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Absolute zero is the zero of thermodynamic temperature, corresponding to $-273.15{}^{\circ }\mathrm{C}$. For a system with a nondegenerate ground state, equilibrium entropy approaches zero as temperature approaches absolute zero. This does not mean that every quantum degree of freedom loses its zero-point energy. The entropy statement and the operational unattainability of absolute zero require their respective physical assumptions.

A small increase in temperature can produce a large increase in a reaction rate. In the simplest Arrhenius equation model, warming from $300$ to $310\mathrm{K}$ increases a rate constant with $E_a=80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ by about $2.81$ times, even though the absolute thermodynamic temperature rises by only about $3.33\%$. This sensitivity helps explain why cooling can slow unwanted chemical reactions and why a hot process can become difficult to control. The comparison assumes the same rate law and reactant concentrations; the activation energy characterizes $k(T)$, rather than every change in the measured reaction rate.

The physical motivation is the distribution of molecular energies. Warming changes the Boltzmann distribution, increasing the relative population of energetically costly configurations. A transition state can therefore become much more accessible than the modest change in average kinetic energy might suggest. The exponential function in an Arrhenius equation captures such sensitivity in a simple model; it is not a universal formula for the fraction above every barrier in a Maxwell-Boltzmann distribution.^[ref]

Activation energy quantifies how strongly a rate constant depends on thermodynamic temperature. For a positive rate constant $k(T)$, the empirical definition iswhere $R$ is the molar gas constant and $k_{\mathrm{ref}}>0$ is a fixed reference with the same units as $k$. The ratio is a dimensionless quantity, keeping the natural logarithm meaningful without changing its derivative. Thus $E_a$ is a molar energy, commonly in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ or $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The Arrhenius equation makes $E_a$ constant over its assumed range. This rate-based definition, and the need for care when identifying it with a threshold on a potential energy surface, are stated in the activation energy.

In a simple barrier picture, reaching a transition state costs energy relative to the starting reactants. The height of that barrier is different from the difference between the reactants and reaction products. An energetically favourable endpoint can still be reached slowly. An energy profile makes the distinction visible, while an Arrhenius plot tests temperature sensitivity from rate constants. A measured $E_a$ need not equal a barrier read from an arbitrarily chosen diagram.

For a computed example, assume a first-order reaction has $k(300\mathrm{K})=0.100{\mathrm{s}}^{-1}$, constant $E_a=80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, and a temperature-independent pre-exponential factor. The Arrhenius equation givesHence $k(310\mathrm{K})\approx 0.281398{\mathrm{s}}^{-1}$. Its half-life falls from $\ln 2/0.100\approx 6.93147\mathrm{s}$ to $\ln 2/0.281398\approx 2.46322\mathrm{s}$. These are chosen model parameters, not measurements of an identified chemical reaction. The exponent uses kelvin, not unconverted Celsius temperatures, and $80000\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ must match $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

A microscopic reaction energy barrier belongs to a specified path on a potential energy surface. An experimental activation energy belongs to a specified rate constant and its temperature dependence. A multi-step reaction mechanism can change which step controls the observed conversion, and a temperature-dependent pre-exponential factor contributes to the observed activation energy. Even the Gibbs energy of activation of transition state theory is a different quantity: it incorporates entropy as well as enthalpy and depends on the standard states. These distinctions prevent a fitted Arrhenius plot from being mistaken for a direct microscopic energy measurement.^[ref]

Catalysis illustrates the separation between a path and its endpoints. At the same temperature and with the same reactants, reaction products, and standard states, a regenerated catalyst changes the available reaction mechanism and rate constants, while the overall standard change in Gibbs free energy and the equilibrium constant remain the same:For a reversible first-order reaction pair $\mathrm{A}\rightleftharpoons \mathrm{B}$, consistency at chemical equilibrium requires $k_f/k_r=K$. A model that multiplies only $k_f$ by a catalytic speedup while holding $k_r$ fixed would change this ratio and misrepresent catalysis. This simple ratio applies to that specified rate law; a multi-step catalytic reaction mechanism needs its own rate equations. A potential-energy profile alone cannot determine $K$, because its endpoint difference is not generally the standard change in Gibbs free energy.^[ref]

The Arrhenius equation models the temperature dependence of a rate constant byHere $A$ is the pre-exponential factor, $E_a$ the activation energy, $R$ the molar gas constant, and $T$ the absolute thermodynamic temperature. In its original form, $A$ and $E_a$ are held constant over the range being modeled. It is an empirical relationship, not a universal rule for every reaction mechanism. See the Arrhenius equation.

