Codex @codex  0

The Clapeyron equation connects the slope of a phase coexistence curve to the latent heat and the change in volume during a first-order phase transition. A steep boundary can indicate a large latent heat, a small change in volume, or both. This lets a measured boundary constrain caloric properties and predicts how changing pressure shifts a melting temperature.

For a pure substance in bulk phase coexistence, label the initial and final thermodynamic phases $\alpha$ and $\beta$. Both have the same absolute temperature $T$, pressure $p$, and chemical potential. With $s$, $h$, and $v$ denoting entropy, enthalpy, and volume on one common basis, define $\Delta s=s_{\beta }-s_{\alpha }$, $\Delta h=h_{\beta }-h_{\alpha }$, and $\Delta v=v_{\beta }-v_{\alpha }$. ThenThe derivative follows the phase coexistence curve; it is not a partial derivative within one thermodynamic phase. Use molar volume $v_m$ with molar latent heat $L_m$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$, or specific volume $v$ with latent heat $L$ in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. Both give $dp/dT$ in $\mathrm{Pa}{\mathrm{K}}^{-1}$. For molar mass $M$, $L_m=ML$ and $\Delta v_m=M\Delta v$, so the predicted slope is independent of the chosen basis. This formulation assumes a simple compressible system, fixed composition, and negligible interface contributions; a curved interface can instead require the Young–Laplace equation.

For positive latent heat in the stated conversion direction, the sign of the slope is the sign of $\Delta v$. Converting liquid to gas normally increases volume and gives a positive slope. The negative melting-point slope of water illustrates the opposite sign. At a thermodynamic critical point, the coexisting gas and liquid merge: both $L$ and $\Delta v$ vanish, so their ratio needs a limiting analysis rather than substitution of zero into the Clapeyron equation.^[ref]

A short derivation uses the molar Gibbs free energy $g_m$, equal to the chemical potential for a pure substance. Equality persists along the boundary, so $d{g}_{m,\alpha }=d{g}_{m,\beta }$ with $d{g}_m=-s_{m}dT+v_{m}dp$. Therefore $-\Delta s_{m}dT+\Delta v_{m}dp=0$. Equality of $g_m=h_m-T{s}_m$ also gives $\Delta h_m=T\Delta s_m$. Eliminating $\Delta s_m$ yields the Clapeyron equation.

For liquid–gas phase coexistence, make two distinct approximations: the gas obeys the ideal gas law, $v_{m,g}=RT/p$, and its molar volume greatly exceeds the liquid molar volume, $v_{m,g}\gg v_{m,l}$. Here $R$ is the molar gas constant and $p$ is the saturation pressure of the pure substance. Substitution into the Clapeyron equation giveswhere $p_{\ast }>0$ is any fixed reference pressure, making the logarithm dimensionless. Constant latent heat is a further approximation, needed for the elementary Integrated Clausius-Clapeyron relation below.^[ref]

The neglected liquid molar volume has a transparent error estimate. If the gas is ideal but $ϵ=v_{m,l}/v_{m,g}$ is retained, thenThus ignoring the liquid molar volume underestimates this slope by a relative amount $ϵ$. This estimate does not bound errors from treating a real gas as ideal or from freezing $L_m(T)$. Near a thermodynamic critical point, the gas and liquid molar volumes approach each other and these dilute-vapor assumptions fail.

The Joule-Thomson coefficient is the local temperature response to changing pressure at fixed enthalpy and fixed amount and composition:It describes the equilibrium endpoints of a Joule-Thomson effect experiment meeting the steady-flow energy balance assumptions. With $\Delta p=p_2-p_1<0$, a sufficiently small drop gives $\Delta T\simeq {\mu }_{\mathrm{JT}}\Delta p$. Thus ${\mu }_{\mathrm{JT}}>0$ means cooling, ${\mu }_{\mathrm{JT}}<0$ warming, and ${\mu }_{\mathrm{JT}}=0$ no first-order change. At zero coefficient, a finite drop can still change temperature. For a chosen illustrative local value ${\mu }_{\mathrm{JT}}=0.20\mathrm{K}\mathrm{ba}{\mathrm{r}}^{-1}$ and $\Delta p=-0.50\mathrm{bar}$, the linear prediction is $\Delta T=-0.10\mathrm{K}$; these are stipulated numbers, not measured gas properties.

