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My Textbook says:

Two states A and B can never lie both on a reversible as will as irreversible adiabatic path. There lies only one unique adiabatic path linkage between two states A and B.

I don't understand what the book means. Why cannot there be both reversible and irreversible adiabatic paths between two states? And if only one path is possible, then how do we know which one?

What I understood is: In adiabatic process: ΔU=W. Now since ΔU is state function, it will mean that Wirreversible=Wreversible, which is not possible. But mathematical proof aside, is there any example/physical explanation to this?

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    $\begingroup$ Wrev <> Wirr. But the question elaboration is better placed to the question. More space and editing preview available. Comments are for a short feedback, not for supplementing additional question elaboration. $\endgroup$
    – Poutnik
    Commented Apr 29, 2021 at 7:48

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For an infinitesimal adiabatic process $dS_\mathrm{surroundings} = 0$ since no heat is exchanged with the surroundings. But $dS_\mathrm{system} = 0$ is only true for a reversible process, since only then $dS_\mathrm{system} = dq/T$. For an irreversible process $dS_\mathrm{system} \neq dq/T =0$. Therefore

$$dS_{\mathrm{irrev}}\neq dS_{\mathrm{rev}}=0$$

This means that the state accessed with a reversible process through an integral change $\int dS_{\mathrm{rev}}=0$ cannot be accessed directly through an irreversible adiabatic process, for which $\Delta S = \int dS_{\mathrm{irrev}} \neq 0$.

Now since S is a state function, the terminal states of the two processes must therefore differ (QED).

As regards the second part (how do we know which is the reversible), you can use the approach just outlined to conclude that for the reversible process $\Delta S =0$.

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  • $\begingroup$ so can we say that irreversible adiabatic process is not directly possible between any two given states? $\endgroup$ Commented Apr 29, 2021 at 11:36
  • $\begingroup$ If you have an irreversible adiabatic process between two end states (i.e., thermodynamic equilibrium states), there is no reversible adiabatic process between these same two end states. $\endgroup$ Commented Apr 29, 2021 at 11:42
  • $\begingroup$ and vice-versa? $\endgroup$ Commented Apr 29, 2021 at 11:56
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    $\begingroup$ Yes, of course. $\endgroup$ Commented Apr 29, 2021 at 13:20
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Suppose that in a given transformation, ${\Delta U}$ is defined, and cannot be changed. But the same ${\Delta U}$ is the sum of two terms $Q$ and $W$ that can be changed and modified arbitrarily. If one of these terms, here the heat $Q$, is chosen to be zero, the other term $W$ cannot vary and stays equal to $W_{rev}$.

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  • $\begingroup$ Sure but why only $Wrev$. why not $Wirreversible$? $\endgroup$ Commented Apr 29, 2021 at 11:27
  • $\begingroup$ It is up to you. It can be reversible work or irreversible work. It does not matter here. The only important thing is that this work is equal to the change of internal energy. $\endgroup$
    – Maurice
    Commented Apr 29, 2021 at 12:54
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Basically, its simple. From first laww of thermodynamics we know that, dU=q+W for an adiabatic process, heat exchange, i.e. q=0 hence change in internal energy is entirely dependent on the amount ofnwork done i.e. ,for an adiabatic process, dU=W
and work done for an irreversible process is different than work done in reversible process i.e., (WD)rev =/= (WD)irrev (since work done is a path function)

but then again dU is a state function and for an adiabatic process entirely dependent on state hence this would imply that to get to the same state T2,V2 from say T1,V1 while following a reversible and an irreversible path dU be different paths be different. That means to achieve the sane final state starting from the same initial sate change in internal energy for 2 different paths be different. Which is not possible for a sate function such as dU. hence this implies that the final state achieved by a gas while undergoing a reversible and irreversible adiabatic process would be different because if not then the change in internal energy would be dependent on path which is simply impossible due to it being a state function

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    $\begingroup$ Do not use unnecessary formatting. It becomes unreadable. Keep your formatting consistent. $\endgroup$ Commented yesterday
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    $\begingroup$ Guides for formatting of chemical/mathematical formulas/expressions/equations: Basics / Detailed / Upright vs Italics / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. $\endgroup$
    – Poutnik
    Commented yesterday
  • $\begingroup$ Reversible and irreversible adiabatic processes simply do not reach the same final state with the same initial one. Unless combined with non adiabatic processes that compensate difference in work by exchanging heat. $\endgroup$
    – Poutnik
    Commented yesterday

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