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Over the plotted range, approximate the surface tension by $\gamma (T)={\gamma }_0-\beta T$ with constant positive $\beta =-{\gamma }^{\prime }$, and take $C_A=70\mathrm{mJ}{\mathrm{K}}^{-1}=0.070\mathrm{J}{\mathrm{K}}^{-1}$. A reversible adiabat hasIntegrating givesThe arbitrary fixed reference temperature $T_{\ast }$ makes the logarithm dimensionless; changing it only shifts the integration constant. This is the dimensionally precise version of writing $C_A\ln T+\beta A$ as constant. Each term has units of entropy. Between two states,If the graph had substantial curvature, the entropy differential would still hold, but this particular constant-$\beta$ exponential would no longer follow.

Equilibrium radiation and cavity frequencies. A blackbody absorbs all incident radiation at every frequency and direction. In thermal equilibrium its emission spectrum depends only on its temperature. Radiation inside an isothermal absorbing cavity, sampled through a small aperture that hardly disturbs equilibrium, realizes black-body radiation to a good approximation. Microscopic absorption and emission establish the equilibrium distribution; a perfectly reflecting wall is the convenient idealization used to count the modes once that distribution is specified.

In a rectangular electromagnetic cavity, the wave equation and perfect conductor wall boundary conditions select component wavenumbers $k_x=\pi l/L_x$, $k_y=\pi m/L_y$, $k_z=\pi n/L_z$. Combining the vacuum dispersion relation $\omega =c|\mathbf{k}|$ and $\nu =\omega /(2\pi )$ givesFor example, electric-field components can have the spatial dependenceFor three positive indices, Gauss's law requires $k_x{A}_x+k_y{A}_y+k_z{A}_z=0$, leaving two independent polarizations of an electromagnetic wave.

Index qualification. The original paper describes the indices as positive integers. That describes the three-nonzero-index family but omits valid electromagnetic modes with exactly one zero index, which have one polarization of an electromagnetic wave. For instance, with $n=0$, $l,m\ge 1$, only $E_z$ above remains; it is normal to the top and bottom faces and vanishes at the side walls. Two zero indices give no nontrivial cavity field. The omitted one-zero-index family does not affect the leading three-dimensional mode count, but is essential to the thin-cavity limit below. Retaining the paper's strict positivity restriction for every mode would instead give a thickness gap and would not yield its requested two-dimensional power law.

Three-dimensional mode count. A lattice cell in positive $\mathbf{k}$ space has volume ${\pi }^3/V$, where $V=L_x{L}_y{L}_z$. In a continuum approximation, a shell of radius $k$ and thickness $dk$ contains its positive octant, multiplied by two polarizations:Using $k=2\pi \nu /c$ givesFor a finite cavity the exact density of states is discrete; this formula is its bulk continuum limit, appropriate when many modes contribute and wall corrections are small.

Mean photon number. A mode is a quantum harmonic oscillator with thermal excitation energies $sh\nu$, $s=0,1,2,\dots$. The zero-point energy does not enter its thermal occupation. Put $q=e^{-h\nu /(k_{B}T)}$. Its partition function and mean occupation areThis is the Bose-Einstein distribution with zero photon chemical potential: equilibrium absorption and emission do not conserve photon number.

Thermal energy. Multiply the density of states by the mean excitation energy $h\nu ⟨s⟩$ and integrate:The substitution $x=h\nu /(k_{B}T)$ isolates all the dependence on cavity dimensions and temperature:Expanding $1/(e^x-1)={\sum }_{r=1}^{\infty }{e}^{-rx}$ gives an integral ${\sum }_{r}6/r^4=6\zeta (4)={\pi }^4/15$. HenceThis is the thermal energy only, excluding the zero-point contribution. The three-dimensional continuum regime requires $k_{B}T$ large compared with the geometric level spacings in all three directions.

Spectral sketch. The spectral energy density per unit ordinary frequency is Planck's law,At small frequency it rises as $8\pi k_{B}T{\nu }^2/c^3$, while at large frequency it decays exponentially. Raising temperature increases the spectrum at every fixed positive frequency, moves the peak to larger frequency in proportion to $T$, and increases the integrated area in proportion to $T^4$. A plot per unit wavelength has a different peak position because the integration variable changes.

