Entropy change for spontaneous/ irreversible gas expansion

[zanick]

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.

 

[Chestermiller]

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.

 

[zanick]

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?

 

[Chestermiller]

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.

 

[zanick]

makes sense. thanks . the temperature is controlled by the isothermal bath due to the energy used in controlling the expansion. makes sense.