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In a regular thermodynamic formulation, a state at absolute zero cannot be reached from positive temperature by finitely many cooling operations with finite control resources. The entropy-difference form of the third law of thermodynamics makes regular entropy curves approach a common limiting entropy. An isentropic process starting above that entropy cannot intersect the common zero-temperature endpoint.

For an illustrative pair of settings with $S-S_0=a_{1}T$ and $S-S_0=a_{2}T$, $0<a_1<a_2$, an isothermal entropy-reduction step followed by reversible adiabatic cooling gives $T_{n+1}=(a_1/a_2)T_n$. Every finite number of stages leaves $T_n>0$. The model illustrates the entropy geometry; a general finite-time impossibility statement additionally needs assumptions about available interactions and cooling rates.

Underdamped Langevin dynamics retains both position $q$ and momentum $p$. For mass $m>0$, temperature $T>0$, friction rate $\gamma >0$ and smooth potential energy $U$, one common normalization isHere $W$ is vector Brownian motion. The noise amplitude is tied to friction by the fluctuation-dissipation relation for a Langevin particle. If the density is normalizable and the dynamics is nonexplosive, its invariant distribution is the Boltzmann distributionThe Hamiltonian flow preserves the total energy and phase-space volume; the momentum Ornstein-Uhlenbeck process preserves a centered normal distribution with coordinate variance $m{k}_{B}T$. These two facts verify the invariant density. In dimensionless units with $k_{B}T=1$ and momentum variance $\eta$, it is $\exp [-U(q)-|p{|}^2/(2\eta )]$, with the corresponding choice $m=\eta$.

Retaining inertia resolves the short-time ballistic-to-diffusive Brownian displacement. The overdamped Langevin dynamics approximation removes momentum relaxation when only much longer times are relevant.

A uniform-density spherical hydrostatic equilibrium model with zero surface pressure has $P(r)=2\pi G{\rho }^2(R^2-r^2)/3$, $M=4\pi \rho R^3/3$ and gravitational energy $-3G{M}^2/(5R)$. A perfect gas can have this mechanical structure with varying temperature and entropy; the imposed equilibrium density does not require incompressible perturbations.

On the stable zero-field branch $M_B=-M_W=M$, put $a=\beta (1-M^2)$. Differentiating the two-sublattice Ising mean-field approximation with respect to the uniform field gives ${\chi }_B=a(1-qJ{\chi }_W)$ and ${\chi }_W=a(1-qJ{\chi }_B)$. Thus ${\chi }_{+}=a/(1+qJa)$ and $\partial M_{-}/\partial h=0$. The true staggered magnetic susceptibility, $\partial M_{-}/\partial g=a/(1-qJa)$, is a different response and diverges at the continuous transition. The factor $1-M^2$ is essential: the uniform response vanishes when the spins saturate at zero temperature.

For a uniform ideal gas sphere, set $\rho =mn$, $p=n{k}_{B}T$ and $M=4\pi \rho R^3/3$. Building the sphere by concentric shells gives gravitational potential energy $W=-3G{M}^2/(5R)$. During a small homologous contraction, released gravitational energy balances the work against gas pressure when $3G{M}^2|dR|/(5R^2)=4\pi R^{2}p|dR|$, or $GMm/R=5k_{B}T$. Eliminating $R$ gives the displayed mass. At the same initial mass density and temperature, larger masses have stronger gravitational contraction than pressure support in this estimate. Its numerical coefficient belongs to the uniform-sphere pressure-work model; the infinite-medium Jeans instability and a pressure-confined isothermal sphere use different equilibrium and boundary assumptions.

An observer at constant proper acceleration in the Minkowski vacuum detects a thermal response at this temperature. The associated Rindler coordinates separate the observer's accessible wedge from a horizon. This is an observer-dependent particle/detector response, not an assertion that inertial observers measure a thermal bath.

The upper part of the terrestrial atmosphere includes regions where ionization makes the response to electromagnetic waves depend strongly on mobile electrons. The ionosphere describes that electrical structure; its boundaries need not coincide with the boundaries defined by temperature in the mesosphere and thermosphere.

The upper endpoint of a two-liquid coexistence region in a temperature-composition diagram. In the ordinary upper-critical case, heating above this temperature removes liquid-liquid demixing.