Thin-cavity radiation. Let $A=L_x{L}_y$. The relevant low-temperature window isNonzero thickness indices are then thermally suppressed, while many in-plane modes are populated. The $n=0$ family has one polarization of an electromagnetic wave. Its allowed in-plane wavenumbers form a lattice of cell area ${\pi }^2/A$; a quarter of a circular shell therefore containsThus the dimensional crossover of thermal cavity radiation givesThe integral is $2\zeta (3)$ by the same geometric-series argument. A model with a different number of independent polarizations of an electromagnetic wave changes the prefactor but not the $T^3$ scaling. For a cavity with fixed finite $L_x,L_y$, cooling still further eventually freezes its lowest in-plane mode as well: the strict $T\to 0$ thermal energy is exponentially suppressed. The requested area law is the intermediate quasi-two-dimensional regime, not an arbitrarily low-temperature continuum limit.

In a long wire away from its ends, the steady temperature is independent of axial position and azimuth. The heat equation reduces toMultiplying by $\rho$ and integrating givesA finite temperature and nonsingular flux at the axis require $C_1=0$. Applying $T(a)=T_0$ determines $C_2=T_0+H{a}^2/(4\kappa )$. ThereforeThis is the steady temperature of a uniformly heated cylindrical wire. Its hottest point is at the centre, with $T(0)-T_0=I^2/(4{\pi }^2{a}^2\kappa \sigma )$. The outward heat flux is $J_Q(\rho )=-\kappa T^{\prime }(\rho )=H\rho /2$, so the power crossing a cylindrical surface per unit length, $2\pi \rho J_Q(\rho )=H\pi {\rho }^2$, equals the heating generated inside it.

Origin of diffusion. In a material near local thermal equilibrium, microscopic carriers transfer energy from hotter regions to colder regions. On distances long compared with their mean free path and times long compared with relaxation times, the local constitutive law is Fourier's law,Combining this temperature-gradient-driven flux with local conservation of energy produces a diffusion equation. This description assumes diffusive thermal conduction; it is not generally valid for ballistic transport, and flow would add advective transport.

Energy balance with heating. Choose any fixed volume $V$, let $c$ be its uniform volumetric heat capacity and $H$ its heat generation rate per unit volume. The rate of increase of internal energy equals heat generation minus outward heat flux:Since $V$ is arbitrary, for constant thermal conductivityThe ratio $\kappa /c$ is the thermal diffusivity in ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Here $c$ is in $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$, $\kappa$ in $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and $H$ in $\mathrm{W}{\mathrm{m}}^{-3}$, so both right-hand terms have units $\mathrm{K}{\mathrm{s}}^{-1}$. If thermal conductivity varies with position or temperature, the divergence $\nabla \cdot (\kappa \nabla T)$ must be retained rather than replacing it by $\kappa {\nabla }^{2}T$.

Replacing the fixed surface temperature by Newton's law of cooling changes the boundary condition but not the radial derivative in steady state. At the surface,HenceHere $\gamma$ is a surface heat transfer coefficient in $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, unrelated to the surface tension symbol used in C10. An independent energy-balance check uses a length $\ell$: the electrical input $I^2\ell /(\sigma \pi a^2)$ must equal the surface loss $2\pi a\ell \gamma [T(a)-T_{\mathrm{air}}]$, which gives the same answer.

The complete temperature profile isThermal conductivity controls the internal gradient, but cancels from the surface-temperature rise because all generated power must eventually cross the prescribed surface. This assumes positive constant $\gamma$, uniform electrical conductivity and negligible end or additional radiative losses; temperature-dependent material properties could make the heating and boundary problem nonlinear.

Entropy at absolute zero. The third law of thermodynamics concerns the low-temperature limit of equilibrium states. In the zero-entropy formulation, the entropy of a perfect crystal with a nondegenerate ground state tends to zero as $T\to 0$. The statistical interpretation is $S=k_B\ln \Omega$: a unique accessible ground state has $\Omega =1$. The entropy-difference formulation of the third law of thermodynamics states that entropy differences associated with regular changes of mechanical parameters tend to zero as absolute zero is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate ground state can carry residual entropy.

For a phase of fixed composition and pressure,A finite limiting entropy constrains this integral. Ordinary equilibrium heat capacities therefore vanish as temperature approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic phonons and $C\propto T$ for a normal degenerate Fermi liquid. A constant positive heat capacity extrapolated to zero would give a logarithmically divergent entropy difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.

Gibbs free energy. For fixed particle numbers,Thus, when $S\to 0$, the Gibbs free energy approaches its zero-temperature value with a horizontal tangent:The equality $G(0,p)=H(0,p)$ uses the enthalpy $H=U+pV$; neither limiting value needs to be zero. The low-temperature Gibbs free-energy expansion makes the approach quantitative. For $C_p=a{T}^r$ with $r>0$ and $S(0)=0$,The second derivative is ${\partial }_T^{2}G=-C_p/T\le 0$, although it need not tend to zero. For a reaction between such equilibrium phases, ${\partial }_T\Delta G=-\Delta S\to 0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine residual entropy $S_0$ is retained instead, $G=G(0,p)-S_{0}T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.

