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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]

For two subsystems with independently specified macrostates, suppose every allowed microstate of $A$ can be paired with every allowed microstate of $B$. Their allowed set is a Cartesian product. With equal probability for each pair, Boltzmann's entropy formula givesFor example, take two independent copies of the toy model, each with two labels choosing between its two boxes. Each copy has four configurations and Boltzmann entropy $k_B\ln 4$. Combining their independent labels gives sixteen configurations and Boltzmann entropy $k_B\ln 16=2k_B\ln 4$. This explains why entropy adds even though microscopic counts multiply. ^[ref]

The product assumption matters. A constraint shared between the subsystems can forbid some pairs; for example, fixing only their combined energy generally requires summing over its possible divisions. See entropy maximization under thermal contact for how thermal equilibrium selects the dominant division. Statistical independence cannot be inferred merely from drawing a boundary between two parts.

ATP is a nucleotide consisting of adenosine, itself adenine joined to ribose, carrying three phosphate groups. From the ribose outward, these groups are labelled $\alpha$, $\beta$, and $\gamma$. ATP connects energy-supplying processes to energy coupling for synthesis and active transport; the same carrier can be regenerated and reused. ^[ref].

The net hydrolysis commonly used in energy coupling isHere ${\mathrm{P}}_i$ denotes inorganic phosphate. At buffered pH $7$, this biochemical shorthand groups the appropriate protonation states into pools and leaves proton exchange with the buffer solution implicit. An explicit, charge-balanced choice is ${\mathrm{ATP}}^{4-}+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{ADP}}^{3-}+{\mathrm{HP}{\mathrm{O}}_4}^{2-}+{\mathrm{H}}^{+}$, without implying these are the only chemical species present. See the IUBMB recommendations for biochemical thermodynamics.

The quoted textbook biochemical standard state value is approximatelyThis convention uses fixed pH $7$, temperature $298.15\mathrm{K}$ ($25{}^{\circ }\mathrm{C}$), reference pressure $100\mathrm{kPa}$ ($1\mathrm{bar}$), unit reference thermodynamic activities for dissolved reactants and reaction products, and water thermodynamic activity approximated by one. The prime indicates the fixed-pH biochemical convention. See OpenStax Biology 2e, 6.4: standard ATP hydrolysis value and Biology 2e, 6.2: biochemical standard conditions. This is a reference value, rather than a universal value inside a biological cell; precise tables must also specify chemical solution conditions and any magnesium ion binding convention. Different reference conventions and data treatments give different tabulated values, as discussed in Li et al., 2010: a database of thermodynamic quantities for biochemical reactions.

An exergonic reaction has a negative change in Gibbs free energy, so the overall ATP hydrolysis is favourable under those reference conditions. Breaking a chemical bond by itself costs energy; forming new chemical bonds can release energy. The net change compares all reactants with all reaction products, including their interactions with water and the entropy contribution. Calling the phosphate connection a "high-energy bond" describes a large favourable change in Gibbs free energy on hydrolysis, rather than energy released by bond breaking alone. See 7.5 Strengths of Ionic and Covalent Bonds and Cooper, The Cell: ATP and metabolic energy.

Composition changes the driving force. At the same fixed pH, temperature, pressure, and binding convention, write the molar reaction change asHere $R$ is the molar gas constant, $T$ the absolute temperature, and $a$ a dimensionless thermodynamic activity for a biochemical pool. Constant water thermodynamic activity is absorbed into the reference term; fixed pH accounts for buffered protons. The shorthand $\Delta G$ means transformed molar reaction Gibbs free energy change, ${\Delta }_r{G}^{\prime }$, rather than a finite sample's total change. The natural logarithm acts on a dimensionless reaction quotient, never on concentrations with units. See the IUBMB biochemical thermodynamics recommendations and the general equilibrium constant relation.

For a stipulated ideal dilute chemical solution, let $a_i=[i]/c^{\circ }$, with $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$. Choose $[\mathrm{ATP}]=5\mathrm{mmol}{\mathrm{L}}^{-1}$, $[\mathrm{ADP}]=0.5\mathrm{mmol}{\mathrm{L}}^{-1}$, and $[{\mathrm{P}}_i]=1\mathrm{mmol}{\mathrm{L}}^{-1}$, ignoring activity coefficient corrections and magnesium binding. These are illustrative inputs, not a claim about a particular biological cell. ThenandIncreasing $Q$ tenfold raises $\Delta G$ by $RT\ln 10\approx 5.708\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at this temperature. Thus the ATP to ADP ratio and available inorganic phosphate matter, even with the same standard state value. These are local molar predictions at the stipulated composition, rather than a constant total yield as an isolated sample reacts.

An adiabat is a curve representing a quasistatic adiabatic change. In a reversible description it is also a curve of constant entropy. For an ideal gas with constant heat-capacity ratio $\gamma$, it obeys $p{V}^{\gamma }=\mathrm{constant}$. A general irreversible adiabatic process need not follow this equilibrium-state curve and can produce entropy.