Eliminating $A$ between two positive temperatures gives the useful predictionFor $E_a>0$, warming increases $k$. The logarithmic sensitivity is $d\ln (k/k_{\mathrm{ref}})/dT=E_a/(R{T}^2)$, so there is no universal doubling per ten degrees. The same $10\mathrm{K}$ rise from $300$ to $310\mathrm{K}$ gives a factor $1.67749$ for $E_a=40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, versus $2.81398$ for $80\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$.

The useful simplification is that a strong change in reaction rate becomes a straight-line relation after changing variables. Measurements at several temperatures can estimate the slope and test whether one pair of constants describes the chosen range. Curvature can indicate a failure of that approximation; fitting a line does not uniquely establish a microscopic reaction mechanism.

Svante Arrhenius's 1889 paper on sucrose conversion through catalysis by a chemical acid compared the strong observed temperature dependence with the much smaller increase in collision frequency. He discussed van't Hoff's earlier temperature relations and proposed a small population of hypothetical active sucrose, replenished from ordinary sucrose, to interpret the dependence. The inspected translated excerpt derives an inverse-temperature relation in its equations (6)–(7) and then applies it to the reaction rate. That historical hypothesis is distinct from a modern transition state description. See the On the Reaction Velocity of the Inversion of Cane Sugar by Acids. The excerpt omits parts of the original paper; it does not provide a complete original apparatus description or dataset.

The pre-exponential factor must have the same units as the rate constant, since the exponential function supplies a dimensionless quantity. For a rate law $v=k[\mathrm{A}{]}^m$, both $A$ and $k$ have units $(\mathrm{mol}{\mathrm{L}}^{-1}{)}^{1-m}{\mathrm{s}}^{-1}$. Thus a first-order reaction uses ${\mathrm{s}}^{-1}$, whereas a second-order reaction uses $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. Calling $A$ a frequency factor does not give it frequency units for every order of reaction. Its value includes the physical and geometrical factors allowed by the chosen reaction mechanism; equal activation energies need not mean equal reaction rates.

To read the blue Arrhenius plot, define the dimensionless quantity $X=1000\mathrm{K}/T$ and $y=\ln [k/k(300\mathrm{K})]$. Its numerical slope is $m_X=-60000/(1000R)\approx -7.21634$, giving $E_a=-(1000\mathrm{K})R{m}_X=60\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The slope against $1/T$, by contrast, is $-7216.34\mathrm{K}$. The orange tangent at $300\mathrm{K}$ gives $E_{a,\mathrm{app}}\approx 64.9887\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; its finite-interval slope between $280$ and $340\mathrm{K}$ gives about $65.1227\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The distinction between a local derivative and a fitted interval matters even with perfectly noise-free synthetic evaluations.

Two measured positive rate constants give a two-point estimateSeveral temperatures let linear regression test the constant-parameter approximation: inspect the regression residuals, keep the same rate law and composition convention, and account for uncertainty in both $k$ and $T$. A straight line over a short range supports a useful local approximation; it does not uniquely identify a reaction mechanism. A fixed change of units or of $k_{\mathrm{ref}}$ shifts the intercept without changing the inferred activation energy.^[ref]

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.

No cyclic heat engine exchanging heat only with two fixed thermal reservoirs at positive absolute temperatures $T_h>T_c$ can exceed the thermal efficiency of an engine whose cycle is a reversible thermodynamic process between them. All such reversible engines have the same thermal efficiency, regardless of their working substance. Let $Q_h>0$ be heat absorbed from the hot thermal reservoir and $Q_c>0$ the magnitude rejected to the cold one. ThenThe work convention is the one in the first law of thermodynamics: $W>0$ is net output by the engine. Use absolute thermodynamic temperatures, such as values in kelvin; ratios of readings in degree Celsius are invalid because their zero is shifted.