Why helium remains a liquid. Helium atoms attract each other weakly and are exceptionally light. Their zero-point energy is large compared with the binding that would localize them in a solid. Both helium-4 and helium-3 can therefore remain liquid down to the lowest temperatures at low pressure; substantial applied pressure is needed to solidify them. This is not a violation of the third law. A quantum liquid can have a unique low-entropy ground state, and zero-point motion does not itself imply thermal entropy. The third law does not require every substance to crystallize at atmospheric pressure.

A helium-4 atom has integer spin zero and is a boson. Below the lambda point near $2.17\mathrm{K}$ at saturated vapor pressure, its liquid is superfluid. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the condensate fraction need not equal the superfluid fraction. The two-fluid model of superfluidity separates an entropy-carrying normal component of a superfluid from the superfluid component, whose entropy is zero. At sufficiently low temperature the dominant thermal excitations are acoustic phonons: their linear dispersion in three dimensions gives $U-U_0\propto V{T}^4$, $C_V\propto V{T}^3$ and $S\propto V{T}^3$. Higher-energy rotons are exponentially suppressed. Thus the normal component of a superfluid and entropy disappear as $T\to 0$, while superfluid flow and quantum motion can persist.

A helium-3 atom has spin $1/2$ and is a fermion. Well below its Fermi temperature, but above its superfluid transition, liquid helium-3 is a degenerate interacting Fermi liquid. Only quasiparticles close to the Fermi surface are thermally excited, giving $C_V\propto T$ and $S\propto T$. At temperatures of order a millikelvin, helium-3 also becomes superfluid through formation of Cooper pairs of fermions. This is not Bose-Einstein condensation of individual helium-3 atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic quasiparticle contribution to heat capacity is exponentially suppressed at temperatures well below the pairing gap. Gapless collective modes can still give thermal terms following a power law. The entropy nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.^[ref]

Helium-3 also illustrates the importance of nuclear spin entropy. Over an intermediate low-temperature range, localized spins in the solid can retain approximately $k_B\ln 2$ of entropy per atom, while the degenerate liquid has much less. Reversible conversion of liquid to the higher-entropy solid then absorbs heat; adiabatic compression along the melting curve can cool the sample. This is Pomeranchuk cooling. At still lower temperature, interactions between the nuclear spins remove this nearly free-spin entropy, so a constant $k_B\ln 2$ cannot be extrapolated to absolute zero.^[ref]

Liquefaction and cooling methods. A gas must be below its liquid-gas critical point to form a distinct liquid phase by compression. Compression above the critical temperature alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression heat, cools the compressed gas in a heat exchanger and uses expansion to lower its temperature further.

A throttling valve operates approximately at constant specific enthalpy when kinetic-energy changes and heat leakage are negligible. The Joule-Thomson coefficient isusing consistent molar quantities, or consistent total quantities for a fixed amount. A pressure drop cools the gas if ${\mu }_{\mathrm{JT}}>0$. An ideal gas has zero Joule-Thomson coefficient; real intermolecular interactions are essential. In particular, room-temperature helium warms on throttling, so it must first be brought into a region of positive Joule-Thomson coefficient. Expansion through an engine or turbine can instead remove work and produce approximately isentropic cooling.

A regenerative liquefier sends the cold returning gas through a counterflow heat exchanger to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces liquid. Modern helium liquefaction commonly combines work-producing expansion with a final throttle. ^[ref]

Once liquid exists, pumping away vapour lowers its saturation pressure and boiling temperature. The escaping vapour carries latent heat away, cooling the remaining liquid. At still lower temperatures, a dilution refrigerator exploits the different entropy carried by helium-3 in concentrated and dilute helium mixtures; adiabatic demagnetization and Pomeranchuk cooling provide other routes.

Approaching zero. The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach absolute zero in finitely many stages. As the entropy available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero entropy cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. Quantum liquids and refrigeration methods can approach $0\mathrm{K}$ while remaining consistent with vanishing equilibrium entropy; they do not provide a finite-operation route to absolute zero.

For a reversible adiabatic process, an ideal gas obeys $T{V}^{\gamma -1}=\text{constant}$. A monatomic gas has heat capacity ratio $\gamma =5/3$. Taking the stated expansion factor to mean $V_f=9V_i$,This assumes the usual quasistatic isentropic process. Heat insulation alone does not determine the final temperature: an ideal-gas free expansion is also adiabatic but leaves the temperature unchanged. The numerical answer uses nine times the original volume, the conventional intended meaning of the source's “nine times larger”.

