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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]

For the insulated gas-only free expansion of an ideal gas, the equilibrium endpoints are $(T,V_1)$ and $(T,V_2)$ with $V_2>V_1$. Because entropy is a state function, calculate their difference along a hypothetical reversible thermodynamic process: an isothermal expansion using a frictionless piston and a thermal reservoir at $T$. This comparison process receives heat and delivers work, even though the actual free expansion does neither. Its reversible heat increment is $\delta Q_{\mathrm{rev}}=pdV=nRTdV/V$, soHere $R$ is the molar gas constant, and the logarithm acts on a dimensionless ratio. The reversible reference gives the correct endpoint entropy difference because entropy is a state function; putting the actual $Q=0$ into $\Delta S=Q/T$ would misuse a relation requiring reversible heat transfer. For the actual isolated gas, all this increase is entropy production. Thus an adiabatic process can increase entropy and need not be an isentropic process. ^[ref]

In the figure's volume doubling, the ideal gas law gives $p_1=99.768\mathrm{kPa}$ and $p_2=49.884\mathrm{kPa}$, while $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. The reference process receives and delivers $nRT\ln 2\simeq 1728.85\mathrm{J}$ as heat and work; its thermal reservoir loses the same entropy that the gas gains. The actual free expansion has $Q_{\mathrm{in}}=W_{\mathrm{out}}=0$ and produces that entropy internally. Reversibly compressing the gas back at $300\mathrm{K}$ would require $1728.85\mathrm{J}$ of work input and rejection of equal heat. Restoring the gas alone therefore changes the surroundings: the actual release is irreversible.

For a real gas the final temperature may differ, so the ideal-gas logarithm formula cannot simply be copied. Evaluate $S(T_2,V_2)-S(T_1,V_1)$ using that gas's equilibrium properties or a suitable reversible comparison path. If the walls and water exchange heat with it, the Second law of thermodynamics constrains the total entropy of the isolated apparatus; the gas alone is then not an isolated system.

A heat engine uses a repeating thermodynamic cycle to turn part of an incoming heat transfer into net outgoing work. In the two-thermal reservoir model, the hot thermal reservoir supplies heat $Q_h>0$ at fixed temperature $T_h$, and the cold thermal reservoir receives heat of magnitude $Q_c>0$ at fixed temperature $T_c<T_h$. A Stirling engine makes the repeated motion visible; the abstract energy budget applies independently of its particular mechanism.^[ref]

Choose the entire cyclic device as a closed system, with no net change in bulk kinetic energy or potential energy. Count heat entering as positive and net work delivered as positive, as in the first law of thermodynamics. Thus $Q_h$ enters with a plus sign and $Q_c$, although a positive magnitude, enters with a minus sign. Assume these two heat transfers and the net work are the only transfers across the boundary. Over one complete thermodynamic cycle, all stored energy, including internal energy $U$, returns to its starting value:The device does not accumulate heat from cycle to cycle. More precisely, heat is a transfer, while internal energy is what can be stored; even a component that temporarily stores internal energy must return to its initial state for this boundary to be cyclic.^[ref]

For engine operation, $Q_h>Q_c$ and $W>0$. Its thermal efficiency isUse net work, after subtracting any work supplied during compression or required by auxiliaries included in the chosen device. For periodic operation, average power satisfies $P_{\mathrm{out}}={\dot{Q}}_h-{\dot{Q}}_c$ and $\eta =P_{\mathrm{out}}/{\dot{Q}}_h$, with the rates averaged over complete thermodynamic cycles.^[ref]

For a stipulated model at $T_h=600\mathrm{K}$ and $T_c=300\mathrm{K}$, let $Q_h=1000\mathrm{J}$ and $Q_c=600\mathrm{J}$ per thermodynamic cycle. Then $W=400\mathrm{J}$ and $\eta =2/5=40\%$. At two thermodynamic cycles per second, these become $2000\mathrm{W}$ incoming heat transfer, $1200\mathrm{W}$ outgoing heat transfer, and $800\mathrm{W}$ net power. These are constructed values, not measurements.

