Why does the entropy increase in free expansion of an ideal gas?

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Discussion Overview

The discussion revolves around the increase of entropy during the free expansion of an ideal gas, exploring the implications of the second law of thermodynamics and the conditions under which entropy is calculated. Participants examine the nature of the process, the applicability of certain equations, and the concept of equilibrium states.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that in free expansion, the heat transfer Q is zero, yet entropy increases, suggesting the process is not quasistatic.
  • Others clarify that free expansion does not pass through a series of equilibrium states, leading to uncertainty about the values of pressure, temperature, and entropy during the process.
  • A participant questions the correctness of the formula $$dS = \frac{dQ}{T}$$, proposing that the correct expression involves a reversible process, $$\Delta S=\int{\frac{dQ_{rev}}{T}}$$.
  • Concerns are raised about whether Q can be treated as a full differential in this context, with some participants expressing uncertainty about the application of the formula.
  • There is a discussion about the work done during free expansion, with conflicting views on whether it is zero or not.
  • Some participants argue that the temperature T cannot be considered constant during the process, as different points in the gas may have varying temperatures, indicating a lack of equilibrium.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the entropy formula and the nature of the free expansion process. There is no consensus on whether the work done is zero or on the correct interpretation of the entropy change in this context.

Contextual Notes

Limitations include the dependence on definitions of equilibrium and the conditions under which the entropy change is calculated. The discussion highlights the complexity of applying thermodynamic principles to irreversible processes.

Mayan Fung
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I learned that
$$ dS = \frac Q T$$
In free expansion of Ideal gas, it is obvious that Q = 0. However, the entropy increases. I guess the reason is that it is because the process is not quasistatic. If I am right, why is this process not quasistatic. If I am not, what's wrong with the formula above. Thanks!
 
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In free expansion the system does not pass through a series of equilibrium states. The P, T and S do not have the same value throughout the extent of the system at every instant. It starts with equilibrium state ends in equilibrium state but in between we are not sure rather we are sure it is not in equilibrium.
 
Chan Pok Fung said:
I learned that
$$ dS = \frac Q T$$
In free expansion of Ideal gas, it is obvious that Q = 0. However, the entropy increases. I guess the reason is that it is because the process is not quasistatic. If I am right, why is this process not quasistatic. If I am not, what's wrong with the formula above. Thanks!
The equation you wrote is incorrect. If you learned it that way, then you were taught incorrectly. The correct formula is $$\Delta S=\int{\frac{dQ_{rev}}{T}}$$
What do you think the subscript "rev" stands for?

Here is a reference to my recent Physics Forums Insights article the provides a cookbook recipe for determining the entropy change in an irreversible process: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
See in particular Example 3

Here is another article on entropy and the second law that should help with your understanding: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/
 
Thanks all for your comments
I am not sure if I am correct. Q may not always be full differential, so I guess we can't always write dQ?
 
Chan Pok Fung said:
Thanks all for your comments
I am not sure if I am correct. Q may not always be full differential, so I guess we can't always write dQ?
As I said, your exact problem is solved in Example 3.
 
But gas is pushing nothing, isn't that the work done = 0?
 
Who said that the work done is not zero? Did you read what I wrote?
 
dq is not perfect differential but [dS =(dq/T)] is perfect differential and T is called integrating factor. One may argue that for infinitesimal dq T can be considered as constant but for the process under consideration at any instant during the process if you were to measure T at different points in the mass of gas you will not get the same value which will convince you that the mass of gas does not have definite value of temperature hence the gas is not in equilibrium at any instant during the process. So formula is not applicable.
 
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Let'sthink said:
dq is not perfect differential but [dS =(dq/T)] is perfect differential and T is called integrating factor. One may argue that for infinitesimal dq T can be considered as constant but for the process under consideration at any instant during the process if you were to measure T at different points in the mass of gas you will not get the same value which will convince you that the mass of gas does not have definite value of temperature hence the gas is not in equilibrium at any instant during the process. So formula is not applicable.

Thanks for your explanation!
 

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