Entropy change for spontaneous/ irreversible gas expansion

In summary: The temperature would change in the reversible case because the gas is doing work in this case. How did you think the gas could be made to expand slowly if the piston were not being controlled? So to prevent the temperature from changing, we need to place the chamber an isothermal bath.makes sense.thanks.the temperature is controlled by the isothermal bath due to the energy used in controlling the expansion.
  • #1
zanick
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When trying to describe why the entropy goes up for a irreversible process, such as gas expanding into a vacuum, it seems fairly easy at a high level. the valve between the two chambers opens, the free expansion occurs, the pressure drops proportional to the volume change and the temp remains constant because no work was done and the energy of the system didnt change. However, using the equation dS=dQ/T, it doesn't seem to apply , because we know the energy (U) didnt change, and we know the temperature (T) didnt change, but we do know the entropy (S) goes up indicated by that pressure drop. does it have something to do with the (Q) heat per unit volume going down or something like that? or is just as simple as using the wrong equation. if so, it still would be interesting to hear the answer by the experts here.
 
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  • #2
The equation dS=dQ/T describes the transfer of entropy across the boundary of a system, but it does not account for entropy generated within the system (due to viscous dissipation and finite temperature gradients) during an irreversible process. Such entropy generation is not present during a reversible process. So, to determine the entropy change taking place for an irreversible process, you must devise an alternate reversible process (that may not resemble the irreversible process very much), and calculate dQ/T for that alternate process between the same two end states. For more details on how this all plays out, see my Physics Forums Insights article https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/. This article presents a simple cookbook recipe on how to determine the entropy change for a reversible process, including worked examples for some typical cases.
 
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  • #3
Can we review your example 3, and step 3 by the breaking out of the conditions of the free expansion example to a mass less piston chamber that is allows the volume to increase by double to end up at the same final state?
in both cases (original vs step 3), the gas is allowed to expand ( instantaneously in the 1st case) into 2x the original volume. the temp doesn't change because no work is done in the 1st step... but in the 3rd step, the piston is moved slowly as well as the entire chamber placed in a isothermal bath. why is that necessary if the temp will not change in the system in either case?
 

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  • #4
zanick said:
Can we review your example 3, and step 3 by the breaking out of the conditions of the free expansion example to a mass less piston chamber that is allows the volume to increase by double to end up at the same final state?
in both cases (original vs step 3), the gas is allowed to expand ( instantaneously in the 1st case) into 2x the original volume. the temp doesn't change because no work is done in the 1st step... but in the 3rd step, the piston is moved slowly as well as the entire chamber placed in a isothermal bath. why is that necessary if the temp will not change in the system in either case?
The temperature would change in the reversible case because the gas is doing work in this case. How did you think the gas could be made to expand slowly if the piston were not being controlled? So to prevent the temperature from changing, we need to place the chamber an isothermal bath.
 
  • #5
makes sense. thanks . the temperature is controlled by the isothermal bath due to the energy used in controlling the expansion. makes sense.
 

1. What is entropy change for spontaneous/ irreversible gas expansion?

Entropy change for spontaneous/ irreversible gas expansion is a measure of the disorder or randomness of a system as it undergoes a spontaneous expansion. It is a thermodynamic property that describes the distribution of energy and particles within a system.

2. How is entropy change related to gas expansion?

Entropy change is directly related to gas expansion because as a gas expands, its particles become more randomly distributed, leading to an increase in disorder and entropy. This is because there are more ways for the particles to be arranged in a larger volume, compared to a smaller volume where they are more confined.

3. What factors affect entropy change for gas expansion?

The main factors that affect entropy change for gas expansion are temperature, volume, and number of particles. As temperature increases, the particles have more energy and can move around more freely, increasing the disorder of the system. An increase in volume also leads to an increase in entropy, as discussed in the previous question. Lastly, the number of particles also plays a role, as more particles means more possible arrangements and therefore a higher entropy.

4. Is entropy change always positive for gas expansion?

No, entropy change is not always positive for gas expansion. It depends on the conditions of the system. If the gas expansion is reversible, meaning it can be reversed without any energy loss, then the entropy change will be zero. However, if the expansion is irreversible, then the entropy change will be positive.

5. How is the second law of thermodynamics related to entropy change for gas expansion?

The second law of thermodynamics states that the total entropy of a closed system can never decrease over time. This means that for a spontaneous gas expansion, the total entropy of the system will always increase. Therefore, the second law of thermodynamics is directly related to entropy change for gas expansion, as it explains why the entropy of a system increases during this process.

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