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In (i), any pair of compatible functional groups can react: monomers, oligomers and long chains all participate. For an equimolar, strictly bifunctional mixture without cyclization in step-growth polymerization, the Carothers equation givesHere $p$ is functional-group conversion and ${\overline{X}}_n$ counts the original monomer molecules. Large degree of polymerization therefore requires nearly complete conversion.

In (ii), growth occurs only at the metal-bound chain end. If initiation is rapid and chain transfer in chain polymerization is negligible, the number of chains is fixed andThis is the linear conversion dependence of living polymerization. Actual zirconocene systems can undergo chain transfer, so linear growth is a limiting model, not a consequence of coordination polymerization alone.

In (iv), a newly initiated chain rapidly attains a large size and then undergoes radical termination. Further conversion mostly creates additional chains. Thus high degree of polymerization can occur at low conversion, without the systematic linear increase of an ideal living system. temperature, initiator depletion and monomer depletion can change that size during the run.

For the ideal equimolar step-growth polymerization, $p=0.5$ givesThis counts individual starting molecules, so it is only roughly one diamine–diester pair per average chain, rather than a long polymer. In a successful radical polymerization, many monomers normally add between initiation and radical termination. The styrene–maleic anhydride copolymer in (iv) therefore has the larger expected degree of polymerization.

Conversion alone does not specify the radical-chain length: the conclusion assumes ordinary polymer-forming conditions, rather than arbitrarily high initiator concentration or extremely strong chain transfer in chain polymerization.

The critical micelle concentration is the approximate total surfactant concentration above which additional surfactant predominantly enters micelles. Below it, most surfactant is present as separate molecules; above it, the free-monomer concentration changes much more slowly while the micelle population increases.

The crossover occurs when aggregation balances the unfavorable exposure of hydrophobic tails to water against head-group repulsion and the loss of molecular translational entropy. It is an aggregation threshold, not the concentration at which every molecule becomes micellar.

A consistent proposed sequence is desilylation, five-membered hemiketal formation, retro hetero-Diels–Alder opening, intramolecular aldol addition and hemiacetal closure. The essential carbon rearrangement is loss of the original ketone–side-chain bond and formation of a bond between its two neighbouring carbons; it is not a simple Michael addition.

For an explicit atom map, label the starting six-membered ring $A-B=C-D-E-O_r-A$, with the ketone at $D$, methyl substituents at $A$ and $B$, the hydroxy group at $A$, and the side chain $E-X-T-\mathrm{Et}$. Here $X$ is the methylene and $T$ is the carbon bearing the protected alcohol. Let $O_D$ be the original ketone oxygen and $O_T$ the protected alcohol oxygen. These labels identify atoms, not new substituents.

First, Fluoride removes the silyl protecting group from $O_T$. The resulting alcohol attacks the protonated ketone at $D$, forming the five-membered hemiketal $D-E-X-T-O_T-D$. The original oxygen $O_D$ is now a hydroxy group; the original pyran ring remains intact at this stage.

Next, The original pyran ring can undergo a retro hetero-Diels-Alder reaction. The three electron-pair movements are $B=C\to C=D$, $D-E\to E=O_r$ and $O_r-A\to A=B$: break $D-E$ and $O_r-A$, and form the three indicated double bonds. The result is the tethered aldehyde/dienol esterThe temporary $D-O_T$ bond keeps both reacting ends within the same molecule. At $D$, both oxygen substituents remain attached, and the new $D-O_T$ tether prevents separation into independent fragments.

Then The $C=D$ enol attacks the tethered aldehyde at $E$. Formation of $C-E$, restoration of $D=O_D$, and proton transfer give the six-membered lactone ring $C-D-O_T-T-X-E-C$, with $E-O_{r}H$. This is an intramolecular aldol reaction, and it supplies the crucial new carbon–carbon bond without losing any carbon atom.

