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For equal masses $M$ of the same liquid with constant specific heat $c_P$, adiabatic constant-pressure mixing gives final temperature $T_f=(T_1+T_2)/2$ by enthalpy conservation. The entropy change is $M{c}_P\ln (T_f/T_1)+M{c}_P\ln (T_f/T_2)=2M{c}_P\ln [(T_1+T_2)/(2\sqrt{T_1{T}_2})]$. The arithmetic-geometric mean inequality proves nonnegativity, with equality only for equal initial temperatures. There is no separate compositional mixing entropy for identical fluids.

When a system at instantaneous temperature $T$ absorbs heat $dQ$ from a reservoir at $T_R$, with no other entropy production in the system, its entropy increases by $dQ/T$ while the reservoir changes by $-dQ/T_R$. The total is positive for spontaneous heat transfer across a finite temperature difference. A quasistatic mechanical path can therefore be thermally irreversible.

With negligible nuclear or external heating, the stellar structure equations give $L_{\mathrm{int}}=-{\int }_0^{M}T\partial s/\partial tdm$. A nearly isentropic convective interior therefore has $\dot{s}=-L_{\mathrm{int}}/\int Tdm<0$ whenever it radiates. Decreasing specific entropy does not require temperature at every depth to decrease immediately: contraction of an ideal gas can raise its central temperature while total energy is lost.

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.

The standard molar entropy difference between the transition state and the starting reactants, at the same temperature and pressure, with specified standard states, in transition state theory.^[ref]

The entropy of mixing is the entropy change between initially separate components and their final mixture, with initial and final temperatures, pressures and component amount of substance values specified. For an ideal solution made from pure liquids at the same temperature and pressure, let $n={\sum }_i{n}_i$. ThenThe right-hand expression is the ideal solution result; the general definition does not require ideality. Each interior mole fraction has a negative logarithm, so mixing distinct components gives positive entropy of mixing. At a pure endpoint, use the limit $x\ln x\to 0$ as $x\to 0^{+}$.

For one mole of the chosen A/B ideal solution with $x_A=0.4$,For the same total amount of substance, equal mole fractions maximize this binary entropy of mixing: the composition function $s(x)=-x\ln x-(1-x)\ln (1-x)$ hasso its unique maximum is $s(1/2)=\ln 2$. This is a comparison between different prepared compositions, not a prediction that a closed sample changes its component inventory to become equimolar.

For an ideal gas with specific internal energy $c_{v}T$ and Fourier's law heat flux $\mathbf{q}=-k\nabla T$, subtraction of the kinetic-energy balance from total energy gives $\rho c_{v}DT/Dt+p\nabla \cdot \mathbf{u}=\nabla \cdot (k\nabla T)$. The thermodynamic entropy satisfies $TDS/Dt=c_{v}DT/Dt+p\nabla \cdot \mathbf{u}/\rho$, soThe divergence describes entropy transport by heat; the final term is local irreversible production. For $T>0$ and $k\ge 0$ it is nonnegative. In a fixed impermeable thermally insulated container, boundary entropy flux vanishes and mass conservation gives $d\int \rho SdV/dt=\int k|\nabla T{|}^2/T^{2}dV$. Insulation conserves total energy while permitting entropy to increase as temperature gradients relax.

For the pressure-density Hugoniot relation for a perfect gas, the change in specific entropy is $[s]/c_v=\log P-\gamma \log D(P)$, with $c_v$ the specific heat capacity at constant volume. Its derivative isIt is positive for a compressive normal shock wave with $P>1$. For a weak shock with $P=1+\delta$, integrating the leading term givesThus the entropy production is cubic in the small pressure jump, even though the mass density and temperature changes already appear at first order.

An enzyme is a biological catalyst: it accelerates a chemical reaction and is regenerated after a successful catalytic cycle. Most enzymes are proteins, but catalytic RNA molecules, called ribozymes, also exist. Many reactions needed by a biological cell would proceed too slowly under its ordinary conditions without enzymes. See Cooper, The Cell: the central role of enzymes as biological catalysts.

The biological beauty is selective speed. A biological cell contains many possible reactants; its enzymes help determine which transformations happen on a useful timescale. In metabolism, one step's reaction product can become another step's substrate, so selective catalysts support an organized flow of material.

