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I'm quite confused about isochoric and isobaric irreversible processes, and, in particular, the doubt is about work W in such processes.

If the processes is reversible (that is I can write $d W= p dV$) I have no problem in saying that

  • In a isochoric process $W=0$, since $dV=0$.
  • In a isobaric process $W=p(V_b-V_a)$ since $\int p dV=p \int dV$.

In other words, in both cases I can be sure that work is a state function (and, in particular, in isochoric is zero).

The problem is: I find in different textbooks that the previous expression for $W$ are used also in the cases of irreversible processes.

But in my view this is not correct because, to reach those conclusions about work it is fundamental to use $d W= p dV$, which holds true iff the process is reversible.

So are the previous relations correctly or incorrectly used in the case of irreveresibility?

If so, can you suggest me any book/source that talks about canonical processes (isobaric, isochoric in particular) without saying that for any isochoric and isobaric process (irreversible or not) the previous ones are the expressions of work?

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  • $\begingroup$ Work is not a state function. For example, if I give you $(p, V, T)$ can you tell me the work? Think about the distinction. Keep in mind that work (and heat) are quantities of energy that enter or leave the system during a process. $\endgroup$
    – garyp
    Commented Aug 3, 2016 at 19:04
  • $\begingroup$ I'm wondering why do you have problem when the process is irreversible. Is there any reason? $\endgroup$
    – user115350
    Commented Aug 3, 2016 at 22:47

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An isochoric process is by definition when $dV=0$ irrespective of being reversible or irreversible. Whether $\delta W = pdV$ depends on what you mean by $p$. If by $p$ you mean the external pressure then it is true irrespective of the nature of the process, and then you can say that $\delta W$ is the external work, that is the work done by external forces on the system that has suffered volume change $dV$. If by $p$ you mean the internal pressure, even if it is homogeneous across the system, then the equation holds when the process is reversible, and does not hold if it is irreversible.

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  • $\begingroup$ Well said!!!!!! $\endgroup$
    – Chet Miller
    Commented Aug 3, 2016 at 23:36
  • $\begingroup$ @hyportnex But only if non-PV work is zero, right? Otherwise $\delta W = p_{ext}dV + \delta W_{nonPV} \ne p_{ext} dV$ $\endgroup$
    – ManRow
    Commented Feb 19 at 6:18
  • $\begingroup$ @ManRow In general, let $p_0$ denote the externally imposed pressure and let $p$ be the internal system pressure. Then $dU=TdS-pdV=\delta q_0 - p_0 dV$, denoting the external work by $p_0dV$. When the process is irreversible then you have both $TdS- \delta q_0 = pdV -p_0dV > 0$, therefore, $p > p_0$ for $dV>0$ (an expansion work) and $p<p_0$ for $dV<0$ (a compression work). If $dV=0$ then all you have is $TdS>\delta q_0$. For some more details read this physics.stackexchange.com/questions/801867/… $\endgroup$
    – hyportnex
    Commented Feb 19 at 10:41

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