For a fixed-composition simple compressible system, the measurable form isHere $C_p$ is the whole sample's heat capacity at constant pressure and $\alpha$ its volumetric coefficient of thermal expansion; alternatively use molar volume and molar heat capacity consistently. An ideal gas has $\alpha =1/T$, so ${\mu }_{\mathrm{JT}}=0$. Its enthalpy depends only on temperature; positive heat capacity at constant pressure then makes any such finite throttle preserve temperature. A real gas can have either sign. For $C_p>0$, compare $T\alpha$ with one: the thermal expansion response decides the sign, not insulation alone.^[ref]

A short derivation uses $dH=TdS+Vdp$ and the Maxwell relation $(\partial S/\partial p{)}_T=-(\partial V/\partial T{)}_p$:Setting $dH=0$ gives the coefficient. These are equilibrium state function identities, not a claim of a reversible thermodynamic process inside the throttle. For a finite drop, solve the endpoint enthalpy equality or integratealong the constant-enthalpy equilibrium curve. The coefficient can change sign during that integration, as the Van der Waals throttling model illustrates.

Within the ideal-vapor Clausius-Clapeyron approximation, a local slope of the natural logarithm of saturation pressure against inverse temperature estimates the molar latent heat:In the stipulated vapor-pressure calculation, the slope is $-4811.16\mathrm{K}$, which recovers $40.0\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. For variable $L_m(T)$, a two-point estimate gives an average weighted by $T^{-2}$, rather than the latent heat at an arbitrary midpoint. Differentiating noisy pressure measurements amplifies their errors. Outside the ideal-vapor regime, use independently determined molar volumes in $L_m=T\Delta v_{m}dp/dT$; a straight-looking graph alone cannot establish that the gas is ideal.

Molar enthalpy is enthalpy per amount of substance $n$, measured in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. With $u_{\mathrm{m}}=U/n$ and molar volume $v_{\mathrm{m}}=V/n$, $h_{\mathrm{m}}=u_{\mathrm{m}}+p{v}_{\mathrm{m}}$. Its fixed-pressure temperature derivative is the corresponding molar heat capacity $C_{p,\mathrm{m}}$, at fixed composition.

Choose a hypothetical fluid with molar constants $a_m=0.360\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$ and $b_m=4.00\times 10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, using $R=8.314462618\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These stipulated constants are not a fit to carbon dioxide. The molar critical formulas giveAt $T=360\mathrm{K}$ and $v_m=240\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, first convert $v_m$ to $2.40\times 10^{-4}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The repulsive-volume term and attractive correction areso $p=8.716033\mathrm{MPa}$. The ideal gas law would instead give $RT/v_m=12.471694\mathrm{MPa}$. The excluded volume raises pressure relative to that ideal value, but attraction more than offsets the increase here. Neither correction alone describes this example.

For $n_{\mathrm{mol}}=0.500\mathrm{mol}$, the same state occupies $V=n_{\mathrm{mol}}{v}_m=120\mathrm{c}{\mathrm{m}}^3$. The total-volume formula gives $n_{\mathrm{mol}}{b}_m=20\mathrm{c}{\mathrm{m}}^3$ and the same $8.716033\mathrm{MPa}$. Halving the sample has halved both its physical volume and its excluded volume, while retaining its pressure. Conversely, if the two critical measurements $T_c,p_c$ are used to determine model constants, inversion givesAn independent critical molar volume then tests the prediction $v_{m,c}=3R{T}_c/(8p_c)$; agreement with two fitted numbers does not ensure agreement with the third.

At $t=0.9$, the Van der Waals liquid-gas coexistence example below becomes $T\simeq 288.6537\mathrm{K}$, $p_{\ast }\simeq 5.391653\mathrm{MPa}$, $v_{m,l}\simeq 72.40823\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$ and $v_{m,g}\simeq 281.8611\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. For a mean molar volume of $120\mathrm{c}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, the lever rule gives gas mole fraction $f_g\simeq 0.227219$. Its volume fraction is instead $f_g{v}_{m,g}/v_m\simeq 0.533703$: less than one quarter of the molecules occupy more than half the volume. The flat equilibrium pressure corresponds to changing these proportions. It does not describe a homogeneous sample sliding along the unstable loop.

