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Gibbs free energy thermodynamics

  • Thread starter mjbm0720
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  • #1
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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
 

Answers and Replies

  • #2
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Last edited by a moderator:
  • #3
siddharth
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Hi mjbm0720, and welcome to the forums.

Hi

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

If it is valid dG=dW
This relation is NOT valid in general at all.

However, assuming that the relation is valid in a specific situation,
dG=-Pdv

P=-dG/dv?
seems to be correct, assuming that the work done is fully resisted.

Is there anything specific you are trying to show?
 
  • #4
4
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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






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?
 

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