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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]

Adiabatic expansion increases the accessible volume without heat transfer. In a loaded piston expansion of the fixed-composition ideal gas considered in adiabatic process, positive outward work comes from internal energy, so the final temperature decreases. This provides a cooling mechanism without removing heat during the stroke. In the Carnot cycle, reversible adiabatic expansion connects two different temperatures between its isothermal processes.^[ref]

For the same constant-$\gamma$ reversible ideal-gas adiabat, put $b=V_2/V_1>1$. The scaling and first law of thermodynamics giveWith the figure's stipulated $\gamma =1.4$, $p_1{V}_1=1\mathrm{kJ}$ and $b=2$, the final ratios are $T_2/T_1\simeq 0.75786$ and $p_2/p_1\simeq 0.37893$. Outward work is $0.60535\mathrm{kJ}$ and $\Delta U=-0.60535\mathrm{kJ}$. Reversing this particular path from its cooled endpoint would require the same work input; the figure's separate compression starts at the original state, so its input is different. Endpoint entropy supplies another independent check:The two logarithm terms cancel using $C_V(\gamma -1)=nR$, confirming the reversible path is an isentropic process. ^[ref]

PASCO's expansion demonstration first compresses trapped air and waits for thermal equilibrium before the rapid expansion. Its manual warns that friction-warmed cylinder regions contact the gas during expansion, compromising quantitative comparison. ^[ref] The hold separates cooling caused by expansion from simply undoing the earlier hot compression. As proposed controls, observe the temperature dip and subsequent warming, compare rapid and slow strokes from matched initial states, and assess wall heat transfer and probe lag before fitting a reversible model.

Cooling is not a universal consequence of adiabatic expansion: free expansion of an ideal gas has no outward work and returns to the same equilibrium temperature. Its entropy increases, so the reversible power laws do not describe it. Reuse free expansion, Joule's free-expansion experiment and temperature sensitivity of a free-expansion experiment for that apparatus, its energy budget and the real-gas qualifications. The useful distinction is the load: expansion cools this ideal gas when it must spend internal energy doing work.

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 ideal Brayton cycle consists of adiabatic compression, constant-pressure heat addition, adiabatic expansion, and constant-pressure heat rejection. Its ideal working substance is an ideal gas with constant heat capacities, and the insulated stages are reversible thermodynamic processes.

For pressure ratio $r_p=p_h/p_l$ and heat capacity ratio $\gamma =C_p/C_V$, a reversible ideal-gas adiabat has $T\propto p^{(\gamma -1)/\gamma }$. Thus $T_2/T_1=T_3/T_4=k=r_p^{(\gamma -1)/\gamma }$, and the constant-pressure heat transfers have $Q_{\mathrm{out}}/Q_{\mathrm{in}}=(T_4-T_1)/(T_3-T_2)=1/k$. The thermal efficiency isThis formula requires constant heat capacities and internally reversible insulated stages. An arbitrary dissipative adiabatic process does not satisfy the same pressure-temperature relation.

A Carnot cycle realizes the reversible bound in Carnot's theorem. Its four steps for an ideal gas working substance are:

The temperature versus entropy plot in the Carnot's theorem figure is therefore a rectangle traversed clockwise. Because $\delta Q_{\mathrm{rev}}=TdS$, its width is $\Delta S=Q_h/T_h$. For the illustrated engine, $\Delta S=5/3\mathrm{J}{\mathrm{K}}^{-1}$: the upper step absorbs $600\times 5/3=1000\mathrm{J}$, the lower step rejects $300\times 5/3=500\mathrm{J}$, and the enclosed area is the $500\mathrm{J}$ net work. Geometry displays the thermal efficiency without solving the gas motion. The two vertical steps are isentropic processes, not constant-temperature steps. ^{[ref] [ref]}

Reversible heat transfer is a limit of vanishing temperature differences; ordinary finite-rate transfer introduces entropy production. The Carnot cycle is a performance benchmark rather than a guarantee of useful power. Running it backward gives the ideal operation of a refrigerator or heat pump, with work supplied to move heat toward the hotter thermal reservoir. ^[ref]

A cloud chamber contains supersaturated vapor. Ionization along a charged particle's path supplies sites for nucleation; the resulting liquid drops scatter light and reveal the trail. The observed objects are drops, not magnified particles. Its beauty is that a spatially distributed microscopic interaction becomes a visible geometrical record.

