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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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  • $\begingroup$ Think longer, and make some calculations. Then come back. $\endgroup$
    – Poutnik
    Apr 29, 2021 at 7:32
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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
    Apr 29, 2021 at 7:48

2 Answers 2

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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? $\endgroup$
    – Vega
    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. $\endgroup$ Apr 29, 2021 at 11:42
  • $\begingroup$ and vice-versa? $\endgroup$
    – Vega
    Apr 29, 2021 at 11:56
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    $\begingroup$ Yes, of course. $\endgroup$ 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 $Wirreversible$? $\endgroup$
    – Vega
    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
    Apr 29, 2021 at 12:54

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