Clausius–Clapeyron relation

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The Clausius–Clapeyron relation, named after Rudolf Clausius and Benoît Paul Émile Clapeyron, is a way of characterizing a discontinuous phase transition between two phases of matter of a single constituent.

Definition

On a pressure–temperature (P–T) diagram, the line separating the two phases is known as the coexistence curve. The Clausius–Clapeyron relation gives the slope of the tangents to this curve. Mathematically,

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} T}}={\frac {L}{T\,\Delta v}}={\frac {\Delta s}{\Delta v}},}$

where ${\displaystyle \mathrm {d} P/\mathrm {d} T}$ is the slope of the tangent to the coexistence curve at any point, ${\displaystyle L}$ is the specific latent heat, ${\displaystyle T}$ is the temperature, ${\displaystyle \Delta v}$ is the specific volume change of the phase transition, and ${\displaystyle \Delta s}$ is the specific entropy change of the phase transition.

Derivations

A typical phase diagram. The dotted green line gives the anomalous behavior of water. The Clausius–Clapeyron relation can be used to find the relationship between pressure and temperature along phase boundaries.

Derivation from state postulate

Using the state postulate, take the specific entropy ${\displaystyle s}$ for a homogeneous substance to be a function of specific volume ${\displaystyle v}$ and temperature ${\displaystyle T}$.

${\displaystyle \mathrm {d} s=\left({\frac {\partial s}{\partial v}}\right)_{T}\,\mathrm {d} v+\left({\frac {\partial s}{\partial T}}\right)_{v}\,\mathrm {d} T.}$

The Clausius–Clapeyron relation characterizes behavior of a closed system during a phase change, during which temperature and pressure are constant by definition. Therefore,

${\displaystyle \mathrm {d} s=\left({\frac {\partial s}{\partial v}}\right)_{T}\,\mathrm {d} v.}$

Using the appropriate Maxwell relation gives

${\displaystyle \mathrm {d} s=\left({\frac {\partial P}{\partial T}}\right)_{v}\,\mathrm {d} v}$

where ${\displaystyle P}$ is the pressure. Since pressure and temperature are constant, by definition the derivative of pressure with respect to temperature does not change. Therefore, the partial derivative of specific entropy may be changed into a total derivative

${\displaystyle \mathrm {d} s={\frac {\mathrm {d} P}{\mathrm {d} T}}\,\mathrm {d} v}$

and the total derivative of pressure with respect to temperature may be factored out when integrating from an initial phase ${\displaystyle \alpha }$ to a final phase ${\displaystyle \beta }$, to obtain

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} T}}={\frac {\Delta s}{\Delta v}}}$

where ${\displaystyle \Delta s\equiv s_{\beta }-s_{\alpha }}$ and ${\displaystyle \Delta v\equiv v_{\beta }-v_{\alpha }}$ are respectively the change in specific entropy and specific volume. Given that a phase change is an internally reversible process, and that our system is closed, the first law of thermodynamics holds

${\displaystyle \mathrm {d} u=\delta q+\delta w=T\;\mathrm {d} s-P\;\mathrm {d} v}$

where ${\displaystyle u}$ is the internal energy of the system. Given constant pressure and temperature (during a phase change) and the definition of specific enthalpy ${\displaystyle h}$, we obtain

${\displaystyle \mathrm {d} h=T\;\mathrm {d} s+v\;\mathrm {d} P}$

${\displaystyle \mathrm {d} h=T\;\mathrm {d} s}$

${\displaystyle \mathrm {d} s={\frac {\mathrm {d} h}{T}}}$

Given constant pressure and temperature (during a phase change), we obtain

${\displaystyle \Delta s={\frac {\Delta h}{T}}}$

Substituting the definition of specific latent heat ${\displaystyle L=\Delta h}$ gives

${\displaystyle \Delta s={\frac {L}{T}}}$

Substituting this result into the pressure derivative given above (${\displaystyle \mathrm {d} P/\mathrm {d} T=\mathrm {\Delta s} /\mathrm {\Delta v} }$), we obtain

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} T}}={\frac {L}{T\,\Delta v}}.}$

This result (also known as the Clapeyron equation) equates the slope of the tangent to the coexistence curve ${\displaystyle \mathrm {d} P/\mathrm {d} T}$, at any given point on the curve, to the function ${\displaystyle L/(T\,\Delta v)}$ of the specific latent heat ${\displaystyle L}$, the temperature ${\displaystyle T}$, and the change in specific volume ${\displaystyle \Delta v}$.

