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If two states $A$ and $ B$ are connected by a reversible path, they can never be connected by an irreversible path during an adiabatic process because:

$\underbrace{\Delta U}_{\text{state function}} = \underbrace{\Delta W}_{\text{path function}}$

And $W_{\text{reversible}}\neq W_{\text{irreversible}}$

Now, Atkins' Physical Chemistry (Section $3.2$) states that:

To calculate the difference in entropy between any two states of a system, we find a reversible path between them, and integrate the energy supplied as heat at each stage of the path divided by the temperature at which heating occurs.

But the problem is that we cannot define a reversible path for an adiabatic irreversible process. If that's the case, how can we calculate the entropy change for an adiabatic irreversible process?

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  • $\begingroup$ This doesn't answer my question because it doesn't discuss this issue. $\endgroup$ – Archer Mar 10 '18 at 13:01
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I agree with you over here.

$\underbrace{\Delta U}_{\text{state function}} = \underbrace{\Delta W}_{\text{path function}}$

And $W_{\text{reversible}}\neq W_{\text{irreversible}}$

But this part seems to be wrong.

If two states $A$ and $ B$ are connected by a reversible path, they can never be connected by an irreversible path during an adiabatic

Just because the process is adiabatic irreversible it doesn't mean that the reversible path needs to be adiabatic as well.

The only requisite for calculating the entropy for an irreversible process is simply that the path taken must be reversible.

So until now we have established that the heat transfer does not have to be 0. For an adiabatic process.

Even though $W_{\text{irrev}} \neq W_{\text{rev}}$ the extra parts will be balanced by the heat transfer in the reversible process.

For example let's take adiabatic expansion.

$$W_{\text{irrev}} > W_{\text{rev}}$$

Numerically speaking in an absolutely hypothetical situation.

\begin{align} \ce{ ∆U &-> 10~J}\\ \ce{ W_{irrev} &-> 10 ~J}\\ \ce{ W_{rev} &-> 5~J}\\ \end{align}

So the possible proposed reversible path will automatically have a transfer of heat into the gas . Whose value will be 5J.
.

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    $\begingroup$ (+1) I agree with this answer. Entropy is a state function. It doesn't matter whether you take an isobaric/adiabatic/blah path from state A to state B. As long as it's between the two same states, the change in entropy will be the same. But, do note that it's only for calculating the change in the system's entropy, i.e., not the surroundings. $\endgroup$ – Gaurang Tandon Mar 10 '18 at 14:57
  • $\begingroup$ (-1) half of your answer is the question itself. Wrt the last two sentences, can you please elaborate? $\endgroup$ – Archer Mar 10 '18 at 14:58
  • $\begingroup$ What part is unclear I'll work on that. $\endgroup$ – Avnish Kabaj Mar 10 '18 at 14:59
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    $\begingroup$ (My vote got locked) but I now realize you didn't answer "they can never be connected by an irreversible path during an adiabatic process" @AvnishKabaj $\endgroup$ – Gaurang Tandon Mar 10 '18 at 15:02
  • $\begingroup$ @Gaurang edited it . Feel free to retract your upvote or downvote it. $\endgroup$ – Avnish Kabaj Mar 10 '18 at 15:16
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The reversible path between the two end states does not have to be adiabatic even though it is adiabatic for the actual adiabatic irreversible path. For the reversible path, we are allowed to add or remove heat from the system reversibly, as long as we start off at the same initial state and arrive at the same final state. And the reversible process does not have to be done using a one-step path. Part of the path can be adiabatic and part of it can be isothermal, for example. Other reversible combinations can also be used. For more details on a cookbook recipe for determining the entropy change for an irreversible process, see this reference: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

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  • $\begingroup$ "the reversible process does not have to be done using a one-step path" don't all reversible processes always have infinite steps? $\endgroup$ – Gaurang Tandon Mar 10 '18 at 15:41
  • $\begingroup$ In Step 1 of Physics Formums insights, it is written: "$\Delta U = nC_v(T_f- T_0)= P_f(V_f-V_o)$" Can you please explain how this is true? $\endgroup$ – Archer Mar 10 '18 at 15:42
  • $\begingroup$ @GaurangTandon By a one-step path, I meant a solely adiabatic path of infinitesimal increments or a solely isothermal path of infinitesimal increments. My point was that the overall reversible process can consist partly of an adiabatic path and partly of an isothermal path. $\endgroup$ – Chet Miller Mar 10 '18 at 15:49
  • $\begingroup$ Which part do you think is not true? (Incidentally, you're missing a minus sign on the right hand side of the equation). $\endgroup$ – Chet Miller Mar 10 '18 at 15:51
  • $\begingroup$ @ChesterMiller How did you obtain it? $nC_vT \ne -P_fV_f$ $\endgroup$ – Archer Mar 10 '18 at 15:53

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