Two ways of making supersaturated vapor must be distinguished. An expansion cloud chamber rapidly increases the volume of moist gas, producing a temporary sensitive interval through cooling by adiabatic expansion. A diffusion cloud chamber maintains a cold base and a warmer vapor supply, producing a continuously replenished sensitive layer through diffusion and a temperature gradient.

Consider an illustrative water–air mixture, initially saturated at $T_0=293.15\mathrm{K}$. Let $\lambda =V/V_0$ be the expansion ratio and approximate the carrier gas by a uniform ideal gas with constant heat capacity ratio $\gamma =1.40$. Before significant condensation, assume fixed numbers of carrier and vapor molecules, negligible incoming heat, and the reversible adiabatic expansion relation. The first law of thermodynamics, $n{C}_{V}dT=-pdV$, with the ideal gas law, givesThe vapor follows the same temperature, so its partial pressure scales as $T/V$. A rapid real piston motion need not be reversible; the formula is an idealized bulk estimate after neglecting flow and losses.

Use a physically motivated Integrated Clausius-Clapeyron relation for the equilibrium water vapor pressure, with stipulated rounded constants $p_0=2339\mathrm{Pa}$ at $T_0$, molar latent heat $L_m=45000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$:These constants define the example, without fitting any historical measurements. Throughout the plotted interval $1\le \lambda \le 1.17$, the temperature stays above the freezing point of water. InitiallyThus the decrease in vapor pressure initially beats dilution. At $\lambda =1.15$, the model gives $T=277.21\mathrm{K}$, $p_v=1923.33\mathrm{Pa}$, $p_{\mathrm{sat}}=809.07\mathrm{Pa}$ and $S=2.3772$.

Why does $S>1$ not instantly make every molecule a visible drop? In the capillarity approximation of classical nucleation theory, a neutral spherical drop of radius $r$ has a positive surface energy cost and a negative bulk Gibbs free energy contribution. For molecular liquid volume $v_l$, surface tension $\sigma$ and Boltzmann constant $k_B$,Differentiating $\Delta G$ shows that the critical nucleus at $r_{\ast }$ is a maximum: smaller neutral drops tend to shrink, whereas larger ones can grow. With stipulated $\sigma =0.075\mathrm{J}{\mathrm{m}}^{-2}$, $v_l=2.99\times 10^{-29}{\mathrm{m}}^3$ and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, the $\lambda =1.15$ example gives $r_{\ast }=1.353\mathrm{nm}$ and $\Delta G_{\ast }/(k_{B}T)=150.3$. A substantial neutral-drop barrier can coexist with supersaturated vapor.

This is deliberately a neutral-drop comparison. An ion changes the cluster free energy through electrical interactions and molecular hydration; the formula does not calculate the ion-assisted nucleation threshold. At nanometre radii, using bulk surface tension is itself approximate. There is no predicted droplet count, growth time or detection efficiency here. Once many drops grow, latent heat release, vapor depletion and heat transfer from the chamber walls invalidate the dry adiabatic expansion trajectory. The example therefore explains the direction of the effect and the need for nucleation, without extrapolating its curves to Wilson's cloud limits.

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]

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]}

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.

For a simple compressible system at fixed particle number, assume the internal energy has continuous second partial derivatives and $T$ is a locally invertible function of entropy at fixed volume. Its exact differential $dU=TdS-pdV$ gives the Maxwell relation $T_V{|}_S=-p_S{|}_V$. Meanwhile the constant-volume heat capacity is $C_V=T{S}_T{|}_V$. The chain rule therefore givesFor one mole of an ideal gas with constant molar $C_V>0$, this becomes $T_V{|}_S=-RT/(C_{V}V)$, whose integral curves satisfy $T{V}^{R/C_V}=\text{constant}$. It expresses cooling on reversible adiabatic expansion in terms of measurable thermal and mechanical derivatives.