Finally, Keto-enol tautomerism at $A=B$ gives the methyl ketone at $A$ and introduces the hydrogen at $B$. The $E-O_{r}H$ alcohol then attacks that ketone, closing the five-membered hemiacetal ring $A-B-C-E-O_r-A$. The resulting fused rings have exactly the product's connectivity: $D$ is the lactone carbonyl, $O_T$ is its ring oxygen, and $O_r$ is the five-membered-ring oxygen.

The stereocentre at $T$ survives throughout this sequence. The configuration initially at $E$ is erased at the aldehyde stage, and those at $A$, $B$, $C$ and $E$ arise during the later cyclization and proton-transfer steps. A folded intramolecular aldol reaction followed by reversible hemiacetal formation is consistent with the drawn cis ring junction, but connectivity and orbital bookkeeping alone do not prove that it is the unique selected diastereomer. In particular, a claim of simple retention at $E$ would contradict this atom map.

A related pyranone-to-fused-lactone rearrangement under aqueous hydrogen fluoride is experimentally reported; that account also explicitly qualifies the relative stereochemistry as assigned on mechanistic grounds. The sequence above is a proposed explanation of the examination structure, rather than a mechanism claimed to have been experimentally established in that paper.^[ref]

Write $a=[\mathrm{A}]$, $b=[\mathrm{B}]$ and $c=[\mathrm{C}]$ for the three concentrations. The letters denote chemical species; the forward arrow represents association of one molecule of each reactant into one complex, and the reverse arrow represents dissociation. The positive constants $k_{+}$ and $k_{-}$ are the respective reaction rate constants. For an elementary chemical reaction, the law of mass action gives forward and reverse reaction ratesEvery forward event removes one molecule of each reactant and adds one complex; a reverse event reverses those changes. Thus the reaction-rate equations areThe units of $k_{+}$ are inverse concentration per unit time, whereas $k_{-}$ has units of inverse time. Adding the equations gives the independent conservation lawsIn particular $a-b$ is constant, but this is a consequence of those two conservation laws. The total $a+b+c$ need not be constant: association changes the number of separate molecules.

For the general stoichiometry, take $p,q$ to be positive integers. The net reaction rate and stoichiometric vector areConsequentlyThe factors $p,q$ outside $R$ count molecules consumed per event; the powers inside $R$ express the law of mass action. The corresponding conservation laws are $a+pc=\text{constant}$ and $b+qc=\text{constant}$. No factorial is required in this deterministic convention for the reaction rate constant.

The molecular formula is $\boxed{{\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2}$, interpreting the vapour density as relative to air. The organic elemental analysis first determines a simpler empirical formula; the vapor-density molar mass measurement then fixes its multiple.

Take 100 mass units of the chemical substance and divide each elemental contribution by its rounded relative atomic mass:The empirical formula is therefore ${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}$, with empirical formula mass approximately 44. Its calculated mass percentages are 54.55% carbon, 9.09% hydrogen and 36.36% oxygen. The small discrepancies from the given organic elemental analysis are consistent with rounded experimental data; they should not be converted into large, meaningless integer subscripts.

For equal volumes of ideal gases at equal temperature and pressure, the density ratio equals the molar mass ratio. On the air convention,Doubling ${\mathrm{C}}_2{\mathrm{H}}_4\mathrm{O}$ gives ${\mathrm{C}}_4{\mathrm{H}}_8{\mathrm{O}}_2$; its rounded relative molecular mass of 88 predicts an air-relative vapor density of approximately $88/28.97=3.04$, close to 3.05. If the stated vapor density were instead relative to hydrogen, it would imply a relative molecular mass near 6.1, smaller than the empirical formula mass and incompatible with the analysis. That interpretation cannot describe a pure, undecomposed molecular vapor with the stated composition.

Under the older equivalent weight convention $\mathrm{C}=6$, $\mathrm{O}=8$, $\mathrm{H}=1$, the same composition and formula weight could be written ${\mathrm{C}}_8{\mathrm{H}}_8{\mathrm{O}}_4$. This is a change of notation, not eight modern carbon chemical atoms in the molecule. The result specifies composition and molar mass; it does not distinguish among structural isomers, such as ethyl acetate and butyric acid.