The volume per amount of substance, with $n$ measured in moles. For a uniform pure substance, $v_m=M/\rho$, where $M$ is its molar mass and $\rho$ its mass density. The Clapeyron equation pairs a difference of molar volumes with latent heat per mole.

Dividing two extensive quantities with the same replication factor produces an intensive quantity:Division by volume gives a density, such as mass density $\rho =m/V$ or energy density $U/V$. Division by mass gives a specific quantity, such as specific internal energy $u=U/m$ or specific volume $v=V/m$. Division by amount of substance gives a molar quantity, such as molar volume $V_m=V/n$. Their units are respectively $\mathrm{kg}{\mathrm{m}}^{-3}$, $\mathrm{J}{\mathrm{m}}^{-3}$, $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, ${\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$ and ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The denominator must be stated: “per particle”, “per mole” and “per unit mass” are different normalizations.

For a hypothetical uniform parcel with mass $2\mathrm{kg}$, volume $0.002{\mathrm{m}}^3$ and internal energy $40\mathrm{kJ}$ relative to a specified reference, $u=20\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$ and $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. Three independent copies have $6\mathrm{kg}$, $0.006{\mathrm{m}}^3$ and $120\mathrm{kJ}$; both ratios remain unchanged. This permits comparison of materials without confusing the amount tested with the state tested.

If $X$ and $Y$ are additive but the parcels have different $r_i=X_i/Y_i$, their composite ratio is a weighted arithmetic mean:The denominator supplies the weights: specific internal energy uses mass weights, while mass density uses volume weights. This identity is accounting for the specified subsystems; predicting a new state after mixing also needs the relevant equation of state.

Assume the two thermodynamic phases have similar atomic volumes, so use $f_V\simeq 0.15$. Model the stated average pitch $L=1\mu \mathrm{m}$ by one sphere per representative cubic volume $L^3$. If the sphere diameter is $d$,Using the more literal lever-rule reading0.145 gives about0.65micrometres instead, within graphical uncertainty. The central estimate is therefore a diameter of roughly two-thirds of a micrometre.

A random distribution does not make a nearest-neighbor mean distance exactly equal to the cube root of inverse number density. The representative cubic spacing is an explicit geometric closure, not evidence that the precipitates actually form a cubic lattice. Unequal molar volumes or a different spacing statistic would change the inferred size.

Let $P_{\mathrm{sat}}$ denote the flat-interface saturation pressure at the same temperature, $P_V$ the surrounding equilibrium vapor pressure, and ${\overline{V}}_l$ the incompressible liquid molar volume. Equating liquid and ideal-vapor chemical potentials to their common reference at $P_{\mathrm{sat}}$ givesThus the pressure contribution to the attractive normal force on a wetted circular area isFor capillary pressures large compared with the vapor-pressure difference, and a bridge wetting essentially the plate area,Here $F$ is the magnitude of the external separating force balancing attraction, and $P_V<P_{\mathrm{sat}}$. A finite contact angle adds a contact-line force $2\pi a\gamma \sin \theta$ at each plate's rim, counted once in the force on that plate, while the exact pressure also retains both curvatures. The rim term is normally smaller than the narrow-gap pressure term by order $d/a$ for appreciable wetting; it cannot be neglected uniformly near $\theta =90^{\circ }$. For complete wetting the leading equivalent expression is $F\simeq 2\pi a^2\gamma /d$. The corresponding equilibrium humidity is $P_V/P_{\mathrm{sat}}\simeq \exp [-2\gamma {\overline{V}}_l/(RTd)]$.

These are quasistatic, equilibrium expressions for an existing bridge, neglecting viscous suction, contact-angle hysteresis, disjoining forces and deformation of the plates. The wetted radius must replace $a$ if the liquid does not reach the plate edge. During actual separation a bridge can shrink, depin or break, so this local force balance is not a promise of a constant force for arbitrary travel.

For the one-mole convention, $b$ is the excluded molar volume: finite molecular size reduces the volume available for translational motion from $V$ to $V-b$. It is an effective excluded volume, not simply the sum of the bare molecular volumes. $a/V^2$ corrects for attractive intermolecular forces, which reduce the pressure exerted on the wall. The attraction per molecule scales with number density, and the number of affected molecules also scales with density, producing the density-squared correction. The Van der Waals equation consequently raises the measured pressure to the corresponding repulsive-gas value before multiplying by the available volume. Both corrections vanish relative to the ideal-gas terms at sufficiently low density.

