Entropy change for non-thermally-isolated system

Click For Summary
SUMMARY

The discussion centers on the concept of entropy change in thermally isolated and non-thermally-isolated systems, referencing Reif's assertion that for a thermally isolated system, the change in entropy (ΔS) equals zero when external parameters are altered quasi-statically. The key point is that in a thermally isolated system, heat transfer (Q) is zero, leading to ΔS = Q/T being zero regardless of temperature changes. Conversely, in non-thermally-isolated systems, heat can enter from surroundings, resulting in entropy gained by the system being offset by entropy lost by the surroundings, maintaining a net entropy change of zero for the universe.

PREREQUISITES
  • Understanding of the second law of thermodynamics
  • Familiarity with the concepts of thermally isolated systems
  • Knowledge of reversible processes and their implications on entropy
  • Basic principles of the first law of thermodynamics
NEXT STEPS
  • Study the implications of the second law of thermodynamics on entropy changes
  • Learn about the characteristics of thermally isolated versus non-thermally-isolated systems
  • Explore the concept of reversible processes in thermodynamics
  • Investigate isothermal processes and their effects on ideal gases
USEFUL FOR

This discussion is beneficial for students of thermodynamics, physicists, and engineers interested in understanding entropy changes in various thermodynamic processes.

wumple
Messages
57
Reaction score
0

Homework Statement


This isn't an actual problem, it's just a question based on my reading. Reif says "If the external parameters of a thermally isolated system are changed quasi-statically by any amount, ΔS = 0."

I don't understand why it has to be thermally isolated. Looking at the entropy relation

dS = dq/T

There isn't any work in there, and the derivation didn't set work = 0. So if a system undergoes a reversible process and dQ is non-zero, it seems that work also has no effect on the entropy change?
 
Physics news on Phys.org
wumple said:

Homework Statement


This isn't an actual problem, it's just a question based on my reading. Reif says "If the external parameters of a thermally isolated system are changed quasi-statically by any amount, ΔS = 0."

I don't understand why it has to be thermally isolated. Looking at the entropy relation

dS = dq/T

There isn't any work in there, and the derivation didn't set work = 0. So if a system undergoes a reversible process and dQ is non-zero, it seems that work also has no effect on the entropy change?

We have the system and we have the surroundings. The two together form the "universe".

If the system is thermally isolated from the surroundings, Q = 0 by definition and so is ΔS = Q/T, no matter how T may change during the process. So ΔS = 0 for the system, surroundings and universe.

If on the other hand the system is not thermally isolated, then heat can, for example, get into the system from the surroundings. If the process is reversible though, entropy gained by the system = entropy lost by the surroundings. Entropy change of the universe = 0.

Example: isothermal expansion of an ideal gas. We supply heat to the gas, keeping T constant, allowing it to expand from V1 to V2 and p1 to p2. ΔU = 0 so 1st law reads Q = W = ∫pdV = nRT∫dp/p from p1 to p2 = nRT*ln(p2/p1). So the gas increases entropy by Q/T. This amount obviously is lost by the surroundings. The universe entropy change is Q/T - Q/T = 0.
 

Similar threads

Replies
14
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
6K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
10
Views
4K