Derivation from Gibbs–Duhem relation

Suppose two phases, ${\displaystyle \alpha }$ and ${\displaystyle \beta }$, are in contact and at equilibrium with each other. Their chemical potentials are related by

${\displaystyle \mu _{\alpha }=\mu _{\beta }.}$

Furthermore, along the coexistence curve,

${\displaystyle \mathrm {d} \mu _{\alpha }=\mathrm {d} \mu _{\beta }.}$

One may therefore use the Gibbs–Duhem relation

${\displaystyle \mathrm {d} \mu =M(-s\,\mathrm {d} T+v\,\mathrm {d} P)}$

(where ${\displaystyle s}$ is the specific entropy, ${\displaystyle v}$ is the specific volume, and ${\displaystyle M}$ is the molar mass) to obtain

${\displaystyle -(s_{\beta }-s_{\alpha })\,\mathrm {d} T+(v_{\beta }-v_{\alpha })\,\mathrm {d} P=0}$

Rearrangement gives

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} T}}={\frac {s_{\beta }-s_{\alpha }}{v_{\beta }-v_{\alpha }}}={\frac {\Delta s}{\Delta v}}}$

from which the derivation of the Clapeyron equation continues as in the previous section.

Ideal gas approximation at low temperatures

When the phase transition of a substance is between a gas phase and a condensed phase (liquid or solid), and occurs at temperatures much lower than the critical temperature of that substance, the specific volume of the gas phase ${\displaystyle v_{\mathrm {g} }}$ greatly exceeds that of the condensed phase ${\displaystyle v_{\mathrm {c} }}$. Therefore, one may approximate

${\displaystyle \Delta v=v_{\mathrm {g} }\left(1-{\tfrac {v_{\mathrm {c} }}{v_{\mathrm {g} }}}\right)\approx v_{\mathrm {g} }}$

at low temperatures. If pressure is also low, the gas may be approximated by the ideal gas law, so that

${\displaystyle v_{\mathrm {g} }={\frac {RT}{P}}}$

where ${\displaystyle P}$ is the pressure, ${\displaystyle R}$ is the specific gas constant, and ${\displaystyle T}$ is the temperature. Substituting into the Clapeyron equation

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} T}}={\frac {L}{T\,\Delta v}}}$

we can obtain the Clausius–Clapeyron equation

${\displaystyle {\frac {\mathrm {d} P}{\mathrm {d} T}}={\frac {PL}{T^{2}R}}}$

for low temperatures and pressures, where ${\displaystyle L}$ is the specific latent heat of the substance.

Let ${\displaystyle (P_{1},T_{1})}$ and ${\displaystyle (P_{2},T_{2})}$ be any two points along the coexistence curve between two phases ${\displaystyle \alpha }$ and ${\displaystyle \beta }$. 

In general, ${\displaystyle L}$ varies between any two such points, as a function of temperature. But if ${\displaystyle L}$ is constant,

${\displaystyle {\frac {\mathrm {d} P}{P}}={\frac {L}{R}}{\frac {\mathrm {d} T}{T^{2}}},}$

${\displaystyle \int _{P_{1}}^{P_{2}}{\frac {\mathrm {d} P}{P}}={\frac {L}{R}}\int _{T_{1}}^{T_{2}}{\frac {\mathrm {d} T}{T^{2}}}}$

${\displaystyle \ln P{\Big |}_{P=P_{1}}^{P_{2}}=-{\frac {L}{R}}\cdot \left.{\frac {1}{T}}\right|_{T=T_{1}}^{T_{2}}}$

or

${\displaystyle \ln {\frac {P_{2}}{P_{1}}}=-{\frac {L}{R}}\left({\frac {1}{T_{2}}}-{\frac {1}{T_{1}}}\right)}$

These last equations are useful because they relate equilibrium or saturation vapor pressure and temperature to the latent heat of the phase change, without requiring specific volume data.

Applications

Chemistry and chemical engineering

For transitions between a gas and a condensed phase with the approximations described above, the expression may be rewritten as

${\displaystyle \ln P=-{\frac {L}{R}}\left({\frac {1}{T}}\right)+c}$

where ${\displaystyle c}$ is a constant. For a liquid-gas transition, ${\displaystyle L}$ is the specific latent heat (or specific enthalpy) of vaporization; for a solid-gas transition, ${\displaystyle L}$ is the specific latent heat of sublimation. If the latent heat is known, then knowledge of one point on the coexistence curve determines the rest of the curve. Conversely, the relationship between ${\displaystyle \ln P}$ and ${\displaystyle 1/T}$ is linear, and so linear regression is used to estimate the latent heat.