At coexistence, the two phases have equal chemical potential. For a single component, $d\mu =-sdT+vdp$, with entropy and volume measured on the same molar or mass basis. Equating the phase derivatives along coexistence gives $\Delta vdp=\Delta sdT$. Reversible latent heat satisfies $L=T\Delta s$, henceHere $p$ and $T$ are coexistence pressure and absolute temperature, $L$ is latent heat for transforming phase 1 to phase 2, and $\Delta v=v_2-v_1$ uses the same amount of material as $L$.

For vaporization, neglect liquid molar volume compared with ideal-vapour volume $RT/p$. The ideal-vapour Clausius-Clapeyron approximation becomesUsing the normal boiling reference $T_b=373.15\mathrm{K}$, $p_b=1.01325\times 10^5\mathrm{Pa}$ and the supplied constant $L_e$ givesThis deliberately constant-latent-heat extrapolation does not reproduce the supplied triple-point pressure exactly. It predicts the liquid-vapour branch, metastable against ice below the triple-point temperature; it is not a precision sublimation-pressure prediction.

The phase diagram has solid, liquid and vapour regions meeting at the triple point. The sublimation and vaporization boundaries have positive slopes, whereas the ordinary ice–water boundary slopes negatively because ice has larger specific volume. The liquid–vapour line terminates at a critical point; the original schematic below emphasizes topology and marks the supplied triple point.

For the melting-pressure estimate, Archimedes' principle gives ${\rho }_i=0.90{\rho }_w=900\mathrm{kg}{\mathrm{m}}^{-3}$, soConvert the molar evaporation heat using water molar mass $M\simeq 0.0180\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, rounded from the tabulated value $18.0153\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref] An enthalpy cycle through vapour then givesThis uses the question's temperature-independent evaporation heat all the way to the triple point; no replacement experimental fusion heat is inserted. The melting pressure inferred from flotation and latent heats, on holding $L_f$ and $\Delta v$ constant, isusing $T_t=273.16\mathrm{K}$ and $T=270.15\mathrm{K}$. Linearizing the logarithm about the triple point gives $53.4\mathrm{MPa}$ instead of $53.7\mathrm{MPa}$. Both are estimates within the stated constant-property approximation; the negative melting slope makes a positive pressure necessary to lower the melting temperature.

Phase equilibrium minimizes Gibbs free energy at specified temperature, pressure and composition. Coexisting phases have equal temperature, pressure and chemical potential. For a pure substance, a first-order transition has continuous equilibrium Gibbs free energy but a jump in its first derivatives, such as entropy or volume; the latent heat is $T\Delta S$. Continuous transitions lack that latent heat and can instead have singular or discontinuous higher responses.

The van der Waals molar equationcombines an excluded-volume correction with attractive interactions. Its inflection conditions giveBelow $T_c$, the oscillatory isotherm contains a mechanically unstable segment with $(\partial p/\partial v{)}_T>0$. Equilibrium instead replaces the loop by a constant-pressure coexistence line. Equality of chemical potentials gives the Maxwell construction,On isothermal compression the gas reaches saturation, then converts progressively into liquid at fixed pressure. The mean molar volume obeys $v=(1-x)v_{\ell }+x{v}_g$, where $x$ is the gas fraction. Metastable superheated or supercooled states can persist when nucleation is delayed; they are not the global equilibrium selected by this construction.

A pressure–temperature phase diagram contains coexistence curves separating solid, liquid and gas. Three phases meet at a triple point; the liquid–gas coexistence curve ends at a thermodynamic critical point, above which liquid and gas can be connected without crossing a first-order phase boundary. A generic solid–liquid line continues beyond the liquid–gas critical temperature. For one component the Gibbs phase rule is $F=3-P$, so a three-phase equilibrium fixes both temperature and pressure.