A molecule is an electrically neutral bound assembly of more than one atom. It can contain identical atoms, as in ${\mathrm{O}}_2$, or different atoms, as in ${\mathrm{H}}_{2}O$. Discrete molecules are not a universal description of matter: an extended ionic crystal does not consist of separate neutral molecules corresponding to each empirical formula unit. The IUPAC definition also specifies binding sufficient to support a vibrational state.^[ref]

Elasticity is a body's tendency to recover its original configuration when the deforming force is removed. Within a linear regime, Hooke's law makes stress proportional to strain, but elasticity does not require exact linearity. The elastic limit matters: permanent plastic deformation is a different response.

Inertia is the property expressed by Newton's first law: without a resultant external force, a body's velocity remains constant in an inertial frame of reference. Mass measures translational inertia. It is not a force acting against motion, and a moving body does not require a continuing forward force merely to keep moving.

A force describes an interaction that changes momentum. In Newtonian mechanics,A force has magnitude and direction; its point of application also matters for the torque and hence the body's rotational response.

The end groups associate reversibly through a complementary hydrogen-bond array. At low concentration, bringing the two ends of the same molecule together avoids the concentration-dependent cost of finding another molecule, so an intramolecular ring is favored.

For A, increasing concentration permits end-to-end intermolecular association to compete successfully with ring closure. It produces long, reversible supramolecular chains. Increasing chain length and entanglement account for the highly viscous solution without requiring covalent polymerization.

B has more favorably positioned attachment points on its rigid aromatic core. Its arms can close a less strained ring, so intramolecular association remains strongly competitive over the stated range. Predominantly discrete rings do not form the same extended chain network, explaining the retained fluidity. Both compounds can make the same local hydrogen bonds; their different balance of ring strain, preorganization and intermolecular association explains the macroscopic difference.

The terminal ureidopyrimidinone groups supply the association sites. Their urea/pyrimidinone NH donors and carbonyl/ring-nitrogen acceptors can form a self-complementary array of four hydrogen bonds between two UPy units. Joining the two terminal units of one molecule by this noncovalent association closes the ring.

The ether-containing arms provide flexibility and spacing; they are not covalently joined together during ring formation. The ring is therefore a supramolecular conformation of the existing molecule, rather than a new covalent ether macrocycle.

Interpret the supplied free energies as standard reaction free energies and use $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$. The dimensionless equilibrium constants areUnder the ideal-solution approximation, $[C]=K_C[L]$ and $[D]=(K_D/c^{\circ })[L{]}^2$. Set $k_D=K_D/c^{\circ }$, with units ${\mathrm{M}}^{-1}$. Conservation counts two original molecules in each dimer:The physical, positive root can be written without subtractive cancellation asThe resulting concentrations are:

For example, the first row satisfies $[L]+[C]+2[D]=0.100\mathrm{M}$; using $[L]+[C]+[D]$ would incorrectly count dimers. At both concentrations almost all molecular units belong to dimers.

dimer formation is favored by sufficiently high concentration because it joins two molecules. Dilution favors the separated species, whereas $[C]/[L]=K_C$ is concentration-independent in this ideal model. More than half the molecular units are dimerized when $2[D]>[L]+[C]$, equivalently $c_0>(1+K_C{)}^2/k_D\simeq 7.36\times 10^{-7}\mathrm{M}$. The supplied free energies at one temperature do not determine a temperature trend without additional enthalpy information. At $1\mathrm{M}$, neglected nonideal activities could be significant; the table is the requested concentration-based model.

The molecule combines three chemically different binding regions.

These assignments identify the preferred local sites, rather than guaranteeing three independent binding constants: occupancy can alter conformation, electrostatics and the affinity of neighboring sites.

