Clausius-Clapeyron relation

The Clausius-Clapeyron relation, named after Rudolf Clausius and Émile Clapeyron, is a way of characterizing the phase transition between two states of matter, such as solid and liquid. 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 this curve. Mathematically,

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

where $d P / d T$ is the slope of the coexistence curve, $L$ is the latent heat, $T$ is the temperature, and $Δ V$ is the volume change of the phase transition.

Disambiguation

The generalized equation given in the opening of this article is sometimes called the Clapeyron equation, while a less general form is sometimes called the Clasius-Clapeyron equation. The less general form neglects the magnitude of the specific volume of the liquid (or solid) state relative to that of the gas state and also approximates the specific volume of the gas state via the ideal gas law.^[1]^:509

Derivation

A typical phase diagram. The dotted line gives the anomalous behavior of water. The Clausius-Clapeyron relation can be used to (numerically) find the relationships between pressure and temperature for the phase change boundaries.

Using the state postulate, take the specific entropy, $s$ , for a homogeneous substance to be a function of specific volume, $v$ , and temperature, $T$ .^[1]^:508

$d s = \frac{\partial s}{\partial v} d v + \frac {\partial s}{\partial T} d T$

During a phase change, the temperature is constant, so^[1]^:508

$d s = \frac{\partial s}{\partial v} d v$ .

Using the appropriate Maxwell relation gives^[1]^:508

$d s = \frac{\partial P}{\partial T} d v$ .

Since temperature and pressure are constant during a phase change, the derivative of pressure with respect to temperature is not a function of the specific volume.^[2]^[3]^{:57, 62 & 671} Thus the partial derivative may be changed into a total derivative and be factored out when taking an integral from one phase to another,^[1]^:508

$s_2 - s_1 = \frac{d P}{d T} (v_2 - v_1)$ ,

$\frac{d P}{d T} = \frac {s_2 - s_1}{v_2 - v_1} = \frac {\Delta s}{\Delta v}$ .

Δ

is used as an operator to represent the change in the variable that follows it—final (2) minus initial (1)

For a closed system undergoing an internally reversible process, the first law is

$d u = \delta q - d w = T d s - P d v\,$ .

Using the definition of specific enthalpy, $h$ , and the fact that the temperature and pressure are constant, we have^[1]^:508

$d u + P d v = d h = T ds \Rightarrow ds = \frac {d h}{T} \Rightarrow \Delta s = \frac {\Delta h}{T}$ .

After substitution of this result into the derivative of the pressure, one finds^[4]^[1]^:508

$\frac{d P}{d T} = \frac {\Delta h}{T \Delta v} = \frac {\Delta H}{T \Delta V} = \frac {L}{T \Delta V}$ ,

where the shift to capital letters indicates a shift to extensive variables. This last equation is called the Clasius-Clapeyron equation, though some thermodynamics texts just call it the Clapeyron equation, possibly to distinguish it from the approximation below.

When the transition is to a gas phase, the final specific volume can be many times the size of the initial specific volume. A natural approximation would be to replace $Δ v$ with $v 2$ . Furthermore, at low pressures, the gas phase may be approximated by the ideal gas law, so that $v 2 = v g a s = R T / P$ , where R is the mass specific gas constant (forcing $h$ and $v$ to be mass specific). Thus,^[1]^:509

$\frac{d P}{d T} = \frac {P \Delta h}{T^2 R}$ .

This leads to a version of the Clasius-Clapeyron equation that is simpler to integrate:^[1]^:509

$\frac {d P}{P} = \frac {\Delta h}{R} \frac {dT}{T^2}$ ,

$\ln P = - \frac {\Delta h}{R} \frac {1}{T} + C$ , or^[3]^:672

$\ln \frac {P_2}{P_1} = \frac {\Delta h}{R} \left ( \frac {1}{T_1} - \frac {1}{T_2} \right )$ .

C

is a constant of integration

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

Other derivation

Suppose two phases, I and II, are in contact and at equilibrium with each other. Then the chemical potentials are related by $μ I = μ I I$ . Along the coexistence curve, we also have $dμ I = dμ I I$ . We now use the Gibbs-Duhem relation $dμ = - s d T + v d P$ , where $s$ and $v$ are, respectively, the entropy and volume per particle, to obtain

$-(s_I-s_{II}) \mathrm{d}T + (v_I-v_{II}) \mathrm{d}P = 0. \,$

Hence, rearranging, we have

$\frac{\mathrm{d}P}{\mathrm{d}T} = \frac{s_I-s_{II}}{v_I-v_{II}}.$

From the relation between heat and change of entropy in a reversible process δQ = T dS, we have that the quantity of heat added in the transformation is

$L= T (s_I-s_{II}). \,$

Combining the last two equations we obtain the standard relation.

