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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.

Boyle's law states that a fixed amount of substance of an ideal gas at constant absolute thermodynamic temperature has constant product of absolute pressure and total volume:Here $p$ is measured relative to a vacuum, not relative to the surrounding atmospheric pressure. Halving the volume doubles the absolute pressure; doubling the volume halves it. A real gas approximately follows Boyle's law when it is sufficiently dilute and far from a phase transition.^[ref]

For an original calculated example, suppose a sealed gas sample has $p_1=100\mathrm{kPa}$ and $V_1=100\mathrm{c}{\mathrm{m}}^3$. After isothermal compression to $V_2=50\mathrm{c}{\mathrm{m}}^3$,Both states have $C=10000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10\mathrm{J}$, since $1\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10^{-3}\mathrm{J}$. The volume ratio is dimensionless, so using the same volume units in numerator and denominator leaves the pressure in kilopascals. Although $pV$ has units of energy, its value is not the compression work: isothermal ideal-gas work involves an integral along the path.

The blue curves show calculated ideal gas states, not measurements. The left panel is a branch of a hyperbola; the right panel plots exactly the same absolute pressures against $1/V$. The dashed line uses gauge pressure with a stipulated constant atmospheric pressure of $100\mathrm{kPa}$. Its displaced intercept explains why absolute pressure in Boyle's law matters.

A short derivation starts from the ideal gas law, $pV=nRT$. Fixing the amount of substance $n$ and absolute thermodynamic temperature $T$ fixes $C=nRT$, where $R$ is the molar gas constant. The microscopic picture in ideal gas pressure from wall collisions is equally appealing: at fixed temperature, the typical particle kinetic energy stays fixed, while compression increases the number density and wall momentum transfer per unit surface area and time. A larger absolute pressure follows without requiring hotter particles.

The constraints are essential. A leak changes the amount of substance. Rapid compression can raise the temperature before heat transfer restores thermal equilibrium, as in the isothermal gas syringe experiment. Molecular size and interactions in a dense real gas invalidate the ideal gas approximation; the Van der Waals equation describes one such correction. Equal endpoint temperatures allow an endpoint test of Boyle's law, but do not establish that every intervening state lies on an isothermal process.

For fixed mass $m$, mass density satisfies $\rho =m/V=mp/C$, so it is proportional to absolute pressure. In a dilute bubbly flow, Boyle's law allows bubble volume and its contribution to buoyancy to increase as hydrostatic pressure decreases, provided gas mass is retained, dissolution is negligible, and the temperature stays approximately fixed. Its usefulness is that a simple ratio predicts a substantial size change without tracking individual molecules.

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]

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]

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]

An isothermal compression reduces the volume of a thermodynamic system at fixed temperature. For a fixed-composition ideal gas compressed reversibly from $V_i$ to $V_f<V_i$, isothermal ideal-gas work givesThe surroundings supply positive work $W_{\mathrm{on}}$, and the gas rejects the same amount of heat to the heat bath. Compressing $1\mathrm{mol}$ at $300\mathrm{K}$ from $20$ to $10\mathrm{L}$ requires approximately $1.729\mathrm{kJ}$ of work input and rejects $1.729\mathrm{kJ}$ of heat. If that heat transfer cannot keep up, the temperature rises: the process departs from isothermal compression toward adiabatic compression. The finite-rate isothermal piston model explains the required small temperature excess.

A gas syringe makes the distinction between a constant-temperature path and equal-temperature endpoints observable. The Ideal Gas Law Apparatus (TD-8596A), Instruction Manual 012-08420B by PASCO scientific specifies a TD-8596A gas syringe with a mechanical stop, an internal thermistor, an attached pressure sensor and data acquisition. Record the stop volume with the coupling disconnected, then set $40\mathrm{mL}$ and seal the coupling. Record pressure and temperature, compress quickly to the stop near $20\mathrm{mL}$, and hold until both readouts settle. The manual gives roughly $30\mathrm{s}$ for settling and about $0.5\mathrm{s}$ for the thermistor response. Releasing the plunger gives a second transient and a final volume to record.^[ref]

The manual describes a pressure overshoot and temperature rise during rapid compression, followed by cooling at the held volume; the settled pressure remains higher than initially. This illustrates approximately adiabatic compression followed by heat rejection, rather than an isothermal compression throughout. Use the settled initial and compressed readings to test Boyle's law, correcting both syringe readings by the unmarked tubing volume $V_d$:Use absolute pressure, not gauge pressure. A transient comparison of $pV/T$ must pair simultaneous pressure and temperature readings: the sensor lag can separate their peaks.^[ref]

For an additional test of the path, repeat with small compression steps, wait for stable temperature and pressure at every step, and repeat in reverse. Keep the sealed amount of substance, connecting tubing and ambient temperature fixed; prevent hand warming of the barrel and check for leaks by watching for continued pressure drift at held volume after thermal equilibrium. These are proposed controls, rather than reported measurements. Agreement of equilibrated points with Boyle's law supports the fixed-temperature ideal gas law approximation, but does not prove that the rapid segments between them were isothermal processes. A single local temperature sensor can miss temperature gradients; plunger friction can produce direction-dependent readings. The pressure and volume data estimate reversible work only insofar as the path approaches a reversible thermodynamic process; they do not directly measure heat. After motion stops, cooling at fixed volume has zero boundary work and $Q=\Delta U<0$, even though the complete equal-endpoint-temperature ideal gas run has $\Delta U=0$.

