Gibbs free energy thermodynamics

[mjbm0720]

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

 

[lalbatros]

read this:

http://en.wikipedia.org/wiki/Gibbs_free_energy

and maybe this too:

http://www.dlr.enge.vt.edu/project_documents/Gibbs%20Free%20Energy-2.pdf [Broken]

 

[siddharth]

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?

 

[mjbm0720]

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?