Change is compatible with natural kinds, provided there is a defensible basis for tracking membership and explaining the changing properties. Requiring every property to remain unchanged would exclude nearly all physical objects. The important distinction is between variation within a kind, a transition out of it, and changes in the population that constitutes it.

Chemical elements offer a relatively sharp case. A carbon atomic nucleus has atomic number six, meaning six protons. Its isotope depends also on its neutron number; the ionization state of a carbon chemical atom can change without changing its chemical element. Temperature, position and velocity can also change. Classifying these objects together is useful because the relevant common constitution supports further inference, not because their entire property lists are identical at every moment.

A nuclear reaction that changes the proton number is different: it can produce a nucleus of another chemical element. This does not refute the original classification. It identifies a process under which membership changes. A theory of natural kinds should permit transformations between kinds rather than require its members to exist forever. The ordinary criterion also has a domain: an arbitrary mixture containing carbon is not itself a single carbon chemical atom.

Water makes a complementary point. A sample can undergo melting or boiling, changing density, rigidity and transport properties while retaining its ordinary chemical identification. The claim must be qualified: a bulk sample contains interacting molecules, and its detailed composition can include dissociation products. It need not consist at every instant exclusively of isolated identical water molecules. Nevertheless, changes of thermodynamic phase are explained through the constitution and organization of the material, rather than by treating every different appearance as a new unrelated substance.

These examples might suggest that some fixed microscopic essence is always required. Species put pressure on that inference. Individuals develop and differ; the distribution of inherited traits can change across generations through natural selection, genetic drift and gene flow. A demand that every member share an unchanging diagnostic appearance would misrepresent this history. It would also make a change in the frequency of one trait automatically create a new species, which is not how the relevant evolutionary processes work.

Homeostatic property cluster theory provides one alternative. Traits can tend to occur together because common development, shared ancestry and interactions within a population sustain their association. The traits need not be universal, and the maintaining processes need not freeze their frequencies. A population can change while retaining enough historical and causal continuity to support useful inference.^[ref] This is not a claim that every species concept gives the same boundaries. Reproductive relations, descent and ecological processes can motivate different criteria for different investigations.

An objection remains: if both members and their traits change, what prevents a natural kind from becoming a mere convenient label? The answer cannot be resemblance alone or an arbitrary chain of small changes. The proposed grouping must track processes that support reliable explanations and suitably qualified predictions. If the processes maintaining a cluster break down, or a lineage divides into independently evolving populations, continuing to use one label may cease to serve the relevant inquiry. Membership can be uncertain near such transitions without making every central case arbitrary.

There is also no requirement that all natural kinds have the same basis of unity. The atomic number criterion and a historically sustained population answer different scientific questions. Imposing the former structure on every biological grouping would mistake one successful form of classification for a universal metaphysical rule.

Changing properties therefore do not by themselves defeat naturalness. The scientific work lies in identifying which variations a kind tolerates, which transformations end membership, and why its remaining unity supports inference. A classification earns its status through that explanatory achievement, not through the fiction that nature is motionless.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

Molecular transport across gradients. A gas can transport momentum, thermal energy and species abundance even when its average bulk velocity is zero. Molecules arriving at a point carry properties acquired near their preceding collisions. A small gradient makes the populations arriving from opposite sides slightly different, producing transport from larger to smaller values of the carried property. A continuum transport law requires a mean free path $\lambda$ small compared with the gradient length, and relaxation times short compared with the observation time.

For a dilute hard-sphere gas,where $d$ is molecular collision diameter. Both speed and collision distance matter: high speed alone does not determine a transport coefficient.

A simple kinetic estimate makes the common structure explicit. For a quantity $a(y)$ carried per particle, a molecule moving at angle $\theta$ to $+y$ brings approximately $a(y-\lambda \cos \theta )$. Expanding to first order and averaging an isotropic velocity distribution givessince $⟨\cos \theta ⟩=0$ and $⟨{\cos }^2\theta ⟩=1/3$. This mean-free-path argument estimates the coefficients; a detailed collision calculation changes their numerical prefactors and distinguishes different relaxation rates.

