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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.

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]

Heat capacity measures the heat supplied per small temperature increase along a specified process:The subscript $X$ records the constraint, such as fixed volume or fixed pressure. The units are $\mathrm{J}{\mathrm{K}}^{-1}$. The symbol $\delta Q$ emphasizes that heat is a transfer along a path, rather than a state function with a differential $dQ$. Consequently, an unspecified heat capacity does not determine the required heat.

For a closed system of fixed composition undergoing a quasistatic process, assume that only pressure-volume work occurs, with no friction and no change in bulk kinetic energy or potential energy. The first law of thermodynamics and $H=U+pV$, where $U$ is internal energy and $H$ is enthalpy, then giveThese partial derivatives define heat capacity at constant volume and heat capacity at constant pressure, with the amount and composition also held fixed. Integrating along the selected constraint gives $Q={\int }_{T_1}^{T_2}{C}_X(T)dT$ within a single thermodynamic phase; $Q=C_X\Delta T$ requires constant $C_X$. ^[ref]

For a computed example, take a monatomic gas in the classical ideal gas regime, with amount of substance $n=1\mathrm{mol}$ and stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The equipartition theorem gives $U=3nRT/2$, and the ideal gas law gives $pV=nRT$. ThusFor the same temperature increase $\Delta T=10\mathrm{K}$, the two processes requireHere work done by the gas is positive. At fixed pressure, the extra heat supplies expansion work; the internal energy increase is the same because internal energy of an ideal gas depends only on temperature. Their final volume values differ. ^[ref]

The practical value of heat capacity is that a measured heat input and temperature response can determine it. Once its constraint and temperature dependence are known, it predicts heating requirements and supplies entropy changes through entropy from heat capacity. ^[ref]

Internal energy $U$ is the energy stored in the microscopic degrees of freedom of a thermodynamic system: random molecular kinetic energy, internal molecular excitations, and interaction potential energy. Its value requires a specified system and an energy reference. In thermodynamic equilibrium, internal energy is a state function; it records the present state, without remembering whether a previous change came from stirring, compression or heat transfer. This makes internal energy the common accounting quantity connecting mechanical experiments to thermal ones.^[ref]

Heat and work describe energy crossing the system boundary during a process. They are not separately stored contents of the system. For a closed system whose bulk kinetic energy and external potential energy have unchanged endpoint values, the first law of thermodynamics givesHere $Q_{\mathrm{in}}$ is positive for incoming heat, and $W_{\mathrm{on}}=-W_{\mathrm{out}}$ is positive for net work done on the system. Two routes between the same equilibrium states have the same $\Delta U$, although their heat and work can differ. The infinitesimal notation $dU=\delta Q_{\mathrm{in}}+\delta W_{\mathrm{on}}$ distinguishes a state function change from path-dependent transfers.^[ref]

For a calculated example, take one mole of a classical monatomic gas in the ideal gas regime, at fixed volume, from $300$ to $310\mathrm{K}$. With stipulated molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, its internal energy change is $3nR\Delta T/2=124.71\mathrm{J}$. A purely thermal route supplies $Q_{\mathrm{in}}=124.71\mathrm{J}$ with $W_{\mathrm{on}}=0$. An ideal insulated stirring route supplies $W_{\mathrm{on}}=124.71\mathrm{J}$ with $Q_{\mathrm{in}}=0$, measured between equilibrium endpoints after the motion dies away. The same stored increment has two different transfer histories; neither history defines a separate kind of stored energy.

The familiar internal energy of an ideal gas is a useful special case. Neglect intermolecular interactions, hold the amount of substance $n$ and composition fixed, and suppose that $f$ independent quadratic terms per particle obey the classical equipartition theorem. Choosing the constant reference contribution as zero givesHere $T$ is absolute temperature, $R$ is the molar gas constant, and $C_V$ is the whole sample's heat capacity at constant volume. A monatomic gas has $f=3$ translational terms. A rigid diatomic molecule has $f=5$ when its two rotational terms are active and vibration is frozen; activating one harmonic oscillator vibration adds two terms, one kinetic energy and one potential energy, giving $f=7$. The counting is of quadratic contributions, so a vibrational coordinate is not just one contribution.^[ref]

This constant-$f$ formula applies only over a regime with those same active terms, classical translational statistics, nonrelativistic motion and unchanged composition. Quantum freezing, quantum degeneracy, chemical reactions or appreciable intermolecular interactions invalidate that simple counting. More generally, at fixed amount and composition an ideal gas can still have $U=U(T)$ with varying heat capacity at constant volume, so use $\Delta U={\int }_{T_1}^{T_2}{C}_V(T)dT$. A nonzero constant reference adds $U_{\ast }$ to $fnRT/2$ without altering $\Delta U$.

Mayer's relation states that the difference between heat capacity at constant pressure and heat capacity at constant volume for a fixed-composition ideal gas isHere both capacities are total heat capacities in $\mathrm{J}{\mathrm{K}}^{-1}$, $n$ is amount of substance, $N$ is the number of particles, $R$ is the molar gas constant and $k_B$ is the Boltzmann constant. In other normalizations the same statement isThe first is a difference of molar heat capacities, in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$; the second uses specific heat capacities, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Writing $C_p-C_V=R$ with total capacities would omit the sample's amount of substance. The relation holds pointwise even when the capacities depend on temperature. ^[ref]

The proof outline is short: enthalpy is $H=U+pV=U+nRT$, and internal energy of an ideal gas is independent of volume at fixed temperature. Differentiating at fixed composition and amount of substance therefore gives $C_p=dH/dT=dU/dT+nR=C_V+nR$. It uses the ideal gas law, rather than assuming any particular number of molecular degrees of freedom. At fixed pressure, the additional heat supplies expansion work, $pdV=nRdT$. The two processes can have the same temperature endpoints and equal $\Delta U$ while requiring different heat inputs.