The ratio of radiogenic heating to heat loss for a stated planetary reservoir, named for Harold Urey. A convective mantle Urey ratio uses mantle radioactive production and heat leaving the mantle; a bulk-silicate or surface ratio uses a different reservoir and heat budget. It must be labeled accordingly. A ratio below one is consistent with drawing on stored heat, but core heat or other inputs must also be included before inferring the sign of the mantle's temperature change.

A crystalline particle in urine. Its presence can depend on concentration, acidity, temperature and storage and does not alone establish a urinary stone.

In the simple semiconductor description, the highest occupied energy band of the unexcited crystal at zero temperature. A missing occupant is an electron hole.

The Van der Waals equation is a simple real gas equation of state that can also describe a liquid-gas critical point. With particle count $N$ fixed, write the volume per particle as $v=V/N$:or equivalently $(p+a{N}^2/V^2)(V-Nb)=N{k}_{B}T$. Here $p$ is pressure, $T$ is absolute temperature, and $k_B$ is the Boltzmann constant. In SI, $v$ and $b$ have units ${\mathrm{m}}^3$ per particle, $a$ has units $\mathrm{Pa}{\mathrm{m}}^6=\mathrm{J}{\mathrm{m}}^3$, and $k_B$ has units $\mathrm{J}{\mathrm{K}}^{-1}$; the particle count is dimensionless. The pole at $v=b$ excludes smaller volume values from this model.

The $b$ correction represents excluded volume; the attraction term lowers pressure at fixed temperature and particle number density $1/v$. This is a mean-field theory, replacing detailed intermolecular structure by two parameters. At $v\gg b$ with $a/v^2\ll k_{B}T/v$, it approaches the ideal gas law $pv=k_{B}T$. It supplies a useful qualitative route from a dilute gas to phase coexistence, without a claim of quantitative accuracy for every fluid.^[ref]

Do not mix this particle convention with the molar volume convention elsewhere in this article. With amount of substance $n_{\mathrm{mol}}=N/N_A$, Avogadro constant $N_A$, and molar gas constant $R=N_A{k}_B$, one has $v_m=N_{A}v$, $b_m=N_{A}b$, and $a_m=N_A^{2}a$, so $p=RT/(v_m-b_m)-a_m/v_m^2$. The molar parameters have units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$, respectively.

Johannes Diderik van der Waals introduced this equation of state in 1873, connecting molecular attraction with departures from the dilute ideal gas law. Its historical significance is the unified treatment of gases and liquids, rather than a universally exact fit.^[ref]

A Van der Waals isotherm is the pressure–particle volume curve at a fixed temperature. Put $x=v/v_c$, $t=T/T_c$, and $q=p/p_c$, using the Van der Waals critical point values below. Then the existing Van der Waals law of corresponding states becomesAll curves in the figure are calculations in this reduced model.

For a uniform thermodynamic phase, the isothermal compressibility isThus the upward-sloping portion of a subcritical Van der Waals isotherm has ${\kappa }_T<0$: compression would reduce pressure, allowing a density disturbance to grow. It is mechanically unstable. The spinodal points bound that portion; setting the derivative $\partial q/\partial x$ to zero gives their curve $t=(3x-1{)}^2/(4x^3)$. Negative slope alone guarantees only local mechanical stability: portions between the spinodal and the phase coexistence endpoints have metastability. Equilibrium replaces the entire interval between those endpoints, including these metastable portions, by a Maxwell construction plateau.^[ref]

The Van der Waals equation becomes independent of its material parameters when pressure, temperature and molecular volume are divided by their critical values: $\bar{p}=8\bar{T}/(3\bar{v}-1)-3/{\bar{v}}^2$. This explains the common reduced isotherms within this model, without claiming exact agreement for all real fluids.

Below $T_c$, the equilibrium liquid and gas have the same temperature, pressure $p_{\ast }$, and per-particle chemical potential. For macroscopic bulk thermodynamic phases, neglect interface contributions. The Maxwell equal-area construction determines $v_l$ and $v_g$ byIt expresses equality of chemical potentials, since $d\mu =vdp$ at fixed temperature. Equivalently, the equilibrium Helmholtz free energy per particle follows a common tangent line between the two endpoint states, lying below the homogeneous continuation between them. The construction uses that mathematical continuation to compare endpoints; it is not a physical compression path through negative isothermal compressibility states.^[ref]

Here is a reproducible calculated example at $t=0.9$. With $q_{\ast }=p_{\ast }/p_c$, the homogeneous intersections solve the cubic equationThe Maxwell construction givesAll three roots of a polynomial satisfy $x>1/3$. Only the smallest and largest are the phase coexistence endpoints; the middle one has negative isothermal compressibility. Equal pressures at three admissible values of volume therefore do not imply three stable thermodynamic phases. An analytic area check isAt full numerical precision, the residual is below $10^{-11}$ in reduced units; independent numerical integration gives the same cancellation. At $t=0.9$, the spinodal points are instead $x\simeq 0.718597189$ and $1.528504964$.