Momentum transport and viscosity. For a shear flow with mean velocity $U_x(y)$, the transported directed momentum per molecule is $m{U}_x$. The net flux of $x$ momentum in the $+y$ direction isHere $\eta$ is dynamic viscosity in $\mathrm{Pa}\mathrm{s}$. The associated shear stress on a material surface has the corresponding traction sign convention; the flux carries momentum from the faster-moving layer towards the slower-moving one. The kinematic viscosity is $\nu =\eta /(nm)$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Viscosity smooths velocity gradients, and the lost ordered mechanical energy becomes heat.

Heat transport. Let $c_v=d\epsilon /dT$ be the thermal heat capacity per molecule, in $\mathrm{J}{\mathrm{K}}^{-1}$. Substituting $a=\epsilon (T)$ gives Fourier's law,Thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a monatomic gas, translational equipartition gives $c_v=3k_B/2$. Rotational or vibrational excitations in other gases can increase the energy carried per molecule, while collisions govern how those modes exchange energy. A temperature gradient can therefore be transported differently from directed momentum even when the dimensional estimates look similar.

Species transport. For a dilute tracer or a mixture with prescribed total density and negligible bulk flow, molecules diffuse down concentration gradients. The first-gradient number flux is Fick's first law,The diffusivity $D$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying number flux by the species mass gives mass flux. More generally one uses diffusion flux relative to the mixture's bulk velocity and specifies mass or mole fractions. The approximation is not a claim that an arbitrary pure-gas pressure gradient can remain at rest while only a Fick flux acts. In a mixture, collision cross-sections and relative molecular speeds, including the reduced mass, determine the species diffusivities.

Dependence on density and temperature. The kinetic estimates of gas transport coefficients predict, for fixed collision diameter and molecular heat capacity,Thus dynamic viscosity and thermal conductivity are approximately density independent in the dilute continuum regime. Reducing density reduces the number of carriers but increases their mean free path in the same proportion. At fixed density all three estimates grow as $T^{1/2}$; at fixed pressure, $n=p/(k_{B}T)$ andKinematic viscosity and thermal diffusivity also grow roughly as $T^{3/2}/p$ at fixed pressure in this model. Real interaction potentials make effective collision cross-sections temperature dependent, and internal excitations change $c_v$. Dense-gas correlations and boundaries invalidate the simple cancellation. In particular, when the mean free path approaches the apparatus scale, transport becomes free molecular; it cannot be described by extending the same local, pressure-independent Fourier and viscosity coefficients indefinitely.

Heat diffusion equation. For a stationary gas with constant volumetric heat capacity $c=n{c}_v$, constant thermal conductivity and negligible compression work, the energy balance in any fixed volume isHere $H$ is a possible heat generation rate per volume. The divergence theorem and Fourier's law yieldThermal diffusivity $\alpha$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is of order $\overline{v}\lambda /3$. In a moving or compressing gas, the material derivative, pressure-volume work and viscous heating must also enter the internal-energy equation. The displayed pure diffusion equation describes the approximation in which those effects are absent or negligible.

Representative solutions. With no source, a steady one-dimensional slab between temperatures $T_a,T_b$ at $x=0,L$ has $T=T_a+(T_b-T_a)x/L$ and uniform heat flux $-\kappa (T_b-T_a)/L$. Thus a constant spatial gradient transports heat without changing local temperature in time when the boundaries continually supply and remove it.

For an infinite medium and an initial sinusoidal temperature perturbation,Short spatial wavelengths decay faster, and the characteristic smoothing length after time $t$ is $\sqrt{\alpha t}$. A finite slab with both ends held at $T_b$ has a sine expansion with factors $e^{-\alpha (n\pi /L{)}^{2}t}$; its slowest relaxation time is $L^2/({\pi }^2\alpha )$.

