Clausius-Clapeyron relation

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The Clausius-Clapeyron relation, in thermodynamics, 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,

<math>\frac{\mathrm{d}P}{\mathrm{d}T} = \frac{L}{T\Delta V} </math>

where <math>\mathrm{d}P/\mathrm{d}T</math> is the slope of the coexistence curve, <math>L</math> is the latent heat, <math>T</math> is the temperature, and <math>\Delta V </math> is the volume change of the phase transition.

1 Derivation
2 Applications
3 Example
4 References
5 Bibliography

[edit] Derivation

Suppose two phases, I and II, are in contact and at equilibrium with each other. Then the chemical potentials are related by <math>\mu_{I} = \mu_{II}</math>. Along the coexistence curve, we also have <math>\mathrm{d}\mu_{I} = \mathrm{d}\mu_{II}</math>. We now use the Gibbs-Duhem relation <math>\mathrm{d}\mu = -s\mathrm{d}T + v\mathrm{d}P</math>, where <math>s</math> and <math>v</math> are, respectively, the entropy and volume per particle, to obtain

<math>-(s_I-s_{II}) \mathrm{d}T + (v_I-v_{II}) \mathrm{d}P = 0 </math>

Hence, rearranging, we have

<math>\frac{\mathrm{d}P}{\mathrm{d}T} = \frac{s_I-s_{II}}{v_I-v_{II}}</math>

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

Combining the last two equations we obtain the standard relation.

[edit] Applications

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:

<math>\ln p^* = -\frac{\Delta \hat H _v}{RT}+B</math>

<math>p^*</math> is the vapor pressure (mmHg)
<math>{\Delta \hat H_v}</math> is the heat of vaporization (kJ/mole)
<math>R</math> is the gas constant
<math>T</math> is the temperature (Kelvin)
<math>B</math> is a variable 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 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

<math> \frac{\mathrm{d}e_s}{\mathrm{d}T} = \frac{L_v e_s}{R_v T^2} </math>

where <math> e_s </math> is saturation water vapor pressure, <math>T</math> is a temperature, <math> L_v </math> is latent heat of evaporation, <math>R_v </math> is water vapor gas constant. One can solve this equation to give<ref>http://dx.doi.org/10.1175/1520-0493(1980)108%3C1046:TCOEPT%3E2.0.CO;2</ref>

<math> e_s(T)= 6.112 \exp \left( \frac{17.67 T}{T+243.5} \right) </math>

where <math> e_s(T) </math> is in hPa, and <math> T </math> is in Celsius. Thus, neglecting the weak variation of (T+243.5) at normal temperatures, one observes that saturation water vapor pressure changes exponentially with <math>T</math>.

[edit] Example

One of the uses of this equation is to work out whether or not a phase transition will occur in a given situation. Consider the question of how much pressure is needed to melt ice at a temperature <math>{\Delta T}</math> below 0°C. We can assume

<math> {\Delta P} = \frac{L}{T\Delta V} {\Delta T} </math>

and substituting in

<math>L = 3.34×10⁵</math> J/kg (latent heat of water),

<math>T = 273</math>K (absolute temperature), and

<math>\Delta V </math> = -9.05×10^-5 m³/kg (change in volume from solid to liquid),

we obtain

<math>\frac{\Delta P}{\Delta T} </math> = 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<ref>http://hypertextbook.com/facts/2000/YanaZorina.shtml</ref>) on a thimble (area = 1 cm²).

[edit] References

[edit] 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