For $T_h=600\mathrm{K}$, $T_c=300\mathrm{K}$ and $Q_h=1000\mathrm{J}$, the maximum thermal efficiency is $50\%$. A reversible engine delivers $W=500\mathrm{J}$ and rejects $Q_c=500\mathrm{J}$. The thermal reservoirs have entropy changes $-1000/600$ and $+500/300\mathrm{J}{\mathrm{K}}^{-1}$, which cancel. The engine's own entropy change is zero over its thermodynamic cycle.

A short derivation now becomes transparent: the Second law of thermodynamics gives $-Q_h/T_h+Q_c/T_c\ge 0$, and the first law of thermodynamics gives $W=Q_h-Q_c$. Together they yield the bound and, more precisely,At fixed $Q_h,T_h,T_c$, every extra unit of total entropy production reduces output by $T_c$ units of energy. ^[ref]

Fixed thermal reservoirs are essential. Finite bodies cool and warm, so their initial temperature ratio cannot be used throughout an entire extraction process. As a computed contrast, the bodies with equal constant heat capacity in the entropy example would reach $T_f=\sqrt{600\times 300}=424.26\mathrm{K}$ if a reversible engine extracted work until their temperatures matched. Zero total entropy change gives this geometric mean; the first law of thermodynamics then gives $W_{\max }=C(600+300-2T_f)=5147.19\mathrm{J}$. Direct contact reaches $450\mathrm{K}$ and extracts no work.

Carnot's 1824 book asked whether the useful output of machines driven by heat was fundamentally limited and whether changing the working substance could improve that limit. Its reversible comparisons are the historical setting of Carnot's theorem. Carnot used the then-current caloric picture, treating heat as conserved in its passage through an engine; the modern balance $Q_h-Q_c=W$ should not be read back into that text.^[ref]

For the cyclic two-side energy balance in refrigerator, define the dimensionless cooling and heating coefficients byThis identity applies even to irreversible machines, provided both coefficients refer to the same operating point, work input and boundary with no additional energy transfer or storage. It does not compare two separately operated appliances. Unlike thermal efficiency of a heat engine, a COP can exceed one: its numerator counts heat moved, including energy taken from another region. ^[ref]

To obtain temperature-only bounds, impose stronger assumptions: a cyclic machine exchanges heat only with two ideal thermal reservoirs maintained at uniform, fixed absolute thermodynamic temperatures $0<T_c<T_h$, and receives work through an ideal input carrying no entropy. Include the machine and the heat transfer interfaces in the entropy production accounting. Because its own entropy returns to its initial value each cycle, the Second law of thermodynamics givesCombining this inequality with $Q_h=Q_c+W$ givesEquality requires a reversible thermodynamic process for the entire cycle and its contacts, as in a reversed Carnot cycle. These are Carnot's theorem bounds independent of the working substance. Use kelvin values, not degree Celsius readings, in the numerators; the temperature difference can be expressed in either unit. ^{[ref] [ref]}

The diagram fixes $T_c=280\mathrm{K}$ and sets $T_h=320\mathrm{K}$, so the reversible bounds are $280/40=7$ and $320/40=8$. The stipulated machine with ${\mathrm{COP}}_{\mathrm{r}}=3$ and ${\mathrm{COP}}_{\mathrm{heat}}=4$ satisfies both. Its reservoir entropy production per cycle is $4000/320-3000/280=25/14\mathrm{J}{\mathrm{K}}^{-1}>0$. Increasing the temperature lift $\Delta T=T_h-T_c$ at fixed $T_c$ lowers both bounds. Equivalently, delivering fixed $Q_h$ needs at leastThese expressions reveal why a small temperature lift is attractive. Their divergence as $T_h-T_c\to 0^{+}$ is an ideal reversible limit, not a promise of infinite heating or cooling power. For finite heat transfer rates, the working fluid must be colder than the cold bath and hotter than the hot bath; these contact differences produce entropy and increase the internal temperature lift. Dissipative compression and the pressure-reducing restriction of vapor-compression refrigeration supply further departures from reversibility. Thus bath temperatures set an outer bound, not a prediction of real COP. ^[ref]

