Free expansion and adiabatic process

Click For Summary
SUMMARY

The discussion centers on the calculation of entropy change (\Delta S) in thermodynamic processes, specifically comparing adiabatic processes and free expansion. In an adiabatic process, where heat transfer (Q) is zero, \Delta S is also zero. However, in free expansion, despite Q and work (W) being zero, \Delta S is not zero due to the irreversible nature of the process. The key takeaway is that the formula for entropy change is only valid for reversible processes, necessitating the use of a reversible path to accurately calculate \Delta S for free expansion.

PREREQUISITES
  • Understanding of thermodynamic concepts, specifically entropy and its calculation.
  • Familiarity with the first and second laws of thermodynamics.
  • Knowledge of reversible and irreversible processes in thermodynamics.
  • Basic proficiency in calculus for integrating thermodynamic equations.
NEXT STEPS
  • Study the principles of reversible and irreversible processes in thermodynamics.
  • Learn how to derive entropy change using the Carnot cycle as a reversible path.
  • Explore the implications of the second law of thermodynamics on entropy in various processes.
  • Investigate real-world applications of free expansion and its impact on system efficiency.
USEFUL FOR

Students and professionals in thermodynamics, physicists, and engineers interested in understanding entropy changes in various thermodynamic processes.

sita1
Messages
2
Reaction score
0
I hope someone in here can help me out of this problem.
I know that for calculating the change in entropy of a system (\Delta S) we can use this formula :

\Delta S=\intδq/T
Well the problem is here, when we calculating ΔS for an adiabatic process we know Q= 0 so ΔS=0 in this process, but why we just can't let Q be equal to zero in a free expansion (while we know that in free expansion W=Q=0) and get to the same result (ΔS=0) for the free expansion process?
I should add that I know ΔS≠ 0 for free expansion process but, you know I just don't know where the problem is in my argument.
 
Science news on Phys.org
The formula you wrote is only valid for a reversible process. And the free expansion is not a reversible process. To calculate the difference in entropy in this case, connect the initial and final state by any reversible path, for which generally you won't have dQ = dL = 0, and use your formula on this path.
 
Thanks, that's really a convincing answer.
 

Similar threads

Undergrad Adiabatic expansion work far exceeds isobaric of same volume, why?
  • · Replies 81 ·
3
Replies
81
Views
6K
Undergrad Irreversible adiabatic processes & entropy change (clarification needed)
  • · Replies 22 ·
Replies
22
Views
6K
Undergrad Adiabatic cooling in this process involving liquid ammonia
  • · Replies 5 ·
Replies
5
Views
3K
Undergrad Graphical difference between adiabatic and isothermal processes.
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Are "isentropic" and "adiabatic" same?
  • · Replies 7 ·
Replies
7
Views
11K
Undergrad Why is the internal energy change of free expansion 0?
  • · Replies 7 ·
Replies
7
Views
5K
Undergrad Confusion between adiabatic and energy
  • · Replies 20 ·
Replies
20
Views
4K
Graduate Why dU may not equals dW in free expansion?
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad The relationship between reversible and irreversible processes
  • · Replies 5 ·
Replies
5
Views
2K
Graduate Euler's equation of thermodynamics in free expansion (Joule expansion)
  • · Replies 4 ·
Replies
4
Views
2K