Different signs describe different tasks. With the same directions of heat transfer, $Q_h=Q_c$ gives $W=0$: a device transmitting heat without net work output. If $Q_c>Q_h$, net work is supplied to that device. A refrigerator or heat pump instead extracts heat from the cold side and delivers heat to the hot side, using positive input work $W_{\mathrm{in}}$ and the balance $Q_h=Q_c+W_{\mathrm{in}}$. Their useful output is heat transfer, measured by a coefficient of performance. Finally, a one-time isothermal expansion can have $Q=W$ and $\Delta U=0$ while leaving the volume changed: it is not a complete thermodynamic cycle. A warming device also fails the cyclic assumption, and a fuel-consuming device requires an accounting of matter crossing its boundary.

For two fixed thermal reservoirs at positive thermodynamic temperatures, the Second law of thermodynamics strengthens the energy balance toEquality requires the overall operation to be a reversible thermodynamic process, including its heat transfer to the thermal reservoirs. The Carnot cycle explains how the reversible benchmark is constructed, and Carnot's theorem explains its universality and the entropy production penalty. Use kelvin values, never ratios of degree Celsius readings. The example above lies below the $50\%$ bound: its total entropy production is $-1000/600+600/300=1/3\mathrm{J}{\mathrm{K}}^{-1}$ per thermodynamic cycle.

This is useful before building a machine: warmer cooling surroundings reduce the possible thermal efficiency, even if the mechanism is unchanged. A small observed thermal efficiency does not by itself test the Second law of thermodynamics; heat leaks, friction, incomplete recovery of work, and the measurement boundary can all reduce the reported ratio. A meaningful comparison requires the actual thermal reservoir temperatures, matched heat and net work measurements, and the hypotheses of Carnot's theorem. Finite bodies whose temperatures change are treated separately there.

The practical question already appears in the 1824 Réflexions sur la puissance motrice du feu et sur les machines propres à développer cette puissance by Sadi Carnot: pp. 6–8 ask whether machines driven by heat have a fundamental limit and whether changing the working substance can improve their output. Pages 9–11 describe heating and cooling in real machines, but use the historical caloric theory, rather than the modern first law of thermodynamics balance. The historical discussion under Carnot's theorem explains that distinction.

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]}

Constant temperature does not generally mean constant internal energy. At fixed amount and composition, within one smooth thermodynamic phase of a simple compressible system, an equilibrium identity isThe bracket vanishes for the ideal gas law $pV=nRT$. It need not vanish for a real gas: changing the volume changes the average interaction potential energy, even when the random-motion contribution stays at the same temperature. The identity follows by writing $dU=TdS-pdV$ and using a Maxwell relation; it does not require the actual process to be reversible. ^[ref]

For a Van der Waals equation model with volume per mole $v=V/n$, $p=RT/(v-b)-a/v^2$ gives $(\partial U/\partial V{)}_{T,n}=a/v^2$. Thus an isothermal expansion from $v_1$ to $v_2>v_1$, at fixed amount of substance $n$, increases internal energy bySeparating attractive particles makes their interaction potential energy less negative. For stipulated $n=1\mathrm{mol}$, $a=0.10\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$, $b=10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, $T=300\mathrm{K}$, $v_1=0.010$ and $v_2=0.020{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, the calculated increase for this one-mole sample is $5\mathrm{J}$, despite $\Delta T=0$. This dilute stable-branch model example is not a measurement of a named gas. ^[ref]

A phase transition gives another reason. During equilibrium vaporization of a pure substance at fixed pressure, the temperature can stay at its phase coexistence value while the phase fraction changes. Molecular separation changes interaction potential energy.^[ref] If the closed system does only pressure-volume work, its latent heat input obeysHere $H=U+pV$ is enthalpy. A vaporization latent heat therefore is not automatically the internal energy increment: some input supplies expansion work. A fixed temperature reading alone cannot tell how much internal energy changed.