The units supplied for 130 indicate specific heat capacity, so the block's total heat capacity is $C=mc=0.5(130)=65\mathrm{J}{\mathrm{K}}^{-1}$. Treat it as constant, and the lake as a thermal reservoir at $T_0=283.15\mathrm{K}$. The initial block temperature is $T_i=373.15\mathrm{K}$.

Compute the block's entropy change using a reversible comparison path, even though the actual cooling is irreversible:The lake receives $Q=C(T_i-T_0)=5850\mathrm{J}$ at constant temperature, soThereforeThis is entropy generated by cooling a finite body in a reservoir. The positive total is consistent with the Second law of thermodynamics, although the cooling block's own entropy decreases.

Initially and finally the block and lake are at the same temperature. After the falling block has come to rest, its lost gravitational potential energy has dissipated into internal energy of the lake and surroundings. In the idealization that all $mgh$ is ultimately thermalized in the lake at $T_0=283.15\mathrm{K}$,The block returns to its original thermal state, so its net entropy change is zero. Mechanical potential energy carries no thermal entropy that must be subtracted. ThusUse of $g=9.80\mathrm{m}{\mathrm{s}}^{-2}$ gives the same three-significant-figure answer. Splashing and any temporary heating do not change this final-state balance if their mechanical energy is eventually dissipated in the same reservoir.

Three phase coexistence curves meet at the triple point, where ice, liquid water and water vapour have equal chemical potential. In the ordinary-pressure region, the low-temperature phase is ice, the high-pressure moderate-temperature phase is liquid, and the low-pressure high-temperature phase is vapour. The triple point is approximatelyBelow $p_t$, warming stable ordinary ice crosses the sublimation boundary directly into vapour; there is no intervening stable liquid. Above $p_t$, melting and then boiling can occur. The liquid–vapour curve terminates at the liquid-gas critical point, approximately $647.1\mathrm{K}$ and $22.06\mathrm{MPa}$, beyond which there is no liquid–vapour phase boundary.

The distinctive feature is the negative slope of the ordinary ice–water melting line. The Clapeyron equation givesMelting increases entropy but reduces volume because ordinary ice is less dense than liquid water, so $\Delta s>0$ and $\Delta v<0$. Increasing pressure consequently lowers the melting temperature locally. Sublimation and vaporization both increase volume and entropy, giving positive slopes. The following original schematic shows these signs; it is not a quantitative fit and omits the additional high-pressure ice polymorphs.

A dielectric responds to an electric field mainly by displacing or orienting bound charges, rather than by sustaining a steady conduction current. The polarization density $\mathbf{P}$ is electric dipole moment per unit volume. It creates bound chargeSeparating these charges from free charge motivatesFor an isotropic linear dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$, soThe electric susceptibility ${\chi }_e$ describes material polarization and is not the magnetic susceptibility. More generally it is a tensor and can depend on frequency, field strength and temperature.

Electronic polarization displaces an electron cloud relative to its nucleus; ionic polarization displaces positive and negative ions; orientational polarization partially aligns permanent molecular dipoles against thermal disorder. For independent permanent moments of magnitude $p$, weak-field orientational response gives ${\chi }_e\simeq n{p}^2/(3{ϵ}_0{k}_{B}T)$ when local-field corrections and interactions are negligible. This explains a common inverse-temperature contribution without asserting that every dielectric follows that law.

At a static interface, $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$ and the tangential electric field is continuous. Bound surface charge is already included through $\mathbf{P}$. For a homogeneous filled capacitor, $C=ϵA/d$, and a reversible linear dielectric stores energy density $u=\mathbf{E}\cdot \mathbf{D}/2$. At fixed voltage, insertion raises the stored field energy but the voltage source also supplies energy; force arguments must include that source. At fixed charge the capacitor energy instead decreases on insertion.

At frequencies too high for a polarization mechanism to follow the field, its contribution weakens. A complex permittivity represents dissipative dielectric loss as well as phase lag; the sign of its imaginary part depends on the time convention. Strong fields can cause nonlinear response or dielectric breakdown, beyond the reversible linear model.

Magnetization and field response. A magnetic material contains microscopic magnetic moments associated with electron spin and orbital motion. The magnetization $\mathbf{M}$ is magnetic dipole moment per unit volume. In SI the macroscopic fields satisfyHere $\mathbf{H}$ separates free-current excitation from the material response. A uniform isotropic linear material has $\mathbf{M}={\chi }_m\mathbf{H}$, givingThis constant scalar magnetic permeability model is useful only within its linear range. Crystals can have a susceptibility tensor; strongly magnetic materials depend on field and magnetic history. For a finite specimen the internal $\mathbf{H}$ also includes its demagnetizing field. In a uniformly magnetized ellipsoid, $H_{\mathrm{int}}=H_{\mathrm{applied}}-N_{d}M$ along a principal axis, so sample shape affects the apparent susceptibility.