The ribosome is a striking biological example: the machinery that builds proteins has RNA at its catalytic centre. The complete ribosome contains both RNA and proteins, but its bond-forming active site is built from RNA. Structural evidence established this distinction in Nissen et al., The structural basis of ribosome activity in peptide bond synthesis, Science 289, 920–930 (2000).

Selection occurs at an active site, the region where substrate binding and catalytic chemistry take place. Shape and chemical interactions favour some substrates over others; binding can also reshape the enzyme, an induced fit. Recognition alone is insufficient: holding a substrate extremely tightly need not make it react quickly. Catalysis requires a favourable route through the transition state, the high-barrier configuration between reactants and reaction products. Positioning reactive groups and stabilizing this configuration relative to the bound substrate can lower that barrier. Specificity is often strong, but need not mean exactly one acceptable substrate. See Cooper, The Cell: mechanisms of enzymatic catalysis.

Read the diagram numerically. Its chosen Gibbs free energies are $0$ for the reactants and $-10\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for the reaction products. The uncatalysed peak is $60\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, whereas the catalysed peak is $40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The forward free-energy barriers are therefore $60$ and $40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; viewed from the reaction products, the reverse barriers are $70$ and $50\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Both decrease by $20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, while the endpoint difference stays $-10\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. These are constructed values, not measurements of a particular enzyme.

This separates chemical kinetics from chemical equilibrium. An enzyme changes how rapidly the system approaches chemical equilibrium; it does not change the equilibrium constant for the same overall chemical reaction under the same conditions. Faster catalysis therefore supplies no extra driving energy. In a biological cell, continuing supply of reactants and removal of reaction products can nevertheless sustain a net flow: the surroundings keep that system away from chemical equilibrium. See IUBMB, Symbolism and terminology in enzyme kinetics: basic definitions.

Some enzymes require a cofactor, an additional nonprotein component such as a metal ion or a small organic molecule. Organic helpers are often called coenzymes; they can assist transfers that the protein alone cannot readily perform. The cofactor may remain bound or exchange with the surroundings, depending on the system. A coenzyme can be chemically changed in one reaction and regenerated by another: regeneration of the enzyme does not mean every helper is unchanged in that individual reaction. Some enzymes need no additional cofactor. See Alberts et al., Molecular Biology of the Cell: small molecules associated with proteins and Alberts et al., activated carriers and their regeneration.

How much faster does adding substrate make a reaction? Enzyme kinetics gives a quantitative answer when a model's assumptions fit. For an illustrative reaction obeying Michaelis-Menten kinetics, use the Michaelis-Menten equation with substrate concentration $s$, limiting reaction rate $V_{\max }$ and Michaelis constant $K_M$. At $s/K_M=0.1$, substitution gives $v/V_{\max }=0.1/1.1=1/11\approx 0.0909$. At $s/K_M=1$, it gives $1/2$; at $s/K_M=10$, it gives $10/11\approx 0.9091$. Thus a hundredfold increase in substrate between the first and last examples yields only a tenfold increase in reaction rate.

For a chosen $V_{\max }=110\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$, those fractions mean reaction rates of $10$, $55$ and $100\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. Doubling substrate from $10K_M$ to $20K_M$ raises the normalized rate only from $10/11$ to $20/21\approx 0.9524$, a relative increase of about $4.76\%$. In this simplest model, the catalytic machinery is already nearly fully occupied. Doubling the amount of active enzyme instead doubles $V_{\max }$ and the predicted reaction rates at fixed free substrate concentration, provided the other conditions are unchanged. This distinction helps decide whether additional substrate or additional catalytic capacity could improve throughput.

The interpretation has limits. The simplest Michaelis-Menten kinetics assumes a single substrate, fixed active enzyme amount, a suitable quasi-steady-state approximation, and initial measurements with negligible reverse reaction from accumulated reaction product. Substantial depletion, interacting binding sites, or inhibition by excess substrate can invalidate this model. Within it, $K_M$ is the substrate concentration giving half the limiting reaction rate; it is not generally the binding dissociation constant $K_d$. See IUBMB, enzyme kinetics: steady-state approximation and substrate dissociation and IUBMB, Michaelis-Menten and non-Michaelis-Menten kinetics. Measuring activity also requires specified temperature, pH and necessary cofactor availability: a weak signal can reflect unsuitable conditions rather than little enzyme. See Brooks et al., Assay Guidance Manual: basics of enzymatic assays for HTS.