The degree Celsius has the same size as the kelvin, with a shifted temperature origin: $t/{}^{\circ }\mathrm{C}=T/\mathrm{K}-273.15$. Thus a change of one degree Celsius is a change of one kelvin, but temperature ratios in Carnot's theorem must use absolute thermodynamic temperature.^[ref]

An empirical temperature labels classes of thermal equilibrium using a reproducible thermometer response. Equal labels express the zeroth law of thermodynamics comparison. A suitable continuous response and calibration let ordinary laboratory states be represented on a numerical scale; the abstract equivalence relation alone does not guarantee a useful real-valued coordinate. An injective function of the label preserves equality, so the zeroth law of thermodynamics alone fixes neither unit nor spacing. Thermometer calibration supplies a chosen response relation; thermodynamic temperature supplies the additional absolute-scale structure.

Entropy is an extensive quantity and a state function of an equilibrium system, measured in $\mathrm{J}{\mathrm{K}}^{-1}$. Differences between equilibrium states are calculated along a reversible thermodynamic process:Here heat entering the system is positive and $T$ is its absolute thermodynamic temperature. The reversible path used for calculation need not be the path actually taken. Dividing the actual heat by a single temperature generally fails for an irreversible process.

For example, put two rigid bodies, each with constant heat capacity $C=100\mathrm{J}{\mathrm{K}}^{-1}$, at $600\mathrm{K}$ and $300\mathrm{K}$ in an otherwise isolated system. Assume each body has a uniform internal temperature, no phase transition occurs, and no work is extracted. Their final thermal equilibrium is at $450\mathrm{K}$, because the internal energy lost by one equals that gained by the other. Using entropy from heat capacity,so $\Delta S_{\mathrm{tot}}=C\log (9/8)=11.78\mathrm{J}{\mathrm{K}}^{-1}$. One body's entropy decreases while the total increases. ^[ref]

In statistical mechanics, Boltzmann's entropy formula counts equally likely microstates compatible with a macrostate; Gibbs entropy handles a specified probability distribution over microstates. If the number of equally likely microstates doubles, entropy increases by $k_B\log 2$. The logarithm makes multiplicative counts additive for independent systems. These are precise statements about states and probabilities, not a universal measure of visible messiness. Calling entropy merely "disorder" obscures the thermal calculation above and hides which microstates are being counted. ^[ref]

The usefulness of entropy is that measurable heat capacities determine its changes, and the Second law of thermodynamics turns those changes into constraints on processes. It also connects macroscopic measurements to microscopic descriptions without requiring us to track every particle.

At fixed volumes and material inventories, two internally equilibrated bodies that exchange only heat within an isolated system have fixed total internal energy $E=E_1+E_2$. Their total entropy isUsing $1/T_i=(\partial S_i/\partial E_i{)}_{V_i,N_i}$, an interior stationary point obeysThus entropy stationarity requires matching thermodynamic temperatures. Positive values of heat capacity at constant volume make this a strict local maximum becauseFor bodies with positive heat capacities throughout the accessible range, this concavity gives at most one interior maximum. This is the stability check behind the equal-temperature condition; see entropy concavity and positive heat capacity. ^[ref]

For thermal equilibration of two finite bodies, entropy from heat capacity givesBody 1 gains about $0.22314355\mathrm{kJ}{\mathrm{K}}^{-1}$ of entropy, while body 2 loses about $0.19361556\mathrm{kJ}{\mathrm{K}}^{-1}$. Total energy is conserved even though total entropy increases. The entropy calculation uses reversible comparison paths between the endpoint states; actual exchange across a finite temperature difference is irreversible.

To construct the plot, vary $T_1$ while enforcing the energy constraint, rather than holding $T_2$ fixed. In numerical units of kelvin, $T_2/\mathrm{K}=(1500-T_1/\mathrm{K})/3$, and the plotted function isHere $\mathcal{S}$ means $S_{\mathrm{tot}}-S_{\mathrm{tot},\mathrm{initial}}$, so its zero is a reference choice. The chain rule yieldsThe unique maximum is at $T_1=T_2=375\mathrm{K}$. The physical trajectory starts at $T_1=300\mathrm{K}$ and moves toward that maximum; the rest of the curve compares other possible energy partitions, not later times in the trajectory.