For one mole of the rigid-rotation diatomic ideal gas model, $C_V=5R/2$ and $C_p=7R/2$. Heating by $10\mathrm{K}$ with $R\simeq 8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ gives $Q_V\simeq 207.86\mathrm{J}$ in a rigid sealed vessel and $Q_p\simeq 291.01\mathrm{J}$ under a frictionless piston at fixed pressure. Both gain $207.86\mathrm{J}$ of internal energy, while the piston delivers approximately $83.14\mathrm{J}$ of work. This is a calculated model budget. If the capacities vary with temperature, the corresponding finite statement is stillunder the fixed-pressure and fixed-volume heating assumptions in heat capacity. Their final volume values generally differ. Combining Mayer's relation with heat capacity ratio also gives $C_V=nR/(\gamma -1)$ and $C_p=\gamma nR/(\gamma -1)$ locally; constant values cannot be extrapolated across intervals in which $\gamma (T)$ changes.

For a smooth stable single thermodynamic phase of a general simple compressible system, the broader Maxwell relation identity iswhere $\alpha$ is the coefficient of thermal expansion and ${\kappa }_T$ the isothermal compressibility, with composition fixed throughout. Substituting the ideal gas law gives $nR$; a real gas need not do so. This distinction matters near a phase transition or when interactions appreciably change the equation of state. It explains why Mayer's relation is useful: thermal measurements under different constraints become linked to a mechanical response, while the difference retains a particularly simple value for an ideal gas.^[ref]

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.

A thermodynamic system is in thermodynamic equilibrium when its macroscopic state has no spontaneous tendency to change through any exchange allowed by its constraints. For an isolated system, stable thermodynamic equilibrium maximizes entropy over the accessible states at fixed total energy and conserved material inventories. A wall can forbid an exchange, so always specify what can cross or move. Thermal, mechanical and chemical conditions answer different questions about that same constrained state. ^[ref]

For example, two compartments separated by a rigid, impermeable wall that transmits heat can reach thermal equilibrium while retaining different pressures and compositions. Allowing an unloaded piston to move adds a mechanical equilibrium condition. Allowing one chemical species to cross adds chemical potential equality for particle exchange; allowing a chemical reaction adds its chemical equilibrium condition. A fixed barrier can therefore support a constrained thermodynamic equilibrium that ceases to be an equilibrium when the barrier is removed.

Uniformity is not the definition of thermodynamic equilibrium. An isothermal atmosphere in hydrostatic equilibrium can have a common temperature while its pressure and mass density vary with height: the pressure gradient balances gravity. In an external field, the exchange condition must include the relevant potential energy; for charged matter this leads to the electrochemical potential. Relativistic gravitational corrections to thermal equilibrium are described separately by the Tolman temperature law.

A steady state only says that the macroscopic description is constant in time. A bar maintained between hot and cold thermal reservoirs has steady heat flux and produces entropy, so it is not in thermodynamic equilibrium. Even matching temperatures at one perfect-contact interface can coexist with heat transfer through the surrounding material; matching those two local values does not establish equilibrium of the whole apparatus. See thermal conduction for that spatial transport problem.

The useful simplification is that equilibrium conditions predict an endpoint without computing every collision or flow during relaxation. The first law of thermodynamics supplies the conserved quantities, and the Second law of thermodynamics supplies the direction of change. For a system held at fixed temperature and volume, the corresponding criterion is minimization of Helmholtz free energy; at fixed temperature and pressure, it is minimization of Gibbs free energy, with the material and exchange constraints still imposed.

If two thermodynamic systems are each in thermal equilibrium with the same third system in the same equilibrium state, they are in thermal equilibrium with each other. Write $A\sim B$ for this relation. The zeroth law of thermodynamics supplies transitivity:Its symmetric relation property and self-equilibrium (reflexivity), together with transitivity, make thermal equilibrium an equivalence relation. Its equivalence classes can be labelled by a common empirical temperature; choosing an absolute numerical scale requires the further structure of thermodynamic temperature.^[ref]

The comparison assumes internally equilibrated macroscopic bodies, unchanged composition and mechanical constraints, and contact permitting only heat exchange. Use rigid, impermeable diathermal walls so that neither matter transfer nor boundary work spoils the test. The same state of $C$ matters: letting $C$ warm while measuring $A$, then cool while measuring $B$, does not supply the premise. Ordinary nonrelativistic laboratory thermodynamics is assumed; the Tolman temperature law treats gravitational qualifications. Equal temperatures establish the thermal condition, without requiring mechanical equilibrium or chemical equilibrium.

This explains why a thermometer can compare unlike materials: it supplies a reusable third system. A temperature is a property of an equilibrium state, while heat is energy transferred because of a temperature difference. Two bodies at the same temperature can have different internal energies and heat capacities. A thermometer therefore does not measure a body's stored amount of heat; the finite-body disturbance and waiting-time calculations in thermometer describe how making the comparison can change the state being measured.^[ref]

The name “zeroth” reflects logical order rather than the date of the first thermometer: this premise about thermal equilibrium precedes the already numbered first law of thermodynamics, Second law of thermodynamics and Third law of thermodynamics. This is the naming explanation given in University Physics Volume 2 by Samuel J. Ling and others (§1.1, “Thermal Equilibrium”).^[ref]

The corresponding Zeroth law of black-hole mechanics states constancy of surface gravity over an equilibrium horizon; Hawking radiation identifies that quantity with physical temperature up to its universal conversion factor.