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Vapor pressure question in chemical thermodynamics

  1. Jan 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Calculate the vapor pressure of iso-propyle alcohol over iso-propyle alcohol in pure liquid state at 298,15K.

    2. Relevant equations

    A possible equation to solve this problem, I think, would be to use this formula put together by two definitions of free energy. [tex]\Delta G = \Delta H - T \cdot \Delta S[/tex] and
    [tex]\Delta G = RT ln \frac{p}{p*}[/tex]

    [tex]p = exp(\frac{\Delta H_{vap}-T \cdot \Delta S_{vap}}{R\cdot T})[/tex]

    And for this we know the following data.
    [tex]\Delta H_{vap} = 44,2 \frac{kJ}{mol}
    [tex]\Delta S_{vap} = 127 \frac{J}{mol \cdot K}

    3. The attempt at a solution

    If we put in the data we get the following solution.

    [tex]p = exp(\frac{44200 \frac{J}{mol}-298,15K \cdot 127 \frac{J}{mol \cdot K}}{8,315\frac{J}{mol \cdot K}\cdot 298,15K}=0,078 atm[/tex]

    The unit of the vapor pressure is given in atm because you always compare a result to its predefined standard. In this case 1 atm. [tex]\frac{p}{p_{ref}}[/tex]

    The result is therefore 0,078 atm, however in my solutions manual for the coursework question, the solution should be 0,18 atm. And this is my problem, I simply can't get to that solution no matter how many times I twist and turn it.

    Sorry if the language is a bit off. If you don't understand my question or some of the midway solutions I come to, please do ask and I will respond as soon as possible.

    Nikolaj Rahmberg
  2. jcsd
  3. Jan 2, 2009 #2
    My error has been found by myself.. The value I had for vapor entropy was wrong. It was really 134,04 J/mol*K..

    But thanks anyway..
  4. Jan 2, 2009 #3


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    Glad you worked it out. Welcome to PF! :smile:
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