Gibbs free energy thermodynamics

  • Thread starter mjbm0720
  • Start date
  • #1
4
0
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

  • #3
siddharth
Homework Helper
Gold Member
1,127
0
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
0
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?
 

Related Threads on Gibbs free energy thermodynamics

Replies
0
Views
2K
Replies
5
Views
248
Replies
7
Views
4K
Replies
14
Views
5K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
2
Views
7K
Replies
1
Views
929
Top