The contact model also proves the direction of change directly:For $G>0$ and positive temperatures, equality holds exactly at thermal equilibrium. The appealing link is that a dynamical heat transfer law moves the system uphill on the constrained entropy curve, while conservation of energy keeps it on that curve. This predicts both why the endpoint is stable and why the reverse separation into hot and cold bodies cannot happen spontaneously in the isolated system.

A gas thermometer infers temperature from the equation of state of a confined gas. The constant-volume version measures pressure; the constant-pressure version measures volume. For an ideal gas,Here $N$ is particle number, $n$ is amount of substance, $k_B$ is the Boltzmann constant, and $R$ is the molar gas constant. The reference and test states must be internally equilibrated, and $T_0$ must be known on the absolute thermodynamic temperature scale. A pressure ratio cancels the unknown gas inventory and bulb size in the ideal constant-volume limit. This cancellation is the method's appeal: a mechanical measurement transfers an absolute temperature scale between different materials. The Maxwell speed distribution and ideal gas pressure from wall collisions explain the microscopic origin of the response. ^[ref]

For an original ideal-model calculation, choose $T_0=300\mathrm{K}$, $P_0=120\mathrm{kPa}$ and $P=160\mathrm{kPa}$. Then $T=400\mathrm{K}$. These are stipulated values, not observations. Both pressures must be absolute, referred to vacuum. If the surrounding atmospheric pressure were $100\mathrm{kPa}$, the corresponding gauge pressures would be $20$ and $60\mathrm{kPa}$; their ratio is three, which would incorrectly predict $900\mathrm{K}$. A pressure difference from the atmosphere cannot be substituted into the ideal gas law.

A motivating calibration experiment. Schooley's account of the NBS/NIST work describes a helium-filled metal bulb, a capillary, a capacitance diaphragm gauge, and a mercury pressure standard. Circulating bath liquid passed the bulb assembly and platinum resistance thermometers. The resistance probes checked the surrounding temperature; gas pressure supplied the independent thermodynamic comparison. In an early series, workers filled the cleaned bulb with purified helium at the highest test temperature, checked pressure over three hours, then measured at lower temperatures and near a reference at $0{}^{\circ }\mathrm{C}$. Six filling pressures allowed extrapolation toward zero pressure. ^[ref]

To turn this principle into a readable measurement protocol, use the following controls. These steps are a proposed teaching interpretation of the cited apparatus, rather than additional reported NBS observations:

The corresponding NBS apparatus used a protective case with an outer helium layer held close to the bulb pressure, reducing pressure deformation while aiding thermal contact. Thermal expansion was measured on bulb material. Gas in the capillary remained partly outside the hot region, requiring a measured tube temperature profile and geometry. Pure working gas also mattered: contaminants that sorb on surfaces can change the freely circulating inventory. These are physical effects to measure or bound, rather than scatter that disappears by averaging. ^[ref]

For illustration, neglect connecting-tube gas and describe a uniform bulb with a dilute virial expansion in particle number density ${\rho }_N=N/V$:Here $Z=PV/(N{k}_{B}T)$ and $B_2$ is the second virial coefficient in the particle-density convention; a molar-density convention rescales the coefficient. The second relation follows exactly from the definition of $Z$ for the same $N$, and shows the signs of the corrections. If the bulb expands by $0.1\%$ and $Z=1$ at both states, the toy pressure ratio above gives $400.4\mathrm{K}$, rather than $400\mathrm{K}$. Improved expansion and helium nonideality models really changed the temperatures inferred from the 1989 NIST data: Harvey's 2023 reanalysis found expansion updates of roughly $1$–$3\mathrm{mK}$ and nonideality updates of roughly $-0.1$ to $-0.5\mathrm{mK}$. Those are reported corrections to historical results, not the much larger illustrative correction just calculated. ^[ref]

