heat transfer occurs and $S_{\mathrm{gen}}$ is internal entropy production. With no heat transfer, this reduces to $\Delta S=S_{\mathrm{gen}}$; reversibility makes $S_{\mathrm{gen}}=0$. ^[ref]

A motivating experiment measures the gas temperature directly during compression and expansion, instead of inferring it from how warm a cylinder feels. PASCO's authored TD-8565 manual describes a sealed piston cylinder, a fine nickel-wire thermometer, an absolute pressure sensor and a position readout for volume. A computer records all three channels together. The wire's electrical resistance changes with temperature; its small size permits a rapid response. ^[ref]

Use ordinary trapped air and the apparatus's designed travel stops; mount the base securely, close the valve during a stroke, and record the actual volume limits. John Hart's university demonstration uses a large transparent cylinder with a low-heat capacity thermocouple, a pressure sensor, a travel gauge and a safety stop. It explicitly keeps the release valve closed during the demonstration. The basic observable is a rise in temperature on compression and a fall on loaded expansion, alongside the corresponding pressure and volume changes. Hart discusses wall thermal conduction, sensor response and leakage as reasons real records differ from the reversible model. ^[ref]

For a proposed rapid-versus-slow comparison, let the air and walls return to the same initial thermal equilibrium before each run, then compare the same starting and finishing volume values with different stroke durations. Read temperature, absolute pressure and position continuously, including the hold after motion stops. A rapid compression gives a temporary temperature rise; holding the smaller volume lets heat leave and the temperature approach the surroundings again. Slow compression in good thermal contact allows that heat transfer during the stroke and approaches an isothermal compression. Loaded expansion reverses the signs: the rapid stroke cools, while heat entering during slow motion reduces the cooling. PASCO explicitly supports rapid nearly adiabatic processes and slow isothermal processes; the matched-run controls here are a proposed extension, not a reported set of measurements. ^[ref]

Insulation suppresses thermal conduction; it does not prove that heat transfer vanishes. As proposed controls, repeat the matched strokes with and without a removable insulating jacket, record room and wall temperatures, measure fixed-position drift, and check valve closure and seal leakage. Calibrate the thermometer and position readout independently, including any connected dead volume; account for the heat capacity of the probe rather than treating it as a passive observer. Hart explicitly identifies the small air sample relative to the probe and walls, sensor inertia and seal leakage as limitations. A fast display cannot eliminate those physical effects. ^[ref]

There are two distinct timing requirements. Let ${\tau }_{\mathrm{eq}}$ describe internal equilibration, ${\tau }_{\mathrm{probe}}$ the thermometer response, $t_{\mathrm{move}}$ the stroke duration and ${\tau }_{\mathrm{heat}}$ the effective heat transfer time constant. A useful sufficient window for a resolved, nearly reversible adiabatic process iswith negligible friction. The left inequality permits a quasistatic process and a usable readout; the right limits heat leakage. A violent stroke may satisfy only the right inequality. PASCO reports a roughly $30$–$50\mathrm{ms}$ lag of its temperature channel behind the other channels, so apparent simultaneous samples require care. If no such window exists, analyze equilibrium endpoints with the first law of thermodynamics, and regard the path as approximate. ^[ref]

The diagram's model is deliberately a hold after the stroke, when boundary work is zero. For constant whole-sample heat capacity at constant volume $C_V$ and effective thermal conductance $K>0$, it assumesThis is Newton's law of cooling with one uniform gas temperature and a fixed surrounding temperature; actual wall and probe dynamics can require more than one time constant. At $t/{\tau }_{\mathrm{heat}}=0.1$, the fraction of the initial temperature anomaly erased is $1-e^{-0.1}\simeq 0.0952$; at one time constant it is about $0.6321$. These are analytical fractions, not experimental tolerances or evidence that the preceding stroke was reversible. Measuring the hold helps estimate heat leakage independently of the motion.

The usefulness of an adiabatic process is that it separates mechanical work from heat transfer. Its reversible limit connects temperature changes, mechanical stiffness and entropy in one model: Rüchardt experiment and Clément-Desormes experiment already use that connection to measure heat capacity ratio. The contrast with free expansion explains the deeper point: suppressing heat does not determine how much work is exchanged or how much entropy is produced.

The Clément-Desormes experiment infers heat capacity ratio from a rapid pressure reduction followed by reheating. Use a rigid bottle, pump or bellows, a quick release valve, and a U-tube pressure gauge connected to the bottle and atmosphere. First pump to a small excess pressure, close the inlet, and wait for thermal equilibrium with the room at $T_0$. Measure $p_1>p_a$, where $p_a$ is local absolute atmospheric pressure. Briefly open the release valve until the bottle reaches $p_a$, then close it promptly. The remaining gas is cooler than the room. As it warms back to $T_0$ at fixed bottle volume, its pressure rises to $p_3$, with $p_a<p_3<p_1$ in the ideal procedure. The rebound reveals cooling during release. ^[ref]