For N independent units with nondegenerate levels 0 and $\Delta >0$, the canonical partition function is $(1+e^{-\beta \Delta }{)}^N$. Its energy is $N\Delta /(1+e^{\beta \Delta })$ and heat capacity is $N{k}_B{x}^2{e}^x/(1+e^x{)}^2$, with $x=\Delta /(k_{B}T)$. Both are activated at low temperature. At high temperature the energy saturates at $N\Delta /2$ and heat capacity decays as $T^{-2}$. The equal-degeneracy Schottky peak is at $x\simeq 2.39936$ and height $0.43923N{k}_B$. A common energy shift leaves heat capacity unchanged; unequal level degeneracies require a different partition function.

An equation of state is a relation among thermodynamic state variables such as pressure, volume, temperature, energy, and particle number.

For noninteracting electrons at zero temperature, define $x=p_F/(m_{e}c)$ with Fermi momentum $p_F=\hslash (3{\pi }^2{n}_e{)}^{1/3}$. The isotropic momentum flux gives electron degeneracy pressureIts nonrelativistic and ultrarelativistic limits arerespectively. In a fully ionized stellar gas, $n_e=\rho /({\mu }_e{m}_u)$, using mean molecular weight per electron ${\mu }_e$. This neglects finite-temperature effects, interactions, and changes in composition.

For a specified chemical reaction and standard state convention, the dimensionless equilibrium constant is the equilibrium value of the reaction quotient $Q={\prod }_i{a}_i^{{\nu }_i}$, where $a_i$ denotes thermodynamic activity and ${\nu }_i$ a signed stoichiometric vector entry. Thermodynamic activities incorporate the reference state and departures from ideal behavior. See IUPAC Gold Book, equilibrium constant.

In the ideal concentration model for A converting to B, use $a_{\mathrm{A}}=[\mathrm{A}]/c^{\circ }$ and $a_{\mathrm{B}}=[\mathrm{B}]/c^{\circ }$ with the same reference $c^{\circ }=1\mathrm{M}$. Their ratio gives $Q=[\mathrm{B}]/[\mathrm{A}]$ and $K=0.75/0.25=3$. Equality $K=k_f/k_r$ applies to this elementary first-order pair; it is not a universal ratio of two measured rate constants for an arbitrary multistep chemical reaction.

At a specified constant temperature, with consistent standard states, the standard molar reaction Gibbs free energy isHere ${\Delta }_{r}G$ is the local slope per mole of extent of reaction at fixed temperature and pressure, rather than the finite total Gibbs free energy change between two mixtures. At equilibrium ${\Delta }_{r}G=0$, while ${\Delta }_r{G}^{\circ }$ generally remains nonzero. For $K=3$ at $298\mathrm{K}$, it is about $-2.72kJ/mol$. For positive activities, $Q<K$ favors forward conversion and $Q>K$ favors reverse conversion; neither sign supplies a reaction rate. See MIT 5.111, Lecture 18, reaction free energy and equilibrium.

An equilibrium stage models a contact whose outgoing liquid and vapor satisfy vapor-liquid equilibrium at the stage's temperature and pressure. It is a composition model, not a claim that each piece of packing reaches thermodynamic equilibrium.^[ref]

For blackbody stars of radii $R_A,R_B$ and temperatures $T_A,T_B$, a distant circumbinary planet with Bond albedo $A_B$ and uniform global reradiation has planetary equilibrium temperatureThis neglects internal cooling, eccentricity, and corrections from finite binary separation. The stars heat the planet through the sum of their absorbed bolometric fluxes, not an average of their temperatures.

The equipartition theorem says that each independent quadratic term in a Hamiltonian described by classical mechanics has mean energy $k_{B}T/2$ in a canonical ensemble. Here $k_B$ is the Boltzmann constant and $T>0$ the absolute temperature. Count terms in the energy, rather than vaguely counting everything that can move: one harmonic oscillator has one position degree of freedom but TWO quadratic terms, one in momentum and one in displacement.