A cooling method that reduces an applied magnetic field while suppressing heat transfer to or from the material. For a suitable material exhibiting paramagnetism, an approximately reversible field reduction can lower its temperature at approximately constant entropy.

The preparation matters as much as the field reduction. First establish thermal equilibrium with a cold heat bath, then raise the magnetic field while maintaining the bath temperature: this is magnetization in an isothermal process. The bath accepts heat and the material loses magnetic entropy. Next open a heat switch, thermally isolating the working material, and lower the magnetic field. If the sweep maintains internal thermal equilibrium with negligible entropy production, this second leg is an isentropic process. An adiabatic process alone does not guarantee constant entropy.

The attraction is a controllable energy scale: changing the magnetic field changes the separation of magnetic energy levels, even across a thermally insulating gap. This lets an already cold solid become colder without needing a colder contacting heat bath. A practical refrigerator also needs a load, a period of absorbing its heat, and a reset that rejects accumulated entropy to the bath. A single field sweep has finite cooling capacity.

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.

For a homogeneous cosmic fluid, the physical volume bounded by fixed comoving coordinates obeys $V\propto a^3$. The cosmological continuity equation therefore gives $d(\rho c^{2}V)=-PdV$. This is the first law of thermodynamics with conserved entropy and no additional chemical work. The internal energy of that volume need not be constant: a component with negative pressure gains internal energy as the universe expands. Local stress-energy conservation does not imply a global conserved energy for an arbitrary expanding spacetime.

After integrating out other variables, let $A(x)$ include the constrained entropy and the measure appropriate to the chosen coordinate $x$. Its equilibrium probability density function is proportional to $e^{-A(x)/(k_B{T}_R)}$. Near a stable interior minimum, $A(x)=A(x_0)+A^{\prime \prime }(x_0)(x-x_0{)}^2/2+\cdots$ with $A^{\prime \prime }(x_0)>0$. The resulting normal distribution has variance $k_B{T}_R/A^{\prime \prime }(x_0)$. This is the leading Gaussian fluctuation approximation; a coordinate-dependent measure contributes to the effective $A$, and higher-order terms matter near a vanishing curvature.

In the constant-parameter battery internal resistance model, the source's available electrical power splits between the electrical load and internal dissipation:The stipulated $3\mathrm{V}$, $1\Omega$, $4\Omega$ calculation gives $P_L=1.44\mathrm{W}$, $P_{\mathrm{int}}=0.36\mathrm{W}$, and $\mathcal{E}I=1.80\mathrm{W}$. A resistive electrical load converts its share to Joule heating; another electrical load might convert some of it to mechanical work or light. The source term is available electrical work per time, not a complete cell heat budget: reaction entropy can also matter.

Define the delivery efficiency relative to the source's available electrical work, rather than to all chemical energy or to a complete charging cycle:Here $\eta =0.80$. The largest load power occurs at $R=r$: since $(R+r{)}^2\ge 4Rr$,Thus maximum load power and maximum delivery efficiency ask different questions. As $R\to \infty$, $V_L\to \mathcal{E}$ and $\eta \to 1$, but $I\to 0$ and $P_L\to 0$: a high nearly unloaded voltage supplies almost no power. As $R\to 0$, the model gives $V_L\to 0$, $P_L\to 0$ and internal power tending to ${\mathcal{E}}^2/r$. Real high-current operation can invalidate constant $r$ and $\mathcal{E}$; the mathematical maximum does not establish a permissible operating point for an actual battery.

Take $n$ independent steps, each $+1$ with probability $p>1/2$ and $-1$ with probability $q=1-p>0$. Reverse a step sequence by reading it backward and changing each sign; use the same biased law for both experiments. With $K$ positive steps, the forward sequence weight is $p^K{q}^{n-K}$ and its reversed weight is $p^{n-K}{q}^K$. Their logarithm ratio isHere $S_n$ is dimensionless entropy production, not an entropy in joules per kelvin. A Markov chain on a three-state ring gives a state-space realization: start uniformly, take a clockwise step with probability $p$ and a counterclockwise step with probability $q$. The uniform stationary distribution makes endpoint weights cancel. A physical interpretation as exchanged heat requires local detailed balance with bath entropy $\pm k_{B}a$ per directed step; the abstract model alone specifies path probabilities.

For an allowed value $s=a(2k-n)$, the binomial distribution givesThis is an exact same-law finite fluctuation theorem, with no long-time approximation. For $p=3/4$, $n=9$ and $k=6$, $s=3\ln 3$ and $p_n(s)/p_n(-s)=27$. The positive mean is $⟨S_n⟩=na(p-q)$, although the all-negative path has nonzero probability $q^n$. Directly, $⟨e^{-S_n}⟩=(p{e}^{-a}+q{e}^a{)}^n=(q+p{)}^n=1$.

The left panel shows exact probability masses for $p=3/4$, with negative values highlighted; the right panel tests the logarithm ratio against $s$. These are model values, without experimental measurements or simulated measurement noise.