Diamagnetic and paramagnetic response. Diamagnetism is a weak negative susceptibility caused by induced orbital response opposing the applied field. It occurs in all matter but can be masked by larger positive contributions. Closed-shell materials such as bismuth or many organic compounds show it clearly. Ordinary diamagnetism is often only weakly temperature dependent; superconducting flux exclusion is a separate collective phenomenon and is not described by an arbitrarily small negative susceptibility.

Paramagnetism arises when available magnetic moments partially align with a field. Thermal disorder prevents complete alignment at weak fields. For independent localized moments of magnitude $m$, the classical weak-field result iswhere the second expression uses $B\simeq {\mu }_{0}H$ in a weakly magnetic sample. Thus localized moments obey a Curie law in this regime. Quantum angular momentum changes the coefficient to the appropriate $J(J+1)$ expression and gives saturation at strong field. Conduction electrons instead exhibit Pauli paramagnetism: only states near the Fermi surface can change spin populations, so a degenerate gas has a nearly temperature-independent response. Its orbital diamagnetism must also be included for a full metallic susceptibility.

Collective magnetic order. Exchange interactions can favor parallel alignment, producing ferromagnetism below the Curie temperature. Iron, cobalt and nickel are familiar examples. A material may contain regions of spontaneous magnetization, or magnetic domains, while having little net external moment because the domains point in different directions. Applied fields move domain walls and rotate moments; at large fields the magnetization approaches saturation. Pinning and energy barriers make this process hysteretic. Remanence is the magnetization left after removing the driving field, and coercivity measures the reverse field needed to bring the chosen response back to zero.

The area of a closed $B$–$H$ loop is the energy dissipated per unit volume per quasistatic cycle, $\oint HdB$, with positive dissipated magnitude. Consequently the magnetic permeability is neither one universal constant nor enough information to describe a ferromagnet: differential magnetic permeability $dB/dH$, saturation and hysteresis all matter. Above the ordering temperature the spontaneous moment disappears, and a Curie–Weiss law is often an appropriate approximate paramagnetic description away from the critical region.

In antiferromagnetism, ordered opposing sublattices have equal moments and cancel macroscopically in the ideal compensated case. The order disappears above the Néel temperature. In ferrimagnetism, opposing sublattices have unequal moments and leave a net magnetization. Ferrites combine ferrimagnetic response with relatively high electrical resistivity, making them valuable when eddy-current losses would be substantial in a metallic core.

Electromagnets and engineering choice. A coil supplies free-current magnetomotive force $NI$. For a simple uniform magnetic path, Ampère's law gives $H\ell \simeq NI$ and flux $\Phi \simeq \mu ANI/\ell$. A high-magnetic permeability soft magnetic core concentrates flux and reduces the current required for a given field. With an air gap $g$,Even a short gap can dominate the magnetic reluctance. The approximate attractive force across one uniform gap is $F\simeq B^{2}A/(2{\mu }_0)$, obtained from the field energy, with fringing neglected.

Soft magnetic materials have small coercivity and narrow hysteresis loops; they suit relays, electrical transformer cores and controllable electromagnets. Hard magnetic materials retain their magnetization and suit permanent magnets. Laminating metallic cores breaks large eddy-current loops; ferrites are useful at higher frequencies because of their resistivity. Increasing current cannot increase $B$ indefinitely through the linear formula: core saturation, coil heating and leakage limit performance. A good design selects material for the field amplitude, switching frequency and acceptable loss, rather than simply choosing the largest quoted magnetic permeability.

Pressure and its physical terms. Solving the molar Van der Waals equation for pressure givesThe excluded-volume parameter $b$ reduces the volume available for translational motion, raising the kinetic pressure relative to $RT/V$. The attraction parameter $a$ lowers the pressure because attractive interactions reduce the momentum transferred to the boundary. This mean-field model treats $a,b$ as temperature-independent material constants. For one mole, $V$ is the sample volume with the molar amount already fixed; $b$ has units of that volume and $a$ of pressure times volume squared.

volume dependence of entropy. The Helmholtz free energy has differential $dF=-SdT-PdV$. Equality of mixed derivatives yields the Maxwell relationFor $a=0$ this derivative is $R/(V-b)$, givingEquivalently, $S=R\ln (V-b)+f(T)$ with a common reference volume understood inside the logarithm. The equation of state fixes this volume dependence, not the otherwise unspecified function of temperature.