The equilibrium plateau changes the phase fraction rather than the endpoint densities. For a gas particle fraction $f_g=N_g/N$ and mean particle volume $\bar{v}=V/N$, the lever rule readsThus $\bar{v}=v_c$ at $t=0.9$ gives $f_g\simeq 0.227219$: this is a particle fraction, not the volume fraction occupied by gas. Differentiating equality of chemical potentials along this model's phase coexistence curve yields the same Clapeyron equation explained below, $d{p}_{\ast }/dT=\ell /[T(v_g-v_l)]$, with latent heat $\ell$ per particle. Both endpoint differences vanish as the Van der Waals critical point is approached; the equation's critical limit must be taken as a ratio.

The Maxwell construction presumes a pure bulk fluid at thermodynamic equilibrium. Its endpoint equality does not specify how rapidly a real compression reaches equilibrium. A homogeneous metastable sample can persist while nucleation is delayed; the spinodal marks loss of local mechanical stability in this mean-field theory, rather than the equilibrium onset of condensation. Surface tension, finite sample size and impurities can affect the observed transition. These qualifications matter when comparing a mathematical Van der Waals isotherm with a visible liquid meniscus.

The molar volume $v_m=V/n_{\mathrm{mol}}$ is the volume per amount of substance, not the total container volume or the volume per particle. In the molar convention of the Van der Waals equation, take constant positive parametersThenor, for any fixed amount of substance,Both terms in the first equation are pressures. The total excluded volume is $n_{\mathrm{mol}}{b}_m$: using $V-b_m$ would subtract quantities with different units. Changing the sample size at fixed molar volume and temperature changes $V$ and $n_{\mathrm{mol}}$ proportionally, while leaving pressure unchanged. This is a useful unit and extensivity check before fitting a real gas.

For the molar Van der Waals critical point, differentiate with respect to molar volume at fixed temperature:Setting both to zero gives $R{T}_c=2a_m(v_{m,c}-b_m{)}^2/v_{m,c}^3$ and $R{T}_c=3a_m(v_{m,c}-b_m{)}^3/v_{m,c}^4$. Equating them and dividing by the nonzero factors yields $2v_{m,c}=3(v_{m,c}-b_m)$, soHere $v_{m,c}$ has units ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, $T_c$ has units $\mathrm{K}$, and $p_c$ has units $\mathrm{Pa}$. This is the molar version of the particle derivation below: $v_{m,c}=N_A{v}_c$, with the same physical $T_c$ and $p_c$. If one writes $v_c=3b$, $T_c=8a/(27Rb)$ and $p_c=a/(27b^2)$ without subscripts, then all of $v_c,b,a$ in those formulas must be molar quantities.

Now define the dimensionless reduced variables explicitly as $x=v_m/v_{m,c}$, $t=T/T_c$ and $q=p/p_c$. Substituting the three critical values cancels $a_m,b_m,R$ and givesBecause $v_m/v_{m,c}=v/v_c$, these are exactly the reduced variables in the Van der Waals isotherm discussion. The Van der Waals law of corresponding states is therefore independent of whether one starts with particle or molar units. It predicts a critical ratio $p_c{v}_{m,c}/(R{T}_c)=3/8$, providing a third check on any proposed parameter fit.

A hypothetical Van der Waals equation model illustrates positive and negative Joule-Thomson coefficient regions and a finite throttle that first warms, then cools. Use the molar convention already distinguished in Van der Waals equation:Here $v_{\mathrm{m}}$ is molar volume, $R$ the molar gas constant, $u_{\mathrm{m}}$ internal energy per mole, and $C_{V,\mathrm{m}}$ the stipulated constant molar heat capacity at fixed volume. The second equation is an additional caloric assumption: an equation of state alone does not specify the temperature dependence of internal energy. It satisfies the equilibrium identity $(\partial u_{\mathrm{m}}/\partial v_{\mathrm{m}}{)}_T=T(\partial p/\partial T{)}_{v_{\mathrm{m}}}-p=a_{\mathrm{m}}/v_{\mathrm{m}}^2$. The arbitrary energy reference is set to zero.