Applications

Chemistry

This equation gives the quantitative dependence of the vapor pressure of a liquid on its temperature. It can be used to predict the temperature at a certain pressure, given the temperature at another pressure, or vice versa. If the corresponding temperature and pressure is known at two points, the enthalpy of vaporization can also be determined from this equation.

$\ln \left(\frac{P_2}{P_1} \right) = \left(\frac{\mathcal 4 H_{vap}}{R} \right) \times \left(\frac{1}{T_1} - \frac{1}{T_2} \right)$

where

$T 1$ and $P 1$ are a corresponding temperature (K) and pressure
$T 2$ and $P 2$ are the corresponding temperature and pressure at another point
$\mathcal 4 H_{vap}$ is the enthalpy of vaporization
$R$ is the gas constant (8.314 J mol^-1K^-1)

Chemical engineering

A specific derivation of the equation is used in chemistry and chemical engineering to estimate the vapor pressure of a substance based on the heat of vaporization of that substance, and on the temperature of the system under consideration. The equation is as follows:

$\ln p^* = -\frac{\Delta \hat H _v}{RT}+B$

where

$p *$ is the vapor pressure (mmHg)
${\Delta \hat H_v}$ is the enthalpy of vaporization (kJ/mole)
$R$ is the gas constant
$T$ is the temperature (kelvins)
$B$ is a constant based on the substance and the system parameters

Meteorology

In meteorology, a specific derivation of the Clausius-Clapeyron equation is used to describe dependence of saturated water vapor pressure on temperature. This is similar to its use in chemistry and chemical engineering.

It plays a crucial role in the current debate on climate change because its solution predicts exponential behavior of saturation water vapor pressure (and, therefore water vapor concentration) as a function of temperature. In turn, because water vapor is a greenhouse gas, it might lead to further increase in the sea surface temperature leading to runaway greenhouse effect. Debate on iris hypothesis and intensity of tropical cyclones dependence on temperature depends in part on “Clausius-Clapeyron” solution.

Clausius-Clapeyron equations is given for typical atmospheric conditions as

$\frac{\mathrm{d}e_s}{\mathrm{d}T} = \frac{L_v e_s}{R_v T^2}$

where:

$e s$ is saturation water vapor pressure
$T$ is a temperature
$L v$ is latent heat of evaporation
$R v$ is water vapor gas constant.

One can solve this equation to give^[5]

$e_s(T)= 6.112 \exp \left( \frac{17.67 T}{T+243.5} \right)$

where:

$e s (T)$ is in hPa (mbar)
$T$ is in degrees Celsius.

Thus, neglecting the weak variation of (T+243.5) at normal temperatures, one observes that saturation water vapor pressure changes exponentially with $T$ .

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 $Δ T$ below 0 °C. We can assume

${\Delta P} = \frac{L}{T\,\Delta V} {\Delta T}$

and substituting in

L

= 3.34×10⁵ J/kg (latent heat of water),

T

= 273 K (absolute temperature), and

Δ V

= -9.05×10^-5 m³/kg (change in volume from solid to liquid),

we obtain

$\frac{\Delta P}{\Delta T}$ = -13.1 MPa/°C.

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^[6]) on a thimble (area = 1 cm²).

References

^ ^a ^b ^c ^d ^e ^f ^g ^h ⁱ Wark, Kenneth [1966] (1988). "Generalized Thermodynamic Relationships", Thermodynamics, 5th (in English), New York, NY: McGraw-Hill, Inc.. ISBN 0-07-068286-0.
^ Carl Rod Nave (2006). PvT Surface for a Substance which Contracts Upon Freezing (English). HyperPhysics. Georgia State University. Retrieved on 2007-10-16.
^ ^a ^b Çengel, Yunus A.; Boles, Michael A. [1989] (1998). Thermodynamics - An Engineering Approach, 3rd, McGraw-Hill Series in Mechanical Engineering (in English), Boston, MA.: McGraw-Hill. ISBN 0-07-011927-9.
^ Salzman, William R. (2001-08-21). Clapeyron and Clausius-Clapeyron Equations (English). Chemical Thermodynamics. University of Arizona. Archived from the original on 2007-07-07. Retrieved on 2007-10-11.
^ American Meteorological Society - The Computation of Equivalent Potential Temperature
^ Mass of a Car - Hypertextbook.com

Bibliography

M.K. Yau and R.R. Rogers, Short Course in Cloud Physics, Third Edition, published by Butterworth-Heinemann, January 1, 1989, 304 pages. EAN 9780750632157 ISBN 0-7506-3215-1

J.V. Iribarne and W.L. Godson, Atmospheric Thermodynamics, published by D. Reidel Publishing Company, Dordrecht, Holland, 1973, 222 pages

H.B. Callen, Thermodynamics and an Introduction to Thermostatistics, published by Wiley, 1985. ISBN 0-471-86256-8

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