A precise sufficient setting is $H(y,\xi )=a{y}^2/2+H_0(\xi )$ with $a>0$, where $y$ is one of the canonical variables, $\xi$ denotes all remaining variables, and the phase space integration measure factors as $dyd\xi$. The range of $y$ is the whole real line, independently of $\xi$; there is no constraint coupling its allowed values to other variables. The partition function must be finite, and the relevant expected values must exist. The Boltzmann factor then givesThe probability density function ${\rho }_y$ is a normal distribution with variance $k_{B}T/a$. A larger coefficient narrows the fluctuations enough to cancel its greater energy cost. This is why the mean contribution is independent of mass or spring constant.^[ref]

For a proof outline, the one-variable Gaussian integral is $Z_y={\int }_{-\infty }^{\infty }{e}^{-\beta a{y}^2/2}dy=\sqrt{2\pi /(\beta a)}$, so $-{\partial }_{\beta }\ln Z_y=1/(2\beta )$. More generally, integration by parts in a canonical variable gives $⟨y{\partial }_{y}H⟩=k_{B}T$ when its boundary term $[y{e}^{-\beta H}]$ vanishes and the integrals converge. For $a{y}^2/2$ this is twice its mean energy. At a finite hard boundary the boundary term need not vanish; hard-wall correction to harmonic equipartition illustrates the resulting departure. A constraint, a nonquadratic potential energy, or a noncanonical integration measure therefore requires its own calculation. The theorem is an ensemble average, not equality of instantaneous energies, and applying it to a recorded trajectory requires equilibration and sufficient sampling.

For a classical one-dimensional harmonic oscillator, with positive mass $m$ and spring constant $k$,The origin is the equilibrium position. Its heat capacity is $k_B$ at fixed $m,k$, not $k_B/2$. Although the two mean energies agree, their measured amplitudes do not: $⟨p^2⟩=m{k}_{B}T$ and $⟨x^2⟩=k_{B}T/k$. In a harmonic optical trap, the position measurement observes the potential energy contribution; it does not measure both contributions by observing $x$ alone.^[ref]

For $N$ particles in a three-dimensional classical monatomic gas, neglect interactions and thermally accessible internal excitations. Its three translational momentum terms give internal energy $U=3N{k}_{B}T/2$, heat capacity at constant volume $C_V=3N{k}_B/2$, and heat capacity at constant pressure $C_p=C_V+N{k}_B=5N{k}_B/2$. Per mole, replace $N{k}_B$ by the molar gas constant $R$. For example, using stipulated $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, the molar heat capacity at fixed volume is $12.471\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These results assume a dilute, nonrelativistic ideal gas with classical translational statistics, rather than a quantum degeneracy regime.

For an ideal gas of diatomic molecules, modeled as two chemical atoms with a bond, there are three translational quadratic terms, two accessible rotational kinetic energy terms for changes of bond orientation, and one bond-stretching harmonic oscillator with two quadratic terms. Rotation about the bond axis is not an additional active mode in this linear-molecule model. The three familiar limits of heat capacity at constant volume areCorresponding molar heat capacities are $3R/2$, $5R/2$ and $7R/2$. Counting the stretching harmonic oscillator only once as $k_{B}T/2$ would give the wrong last value.^[ref]

The frozen regimes require quantum mechanics. In the simple molecular rotation model, $E_j=k_B{\theta }_{r}j(j+1)$ with state degeneracy $2j+1$, ${\theta }_r={\hslash }^2/(2I{k}_B)$, and moment of inertia $I$. For the bond-stretching quantum harmonic oscillator, ${\theta }_v=\hslash \omega /k_B$ with angular frequency $\omega$ and $\hslash =h/(2\pi )$ from the Planck constant. When ${\theta }_r\ll {\theta }_v$, the three limits above correspond to $T\ll {\theta }_r$, ${\theta }_r\ll T\ll {\theta }_v$, and $T\gg {\theta }_v$, provided translational motion remains classical. They are smooth crossovers, not abrupt switches. Real molecules can have different allowed rotational populations; at high temperature, thermal dissociation, internal excitations and nonharmonic bond motion can invalidate the fixed-molecule model before its last limit is reached.^[ref]

For a quantum harmonic oscillator with fixed $\omega$, put $z={\theta }_v/T$. Its exact mean energy and heat capacity areAt $T\gg {\theta }_v$, the mean energy approaches $k_{B}T$ and $C_{\mathrm{osc}}$ approaches $k_B$. At $T\ll {\theta }_v$, thermal excitation is exponentially suppressed and $C_{\mathrm{osc}}$ tends to zero, although the zero-point energy $\hslash \omega /2$ survives. Frozen thus means little thermal response, not zero motion or zero energy.^[ref]