Isothermal compressibility is the compressibility measured at fixed temperature, particle number and composition. A slowly compressed sealed piston in contact with a thermal reservoir can approach this constraint. For an ideal gas, differentiating $V=N{k}_{B}T/P$ at fixed temperature gives $(\partial V/\partial P{)}_{T,N}=-V/P$, hence ${\kappa }_T=1/P$. A small pressure increase $\Delta P$ therefore gives $\Delta V/V\simeq -\Delta P/P$. For a finite change, use the whole isothermal process curve $V_2/V_1=P_1/P_2$ instead.

Take one mole of classical monatomic gas, with stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, initially at $T_0=300\mathrm{K}$ and $P_0=100\mathrm{kPa}$. The initial volume is $V_0=nR{T}_0/P_0=0.024942{\mathrm{m}}^3$. Compressing to $V_0/2$ givesThe isothermal compression releases heat to the thermal reservoir; the reversible adiabatic compression retains the input work as internal energy. Reversing either curve gives the corresponding expansion, with adiabatic expansion cooling the gas.

The equilibrium curvature of Gibbs free energy provides another measurement of isothermal compressibility:This follows from $V=(\partial G/\partial P{)}_{T,N}$, with $G$ denoting Gibbs free energy. Within a stable single thermodynamic phase, increasing pressure reduces volume and makes this curvature negative. At phase coexistence, changing the phase fraction can change the volume at constant temperature and pressure; the usual finite single-phase isothermal compressibility then need not exist. Near a critical point, it can diverge. ^[ref]

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.

The ideal Otto cycle consists of reversible adiabatic compression, constant-volume heat addition, reversible adiabatic expansion and constant-volume heat rejection. For a fixed amount of ideal gas with constant heat capacity ratio $\gamma =C_p/C_V>1$ and compression ratio $r=V_1/V_2>1$, its thermal efficiency isThe reversible adiabats are isentropic processes; merely preventing heat exchange would not make an irreversible compression obey the same relations.

To derive the formula, number the four states in cycle order. The adiabats giveWith constant whole-sample heat capacity at constant volume $C_V$, the absorbed and rejected heat magnitudes areSubstitution into $\eta =1-Q_{\mathrm{out}}/Q_{\mathrm{in}}$ cancels the heat-addition temperature difference and yields the stated result.

Within this model, larger $r$ increases thermal efficiency. Real engines also depend on combustion, heat transfer, friction and the temperature dependence of heat capacities; the formula does not predict those losses.

An adiabatic change of box size preserves the mode labels, whose momenta scale as $p\propto L^{-1}$. Since $V=L^3$, differentiation gives $dp/dV=-p/(3V)$. Occupation numbers are fixed under this slow mechanical change, so summing the changes in the individual particle energies givesComparison with $dE=-PdV$ at fixed entropy and particle number yieldsThe integration notation means summing over the original momentum-mode labels; it does not assume that the density in momentum space is unchanged after rescaling. In the ultrarelativistic limit $E(p)=cp$, so $p{E}^{\prime }(p)=E(p)$ and $P=E/(3V)=\rho c^2/3$.

Adiabatic radiation obeys $d(E{a}^3)=-Pd(a^3)$, giving energy density proportional to $a^{-4}$. Equilibrium photon energy density is proportional to $T^4$, so $T\propto a^{-1}$. Therefore the photon number in a comoving volume, $n_{\gamma }{a}^3\propto T^3{a}^3$, is constant. Its entropy is also constant: photon entropy density is $({ϵ}_{\gamma }+P)/T\propto T^3$. Photon number is not a microscopic conserved charge, but equilibrium processes maintaining the spectrum do not alter this comoving-number scaling in adiabatic expansion; entropy-producing processes would invalidate the assumptions.