Excess entropy at fixed volume and temperature. The ideal gas has $(\partial S_{\mathrm{id}}/\partial V{)}_T=R/V$. ThereforeIntegrate from infinite volume, where the entropy difference is specified to vanish:It is negative because excluding configurations reduces the accessible state count. It tends to zero as $V\to \infty$ and has leading dilute limit $-Rb/V$. This is the entropy excess of a van der Waals gas.

Effect of attraction. Restoring a temperature-independent $a$ changes the pressure by $-a/V^2$, but not its derivative with respect to temperature. With the same infinite-volume entropy reference, it therefore does not change $\Delta S^{\mathrm{ex}}$ at fixed $V,T$. This can also be seen fromThe attraction affects internal energy and the equilibrium volume, so it can affect entropy indirectly when different state points are compared. If $a$ were temperature dependent, its derivative would contribute directly to the entropy; that is outside the stipulated constant-parameter model.

Coexistence isotherm. Below the Van der Waals critical point, the formal homogeneous isotherm has a local minimum and maximum. Its middle branch has $\partial P/\partial V>0$, hence negative isothermal compressibility, and is mechanically unstable. Portions of the negative-slope branches can be metastable rather than globally stable. The equilibrium isotherm replaces the loop between liquid volume $V_l$ and vapour volume $V_g$ by a constant saturation pressure $P_s$. The two volume are fixed byThe Maxwell equal-area construction expresses equality of the two phases' chemical potentials. Intermediate total volume are mixtures: changing the vapour fraction changes total volume at the same pressure, rather than following the unstable homogeneous curve. The spinodals are the turning points of the formal isotherm; they are distinct from the coexistence endpoints.

Dense-phase volume at low temperature. At a general subcritical temperature, the two coexistence equations above must be solved together. For this equation of state the area condition is explicitlyThis supplies an exact implicit determination of the liquid volume even when a simple closed form is unavailable.

For the asymptotically low-temperature limit, put $\tau =RTb/a\ll 1$. The liquid is dense, $V_l\simeq b$, while the vapour is dilute, $P_s{V}_g\simeq RT$ and $V_g\gg V_l$. The area condition then givesThus the vapour volume grows exponentially in $1/\tau$, and the saturation pressure is exponentially small. It is consequently consistent to neglect $P_s$ in the liquid equation to algebraic order in $\tau$:Choose the root near $b$, not the large-volume zero-pressure root:Expanding gives the leading requested dense-phase behavior,This is the low-temperature liquid volume of a van der Waals gas. Setting $P_s=0$ is an asymptotic approximation, not an exact coexistence condition at every temperature below $T_c$. The figure illustrates generic subcritical coexistence; its finite-temperature liquid volume is determined by the full equal-area equations. The van der Waals liquid also ceases to be a realistic description at sufficiently low physical temperature, where other phases or quantum effects can matter.

Two equivalent second-law statements. The Clausius statement of the second law forbids a cyclic device whose sole net effect is transferring heat from a colder reservoir to a hotter one. A refrigerator can make that transfer when supplied with work, so the sole-effect qualification matters. The Kelvin-Planck statement of the second law forbids a cyclic device whose sole net effect is drawing heat from a single reservoir and converting all of it into work. A heat engine can produce work when it also rejects heat to a colder reservoir.

The two statements are equivalent by combining a supposed violating device with an ordinary reversible engine or refrigerator: one can cancel one reservoir exchange and obtain a violation of the other statement. The following original schematic shows the forbidden sole effects, with arrows indicating energy flow.

Microscopic fluctuations. The second law is a macroscopic statistical law. A finite fluctuation can transfer a small heat amount $Q$ from cold to hot, with reservoir entropy change $Q/T_h-Q/T_c<0$. Such an individual event does not imply that a cyclic engine can extract systematic useful work from equilibrium. Entropy-decreasing trajectories become exceedingly improbable when the entropy involved is large compared with $k_B$. Under suitable microscopic reversibility conditions, fluctuation relations quantify the relative probabilities of positive and negative entropy production and give nonnegative mean total entropy production. Thus a deterministic no-fluctuation reading has a limit at small scales or short times; the statistical second law remains applicable.

Maximum heating rate. Interpret the quoted $0.5\mathrm{kW}$ as the maximum work-input rate ${\dot{W}}_{\max }$ to an ideal reversible heat pump. This interpretation is needed to obtain a heating rate larger than its electrical input; the wording does not explicitly distinguish input power from delivered heat. A reversible cycle satisfiesTherefore the heating coefficient of performance isWith $T_h=293\mathrm{K}$ and $T_c=273\mathrm{K}$,The additional heat comes from the outdoor reservoir; it is not produced by converting work into more energy than was supplied.