A motivating measurement is thermal calibration of an optical trap. A focused laser through a microscope objective confines a bead in water; imaging or a quadrant photodetector records a position time series. Gieseler and colleagues describe a polystyrene bead of radius $1.03\mu \mathrm{m}$ at $293.15\mathrm{K}$, a $1064\mathrm{nm}$ trapping laser, and photodetector acquisition at $10^5\mathrm{Hz}$. Their recorded bead continues undergoing Brownian motion; higher trapping power produces a narrower equilibrium position distribution. Infer the centered variance, not a squared distance from an arbitrary camera origin:The normal distribution position histogram and agreement with dynamical calibration methods support the local harmonic optical trap interpretation. For a stipulated example, $T=300\mathrm{K}$ and ${\sigma }_x=20\mathrm{nm}$ imply $k\simeq 1.04\times 10^{-5}N/m=0.0104pN/nm$. This calculated scale is not an experimental reading.^[ref]

Position calibration is essential: if detector output $v$ has conversion $x=s(v-v_0)$, its physical variance is $s^2\mathrm{Var}(v)$. An unknown scale $s$ cannot be determined together with $k$ from that one variance. A complementary calibration demonstrated by Tolić-Nørrelykke and colleagues combines the power spectrum of thermal motion with the bead response to known sinusoidal motion of a translation stage. The driven response supplies an additional length scale, and their method does not require bead size, liquid dynamic viscosity or wall distance as inputs. It still uses a thermal dynamical model; agreement between methods is a consistency check, not a theory-free proof of equipartition theorem.^[ref]

The readout itself can hide fluctuations. Wong and Halvorsen used a $1064\mathrm{nm}$ optical trap and adjustable-exposure CCD imaging of a bead in water. They varied exposure and trapping power, held the bead away from the nearest wall, adjusted illumination to maintain comparable images, and measured tracking noise using a stuck bead. The measured variance fell as exposure increased; corrected results agreed with their time-averaging model. For a stationary overdamped harmonic optical trap, with optical trap relaxation time $\tau$, an exposure of length $W$ measures ${\bar{x}}_W=W^{-1}{\int }_0^{W}x(t)dt$. Its ideal noise-free variance isFor $\alpha >0$, $S(\alpha )<1$: using the blurred variance in the uncorrected formula overestimates $k$ by $1/S(\alpha )$. At $W=\tau$, the calculated factor is $e/2\simeq 1.36$. Additive tracking noise instead raises the measured variance and biases the inferred $k$ downward if left uncorrected.^[ref]

Useful controls are to record over many optical trap relaxation times, compare blocks for drift, calibrate the position scale, measure fixed-bead noise, vary exposure, and check the position histogram for departures from a normal distribution. A sparse sequence of accurate instantaneous positions can estimate an equilibrium variance; a long exposure averages away motion even if many frames are collected. Flow, appreciable laser heating, escape from the harmonic region or nonconservative forces can break the assumed thermal equilibrium. Inferring $k$ by imposing equipartition theorem is a calibration, whereas testing the relation requires an independently established stiffness and position scale. The beautiful practical connection is that microscopic randomness can set a force scale: once calibrated, a small constant applied force displaces the mean by $f/k$. The same quadratic-term counting predicts macroscopic heat capacity and makes microscopic fluctuations useful measurements.

The equivalence point is the composition at which added titrant and the initial sample inventory match the specified chemical reaction's stoichiometry. It is a theoretical inventory condition; the titration endpoint is an observed stopping criterion. For a monoprotic acid consuming one hydroxide per acid unit, initial acid molar concentration $c_A$, sample volume $V_A$, and base titrant molar concentration $c_B$ giveThe proof is amount of substance conservation with one-to-one neutralization; a different stoichiometry requires its coefficients. All volumes must use compatible units. Since $1\mathrm{mL}=10^{-3}\mathrm{L}$, $0.100\mathrm{mol}{\mathrm{L}}^{-1}\times 10.00\mathrm{mL}=0.001000\mathrm{mol}$, not $1\mathrm{mol}$.

An equivalence point does not prescribe pH $7$. A weak acid titrated with a strong base leaves its conjugate base, which reacts with water; the corresponding chemical equilibrium can make the chemical solution alkaline. The neutral strong acid/strong base case additionally depends on temperature, through self-ionization of water.^[ref]

A physicist who studied ultracold monatomic gases at JILA.^[ref]

A physicist.^[ref]