An isothermal expansion increases the volume of a thermodynamic system at fixed temperature. For a reversible ideal gas path, the gas delivers positive work and absorbs positive heat, with $Q=W_{\mathrm{by},\mathrm{rev}}=nRT\ln (V_f/V_i)$ and $\Delta U=0$. The heat bath supplies the expansion energy; without sufficient heat transfer, doing work instead lowers the temperature. This transfer is the hot step of a Carnot cycle, and it is why an expanding gas need not cool when its surroundings continuously replenish its internal energy.

An insulated free expansion into an evacuated compartment is different. In the idealization with no heat exchange, no external work and no container energy change, $Q=W_{\mathrm{by}}=0$. Since internal energy of an ideal gas depends only on temperature, its final equilibrium temperature equals its initial value. The transient need not have a uniform temperature or pressure, so equal endpoint temperatures do not make it a reversible isotherm. Do not apply the logarithmic isothermal ideal-gas work formula to this irreversible process. Its entropy still increases by $\Delta S=nR\ln (V_f/V_i)$; entropy change in free expansion evaluates this state function difference using a hypothetical reversible path.^[ref]

For a real gas, even equal-temperature equilibrium endpoints need not have equal internal energy. An insulated free expansion can therefore change the temperature, while a controlled isothermal expansion must provide the heat needed for both its internal energy change and its delivered work. The internal energy at constant temperature article gives the Van der Waals equation example.^[ref]

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.

No cyclic device can have as its sole effect the extraction of heat from one thermal reservoir and its complete conversion into net work. A putative device absorbing $Q_h>0$ and returning to its starting state with $W=Q_h$ and no other transfer would conserve energy, yet violate the Second law of thermodynamics. Rejecting heat to colder surroundings is an essential part of the ordinary two-thermal reservoir heat engine budget.^[ref]

The cyclic and sole-effect qualifications matter. A single isothermal expansion of an ideal gas can absorb heat and deliver equal work while leaving its volume changed; restoring the initial state must also be counted. Likewise, a device discharging stored internal energy or consuming fuel has another effect. Converting supplied electrical work completely to heat through Joule heating reverses the proposed conversion and is allowed. The statement also allows cyclic devices that exchange heat with one thermal reservoir while consuming net work; it forbids positive net work output with that single heat source as the only external change.

Conversely, a hypothetical complete-conversion device could take $100\mathrm{J}$ from the hot thermal reservoir and supply $100\mathrm{J}$ of work to an ideal refrigerator that extracts $100\mathrm{J}$ from the cold thermal reservoir and delivers $200\mathrm{J}$ to the hot one. At $600\mathrm{K}$ and $300\mathrm{K}$, that refrigerator is itself admissible: its reservoir entropy changes cancel. The composite would consume no external work and move a net $100\mathrm{J}$ from cold to hot, violating the Clausius statement of the second law. Together with the construction there, this proves the equivalence of the two impossibility statements.^[ref]

For the two fixed positive-temperature thermal reservoir hypotheses, Carnot's theorem gives the sharper restriction $\eta \le 1-T_c/T_h<1$. A reversible thermodynamic process reaches that bound in the ideal Carnot cycle; an ordinary device's smaller thermal efficiency reflects its entropy production and measurement boundary. An inefficient observed engine is compatible with the Second law of thermodynamics, rather than evidence that the bound has been experimentally determined.

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.

For a reversible isothermal expansion of one mole of an ideal gas, internal energy depends only on temperature, so $\Delta U=0$. The first law of thermodynamics givesThis uses quasistatic expansion against the gas pressure. Specifying only the initial and final volumes and constant temperature would not fix heat for an arbitrary irreversible process: an insulated ideal-gas free expansion has $Q=W=0$.

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.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.