An instantaneous heat input $Q_A$ per unit transverse area at $x=0$ in an otherwise uniform infinite medium gives the Gaussian solutionIts integral $\int c(T-T_b)dx=Q_A$ conserves the deposited heat. The peak falls as $t^{-1/2}$ while its width grows as $\sqrt{t}$; diffusion redistributes energy instead of transporting a sharp front at a single velocity.

Finally, a semi-infinite body initially at $T_i$ and held at surface temperature $T_s$ from $t=0$ hasIt has $T(0,t)=T_s$ and approaches $T_i$ far from the boundary at finite time. These solutions demonstrate steady conduction, mode decay, spreading of localized heat and boundary-driven diffusion. Their idealized behaviour at extremely short times or arbitrarily long distances must be interpreted within the continuum regime rather than as an assertion of instantaneous microscopic communication.

Model air as a dilute ideal gas of molecules described by the hard sphere model. The number density and mean free path areThe factor $\sqrt{2}$ accounts for the relative motion of the colliding molecules. A molecular collision diameter has to be estimated because none is supplied. Taking $d\simeq 0.37\mathrm{nm}$, appropriate as an order-of-magnitude size for the main constituents of air, givesThus the mean free path is about $7\times 10^{-8}\mathrm{m}$, or $70\mathrm{nm}$. A slightly smaller assumed diameter $0.35\mathrm{nm}$ gives about $76\mathrm{nm}$, illustrating the intended precision of the estimate.

From molecular dipoles to macroscopic fields. A dielectric contains charges which remain bound to its atoms or molecules. An applied electric field displaces positive and negative charge centres and can orient permanent electric dipoles. The electric polarization $\mathbf{P}$ is electric dipole moment per unit volume, in $\mathrm{C}{\mathrm{m}}^{-2}$. Its associated bound charge isat the boundary of the polarized material, where $\hat{\mathbf{n}}$ points outward. A uniform electric polarization has no volume bound charge, but generally has surface bound charge. The electric field remains the force per unit test charge; it includes the fields of both free and bound charges.

Introducing the electric displacement field $\mathbf{D}={ϵ}_0\mathbf{E}+\mathbf{P}$ packages the free-charge source separately:For a homogeneous linear isotropic dielectric,The electric susceptibility ${\chi }_e$ is dimensionless; the permittivity $ϵ$ has units $\mathrm{F}{\mathrm{m}}^{-1}$. This is a constitutive approximation. An anisotropic dielectric requires a permittivity tensor, while a nonlinear dielectric need not have electric polarization proportional to electric field. The field in these equations is the macroscopic field, not necessarily the local field at an individual molecule.

Interfaces. Let $\hat{\mathbf{n}}$ point from medium 1 into medium 2. Integrating Gauss's law over a thin pillbox, and the electrostatic curl equation around a thin loop, gives the electrostatic interface boundary conditionsThus normal electric displacement is continuous when there is no free surface charge, whereas tangential electric field is continuous even if the permittivity changes. Normal electric field generally jumps because the interface carries bound charge. The electric potential is continuous for an ordinary interface without an ideal dipole sheet. These conditions, together with Laplace's equation in each homogeneous charge-free region, determine the field.

Energy and electrical sources. Reversibly assembling free charge requires incremental work $\delta W=\int \mathbf{E}\cdot \delta \mathbf{D}dV$. For a linear dielectric whose response can be reversibly integrated from zero field, the electrostatic energy isThis includes the material polarization response, rather than simply using ${ϵ}_0{E}^2/2$ everywhere. In a capacitor, $W=Q^2/(2C)=C{V}^2/2$. Inserting a dielectric increases capacitance. At fixed charge the stored energy decreases. At fixed voltage the stored field energy increases, but the voltage source supplies $VdQ$; the appropriate mechanical energy including that source changes by $-V^{2}dC/2$. Both cases predict attraction of dielectric into the capacitor. A force calculation must state which electrical quantity is fixed. A lossy or hysteretic dielectric also dissipates energy, so the reversible quadratic expression does not describe its full history.

