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Entropy of vaporisation

  1. May 3, 2017 #1
    1. The problem statement, all variables and given/known data

    5 moles of liquid argon undergoes vaporization at its normal boiling point (87.5 K) and the resulting argon gas is subsequently heated to 150 K under constant volume conditions. Calculate the change of entropy for this process. The standard enthalpy change of vaporization, ∆ ⊖ is +6.5 kJ mol-1 for argon. Treat gaseous argon as an ideal gas in your calculations.
    2. Relevant equations
    ΔS= ΔvapH/ T
    ΔS= q/T
    ΔS= nCvln(Tf/Ti)
    3. The attempt at a solution
    For entropy of vaporisation at 87.5 K
    ΔvapH= 6.5 kJ mol-1 for one mole ∴ 5 moles = (6.5⋅5)⋅1000= 32500 J mol-1
    Using Trouton's Rule ΔS1= 32500/87.5 = 371.43 J K-1

    Entropy from 87.5 K to 150 K
    ΔS2= 5⋅ 3/2R⋅ ln(150/87.5)
    ΔS2= 33.61 J K-1

    Total entropy change= ΔSsys= ΔS1 + ΔS2
    ΔSsys= 371.43 + 33.61 = 405.04 J K-1
  2. jcsd
  3. May 3, 2017 #2
    Your methodology looks correct. I haven't checked your arithmetic.
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