Lowest outdoor temperature. Let $\Delta T=T_h-T_c>0$. The largest available heating rate is ${\dot{W}}_{\max }{T}_h/\Delta T$, while the house loses $\lambda \Delta T$. Equilibrium can be maintained whenAt the limiting outdoor temperature,HenceThis is the indoor-temperature limit of a Carnot heat pump. It assumes the ideal pump retains its stated input-power capability over the operating range. Real heat exchangers require finite temperature differences and incur irreversibility, so the reversible result is an ideal upper performance bound.

Conventional heater. At this limiting outdoor temperature, the required delivered heating power is simply the house's heat loss:This is heat delivered to the house. The fuel-input power would be $3.83/\eta \mathrm{kW}$ for heater efficiency $\eta$. At the earlier outdoor temperature of $273\mathrm{K}$, the required delivered heat would instead be $0.1(293-273)=2.00\mathrm{kW}$; the final question's reference temperature must be kept clear.

If $0.5\mathrm{kW}$ is read literally as a maximum delivered heat rate rather than a work-input rating, the alternative answers are ${\dot{Q}}_{h,\max }=0.5\mathrm{kW}$, $T_{c,\min }=293-0.5/0.1=288\mathrm{K}$, and a conventional heater delivering $0.5\mathrm{kW}$ at that limit. Such a pump could not balance the stated house loss at $273\mathrm{K}$. Both readings follow consistently once the ambiguous power rating is defined; the Carnot calculation above uses the work-input reading.

Configuration count. Choose the end-to-end direction as positive and let $N_R,N_L$ count rightward and leftward links. ThenA configuration is determined by choosing the positions of its leftward links, soThe counts must be nonnegative integers; in particular, allowed signed extensions occur in steps of $2a$. If distance means $|L|$ with neither direction selected, there are two corresponding sets for $L>0$, adding a factor of two. That constant does not affect the tension. There is no excluded-volume penalty in this one-dimensional idealized count.

Entropy from the multiplicity. Put $x=L/(Na)$, so $|x|<1$. Applying the leading Stirling's approximation to all three large factorials givesTerms subleading in $N$, including logarithmic factorial prefactors, have been omitted consistently. Expanding near zero extension givesStretching reduces the number of available configurations and hence the entropy.

Tension and the range of Hooke's law. The model assigns no extension-dependent internal energy. At fixed temperature, the externally applied tension is $\mathcal{F}=(\partial F/\partial L{)}_T=-T(\partial S/\partial L{)}_T$. Differentiating the full leading entropy givesIt is the external force needed to oppose the chain's restoring entropic force. Only when $|L|\ll Na$ can one put $\mathrm{artanh}x\simeq x$, yieldingThe paper's displayed linear law therefore requires this small-extension approximation, which is not explicitly stated in its premise $L<L_0$. It is not the exact force law up to full extension. The entropic spring constant grows with temperature because stronger thermal disorder makes a given reduction in configurational entropy cost more free energy. This is the one-dimensional entropic chain force-extension relation.

A sparse number of backward links. Write $q=N_L\ll N$, so $N_R=N-q$ and $L=a(N-2q)$. Exactly,For fixed $q$ and large $N$, each numerator factor approaches $N$, givingMore precisely, the logarithm of the product is $-q(q-1)/(2N)+O(q^3/N^2)$. A small relative error in the multiplicity requires $q^2/N\ll 1$; merely $q/N\ll 1$ need not make the omitted product close to one. The weaker sparse-link condition does suffice for the leading entropy density and leading high-extension tension. These are sparse backward links near full chain extension.

Tension in that limit. In the overlap regime $1\ll q\ll N$, use Stirling also for $q!$:ThusThe full leading entropy would give $(k_{B}T/(2a))\ln [(N-N_L)/N_L]$, which has the same dominant logarithm when $N_L/N\to 0$.

Approaching full extension. The continuum force law diverges logarithmically as $N_L\to 0$, or $L\to Na$. One must not literally apply Stirling to a zero factorial: at the discrete endpoint there is exactly one directed configuration, $\Omega =1$, and its entropy is zero. A continuous fixed-extension derivative is not defined at that endpoint.

An independent force-ensemble calculation makes the limiting physical statement precise. For applied tension $\mathcal{F}$, a link has partition function $z=2\cosh (\beta \mathcal{F}a)$, giving the exact mean extensionIt approaches $Na$ only as $\mathcal{F}\to \infty$ at positive temperature, and inversion reproduces the logarithmic force law for mean extension. In a real chain, bond extension or failure eventually replaces the ideal inextensible-link model.