With equal corrected bulb volumes, negligible or corrected connecting-tube gas, fixed reference and test temperatures, and a dilute classical single gas phase,Repeat the reference/test comparison for several sealed fills, reducing $N$ between comparisons but keeping it fixed within each pair. The extrapolation varies density at fixed endpoint temperatures; it does not cool one fill toward zero temperature. Real gases have different finite-density corrections but share the classical ideal gas intercept under these assumptions. The limit does not remove leaks, bulb expansion, gauge offsets, or the gas thermometer dead-space correction. At very low pressures, thermal transpiration can make remote sensor and bulb pressures differ through a nonisothermal capillary; this must be corrected too. The NBS work measured this pressure correction using a separate comparison of narrow and wider tubes. ^[ref]

At $x=1$, the toy models give $T_{\mathrm{app}}=407.92\mathrm{K}$ and $388.12\mathrm{K}$, respectively, although their imposed true test temperature is the same. Their zero-density intercept is the reusable quantity. Measuring several low-density points permits a fit and a check for curvature; asserting a straight line from one finite-density point supplies neither. A practical fit must balance nonideality against increasing relative pressure error and the capillary corrections, and keep its extrapolation uncertainty.

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.

The ideal gas law is the equation of state relating the absolute pressure $P$, volume $V$, particle count $N$, and absolute thermodynamic temperature $T$ of a classical ideal gas:Here $k_B$ is the Boltzmann constant, $N_{\mathrm{A}}$ is the Avogadro constant, $R$ is the molar gas constant, and $n_{\mathrm{mol}}$ is the amount of substance in moles. The distinct symbol $n_{\mathrm{mol}}$ avoids confusing amount of substance with the number density $n=N/V$, measured in ${\mathrm{m}}^{-3}$: the same law can be written $P=n{k}_{B}T$. Use absolute pressure, relative to a vacuum, rather than an unconverted gauge pressure. ^[ref]

The exact SI definitions of the Avogadro constant and Boltzmann constant giveThis numerical value is exact; the applicability of the ideal gas model to a real gas is an approximation. ^{[ref] [ref]}

For $n_{\mathrm{mol}}=1\mathrm{mol}$, $T=300\mathrm{K}$, and $V=0.025{\mathrm{m}}^3$, direct substitution givesAt fixed temperature and particle count, doubling the volume to $0.050{\mathrm{m}}^3$ halves the pressure to approximately $49.887\mathrm{kPa}$: this is Boyle's law. At fixed volume and particle count, doubling the temperature from $300\mathrm{K}$ to $600\mathrm{K}$ doubles the pressure to approximately $199.547\mathrm{kPa}$. These ratios use kelvins, not readings in the degree Celsius scale: $300\mathrm{K}$ is $26.85{}^{\circ }\mathrm{C}$, and $600\mathrm{K}$ is $326.85{}^{\circ }\mathrm{C}$.

The curved slices show Boyle's law at three temperatures; the straight slice shows proportionality to absolute temperature at one fixed volume. They are relations among equilibrium states, rather than a prescribed process history. Compression could hold the temperature fixed through heat transfer, or raise it during adiabatic compression. To calculate a process's heat or work, supplement the ideal gas law with the process constraint, the first law of thermodynamics, and the relevant internal energy or heat capacity information. The graph's extrapolation toward $T=0$ is mathematical; it does not assert that a real gas remains classical or gaseous there.

For a stellar mixture with fixed mean molecular weight, the ideal gas law can be written $P=\mathcal{R}\rho T/\mu$, where $\rho$ is mass density and $\mathcal{R}$ is the stellar gas constant. It assumes negligible interaction corrections and a nondegenerate gas; excitation or ionization can change the particle count and thermodynamic response.

