Irreversible adiabatic process - is the entropy change zero?

In summary, the question asks for the entropy change per mole when a partition separating an ideal gas and a vacuum in a well-insulated container is removed and the gas expands. The entropy change is calculated using the equation dS = dQrev/T, and the result is expected to be non-zero due to the irreversible nature of the process. However, after performing the calculation, the final answer is zero. To resolve this confusion, it is necessary to forget about the actual irreversible path and devise a reversible path, such as an isothermal expansion, in order to calculate the integral of dQrev/T and obtain the correct value for the entropy change.
  • #1
Ortanul
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Homework Statement


A well insulated container consists of two equal volumes separated by a partition - one half an ideal gas while the other is a vacuum. The partition is removed, and the gas expands. What is the entropy change per mole?

Homework Equations


dS = dQrev/T
S/R = Cp/R dT/T - ln(P/P0)

The Attempt at a Solution


As the process is irreversible, the entropy change should be non-zero even though there is no change in heat. However, after I perform the calculation with the aforementioned equation, I got zero as my final answer.
1.pic.jpg

I'm extremely confused about my result because I can't find any mistake in my calculation, while I do believe that the change should be non-zero.
Any help would be highly appreciated!
 
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  • #2
Ortanul said:

Homework Statement


A well insulated container consists of two equal volumes separated by a partition - one half an ideal gas while the other is a vacuum. The partition is removed, and the gas expands. What is the entropy change per mole?

Homework Equations


dS = dQrev/T
S/R = Cp/R dT/T - ln(P/P0)

The Attempt at a Solution


As the process is irreversible, the entropy change should be non-zero even though there is no change in heat. However, after I perform the calculation with the aforementioned equation, I got zero as my final answer.
View attachment 96098
I'm extremely confused about my result because I can't find any mistake in my calculation, while I do believe that the change should be non-zero.
Any help would be highly appreciated!
For the irreversible process you are analyzing, you should have found that the change in temperature is zero, and the final pressure is half the original pressure. When you calculate the entropy change between the initial and final states, you then follow the following procedure:

1. Completely forget about the actual irreversible path between the initial and final states.
2. Devise a reversible path between these same two ends states. This reversible path does not need to bear any resemblance whatsoever to the actual process path. So, in your problem, for example, it doesn't have to be adiabatic. Any reversible path will do, since they will all give exactly the same result for the entropy change. For your problem, a reversible isothermal expansion might be a good choice.
3. Calculate the integral of dQ/T for the reversible path that you have identified. This will be ##\Delta S##

This is what they mean when they say ##\Delta S## is equal to the integral of dQrev/T.

Chet
 
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1. What is an irreversible adiabatic process?

An irreversible adiabatic process is a type of thermodynamic process in which no heat is exchanged between the system and its surroundings, and the process is not reversible, meaning it cannot be reversed to its original state without causing an increase in entropy.

2. How does an irreversible adiabatic process differ from a reversible adiabatic process?

In a reversible adiabatic process, the system is able to return to its original state without any change in entropy, while in an irreversible adiabatic process, the system cannot be returned to its original state without an increase in entropy.

3. Is the entropy change always zero in an irreversible adiabatic process?

No, the entropy change is only zero in a reversible adiabatic process. In an irreversible adiabatic process, there is always an increase in entropy.

4. What is the significance of the entropy change being zero in a reversible adiabatic process?

The fact that the entropy change is zero in a reversible adiabatic process means that the process is thermodynamically efficient, as no energy is lost to the surroundings. This is often used as a benchmark for the efficiency of real-world processes.

5. Can an irreversible adiabatic process ever be reversed?

No, an irreversible adiabatic process cannot be reversed without causing an increase in entropy. This is due to the second law of thermodynamics, which states that the total entropy of a closed system can never decrease over time.

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