Differentiate $E=\rho c^2{a}^3{V}_0$ and use $dE=-PdV$, with $V=a^3{V}_0$. Division by $c^2{a}^3{V}_0$ givesThis is homogeneous adiabatic expansion without entropy production or exchange across the comoving element.

The photon number density is obtained by integrating the Planck photon distribution. With $x=h\nu /(kT)$,The integral converges, since its integrand is of order $x$ near zero and decays exponentially at infinity. Hence $n=\alpha T^3$, with $\alpha$ independent of temperature.

For equilibrium radiation undergoing adiabatic expansion, the total entropy in a comoving volume is constant. Since the entropy density is $\beta T^3$, this means $\beta T^{3}V$ is constant. A comoving volume scales as the cube of the scale factor, soThis agrees with the cosmological redshift of each photon frequency. The conclusion uses entropy conservation during the expansion.

A clockwise engine Carnot cycle has four reversible steps. At hot temperature $T_h$, isothermal expansion absorbs $Q_h=T_h\Delta S>0$ and does positive work by the gas. Adiabatic expansion then cools it to $T_c$, with zero heat and positive work by the gas. Isothermal compression at $T_c$ emits $Q_c=-T_c\Delta S$ and requires work on the gas. Finally adiabatic compression returns it to $T_h$, with zero heat and work on the gas.

For an ideal gas, the $p,V$ isotherms obey $pV=\text{constant}$ and the adiabats $p{V}^{\gamma }=\text{constant}$, the latter being steeper. In the $T,S$ plane, the isotherms are horizontal and the adiabats vertical, forming a clockwise rectangle. Heat occurs only on the horizontal legs, while work occurs on all four $p,V$ legs.Reversing the arrows gives a refrigerator, reversing the heat and work signs.

For a homogeneous canonical scalar field, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Potential-energy domination gives $P_{\varphi }\simeq -{\rho }_{\varphi }$, accelerated expansion and a decreasing comoving Hubble radius. Hubble friction can permit slow roll down a sufficiently flat potential. Enough inflation stretches an initially small causally connected region to encompass our observable universe, suppresses the relative spatial-curvature term and dilutes unwanted massive relics. quantum fluctuations supply possible seeds for cosmological structure; a successful model must also end inflation and create the hot post-inflationary universe. These statements do not imply that arbitrary initial conditions inevitably inflate.

Use the stated reduced-Planck units throughout the field calculation. Differentiating the Friedmann constraint and substituting the field equation givesDivision assumes the nontrivial expanding branch $H>0$; the vacuum endpoint follows continuously.

During rapid oscillations, require $m\gg H$ and $|\dot{A}|\ll mA$. To leading order, $\dot{\varphi }\simeq -mA\sin mt$, while the Friedmann constraint gives $H_m^2\simeq m^2{A}^2/6$. ThusAveraging over many oscillations on a timescale still short compared with $H^{-1}$ givesThe quadratic-inflaton Hubble oscillations obey, to leading order in $H_m/m$,Differentiating this expression reproduces the cosine term, with smaller corrections from the slowly varying amplitude. Instantaneously $H$ remains nonincreasing because its exact derivative is $-{\dot{\varphi }}^2/2$; an oscillatory correction does not mean oscillatory growth of the total Hubble parameter.

