Free Expansion is non-spontaneous?

Click For Summary

Discussion Overview

The discussion revolves around the concept of free expansion in thermodynamics, particularly focusing on the relationship between internal energy, work done, and entropy changes. Participants explore whether free expansion is spontaneous and the implications of entropy as a state variable in this context.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants assert that internal energy change and work done are zero during free expansion, leading to the conclusion that no heat is absorbed and thus entropy is zero, questioning the spontaneity of the process.
  • Others argue that the entropy change is not zero; it actually increases as the gas expands, challenging the initial claim.
  • There is a discussion about the definition of entropy change, with references to the equation dS = δQ/T, which is noted to apply only to reversible processes.
  • Some participants clarify that while free expansion is not reversible, one can compute the change in entropy using a reversible isothermal process that connects the same initial and final states.
  • One participant raises a question about the change in entropy being calculated as nR * ln(V_2/V_1) if given the initial and final volumes and moles of gas, and inquires about the entropy change of the surroundings.
  • There is a query regarding the use of irreversible processes for calculating entropy changes and the reasoning behind the restriction of dS = dQ/T to reversible processes.

Areas of Agreement / Disagreement

Participants express differing views on the nature of entropy change during free expansion, with some asserting it is zero and others stating it increases. The discussion remains unresolved regarding the implications of these differing perspectives on spontaneity and the applicability of certain thermodynamic equations.

Contextual Notes

Participants note that the definitions and calculations related to entropy depend on the path taken during the process, highlighting the distinction between state variables and path-dependent quantities. There is also an acknowledgment of the complexities involved in understanding thermodynamic principles.

MacNCheese
Messages
7
Reaction score
0
I searched in the fora, and I did find https://www.physicsforums.com/showthread.php?t=292278" asking pretty much the same thing.

Internal energy change and work done are zero, so no heat is absorbed. Thus entropy is zero, hence it isn't spontaneous. What am I doing wrong?

Could you also explain why internal energy changes/work are zero?
 
Last edited by a moderator:
Science news on Phys.org
MacNCheese said:
Internal energy change and work done are zero, so no heat is absorbed. Thus entropy is zero, hence it isn't spontaneous. What am I doing wrong?
The entropy change is not zero--it increases as the gas expands.
Could you also explain why internal energy changes/work are zero?
No heat flows and no work is done. (Nothing pushes on it.)
 
But isn't entropy change equal to heat supplied divided by the temperature?
 
MacNCheese said:
But isn't entropy change equal to heat supplied divided by the temperature?
You are talking about dS = δQ/T. This expression is true only for a reversible process. Free expansion is not a reversible process.

That said, you can use this expression, dS = δQ/T, to compute the change in entropy due to free expansion. Entropy is a "state variable": The entropy of a system is a function of the system's state only. Compare that to work and heat transfer. The amount of work done by a system in getting from some initial state to a final state, along with the heat transferred to a system, depend on the path taken as the system changes from that initial state to the final state.

Because entropy is path-independent, if you can find a path along which you can compute the change in entropy then the computed entropy will be the entropy of the system no matter what path is taken.

The final temperature of the gas in free expansion is the same as the initial temperature: It is an isothermal process. So, find a reversible isothermal process that has the same initial and final states as does the free expansion, compute the change in entropy for this process, and you will have the change in entropy for free expansion.
 
MacNCheese said:
But isn't entropy change equal to heat supplied divided by the temperature?
For a reversible process, yes. But free expansion is not reversible.

If you can imagine a reversible process taking the gas from one volume to another, you can use it to calculate the change in entropy. (Entropy, being a state variable, doesn't depend on how you go from initial to final state.)

Edit: D H beat me to it!
 
Just to add, the entropy change determines if a process is *reversible*. The (Gibbs) free energy change determines if a process is *spontaneous*.
 
Wow, thanks a bunch. That helped clear my concept of entropy a lot.

Does this mean that if I'm given the initial and final volumes and the moles of the gas, the change in entropy will be nR * ln(V_2/V_1)?

Also, is the change in entropy of the surroundings zero?

EDIT: Why can't I use an irreversible process? As long it gets to and from the same (corresponding) states, it should work fine, right? And why is dS = dQ/T defined for only reversible processes?

Sorry, I just cannot wrap my mind around most of thermodynamics. Wish I'd listened in class :P
 
Last edited:

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 22 ·
Replies
22
Views
6K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
5
Views
1K