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Clausius-Clapeyron equation

  1. Mar 3, 2005 #1
    The Clausius-Clapeyron equation is given by:

    [tex]\frac{dP}{dT}=\frac{LJ}{T(V_2-V_1)}[/tex]

    [itex]dP[/itex] is the change in pressure
    [itex]dT[/itex] is the change in temperature
    [itex]L[/itex] is the Latent heat of fusion/vaporisation
    [itex]T[/itex] absolute temperature of substance
    [itex]V_2-V_1 [/itex] is the change in volume
    [itex]J[/itex] is heat constant

    What is the physical significance of this equation?
     
  2. jcsd
  3. Mar 4, 2005 #2
    What, no replies so far!
     
  4. Mar 4, 2005 #3

    Andrew Mason

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    When a liquid and vapour are in thermal equilibrium in a closed volume at a certain temperature, and the volume is increased slightly, some more liquid turns to vapour. Since the liquid has a latent heat (energy just required to break molecular bonds of liquid) the heat has to be drawn from the liquid so the temperature of the liquid/vapour decreases. (the opposite effect when the volume decreases - some vapour turns to liquid and the temperature increases). The Clausius-Clapeyron equation shows the relationship between a change in volume and the rate of change of vapour pressure with temperature.

    AM
     
    Last edited: Mar 4, 2005
  5. Mar 5, 2005 #4
    Thank you very much, sir!
     
  6. Aug 31, 2005 #5
    Example

    Anderw has described it correctly. Here is an exmple from a lab experience. Those who work in lowtemperature physics employ Liq.He and N2 and make use of this principle to reach temp. below the (normal) boiling point of these liquids. For example. Liq. He4 boils at 4.2K and by pumping over He4 vapour, one can attain temperature as low as 1K ! Said in a crude fashion, this means the He4 molecules having the highest Kin.Energy leavs the liquid surface making the total temp to decrease.
     
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