The mean equation integrates toThe shift $t_{\ast }$ is set by matching to the earlier evolution, not necessarily the Big Bang origin of the full solution. Averaged kinetic and potential energies are equal, so $⟨P_{\varphi }⟩\simeq 0$, $⟨{\rho }_{\varphi }⟩\propto a^{-3}$ and $A\propto a^{-3/2}$. The coherent quadratic oscillations behave as pressureless matter, not radiation, even though they are a scalar field rather than a gas of cold particles. radiation would instead give $a\propto t^{1/2}$. The averaged approximation is controlled only after oscillations are fast compared with expansion.

reheating requires interactions beyond the isolated field equations above. The oscillating inflaton can decay into other particles, whose scattering and further decays establish a thermal bath. The perturbative inflaton reheating equations areExpansion redshifts the components differently; transfer becomes efficient around $H\sim \Gamma$, and radiation domination follows when the transferred relativistic energy dominates. For suitable couplings, nonperturbative preheating by parametric resonance can precede thermalization, with backreaction and rescattering affecting its completion. Neither the decay rate nor thermalization is specified by $m$ and $V$ alone. Instantaneous reheating idealizes all this as immediate conversion of the end-of-inflation energy to radiation.^[ref]

For instantaneous-reheating horizon matching, specify the comparison: take the initial patch to be of order $H_i^{-1}$, compare the same-shaped inflationary history with approximately fixed $H_i/H_{\mathrm{end}}$, and assume instantaneous reheating followed by adiabatic expansion. Ignoring changes in relativistic degrees of freedom, $a_0/a_{\mathrm{reh}}\propto T_{\mathrm{reh}}$, while energy conservation and the Friedmann equation give $H_{\mathrm{end}}\propto T_{\mathrm{reh}}^2$. The present size of this inflated patch therefore scales asAt fixed present Hubble radius, a higher instantaneous reheating scale requires more e-folds in this standard comparison, approximately $\Delta N\simeq \ln (T_2/T_1)$, up to slow scale and entropy-degree corrections. Hotter reheating gives more subsequent expansion, but the associated initial Hubble patch is smaller by a stronger power of temperature. If instead one artificially holds the initial physical patch or $H_i$ fixed while varying the end energy independently, the post-inflationary geometric effect alone would give the opposite trend. temperature without an inflation-scale comparison convention does not fix that alternative problem; this is why the matching assumptions must be stated.^[ref]

In the Carnot cycle, let $1\to 2$ be reversible isothermal expansion at $T_h$, $2\to 3$ reversible adiabatic expansion, $3\to 4$ isothermal compression at $T_c$, and $4\to 1$ adiabatic compression. The pressure–volume curve is clockwise. The entropy–temperature figure below uses temperature on the horizontal axis and entropy on the vertical axis, literally an $S$–$T$ plot: isotherms are vertical and adiabats horizontal. The hot isotherm increases entropy by $\Delta S$ and absorbs $Q_h=T_h\Delta S$; the cold one rejects $Q_c=T_c\Delta S$. Thus $W=(T_h-T_c)\Delta S$ and ${\eta }_C=1-T_c/T_h$.

For the second engine, choose the working direction with heat addition during $A\to B$ at $p_1$ and rejection during $C\to D$ at $p_2$. A reversible ideal-gas adiabat obeys $T\propto p^{(\gamma -1)/\gamma }$. DefineThe two constant-pressure heat transfers are $Q_{\mathrm{in}}=c_p(T_B-T_A)$ and $Q_{\mathrm{out}}=c_p(T_C-T_D)=Q_{\mathrm{in}}/r$, where rejected heat is quoted positive. Therefore the Brayton cycle efficiency isfor $\gamma =5/3$. All heat-capacity amount conventions cancel in the ratio.

Since $T_B>T_A$ and $T_C=T_B/r$, $T_D=T_A/r$, the maximum is at B and the minimum at D. The Brayton-Carnot efficiency comparison isThe inequality is strict for a nondegenerate cycle with positive heat input. Although each process can be reversible, the isobaric heat transfer spans a range of temperatures; making the whole cycle reversible requires a succession of matched-temperature reservoirs, rather than just the hottest and coldest reservoirs.

For an ideal diatomic gas with active rotation and frozen vibration, $\gamma =7/5$ andHere $0<p_2/p_1<1$. This comparison holds at the same pressure ratio and assumes temperature-independent heat capacities in the stated mode regime. Frozen rotation or excited vibration would require their actual heat capacities.