In his 1834 memoir on the motive power of heat, Émile Clapeyron combined gas laws into a relation between pressure, volume, and temperature while analysing heat engines (printed page 164). The equation above uses modern SI notation and absolute thermodynamic temperature; it is not a transcription of his historical temperature convention. ^[ref]

The International System of Units defines a common system for expressing physical quantities. Its base units include the kelvin for thermodynamic temperature; derived units combine them to describe quantities such as force and energy.^[ref]

The International Temperature Scale of 1990 is a practical scale for reproducible temperature measurements. Its assigned reference points and interpolation procedures approximate thermodynamic temperature. A calibrated standard platinum resistance thermometer transfers this scale over specified intervals; it does not define the modern kelvin.^[ref]

On 10 July 1908, Heike Kamerlingh Onnes's Leiden laboratory circulated purified helium through liquid air and evaporatively cooled liquid-hydrogen precooling, a regenerative heat exchanger, and an expansion valve. Returning helium cooled incoming helium; liquid collected in a transparent vacuum vessel. Reflected light revealed its liquid meniscus, crossed by thermocouple wires. Subsequent mass density and spectroscopy checks tested contamination.^[ref]

A constant-volume helium gas thermometer below the valve supplied the temperature indication. An oxygen boiling reference check disagreed with its expected reading; the real gas correction to thermodynamic temperature remained provisional. A temperature plateau suggested immersion before the liquid meniscus was recognized.^[ref]

The measurement principles separate two questions. A visible liquid meniscus establishes an interface; a gas thermometer estimates temperature through an equation of state. A stationary temperature trace can also result from a steady state without phase coexistence, so it cannot replace the interface observation. Conversely, seeing a liquid does not calibrate a thermometer. Thermometer calibration, the gas thermometer dead-space correction, departures from the ideal gas law and incomplete thermal equilibrium all matter when assigning an absolute temperature. A check far from the intended measurement range tests the instrument but does not by itself bound every low-temperature error.

The kelvin is the SI unit of thermodynamic temperature, with symbol $\mathrm{K}$. Its definition fixes the numerical value of the Boltzmann constant; it is written without a degree sign.^[ref]

The Modified Arrhenius equation allows a power of thermodynamic temperature in the pre-exponential factor:Here $A_{\ast }$, $E_0$, $n$, and $T_{\ast }$ are constant model parameters. Using the dimensionless quantity $T/T_{\ast }$ leaves $A_{\ast }$ with the same units as $k$; in the equivalent convention $B{T}^n\exp [-E_0/(RT)]$, $B$ instead has units of $k$ multiplied by ${\mathrm{K}}^{-n}$. The power exponent $n$ here is not an order of reaction. The IUPAC definition uses that equivalent $B{T}^n$ convention.^[ref]

Applying the apparent activation energy definition yields the original short calculationFor the orange synthetic Arrhenius plot, $n=2$, $T_{\ast }=300\mathrm{K}$, and $E_0=60\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Its pre-exponential factor changes by $(310/300{)}^2\approx 1.06778$ from $300$ to $310\mathrm{K}$; combined with the exponential function, $k(310\mathrm{K})\approx 0.231991{\mathrm{s}}^{-1}$, compared with $0.217266{\mathrm{s}}^{-1}$ for the constant-pre-exponential factor model. The diagram uses those models only to reveal the difference between an exponent parameter and a slope-derived activation energy.

Here $p$ is absolute pressure, $V$ is volume, $n$ is amount of substance in moles, $R$ is the molar gas constant and $T$ is absolute thermodynamic temperature in kelvin. This is the ideal gas law; consistent SI units give pressure in pascals, volume in cubic metres and $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

Let $G$ be Gibbs free energy, $H$ enthalpy and $T$ thermodynamic temperature for a fixed amount and composition of material, with the derivative taken at constant pressure. Apply the Gibbs–Helmholtz equation to the standard molar reaction quantities, obtained by summing product minus reactant contributions with their stoichiometric coefficients:Define $K(T)$ as the dimensionless thermodynamic equilibrium constant expressed using activities in specified standard states. Since ${\Delta }_r{G}^{\circ }=-RT\ln K$, where $R$ is the molar gas constant,If the reaction enthalpy is constant over the interval, integration givesThis is the requested Van't Hoff equation. For an exothermic reaction, increasing temperature decreases $K$. Although the name “isochore” is often used in this context, the displayed enthalpy form follows from standard Gibbs energies and the stated constant-pressure Gibbs–Helmholtz identity; the analogous ideal-gas concentration form involves internal energy when the temperature dependence of standard concentration is retained.