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Second Law of Thermodynamics

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that the second law of thermodynamics predicts the spontaneous freezing of liquid water at -5[tex]^{o}[/tex]C under 1 bar of constant pressure. Assume that Cp is temperature independent.

    Standard water fusion enthalpy = 6.008 kJ/mol @ 273.15 K

    [tex]\geq[/tex] has to be interpreted as greater than, not greater or equal, in this example.


    2. Relevant equations

    A: [tex]dS[/tex] [tex]\geq[/tex] [tex]\frac{\partial q}{T}[/tex]

    B: [tex]\Delta[/tex][tex]S_{transition}[/tex] = [tex]\frac{\Delta H_{transition}}{T_{transition}}[/tex]

    C: [tex]\Delta S = \int \frac{Cp}{T} dT[/tex]

    D: [tex]\Delta H = q[/tex] in an isobaric process

    3. The attempt at a solution

    After having wasted an hour or so on this presumably easy problem, I cannot get the second law to predict the freezing.

    [tex]Cp\ ln(T) \geq \frac{\Delta H_{transition}}{T_{transition}}[/tex] ?

    Some advice would be appreciated. I know the proof is simple. I'm having some brain fog atm though.

    Thanks!
     
  2. jcsd
  3. Mar 24, 2010 #2

    Mapes

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    Hi Aeon, welcome to PF. What thermodynamic potential stays constant for phase changes at constant temperature and pressure? And using this, can you find the entropy of freezing from the enthalpy of freezing?
     
  4. Mar 24, 2010 #3
    I have been reading on the forums for some time. I never had to ask a question yet though.
    Thanks!

    On a phase transition diagram, Gibbs free energy of one phase is equal to G of another phase when you follow the phase transition curves.

    Is the enthalpy of freezing opposite of the enthalpy of fusion?
     
  5. Mar 25, 2010 #4

    Mapes

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    Exactly. So by using the definition of G, the entropy of fusion can be determined. Now you can calculate the difference in G between the phases - and a spontaneous process will tend to minimize G.

    Yes, and that was a typo; I meant fusion! Enthalpy and entropy of freezing would have the opposite signs.
     
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