The virial expansion describes the first departures from the ideal gas law as a power series in number density. For a homogeneous, single-species gas in three dimensions at thermal equilibrium, write $n=N/V$, with $V$ the volume, and define the dimensionless compressibility factor $Z$ byHere $p$ is absolute pressure, $T$ is absolute thermodynamic temperature, and $k_B$ is the Boltzmann constant. The second virial coefficient determines the initial slope away from $Z=1$; higher coefficients describe further corrections. The symbol $Z$ here is an equation of state ratio, not a canonical partition function. ^[ref]

The left panel sketches the measurement principle; the right panel uses deliberately chosen polynomials, $Z=1+bx+6x^2$ with $x=n{\ell }^3$ and $b=\pm 2$, where $\ell$ is a reference length. These curves are not measured data. Their initial slopes have opposite signs, but both bend because of the quadratic term. Using a finite secant from $x=0$ to $x=h$ gives $(Z(h)-1)/h=b+6h$, rather than $b$: reducing the fitting range reduces this particular truncation bias. This algebra explains why measuring one nonideal pressure is insufficient to extract a reliable second virial coefficient.

Waxman and Hastings's 1971 Burnett experiment used sample and expansion chambers in a thermostatic bath, with a valve between them and a pressure transducer linked to a piston gauge. They measured successive isothermal expansions: evacuate the expansion chamber, open the valve, record the equilibrated pressure, isolate and evacuate that chamber again, then repeat. This avoids separately measuring the absolute gas amount and chamber volumes. For carbon dioxide at $50{}^{\circ }\mathrm{C}$, their measurements below $35$ bar gave a molar second virial coefficient of approximately $-102.2\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. ^[ref]

An idealized reduction makes the signal explicit. Let $q=(V_A+V_B)/V_A$ be the calibrated chamber volume ratio, $p_i$ the pre-expansion pressure, and $p_f$ the post-expansion pressure. Particle conservation during one valve opening gives $n_f=n_i/q$. At the same temperature, division of the two equations of state yieldsThus the intercept determines $q$, while the dilute-limit slope determines the second virial coefficient. In the actual experiment, helium runs calibrated the cell constant and chamber deformation corrections were included. Agreement of the carbon dioxide low-pressure intercept with the helium calibration, and the absence of an anomalous low-pressure upturn, checked for gas adsorption on chamber walls. ^[ref]

For a limited dilute-gas fit, hold temperature fixed, allow thermal equilibrium after every expansion, calibrate absolute pressure and the chamber ratio, and check leaks and sample composition. Repeat the analysis with a smaller upper pressure cutoff and an added curvature term; compare the resulting second virial coefficients with their uncertainties. Too wide a range mixes in higher coefficients; too narrow a range makes the nonideal signal hard to distinguish from instrument error. The displayed chamber equation neglects deformation, dead spaces and adsorption, so it is a teaching reduction rather than the complete experimental analysis.

Units matter. With number density in ${\mathrm{m}}^{-3}$, $B_2$ has units ${\mathrm{m}}^3$ and $B_3$ has units ${\mathrm{m}}^6$. For molar density $c=n/N_{\mathrm{A}}$ in $\mathrm{mol}{\mathrm{m}}^{-3}$, the same equation of state isHere $N_{\mathrm{A}}$ is the Avogadro constant and $R=N_{\mathrm{A}}{k}_B$ is the molar gas constant; the molar coefficients have units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and ${\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$. A coefficient tabulated per mole must therefore multiply molar density, not number density.

For the classical formulas below, use a nondegenerate gas with thermal de Broglie wavelength ${\lambda }_{\mathrm{th}}$ satisfying $n{\lambda }_{\mathrm{th}}^3\ll 1$, negligible quantum corrections to relative motion, and a stable, sufficiently short-ranged pair potential. Diluteness must also control the omitted interaction terms: $|B_{2}n|\ll 1$ alone is not enough when positive and negative contributions cancel. A finite truncation is a local low-number density approximation, not an equation of state for a dense liquid or across a phase transition. ^{[ref] [ref]}