Is dG Always Equal to Maximum Work in Thermodynamics?

Click For Summary
SUMMARY

The discussion centers on the relationship between Gibbs free energy (dG) and maximum work (dW) in thermodynamics. It is established that the equation dG = dW is not universally valid, although it may hold true under specific conditions. The correct expression for Gibbs energy change in terms of pressure and volume is dG = -Pdv, leading to the conclusion that P = -dG/dv when work is fully resisted. Participants emphasize the importance of context in applying these thermodynamic principles.

PREREQUISITES
  • Understanding of Gibbs free energy (G) and its significance in thermodynamics.
  • Familiarity with the concept of maximum work (W) in thermodynamic processes.
  • Knowledge of pressure (P) and volume (V) relationships in thermodynamic systems.
  • Basic principles of surface tension and its relation to Gibbs energy.
NEXT STEPS
  • Study the derivation of Gibbs free energy equations in thermodynamics.
  • Learn about the conditions under which dG = dW is applicable.
  • Explore the implications of surface tension in thermodynamic systems.
  • Investigate the role of external pressure in work done during thermodynamic processes.
USEFUL FOR

This discussion is beneficial for students and professionals in thermodynamics, particularly those studying Gibbs free energy, maximum work calculations, and related physical chemistry concepts.

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