Entropy Change in Reversible vs Irreversible Gas Expansions

In summary: Hello! I think I formulated my statement in a wrong way, but this is also what I said. There are was to make just the system reversible. Shouldn't the term "reversible" refer to both the system and the surroundings (and another term for just...the system)?Yes, the term "reversible" refers to both the system and the surroundings.
  • #1
Silviu
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Hello! I have this GRE question:
In process 1, a monoatomic ideal gas is heated from temperature T to temperature 2T reversibly and at constant temperature. In process 2, a monoatomic ideal gas freely expands from V to 2V. Which is the correct relationship between the change in entropy ##\Delta S_1## in process 1 and the change in entropy ##\Delta S_2## in process 2?

So, I am not sure in the first process what do they mean by, at constant temperature, if the temperature goes from T to 2T, but leaving that aside the correct answer is: ##0 < \Delta S_2 < \Delta S_1 ##. However, another option is ##0 =\Delta S_1 < \Delta S_2##. My main question is: isn't the change in entropy always 0 in a reversible process? So, shouldn't be this answer the right answer (being also the only one with ##0 =\Delta S_1##)?

Thank you!
 
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  • #2
Silviu said:
In process 1, a monoatomic ideal gas is heated from temperature T to temperature 2T
I think it is a misprint. Probably, the question meant to to say "from volume V to volume 2V reversibly and at constant temperature" the idea being to compare the entropy change for the two processes when the volume changes are identical.
 
  • #3
kuruman said:
I think it is a misprint. Probably, the question meant to to say "from volume V to volume 2V reversibly and at constant temperature" the idea being to compare the entropy change for the two processes when the volume changes are identical.
But still the change in entropy should be 0, as the process is reversible, right?
 
  • #4
Silviu said:
But still the change in entropy should be 0, as the process is reversible, right?
Not necessarily. It depends on whether you are talking about the system or about the system plus surroundings.
 
  • #5
Chestermiller said:
Not necessarily. It depends on whether you are talking about the system or about the system plus surroundings.
Hello! I am a bit confused. When you say reversible process, don't you always talk about system plus surroundings? If you talk only about the system, then basically any process that bring the system to the initial state would be reversible, as the change in entropy of the system would be zero (but of course the change in entropy of the universe would increase). So isn't somehow pointless to define reversibility only for the system?
 
  • #6
Silviu said:
Hello! I am a bit confused. When you say reversible process, don't you always talk about system plus surroundings? If you talk only about the system, then basically any process that bring the system to the initial state would be reversible, as the change in entropy of the system would be zero (but of course the change in entropy of the universe would increase). So isn't somehow pointless to define reversibility only for the system?
No. If the system undergoes the exact same path it would have in a reversible process (i.e., a continuous sequence of thermodynamic equilibrium states), even if the surroundings did not undergo a corresponding continuous sequence of thermodynamic equilibrium states (i.e., it underwent an irreversible path), the process is still considered reversible for the system but not for the surroundings. An example of this would be where you manually (by hand) cause a gas in a cylinder/piston arrangement to undergo a reversible adiabatic compression (or expansion) by controlling the history of the force that you apply to the piston. Certainly the biological and mechanical processes taking place in your body during this process are not reversible. So the gas has undergone a reversible process, but not the surroundings. In Fundamentals of Engineering Thermodynamics by Moran et al, this is referred to as an internally reversible process.
 
  • #7
Chestermiller said:
No. If the system undergoes the exact same path it would have in a reversible process (i.e., a continuous sequence of thermodynamic equilibrium states), even if the surroundings did not undergo a corresponding continuous sequence of thermodynamic equilibrium states (i.e., it underwent an irreversible path), the process is still considered reversible for the system but not for the surroundings. An example of this would be where you manually (by hand) cause a gas in a cylinder/piston arrangement to undergo a reversible adiabatic compression (or expansion) by controlling the history of the force that you apply to the piston. Certainly the biological and mechanical processes taking place in your body during this process are not reversible. So the gas has undergone a reversible process, but not the surroundings. In Fundamentals of Engineering Thermodynamics by Moran et al, this is referred to as an internally reversible process.
Hello! I think I formulated my statement in a wrong way, but this is also what I said. There are was to make just the system reversible. Shouldn't the term "reversible" refer to both the system and the surroundings (and another term for just the system - like you said)? But in any case, in the problem that I posed initially, doesn't reversible means a 0 change in entropy for the system they are talking about?
 
  • #8
Silviu said:
Hello! I think I formulated my statement in a wrong way, but this is also what I said. There are was to make just the system reversible. Shouldn't the term "reversible" refer to both the system and the surroundings (and another term for just the system - like you said)? But in any case, in the problem that I posed initially, doesn't reversible means a 0 change in entropy for the system they are talking about?
In both process 1 (with the typo corrected to go from V to 2V, rather than T to 2T) and process 2, the change in entropy of the system is greater than zero, and is the same for both processes. It has to be the same for both processes because they both start in the same state and both end in the same state. Here are a couple of Physics Forums Insights articles I wrote which may help to supplement your understanding:

https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

In the second article, pay particular attention to Example 3.

Chet
 

1. What is entropy change?

Entropy change refers to the measure of the disorder or randomness of a system. In thermodynamics, it is a measure of the heat energy that is converted into work during a process.

2. What is reversible gas expansion?

Reversible gas expansion is a process where the gas expands and does work on its surroundings in a way that the system remains in equilibrium throughout the process. This means that the system can be brought back to its original state without any change in the environment.

3. What is irreversible gas expansion?

Irreversible gas expansion is a process where the gas expands and does work on its surroundings, but the system does not remain in equilibrium throughout the process. This means that the system cannot be brought back to its original state without causing changes in the environment.

4. How does entropy change in reversible gas expansion differ from irreversible gas expansion?

In reversible gas expansion, the entropy change is zero because the system remains in equilibrium throughout the process. However, in irreversible gas expansion, the entropy change is positive because the system does not remain in equilibrium and there is an increase in disorder.

5. Why is reversible gas expansion considered to be more efficient than irreversible gas expansion?

Reversible gas expansion is considered more efficient because it does not result in an increase in entropy, meaning that all the energy used during the process is converted into work. On the other hand, irreversible gas expansion results in an increase in entropy, meaning that some of the energy is lost as heat and cannot be converted into work.

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