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Enthalpy change of a vaporization rxn

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data

    A certain liquid has a standard enthalpy of vaporization of 32 KJ/mol. Calculate q, w, delta H and delta U when 0.75 mol is vaporized at 260 K and 765 Torr.



    2. Relevant equations



    3. The attempt at a solution

    I was able to get delta H, but the solution states that q is equal to delta H. If delta H is equal to q at constant pressure, I don't know how I'm supposed to know when to make that assumption. Also, why does work not equal zero if the pressure is constant, is this additional work?
     
  2. jcsd
  3. Sep 21, 2008 #2

    Mapes

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    The amount of work is nonzero because the volume increases and the system therefore does work on its surroundings. This is the big difference between energy and enthalpy; enthalpy takes into account that a system's volume tends to vary during processes such as heating and vaporization.
     
  4. Jun 11, 2009 #3
    I have a similar question, just wondering how you solved ∆H? I can do the rest.

    Thanks
     
  5. Jun 12, 2009 #4

    Mapes

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    Hi hinglis, welcome to PF. ∆H is the enthalpy of vaporization, adjusted to the number of moles in the system.
     
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