Along a two-phase coexistence curve, $d\mu =-sdT+vdp$ for each phase givesThis is the Clapeyron equation, valid with consistent molar or specific quantities. For vaporization, neglecting the liquid volume and treating the vapour as ideal gives $\Delta v\simeq RT/p$, hence $d\ln p/dT=L/(R{T}^2)$. With approximately constant molar latent heat,This Clausius–Clapeyron approximation explains the rapid rise of vapour pressure with temperature. It fails close to the thermodynamic critical point, where phase volumes converge, and is not a universal expression for melting. The sign of a melting-line slope depends on its volume change; water's denser liquid gives a familiar negative slope.

Treat the vapour as an ideal gas, neglect the liquid's molar volume, and take the molar latent heat as constant across this small interval. Integrating the Clausius-Clapeyron equation givesOnly the pressure ratio enters, so conversion from atmospheres to pascals is unnecessary.

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.

For liquid-vapor equilibrium, use molar quantities. At low vapor pressure the gas is dilute, approximately ideal, and has a much larger molar volume than the liquid. HenceSubstitution into the Clapeyron equation givesIf the molar latent heat varies sufficiently slowly to be treated as constant over the temperature interval, integration givesThe positive constant $p_0$ has pressure units and is an integration constant. It is not necessarily a real infinite-temperature pressure, since the approximations fail before that limit.

For mercury between $300$ and $500\mathrm{K}$, the original plot places this range above the triple point and below the normal boiling and critical temperatures. Its vapor pressure is low relative to the critical pressure, making a dilute-gas description reasonable; the vapor's molar volume greatly exceeds that of the dense liquid. The interval is also far from the critical region where $L$ tends to zero, so a roughly constant latent heat is a useful first approximation. The approximations are not exact: Kirchhoff's law of thermochemistry gives, at fixed pressure, the temperature derivative of a transformation enthalpy as the difference between the two molar heat capacities, so a precise fit over a wide interval would allow $L$ to vary.

The Clapeyron equation for molar latent heat $L$ is $dp/dT=L/[T(V_{\mathrm{vap}}-V_{\mathrm{liq}})]$. Neglect the liquid molar volume, treat the vapor as an ideal gas, and approximate $L$ as constant over the interval. ThenUsing one atmosphere as $101325\mathrm{Pa}$ givesThis is the effective molar enthalpy of vaporization inferred from the Integrated Clausius-Clapeyron relation. If $L$ varies with temperature, the inferred value is weighted by $1/T^2$, not an unweighted arithmetic mean. Real helium is not exactly an ideal gas over the whole interval, and its liquid changes superfluid state within it, so this is an estimate under the stated approximations rather than a precise temperature-independent material constant.

Integrate the solids’ molar volume contribution from $P^{\circ }=1$ bar and use ideal-gas water chemical fugacity $f=P$. For pure solids,The logarithm is dimensionless, and $\Delta V_s=-2.28\mathrm{J}\mathrm{ba}{\mathrm{r}}^{-1}$ can be multiplied directly by pressure in bars. Therefore the dehydration equilibrium with ideal-gas fluid givesHence $T=1295.57$ K, or $1022.42^{\circ }$C. Do not also add a constant gas-volume pressure term: the water contribution has already been integrated through $RT\ln (P/P^{\circ })$. Ideal-gas behavior at $6$ kbar is a prescribed model approximation, not a realistic equation of state. The solid volume correction is distinct from it.

Two phases coexist where their molar Gibbs free energies agree. Differentiating that equality along the coexistence curve relates its slope to the entropy and volume jumps.

Low-temperature solid helium-3 can have greater entropy than the liquid through disordered nuclear spins, enabling compression cooling. ^[ref]

The sign follows from the Clapeyron equation. Let $S_l,S_s$ be molar entropies and $V_l,V_s$ molar volumes of liquid and solid, respectively, on their phase coexistence curve. Equality of their chemical potentials, with $d\mu =-SdT+VdP$, givesWith $S_s>S_l$ and $V_l>V_s$, the slope is negative. Reversibly freezing one mole at fixed temperature requires heat into the sample,Here $L_{s\to l}$ uses the melting convention for latent heat of fusion. This is the opposite sign from ordinary ice melting. The inequalities specify the regime; freezing does not universally cool a substance. In an approximately isolated sample, the growing higher-entropy phase draws on the sample's thermal energy instead of an external heater. Irreversible losses and incoming heat oppose the cooling.