Shape and depolarization. Suppose the dielectric is surrounded by vacuum and the distant imposed electric field is ${\mathbf{E}}_0$. Bound surface charge produces a depolarization field. For a uniformly polarized ellipsoid, along a principal axis,Here $N_d$ is the depolarization factor of a dielectric ellipsoid. This self-consistent field explains why the same material responds differently in different shapes.

An infinite slab with electric field normal to its faces has $N_d=1$, giving $E_{\mathrm{in}}=E_0/{ϵ}_r$; its opposite surface bound charges reduce the internal field. For electric field parallel to the faces, $N_d=0$, giving $E_{\mathrm{in}}=E_0$. End effects are neglected.

A very long circular rod in an axial electric field has negligible depolarization away from its ends, so $E_{\mathrm{in}}\simeq E_0$. In a transverse electric field the infinite-cylinder solution has $N_d=1/2$, giving $E_{\mathrm{in}}=2E_0/({ϵ}_r+1)$. To see the latter directly, take the applied field along $x$. The interior and exterior electric potentials can be written ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +A\cos \theta /r$. Continuity of potential and normal electric displacement at $r=a$ gives $B=2E_0/({ϵ}_r+1)$.

For a sphere of radius $a$, $N_d=1/3$ andFor example, write ${\varphi }_{\mathrm{in}}=-Br\cos \theta$ and ${\varphi }_{\mathrm{out}}=-E_{0}r\cos \theta +C\cos \theta /r^2$. Continuity of potential gives $B=E_0-C/a^3$, and continuity of normal electric displacement gives $ϵB={ϵ}_0(E_0+2C/a^3)$. Solving yields $B=3E_0/({ϵ}_r+2)$ and $C=a^3{E}_0({ϵ}_r-1)/({ϵ}_r+2)$. The $1/r^2$ potential is that of the electric dipole moment above. For ${ϵ}_r=1$ all these induced effects vanish; as ${ϵ}_r$ becomes very large the normal-field sphere and slab approach the corresponding conductor field exclusion. The ideal infinitely long rod in an exactly axial imposed field has a different, singular end-free limit.

This is the one-mole, or equivalently molar volume, form of the Van der Waals equation, an approximate equation of state for a real gas:Here $p$ is the gas pressure, $T$ its absolute temperature, $R$ the molar gas constant, and $V$ the volume occupied by one mole. If $V$ denotes molar volume explicitly, its units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $b$ represents excluded volume arising from the finite size and short-range repulsion of the molecules. The volume available for translational motion is reduced from $V$ to $V-b$. This is an effective excluded volume, not simply the sum of the molecules' geometric volumes. Its molar units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $a$ measures intermolecular attraction in a mean-field approximation. Attractions reduce the force exerted on the walls, lowering the measured pressure by $a/V^2$ relative to the repulsion-only expression. Thus $p+a/V^2$ is the pressure corrected for this attraction. The squared inverse-volume dependence reflects a contribution proportional to the square of the molecular number density. The molar units of $a$ are $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$.

For an arbitrary amount $n$ measured in moles, with total container volume $V_{\mathrm{tot}}$, the corresponding formula isAt low density, when $b\ll V$ and $a/V^2\ll RT/V$, it reduces to the ideal gas law. The model captures qualitative attraction, repulsion and liquid-gas behavior, but does not supply an exact equation of state for every material.

At low temperature, $T\ll {\theta }_r$, retain the ground level and the first excited level. Their energies are $0$ and ${\hslash }^2/I$, with degeneracies $1$ and $3$. Thereforeso in the notation of the question,$A$ is dimensionless and $B$ has units of kelvins. The first omitted level has three times the first excitation energy, explaining the exponent in the error term.

At high temperature, $T\gg {\theta }_r$, many rotational levels are occupied and the sum can be replaced to leading order by an integral:Set $u=J(J+1)$, so that $du=(2J+1)dJ$. ThenThusThe high-temperature approximation here concerns the rotational spectrum. For a real molecule it should still be used below temperatures where neglected vibrational excitation or dissociation invalidates the rigid-rotor model.

