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Jun 29, 2022 at 19:43 vote accept Boson
Jun 29, 2022 at 17:06 answer added Bob D timeline score: 1
Jun 29, 2022 at 7:19 comment added Miyase I know this sort of graph, but it's technically wrong unless there's a quasi-static assumption. Without this assumption, pressure $P$ is undefined during the process, so the graph is meaningless. With this assumption, $P$ is defined. Almost automatically, you can add a mechanic equilibrium assumption, so $P=P_\text{ext}$. If your graph is in this case, then yes $P$ is constant, so $P_\text{ext}$ is constant too. But it's completely unrelated to the fact that it's irreversible.
Jun 29, 2022 at 7:01 comment added Boson Also, what would be the graphs for irreversible isochoric, isothermal and adiabatic? Would they differ from reversible graphs?Could you write an answer based on this?
Jun 29, 2022 at 7:00 comment added Boson @Miyase Ok. so is the above graph only for isobaric irreversible process?
Jun 29, 2022 at 6:55 comment added Miyase @Boson This is incorrect. That a process is irreversible says nothing about $P_\text{ext}$, it only tells you than the created entropy term is strictly positive. There's no relation between the two.
Jun 29, 2022 at 2:49 history edited Boson CC BY-SA 4.0
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Jun 29, 2022 at 2:25 comment added Boson @BobD Because Pext is constant in all irreversible processes(afaik) which is the condition to be an isobaric process.
Jun 28, 2022 at 21:29 comment added Bob D "Why aren't all irreversible processes isobaric?" Why would you think that all irreversible processes are isobaric? I intend to to post an answer, but first I need to know why you would think that.
Jun 28, 2022 at 17:15 history edited Qmechanic
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Jun 28, 2022 at 17:13 history edited Boson CC BY-SA 4.0
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Jun 28, 2022 at 17:13 history edited Boson CC BY-SA 4.0
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S Jun 28, 2022 at 17:11 review First questions
Jun 28, 2022 at 19:16
S Jun 28, 2022 at 17:11 history asked Boson CC BY-SA 4.0