Meteorology and climatology

Atmospheric water vapor drives many important meteorologic phenomena (notably precipitation), motivating interest in its dynamics. The Clausius–Clapeyron equation for water vapor under typical atmospheric conditions (near standard temperature and pressure) is

${\displaystyle {\frac {\mathrm {d} e_{s}}{\mathrm {d} T}}={\frac {L_{v}(T)e_{s}}{R_{v}T^{2}}}}$

where:

• ${\displaystyle e_{s}}$ is saturation vapor pressure
• ${\displaystyle T}$ is temperature
• ${\displaystyle L_{v}}$ is the specific latent heat of evaporation of water
• ${\displaystyle R_{v}}$ is the gas constant of water vapor

The temperature dependence of the latent heat ${\displaystyle L_{v}(T)}$ (and of the saturation vapor pressure ${\displaystyle e_{s}}$) cannot be neglected in this application. Fortunately, the August–Roche–Magnus formula provides a very good approximation:

${\displaystyle e_{s}(T)=6.1094\exp \left({\frac {17.625T}{T+243.04}}\right)}$

In the above expression, ${\displaystyle e_{s}}$ is in hPa and ${\displaystyle T}$ is in Celsius, whereas everywhere else on this page, ${\displaystyle T}$ is an absolute temperature (e.g. in Kelvin). (This is also sometimes called the Magnus or Magnus–Tetens approximation, though this attribution is historically inaccurate.) But see also this discussion of the accuracy of different approximating formulae for saturation vapour pressure of water.

Under typical atmospheric conditions, the denominator of the exponent depends weakly on ${\displaystyle T}$ (for which the unit is Celsius). Therefore, the August–Roche–Magnus equation implies that saturation water vapor pressure changes approximately exponentially with temperature under typical atmospheric conditions, and hence the water-holding capacity of the atmosphere increases by about 7% for every 1 °C rise in temperature.

Example

One of the uses of this equation is to determine if a phase transition will occur in a given situation. Consider the question of how much pressure is needed to melt ice at a temperature ${\displaystyle {\Delta T}}$ below 0 °C. Note that water is unusual in that its change in volume upon melting is negative. We can assume

${\displaystyle \Delta P={\frac {L}{T\,\Delta v}}\,\Delta T}$

and substituting in

${\displaystyle L=3.34\times 10^{5}{\text{ J}}/{\text{kg}}}$ (latent heat of fusion for water),

${\displaystyle T=273}$ K (absolute temperature), and

${\displaystyle \Delta v=-9.05\times 10^{-5}{\text{ m}}^{3}/{\text{kg}}}$ (change in specific volume from solid to liquid),

we obtain

${\displaystyle {\frac {\Delta P}{\Delta T}}=-13.5{\text{ MPa}}/{\text{K}}.}$

To provide a rough example of how much pressure this is, to melt ice at −7 °C (the temperature many ice skating rinks are set at) would require balancing a small car (mass = 1000 kg) on a thimble (area = 1 cm²).

Second derivative

While the Clausius–Clapeyron relation gives the slope of the coexistence curve, it does not provide any information about its curvature or second derivative. The second derivative of the coexistence curve of phases 1 and 2 is given by 

${\displaystyle {\begin{aligned}{\frac {\mathrm {d} ^{2}P}{\mathrm {d} T^{2}}}={\frac {1}{v_{2}-v_{1}}}\left[{\frac {c_{p2}-c_{p1}}{T}}-2(v_{2}\alpha _{2}-v_{1}\alpha _{1}){\frac {\mathrm {d} P}{\mathrm {d} T}}\right]+{}\\{\frac {1}{v_{2}-v_{1}}}\left[(v_{2}\kappa _{T2}-v_{1}\kappa _{T1})\left({\frac {\mathrm {d} P}{\mathrm {d} T}}\right)^{2}\right],\end{aligned}}}$

where subscripts 1 and 2 denote the different phases, ${\displaystyle c_{p}}$ is the specific heat capacity at constant pressure, ${\displaystyle \alpha =(1/v)(\mathrm {d} v/\mathrm {d} T)_{P}}$ is the thermal expansion coefficient, and ${\displaystyle \kappa _{T}=-(1/v)(\mathrm {d} v/\mathrm {d} P)_{T}}$ is the isothermal compressibility.