Define dimensionless variables $r=v_{\mathrm{m}}/b_{\mathrm{m}}$, $\theta =R{b}_{\mathrm{m}}T/a_{\mathrm{m}}$, $\pi =b_{\mathrm{m}}^{2}p/a_{\mathrm{m}}$ and $\eta =b_{\mathrm{m}}{h}_{\mathrm{m}}/a_{\mathrm{m}}$, where $h_{\mathrm{m}}$ is molar enthalpy. These differ from the critical-value scaling in Van der Waals law of corresponding states. The throttle calculation reduces toThus a prescribed inlet fixes $\eta$, and solving the first and last equations at the outlet pressure gives the final temperature. For an inlet $({\theta }_1,{\pi }_1)=(0.8,0.12)$ and outlet ${\pi }_2=0.025$, the model gives ${\theta }_2\simeq 0.752287$, a decrease of $0.047713$. The inlet $(2.2,0.12)$ instead gives ${\theta }_2\simeq 2.205786$, an increase of $0.005786$. The purple inlet $(1.45,0.30)$ initially warms but ends at ${\theta }_2\simeq 1.437585$ for the same outlet pressure. These are reduced toy values, not kelvin readings or experimental evidence. No physical gas parameters were fitted.

For a short inversion derivation, implicit differentiation of the Van der Waals equation at fixed pressure givesInsert this into $T(\partial v_{\mathrm{m}}/\partial T{)}_p=v_{\mathrm{m}}$ and substitute back into the equation of state. The Joule-Thomson inversion curve isPositive inversion pressure requires $r>3/2$. Its maximum pressure is $\pi =1/3$ at $r=3$, while the upper dilute limit is ${\theta }_{\mathrm{inv}}\to 2$. Using Van der Waals critical point values, the latter is $T_{\mathrm{inv},\max }/T_c=27/4$. The inversion locus depends on the equation of state; the chosen molar heat capacity additionally determines how much temperature changes along each finite throttle.^[ref]

The displayed region has $\theta >8/27$, above the model's critical temperature, and uses only $r>1$, $p>0$, $(\partial p/\partial v_{\mathrm{m}}{)}_T<0$ and positive $C_{p,\mathrm{m}}$. Subcritical calculations require attention to phase coexistence, metastability and the Maxwell construction; an unstable homogeneous continuation is not an equilibrium throttle endpoint. Even on a stable branch, this mean-field theory with constant $C_{V,\mathrm{m}}$ is a qualitative model, not an accurate inversion map for a particular real gas. Phase transitions, changing composition, appreciable heat leakage or bulk kinetic energy changes require a different property calculation or steady-flow energy balance. At phase coexistence, the smooth single-thermodynamic phase coefficient formula cannot replace a balance involving the phase fraction.

For an ideal Diesel cycle with compression ratio $r=V_1/V_2>1$, cutoff ratio $\rho =V_3/V_2>1$ and constant heat-capacity ratio $\gamma$, its thermal efficiency is $\eta =1-({\rho }^{\gamma }-1)/[\gamma r^{\gamma -1}(\rho -1)]$. As $\rho \to 1^{+}$ the quotient tends to $\gamma$, giving $\eta \to 1-r^{1-\gamma }$. The maximum-to-minimum temperature ratio tends to $r^{\gamma -1}$, so the limiting efficiency equals that of a Carnot cycle between those temperatures. Both heat input and net work vanish in this limit: the result is a limit of their ratio, not a finite-work engine at exactly zero cutoff interval.

In the free-slip Stokes flow temperature model, let the critical temperature be $g=\sin \pi z$ and vertical velocity $f=sg$, $s=k_c^2+{\pi }^2$. A forced positive second harmonic $(W,\Theta )e^{2i{k}_{c}x}$ couples to the negative critical harmonic. Its temperature solvability condition has integrandThe identity follows by substituting $f=sg$ and differentiating. Its integral is zero because $g$ vanishes on both plates, regardless of the forced boundary value of $\Theta$. Thus the first resonant forcing coefficient vanishes. A generic nonzero conjugate-amplitude term cannot be inferred from wave-number matching alone.

The temperature relation $\mathrm{d}\ln K/\mathrm{d}T={\Delta }_r{H}^{\circ }/(R{T}^2)$ for a dimensionless thermodynamic equilibrium constant. With constant reaction enthalpy, integration gives the difference of reciprocal temperatures.