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Standard Entropy of a liquid at melting point with no S(298.15) given

  1. May 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Vapour pressures of a liquid have been measured and fit to the following equation:
    Log10 P (mmHg) = -3571/T + 8.999
    The melting point has been determined to be 392.7 K.
    A Cp value given for the liquid is 250 J/mol K
    and the ΔSvap is 117.20 J/mol K

    2. Relevant equations
    Clausius Clapeyron:

    ln(P1/P2)= ΔH/R*(1/T2 - 1/T1)

    3. The attempt at a solution

    T1= 392.7 K P1 = 0.8045 mmHg (from the equation, giving the vapour pressure of the LIQUID at the melting point).
    T2= 298.15 K P2= 1.052E-3 mmHg

    From the Clausius Clapeyron:

    ln(0.8045 mmHg/1.052E-3 mmHg)= ΔH/R*(1/298.15 K - 1/392.7 K)

    ΔH/R = 8221.89 K-1
    ΔH = 68361 J/mol

    ΔH/T = ΔS = 68361/298.15 = 229.29 J/mol K

    However the answer is 350.60 J/mol K.

    I tried adding the ΔSvap to the derived value (229.29+117.2) , however that only gives me 346.5 J/mol K
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 24, 2014 #2
    Maybe this relation might be helpful: ##L = T\Delta S = T(S_2 - S_1)##.
     
  4. May 26, 2014 #3
    I would use L= TΔS = 68361 J/mol = 229.29 J/mol K x 392.7 K
    = 392.7(S2-S1),
    At this point I am unsure what to use for S2 to solve for S1 or vice versa.
     
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