Temperature as an equilibrium relation. Classical thermodynamics begins with the zeroth law of thermodynamics: thermal equilibrium is transitive. A thermometer can therefore assign a common empirical temperature to systems in mutual thermal equilibrium. Different thermometer properties initially give different monotone empirical scales; a resistance reading by itself does not define the absolute scale.

The Second law of thermodynamics provides a substance-independent thermodynamic scale. A reversible engine operating between reservoirs has a heat ratio depending only on their temperatures. One chooses absolute temperature so that $Q_h/Q_c=T_h/T_c$, with $Q_h,Q_c$ the positive heat magnitudes. Its reversible efficiency is $1-T_c/T_h$. Equivalently, reversible heat defines entropy through $dS=\delta Q_{\mathrm{rev}}/T$. For an equilibrium simple compressible system,Classical thermodynamics uses this relation without needing a microscopic account of entropy. A dilute-gas thermometer realizes an empirical scale approaching the thermodynamic scale after corrections for nonideal behavior.

Statistical meaning of the same derivative. In the microcanonical description, an equilibrium energy shell containing $\Omega (U,V,N)$ accessible states has $S=k_B\ln \Omega$. Its statistical temperature is consequentlyTo see why this is an equilibrium temperature, let two weakly coupled systems exchange a small energy while their total energy stays fixed. The combined multiplicity is ${\Omega }_A(U_A){\Omega }_B(U_{\mathrm{tot}}-U_A)$. Maximizing its logarithm giveswhich is equality of the two inverse temperatures. Stability requires the total entropy maximum, not only stationarity. The overwhelmingly dominant macrostate in a large system therefore realizes the thermodynamic equilibrium condition.

A large reservoir gives a small subsystem the canonical probabilities $p_i\propto e^{-\beta E_i}$, with $\beta =1/(k_{B}T)$. In that ensemble, $F=-k_{B}T\ln Z$ and $S=-k_B{\sum }_i{p}_i\ln p_i$ agree with macroscopic thermodynamic relations in their usual equilibrium regime. The equipartition identity for a classical dilute monatomic gas, $⟨p^2/(2m)⟩=3k_{B}T/2$, is a useful operational connection between microscopic motion and temperature, but is not a universal definition: quantum modes need not obey equipartition, and a system away from equilibrium need not have one temperature at all.

For finite systems one must specify the entropy convention and energy resolution; surface and fluctuation corrections can distinguish definitions that agree in the ordinary thermodynamic limit. Bounded-energy population-inverted systems can also have a negative entropy derivative and hence a negative statistical temperature. That possibility lies outside an ordinary positive-temperature ideal-gas thermometer; it does not mean temperature is merely an arbitrary average kinetic energy.

Obtaining the energy–temperature conversion. The Boltzmann constant connects entropy measured in $\mathrm{J}{\mathrm{K}}^{-1}$ with the dimensionless logarithm of a state count. Historically it could be determined from the ideal gas law $pV=N{k}_{B}T$: measure the molar gas constant $R$ and independently determine the Avogadro constant $N_A$, then obtain $k_B=R/N_A$. The independent particle-number determination matters; using $R$ alone would not fix energy per particle.

An electrical method uses Johnson-Nyquist noise. A resistor of resistance $R_e$ at known temperature has, in the classical range $hf\ll k_{B}T$, a one-sided open-circuit voltage-noise spectrum $S_V(f)=4k_{B}T{R}_e$. Over a calibrated flat bandwidth $\Delta f$,A practical measurement uses an equilibrium resistor, a calibrated low-noise amplifier and a known transfer function, subtracts or independently characterizes amplifier noise, and replaces a nominal bandwidth by the actual effective noise bandwidth. Varying temperature and fitting the mean-square voltage is more informative than treating one amplified voltage sample as the thermal variance. Other primary methods relate temperature to sound speed in a dilute gas or to calibrated particle diffusion and mobility.

There is a historically relevant unit-definition qualification: by the date of this exam, the revised SI had already taken effect on 20 May 2019. The kelvin is now defined by fixingThe same experiments now realize or compare thermodynamic temperatures using this fixed constant, whereas historically they measured the conversion constant on an independently defined temperature scale. The physical distinction between equilibrium temperature and microscopic state counting is unchanged.^[ref]

For the intended reversible adiabatic process, an ideal gas obeys $T{V}^{\gamma -1}=\text{constant}$. A monatomic gas has heat capacity ratio $\gamma =5/3$, soThe required extra assumption is that “adiabatic” denotes the usual quasistatic isentropic process. Absence of heat transfer alone would not fix the final temperature: an ideal gas undergoing free expansion into vacuum is also adiabatic but leaves its temperature unchanged.