Timeline for Is there a quasistatic process that is not reversible?
Current License: CC BY-SA 3.0
13 events
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| Mar 19, 2020 at 6:51 | comment | added | Nicolas | @march good point ! I think your example is not quasistatic even in the limit of infinitely small $dV$ for each step though. The reason is that during each step, the pressure varies a finite amount (between $P$ and $0$ and back again to $P-dP$) for at least part of the system. | |
| Jul 4, 2017 at 23:04 | comment | added | march | @joshphysics. (cont.) In this new case, entropy should also increase, but it seems like there should be a corresponding decrease in entropy elsewhere, because it seems like this should be reversible, but where would the entropy decrease be? I'm not sure. | |
| Jul 4, 2017 at 23:04 | comment | added | march | @joshphysics. I will have to think about this. In the case of the "free" expansion of the gas into a sequence of infinitesimal volumes, at first I was inclined to say that that was quasi-static, exactly because the system would move through a sequence of equilibrium states. But if I imagine how to accomplish this: you remove a sequence of infinitely thin partitions, in which case the system does no work, and so the final state must be the same as in the true free expansion case. It feels like there should be a contradiction there: in the free case, entropy increases. (cont.) | |
| Jul 1, 2017 at 16:24 | comment | added | joshphysics | (contd.) some reasonably smooth curve in the thermodynamic state space of the system, and this curve can be used to correctly compute any thermodynamic quantity one chooses for the process by integrating an appropriate differential form, e.g. heat or work, along it?" | |
| Jul 1, 2017 at 16:23 | comment | added | joshphysics | I'm concerned that your definition of quasistatic might be misinterpreted the way it's stated. Consider, for example, the adiabatic free expansion of an ideal gas into a sequence of extremely small extra volumes. If we let the gas come to equilibrium before allowing it to expand into the next small compartment, we might be tempted to call this process quasistatic based on your definition, but I personally think that would violate the spirit of the term. Shouldn't one add something like "as the process becomes slower/more incremental, the successive equilibrium states become more dense on | |
| Jun 22, 2017 at 5:07 | comment | added | joshphysics | This is an interesting description as I think it gets at a the heart of an issue brought up in the comments to an answer of mine here: physics.stackexchange.com/a/78443/19976 | |
| Dec 21, 2016 at 14:42 | vote | accept | Dimitri | ||
| Dec 21, 2016 at 14:42 | history | bounty ended | Dimitri | ||
| Dec 21, 2016 at 14:41 | comment | added | Dimitri | I think I get your point, thanks for the clarification. | |
| Dec 14, 2016 at 16:56 | comment | added | march | the combined system is not undergoing a quasi-static process, because it's not in thermal equilibrium, as you said. This is exactly what I meant by the difference between the two terms: you get to choose the system you're talking about when you use the term quasi-static, but you don't when you use the term reversible. | |
| Dec 14, 2016 at 16:56 | comment | added | march | @Dimitri. I don't think I agree with your assessment. Imagine an idealized situation where the two systems in contact have very large specific heats and large thermal conductivities so that the temperature necessarily changes very slowly, allowing each system to equilibrate very quickly when it gains a little bit of thermal energy. This is really what we mean by quasi-static, that the system's behavior approaches that of being in equilibrium at all times. In this way, each system is undergoing a quasi-static process. However, since the two systems are at different temperatures, ... | |
| Dec 13, 2016 at 10:17 | comment | added | Dimitri | Thanks for the clarification between single and multiple systems. However I disagree with your example, if there is a finite temperature difference then the process is not quasistatic because there is no thermal equilibrium at all times. | |
| Dec 9, 2016 at 1:01 | history | answered | march | CC BY-SA 3.0 |