Is dG Always Equal to Maximum Work in Thermodynamics?

Click For Summary

Homework Help Overview

The discussion revolves around the relationship between Gibbs free energy (dG) and maximum work (dW) in thermodynamics. Participants are exploring whether the equation dG = dW holds true in general and are examining specific conditions under which it might apply.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the validity of the relation dG = dW and discussing its applicability in specific scenarios. There are attempts to derive relationships involving pressure and volume changes, as well as surface tension, in the context of Gibbs energy.

Discussion Status

Some participants have provided references for further reading, while others are clarifying the assumptions behind the equations being discussed. There is an acknowledgment that the relation dG = dW is not universally valid, and the conversation is exploring the nuances of its application.

Contextual Notes

There is a mention of specific conditions under which the relation might hold, such as when work is fully resisted. Participants are also reflecting on potential mistakes in their reasoning regarding the relationships between pressure, volume, and Gibbs energy.

mjbm0720
Messages
4
Reaction score
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
 
Physics news on Phys.org
Last edited by a moderator:
Hi mjbm0720, and welcome to the forums.

mjbm0720 said:
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?
 
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






siddharth said:
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?
 

Similar threads

Is Gibbs Energy Truly Equivalent to Work?
  • · Replies 1 ·
Replies
1
Views
2K
DeltaG and DeltaA calculation for heating a gas at constant volume
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Minimization of the Gibbs energy when number of molecules varies
  • · Replies 16 ·
Replies
16
Views
3K
Maxwell Relation, Gibbs Free Energy, Thermal Expansion Coefficient
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Gibbs Free Energy Correction Term
  • · Replies 5 ·
Replies
5
Views
1K
Gibbs free energy for superconductor in intermediate state
  • · Replies 1 ·
Replies
1
Views
1K
Chemistry How to derive 2nd law extremum principles for U, A, G, and H?
  • · Replies 1 ·
Replies
1
Views
3K
High School Change in entropy of reversible isothermal process
  • · Replies 11 ·
Replies
11
Views
3K
Thermodynamics: Gibbs free energy from this battery reaction?
  • · Replies 7 ·
Replies
7
Views
6K
How Does Temperature and Pressure Affect Gibbs Free Energy in Water?
  • · Replies 4 ·
Replies
4
Views
4K