For a reversible thermodynamic process that changes the volume of a fixed amount of ideal gas at absolute temperature $T$, the work done by the gas isHere $n$ is the amount of substance, $R$ the molar gas constant, $p$ the pressure, and $\ln$ the natural logarithm. The ideal gas law supplies $p=nRT/V$; the integral is the signed area under the pressure-volume curve. Isothermal expansion gives positive work; isothermal compression gives negative work. A convention counting work done on the gas instead has $W_{\mathrm{on}}=-W_{\mathrm{by}}$.^[ref]

For a calculated example, let $n=1\mathrm{mol}$, $T=300\mathrm{K}$ and $V_i=10\mathrm{L}$, $V_f=20\mathrm{L}$. The pressure falls from approximately $249.4$ to $124.7\mathrm{kPa}$, andTraversing the same isothermal process backward changes both transfer signs: $W_{\mathrm{by},\mathrm{rev}}=Q\approx -1.729\mathrm{kJ}$. Neither direction changes internal energy of an ideal gas. These stipulated values illustrate the energy balance; they are not experimental measurements.

The shaded geometric area is positive in both panels; the direction of the integral supplies the sign of work. Knowing only the two equilibrium endpoints does not justify shading this curve for an irreversible process. For an idealized piston moving against a specified external resisting pressure, boundary work is $W_{\mathrm{by}}=\int p_{\mathrm{ext}}dV$. Substitution of the equilibrium gas pressure requires the appropriate reversible limit. In free expansion into vacuum, $p_{\mathrm{ext}}=0$ and this boundary work is zero.

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]}

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.

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.

Use an ideal gas to illustrate the Carnot cycle, with hot and cold heat baths at $T_h=T_2$ and $T_c=T_1$. The four processes are all reversible thermodynamic processes. Lettered states below belong only to this illustration; the engine later in the question uses numbered states.

Starting at $A$, perform the following four stages:

The clockwise enclosed area in the pressure-volume diagram is the positive net work per cycle, $W=\oint pdV=Q_h-Q_c$.

Sadi Carnot's ideal engine was deeply indebted to practical heat engines as a source of questions and physical constraints, while its decisive result depended on deliberate idealization. In 1824 steam power was an established technology, and variations in engine economy made the conversion of heat into useful work a problem with immediate industrial meaning. Comparing different working substances and operating temperatures required a theory more general than the description of one machine.

Real engines supplied the essential architecture: a working fluid, heating from a hot source, expansion producing work, cooling or condensation, and return to an initial condition. Their losses also suggested what an ideal comparison had to remove. Friction, leakage, finite temperature differences during heat exchange and incomplete expansion waste useful performance. The idea of obtaining more work by using expansion rather than simply discarding pressure connected the argument directly to engineering experience. Sadi Carnot's opening discussion explicitly places the inquiry in the context of the steam engine's practical power.^[ref]

The theoretical advance was to replace those irreversible operations by a closed thermodynamic cycle. In its modern description the Carnot cycle contains an isothermal expansion at $T_h$, an adiabatic expansion, an isothermal compression at $T_c$, and an adiabatic compression. The fluid returns to its initial state, so the result can be compared repeatedly rather than consuming a hidden initial store. Reversibility permits an engine to be run backwards and paired with another engine, providing a powerful limiting argument independent of detailed construction.

It is essential not to attribute the whole modern interpretation to 1824. Sadi Carnot reasoned within a caloric theory in which heat was conserved and motive power was associated with its passage from hot to cold, rather like a fall between levels. The modern first law of thermodynamics instead gives $W=Q_h-Q_c$, and the reversible entropy balance gives $Q_h/T_h=Q_c/T_c$. Thus the modern resultexpresses a lasting structure of his argument in a later thermodynamic language; it should not be presented as his original derivation with contemporary concepts.

Actual engines therefore did more than decorate an abstract proof: they provided the phenomenon to explain, the operating ingredients and the direction in which improvements were sought. Yet no real engine executes a perfectly reversible Carnot cycle; finite power requires finite rates, and practical machines face material and transfer constraints. The engine is most valuable as a comparative upper limit and an argument for universality. Its historical interest lies in extracting such a general limit from engineering knowledge using an initially mistaken theory of heat, not in either pure deduction without machinery or mere generalization of workshop measurements.