Clausius-clapeyron Relation Website Links For
Relation
 

Information About

Clausius-clapeyron Relation




: rac{dP}{dT} = rac{L}{T\Delta V}
where ''dP''/''dT'' is the slope of the coexistence curve, ''L'' is the Latent Heat , ''T'' is the Temperature , and \Delta V is the Volume .


DERIVATION


Suppose two phases, I and II, are in contact and at equillibrium with each other. Then the chemical potentials are related by μI = μII. Along the coexistence curve, we also have dμI = dμII. We now use the Gibbs-Duhem relation
d\mu = -sdT + vdP, where s and v are, respectively, the entropy and volume per particle, to obtain

:(s_I-s_{II}) dT + (v_I-v_{II}) dP = 0

Hence, rearranging, we have

: rac{dP}{dT} = rac{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.


APPLICATION


This equation can be used to work out whether or not a phase transition will occur. For example, the Clausius-Clapeyron relation is often invoked to explain Ice Skating : the increased pressure of the skater on the skates causes the ice to melt. Does this explanation work? If ''T'' = −2 °C, we can use the Clausius-Clapeyron relation to work out how much pressure is needed to melt ice. We can assume

: {\Delta P} = rac{L}{T\Delta V} {\Delta T}

and substituting in

  • 105 J/kg, T=271K, \Delta V = -9.05 ---10-5m3/kg,


and

:\Delta T = 2K,

we get

:\Delta P = 27.2 MPa.

This is an equivalent pressure to a Sumo Wrestler (mass = 150 kg) standing on a Stiletto Heel (area = 0.5 cm2)! Evidently, this is not how ice skating works.