Gibbs–Helmholtz equation

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The Gibbs–Helmholtz equation is a thermodynamic equation used for calculating changes in the Gibbs energy of a system as a function of temperature. It is named after Josiah Willard Gibbs and Hermann von Helmholtz.

The equation is:

${\displaystyle \left({\frac {\partial \left({\frac {G}{T}}\right)}{\partial T}}\right)_{p}=-{\frac {H}{T^{2}}},}$

where H is the enthalpy, T the absolute temperature and G the Gibbs free energy of the system, all at constant pressure p. The equation states that the change in the G/T ratio at constant pressure as a result of an infinitesimally small change in temperature is a factor H/T².

Chemical reactions

Main article: Thermochemistry

The typical applications are to chemical reactions. The equation reads:

${\displaystyle \left({\frac {\partial (\Delta G^{\ominus }/T)}{\partial T}}\right)_{p}=-{\frac {\Delta H}{T^{2}}}}$

with ΔG as the change in Gibbs energy and ΔH as the enthalpy change (considered independent of temperature). The o denotes standard pressure (1 bar).

Integrating with respect to T (again p is constant) it becomes:

${\displaystyle {\frac {\Delta G^{\ominus }(T_{2})}{T_{2}}}-{\frac {\Delta G^{\ominus }(T_{1})}{T_{1}}}=\Delta H^{\ominus }(p)\left({\frac {1}{T_{2}}}-{\frac {1}{T_{1}}}\right)}$

This equation quickly enables the calculation of the Gibbs free energy change for a chemical reaction at any temperature T₂ with knowledge of just the standard Gibbs free energy change of formation and the standard enthalpy change of formation for the individual components.

Also, using the reaction isotherm equation, that is

${\displaystyle {\frac {\Delta G^{\ominus }}{T}}=-R\ln K}$

which relates the Gibbs energy to a chemical equilibrium constant, the van 't Hoff equation can be derived.

Derivation

Background

The definition of the Gibbs function is

${\displaystyle H=G+ST\,\!}$

where H is the enthalpy defined by:

${\displaystyle H=U+pV\,\!}$

Taking differentials of each definition to find dH and dG, then using the fundamental thermodynamic relation (always true for reversible or irreversible processes):

${\displaystyle dU=T\,dS-p\,dV\,\!}$

where S is the entropy, V is volume, (minus sign due to reversibility, in which dU = 0: work other than pressure-volume may be done and is equal to −pV) leads to the "reversed" form of the initial fundamental relation into a new master equation:

${\displaystyle dG=-S\,dT+V\,dp\,\!}$

This is the Gibbs free energy for a closed system. The Gibbs–Helmholtz equation can be derived by this second master equation, and the chain rule for partial derivatives.

Derivation

Starting from the equation ${\displaystyle dG=-S\,dT+V\,dp\,\!}$ for the differential of G, and remembering ${\displaystyle H=G+T\,S\,\!}$ one computes the differential of the ratio G/T by applying the product rule of differentiation in the version for differentials: ${\displaystyle {\begin{aligned}d\left({\frac {G}{T}}\right)&={\frac {T\,dG-G\,dT}{T^{2}}}={\frac {T\,(-S\,dT+V\,dp)-G\,dT}{T^{2}}}\\&={\frac {-T\,S\,dT-G\,dT+T\,V\,dp}{T^{2}}}={\frac {-(G+T\,S)\,dT+T\,V\,dp}{T^{2}}}\\&={\frac {-H\,dT+T\,V\,dp}{T^{2}}}\end{aligned}}\,\!}$ Therefore, ${\displaystyle d\left({\frac {G}{T}}\right)=-{\frac {H}{T^{2}}}\,dT+{\frac {V}{T}}\,dp\,\!}$ A comparison with the general expression for a total differential ${\displaystyle d\left({\frac {G}{T}}\right)=\left({\frac {\partial (G/T)}{\partial T}}\right)_{p}\,dT+\left({\frac {\partial (G/T)}{\partial p}}\right)_{T}\,dp\,\!}$ gives the change of G/T with respect to T at constant pressure (i.e. when dp = 0), the Gibbs-Helmholtz equation: ${\displaystyle \left({\frac {\partial (G/T)}{\partial T}}\right)_{p}=-{\frac {H}{T^{2}}}\,\!}$