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Gibbs free energy

  1. Feb 11, 2007 #1
    Hi

    I am new in thermodynamics , and I thought you can help me

    If it is valid dG=dW

    May I say

    dG=-Pdv

    P=-dG/dv?


    thanks in advance,Mariana



    P=external pressure
    G=Gibbs energy
    W=maximum work
     
  2. jcsd
  3. Feb 11, 2007 #2
  4. Feb 11, 2007 #3

    siddharth

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    Homework Helper
    Gold Member

    Hi mjbm0720, and welcome to the forums.

    This relation is NOT valid in general at all.

    However, assuming that the relation is valid in a specific situation,
    seems to be correct, assuming that the work done is fully resisted.

    Is there anything specific you are trying to show?
     
  5. Feb 11, 2007 #4
    Thank you so much


    I was trying to show this:

    Because of dG=dWmax and dw=surface tension por dA

    Surface tension=dG/dA

    I was thinking because dw=pdV
    I could say
    p=-dG/dV
    But I see that it is a mistake






     
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