An electrical insulator has no freely conducting bulk charge in the ideal static description, but an applied electric field can displace bound positive and negative charge relative to one another. The resulting electric dipole moments per unit volume define the polarization density $\mathbf{P}$. Distortion of electronic charge clouds, relative displacement of ions, and partial alignment of permanent molecular electric dipoles can all contribute. Thermal agitation opposes the last mechanism.

For a weak field in a linear isotropic dielectric, the constitutive law isHere ${\chi }_e$ is the electric susceptibility, $ϵ$ is the permittivity, and $\mathbf{D}$ is the electric displacement field. In an anisotropic material, ${\chi }_e$ is a tensor and $\mathbf{P}$ need not be parallel to $\mathbf{E}$. These response laws describe the macroscopic field, not necessarily the microscopic field acting on one molecule.

Spatial variation and surface termination of the polarization density give bound charge and bound current through the bound charge densitieswhere $\hat{\mathbf{n}}$ is outward from the material. Thus uniform polarization density has no bulk bound charge, but generally has opposite bound charges on opposing surfaces. Gauss's law can be rewritten asWith no free surface charge, the normal component of $\mathbf{D}$ is continuous across an interface; the tangential electric field is continuous in electrostatics.

For example, fill a parallel-plate capacitor of plate area $A$ and gap $d$ with a uniform dielectric, neglecting fringing. At fixed free plate charge $Q$, $D=Q/A$, soThe induced bound surface charges oppose the original field, reducing the electric field and potential difference at fixed $Q$. At fixed applied potential difference, the field remains $V/d$ and the battery instead supplies more free charge. The electric displacement field separates these free-charge and bound-charge roles.

An electric dipole is a localized pair of opposite electric charges, or more generally a neutral electric charge distribution with nonzero first electric moment. For charges $-q$ and $+q$ separated by displacement $\mathbf{d}$ directed from negative to positive, the electric dipole moment isA neutral molecule can have a permanent electric dipole moment when its positive and negative charge centres do not coincide. An applied electric field can also produce an induced electric dipole moment by distorting an initially nonpolar electric charge distribution.

At distance $r\gg |\mathbf{d}|$ in vacuum, expand the difference of the two point-charge potentials to first order in the separation. The leading electric potential isThe monopole term cancels because the net electric charge is zero. Taking $\mathbf{E}=-\nabla \Phi$ gives the electric dipole fieldOn the dipole axis its magnitude is $2p/(4\pi {\epsilon }_0{r}^3)$ and its direction is along $\mathbf{p}$; in the equatorial plane the magnitude is $p/(4\pi {\epsilon }_0{r}^3)$ and the field points opposite to $\mathbf{p}$. The $r^{-3}$ decay is faster than the $r^{-2}$ field of an isolated point charge.

In a uniform external electric field, the opposite forces on the two charges cancel in their resultant but form a couple. Its torque isFor a fixed permanent electric dipole moment, the interaction potential energy isDifferentiating with respect to orientation recovers the torque. Parallel alignment minimizes $U$ and is stable against a small rotation; antiparallel alignment maximizes $U$ and is unstable. Rotating a fixed dipole slowly from parallel to antiparallel requires external work $2pE$.

A nonuniform electrostatic electric field can also exert a net force. Expanding the two charge forces gives $\mathbf{F}\simeq (\mathbf{p}\cdot \nabla )\mathbf{E}$, which equals $\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed dipole in a curl-free electrostatic field. Thus a suitably aligned dipole is attracted toward stronger field. An induced electric dipole moment requires care in computing energy: when $\mathbf{p}=\alpha \mathbf{E}$ for constant isotropic electric polarizability, building the dipole costs energy and the effective interaction is $U=-\alpha E^2/2$, rather than substituting directly into the fixed-dipole formula. These field, torque and energy effects explain molecular orientation and electric polarization.