For independently labelled particles with nondegenerate levels, choosing which $N_{+}$ particles occupy the upper level gives a binomial multiplicity. With $x=(N_{+}-N_{-})/N$, its large-number entropy is $N{k}_B[\ln 2-(1+x)\ln (1+x)/2-(1-x)\ln (1-x)/2]$. This counting assumes distinguishable or localized degrees of freedom, not indistinguishable particles sharing one pair of quantum orbitals.

In Boltzmann entropy of a two-box occupancy, impose an initial restriction that all labels are left. There is one allowed coarse microstate. Remove that restriction, allowing all occupancy values: the allowed set is now the union of every occupancy macrostate, containing $2^N$ strings. If the final description assigns equal probability to this whole set, its Boltzmann entropy change isThe counting identity is the binomial theorem. For twenty labels, $\Delta S=20k_B\ln 2\approx 13.862944k_B$. This final description fixes neither box occupancy; it is coarser than the particular macrostate $k=10$, whose Boltzmann entropy is $k_B\ln 184756$. The two counts answer different questions, so substituting the balanced macrostate for the full allowed set would change the calculation.

For four labels, the initial all-left string is 0000; removing the restriction makes all sixteen strings available. The $k=2$ column has only six strings. Accordingly, the whole allowed set has Boltzmann entropy $k_B\ln 16=4k_B\ln 2$, while the balanced macrostate has Boltzmann entropy $k_B\ln 6$.

Removing a barrier changes the allowed set immediately, but it does not instantaneously randomize the actual probability distribution. Motion and equilibration must make those configurations accessible. The counting result illustrates the configurational contribution to free expansion; it does not describe the relaxation time or every contribution to the entropy of a real gas.

The integral uses $0\log 0=0$ when it is defined. For the Boltzmann equation, the Boltzmann H theorem makes it nonincreasing under suitable regularity and integrability. This functional is distinct from the microstate-counting Boltzmann entropy; the physical kinetic entropy has the opposite sign.

Why does a gas spread through an available container? One part of the answer is counting: a nearly uniform macrostate can be realized by vastly more microstates than a macrostate with every particle crowded into one corner. Boltzmann's entropy formula turns that multiplicity into entropy. The useful insight is that a large entropy measures how many microscopic realizations fit a specified macroscopic description, rather than how untidy a picture looks.

For a specified macrostate $M$ containing $\Omega (M)$ accessible, equally likely microstates, writeHere $k_B$ is the Boltzmann constant, and $\ln$ is the natural logarithm, also written $\log$ in this article's title. The count $\Omega (M)$ is dimensionless; entropy has units of energy divided by temperature. Always specify the macrostate, the constraints and the model's microscopic resolution before counting. A microcanonical ensemble provides a familiar equilibrium setting for the equal-probability assumption. ^[ref]

The natural logarithm compresses an enormous count into a manageable number and makes products of counts into sums of entropies; see additivity of Boltzmann entropy. With this counting convention, one allowed microstate has zero Boltzmann entropy. This zero refers to the chosen model and resolution; it does not by itself assign an absolute entropy to a real material.

A product trial distribution with actual mean Ising spin $m$ givesIts entropy is the sum of independent-spin entropies, with $0\log 0=0$. Differentiation gives $k_{B}T\mathrm{arctanh}m=qJm+h$, the self-consistency equation. The variational mean parameter is distinct from the assumed field parameter of the Ising auxiliary mean-field free energy until the equilibrium condition is imposed.

For $Z={\sum }_i{e}^{-\beta E_i}$ and $U=⟨E⟩$, the entropy is $S=k_B\ln Z+U/T$. It follows directly by substituting canonical probabilities into $S=-k_B{\sum }_i{p}_i\ln p_i$.

The partition function $Z(\beta )={\sum }_s{e}^{-\beta E_s}$ is the sum of the Boltzmann weights over individual microstates. Dividing by it makes their probabilities sum to one. When summing distinct energies instead, include the state degeneracy: $Z={\sum }_{E}g(E)e^{-\beta E}$. For two independent distinguishable two-level systems,The middle coefficient counts two microstates, exactly as in the preceding figure. For $N$ such units, $Z_N=Z_1^N$; this factorization requires additive energy and independence, and generally fails with interactions.

The usefulness goes beyond normalization. For fixed energies, derivatives of the natural logarithm of the partition function yield measurable quantities:Here $U$ is internal energy, the second expression is the energy variance, and $S$ is Gibbs entropy. For one two-level system at fixed $\Delta$, its heat capacity $C=\partial U/\partial T$ and entropy areAt $\beta \Delta =\ln 3$, these give $U/\Delta =0.25$, $C/k_B\simeq 0.2263$ and $S/k_B\simeq 0.5623$. For independent units, all three quantities add.

The figure shows why heat capacity peaks. At low temperature, the excited energy is rarely occupied; at high temperature, the probabilities are almost equal and further warming barely changes internal energy. The largest response lies between these limits, at $k_{B}T/\Delta \simeq 0.4168$, where $C/k_B\simeq 0.4392$. Meanwhile entropy rises from zero toward $k_B\ln 2$, reflecting two equally likely alternatives. This high-temperature limit belongs to the stipulated two-level system; extra accessible energy eigenstates would change it.

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]