The bottle loses molecules while the valve is open: it is not a closed system whose geometrical volume expands. In the well-mixed, negligible-heat transfer discharge model with constant heat capacity ratio, the remaining gas obeys $T_2/T_0=(p_a/p_1{)}^{(\gamma -1)/\gamma }$. Subsequent heating at fixed amount of substance and bottle volume gives $p_3/p_a=T_0/T_2$ by the ideal gas law. Eliminating the unmeasured $T_2$ yieldsThe logarithms use absolute pressure ratios. The approximation requires small gauge pressures $\Delta p_i\ll p_a$; $h_i$ are signed height differences between the two liquid surfaces, and the final equality uses the same liquid in both readings with hydrostatic equilibrium. It is not an instruction to insert gauge readings into the logarithmic formula. For synthetic states generated with $p_a=100000\mathrm{Pa}$, $p_1=102000\mathrm{Pa}$ and $\gamma =1.4$, the model gives $p_3\simeq 100567.39\mathrm{Pa}$. The logarithmic inference recovers $1.4$; the small-pressure approximation gives approximately $1.3961$.

Wait for a stable initial pressure after pumping and for a stable final pressure after closing. Vary the valve-open interval and initial excess pressure across repeated trials: closing too early leaves residual excess pressure, while leaving the valve open during warming lets more gas escape and suppresses the rebound. Test the seal for leaks, and read both liquid levels consistently. In a variant with incomplete release, measure the actual immediate post-release absolute pressure $p_2$ and replace the numerator by $\ln (p_1/p_2)$. The Zurich laboratory manual explains that liquid-column oscillations can delay this reading until appreciable reheating has occurred; its proposed backward extrapolation assumes an exponential thermal relaxation and must be checked against the recorded readings. These controls separate a thermal signal from valve timing, instrumental lag and leakage. ^[ref]

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]

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.

Isentropic compressibility is the compressibility at fixed entropy, particle number and composition. The conventional name adiabatic compressibility refers to a reversible adiabatic process; an irreversible adiabatic compression need not keep entropy fixed. For an ideal gas, it is ${\kappa }_S=1/(\gamma P)$, where $\gamma =C_P/C_V$ is the heat capacity ratio. It describes a steeper local pressure response than isothermal compressibility when $\gamma >1$.

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]

Normal shock tables tabulate upstream-to-downstream property ratios for a normal shock wave, typically as functions of upstream Mach number and heat capacity ratio. They encode the Rankine-Hugoniot conditions for a perfect gas rather than defining the discontinuity itself.

The Rüchardt experiment infers heat capacity ratio from the period of an oscillation of a piston supported by trapped gas. Use a vertical precision cylinder with a close-fitting movable piston, known moving mass $M$, face area $A$, trapped equilibrium volume $V_0$, and local atmospheric pressure $p_{\mathrm{atm}}$. With negligible static friction, force balance gives the absolute gas pressure $p_0=p_{\mathrm{atm}}+Mg/A$, where $g$ is gravitational acceleration. Displace the piston slightly and release it: compression raises pressure, and expansion lowers it, producing a restoring force. The oscillation is a mechanical probe of thermal response, without separately measuring two heat inputs. ^[ref]

Let $x$ be upward displacement from equilibrium, so $V=V_0+Ax$. For constant heat capacity ratio and a reversible adiabatic process, the net upward force, including the ambient load and weight, isThe linear approximation requires $|Ax|/V_0\ll 1$. Applying Newton's second law gives simple harmonic motion withHere ${\omega }_0$ is the undamped angular frequency and $\mathcal{T}$ the undamped oscillation period, rather than temperature. Record repeated cycles, measure $M,A,V_0,p_{\mathrm{atm}}$ independently, and include all connected trapped volume. An illustrative model with $M=0.020\mathrm{kg}$, $A=2.0\times 10^{-4}{\mathrm{m}}^2$, $V_0=0.010{\mathrm{m}}^3$, $p_{\mathrm{atm}}=100000\mathrm{Pa}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\gamma =1.4$ gives $p_0=100981\mathrm{Pa}$ and $\mathcal{T}\simeq 1.182\mathrm{s}$. These stipulated values are not measurements.

In Shearwood and Sloan's implementation, piston-mounted magnets induce an electromotive force in a stationary coil through Faraday's law of induction; an oscilloscope records the signal. They observed faster oscillations at smaller trapped volume values and stronger damping at small cylinder volume values. The signal tracks velocity locally, so its amplitude is not directly piston displacement. Calibration also identified inlet dead space beyond the graduated cylinder volume. ^[ref]

Repeat at smaller initial amplitudes to test the linear approximation, and compare several $V_0$ values. Heat transfer weakens the restoring force toward the isothermal process limit, biasing the inference toward $1$. Check leakage through resting-piston drift and before-and-after trapped volume readings. For viscous damping with envelope $e^{-t/\tau }$, the measured angular frequency obeys ${\omega }_0^2={\omega }_d^2+{\tau }^{-2}$; this correction does not remove thermal leakage. Internal equilibration must be faster than an oscillation period, with heat transfer and leakage slow over a run. Large amplitude or sticking friction require a richer model. ^[ref]