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Clausius-Clapeyron equation and water

  1. Jul 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Using the Clausius-Clapeyron equation, determine the slope of dp/dT for water-ice I equilibrium and explain why you can skate on ice but not on solid argon.

    2. Relevant equations

    Clausius-Clapeyron equation

    3. The attempt at a solution

    I know what the clausius-clapeyron equation is but I have no idea how I'm supposed to determine the slope for water - there is no information in the book so far on water so I really have no idea how I could write the change in entropy and volume as functions of temperature..

    I'm completely at a loss here (and most likely overlooking something simple) :confused:

    (I wasn't sure if this should be in introductory or advanced physics, the textbook I'm using doesn't really mention what level it's geared at)
     
  2. jcsd
  3. Jul 13, 2012 #2

    TSny

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    Consider the melting of 1 gram of H2O.

    Note ΔS= Q/T where Q is the heat required to melt 1 gram of ice (79.7 cal = 333 J) and T is the temperature corresponding to equilibrium between ice and water.

    ΔV = Vf-Vi where Vi is the volume of 1 gram of ice and Vf is the volume of 1 gram of water. You can calculate these volumes from the known densities of ice and water at the melting point temperature (ice: 0.917 g/cm3, water: 1.000 g/cm3).

    Watch your units.
     
  4. Jul 14, 2012 #3
    Ah, thank you! :biggrin:
    I knew it'd turn out to be something simple
     
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