0
$\begingroup$

let's say we've got a carnot-cycle in a pressure volume diagram with the following processes:

1 -> 2: reversible isothermal

2 -> 3: reversible adiabatic

3 -> 4: reversible isothermal

4 -> 1: irreversible adiabatic

the diagram looks like this:

enter image description here

Now the question: How would the diagram look like if the last process was

4 -> 1: reversible adiabatic?

$\endgroup$
  • $\begingroup$ What if I told you that, in the irreversible compression, the pressure is not spatially uniform within the system? What value of pressure would you plot on your diagram? What if I also told you that, in an irreversible compression, the average pressure of the gas depends not only on the volume but also on the rate of change of volume. How would you take this into account on your diagram. $\endgroup$ – Chet Miller Nov 24 '16 at 1:59
  • 1
    $\begingroup$ As @ChesterMiller has pointed out, you cannot plot an irreversible path on a thermodynamic diagram. Usually such processes are shown schematically by a broken line, which has no meaning besides indicating the fact that an irreversible process occurred between two equilibrium states. So when 4 to 1 process becomes reversible, what happens to your diagram is that a solid line will replace the broken line! $\endgroup$ – Deep Nov 24 '16 at 6:24
  • $\begingroup$ In addition to what @Zero said, I should mention that it is not possible to have and adiabatic reversible path and an adiabatic irreversible path between the same to end states. $\endgroup$ – Chet Miller Nov 24 '16 at 12:41
1
$\begingroup$

The diagram looks exactly how it should for a reversible process $4\to 1$. If the process is irreversible, on the other hand, the smooth solid line $4\to 1$ is deceptive, for it suggests that the system is passing through a sequence of equilibrium states in the process $4\to 1$. This is not the correct way to represent an irreversible process.

To clarify, for sure the system might possibly have at any time a definite pressure and volume in an irreversible process (although I fail to imagine at the present moment an irreversible and adiabatic process in which the $PVT$ system is also in mechanical equilibrium). But since a general irreversible path without any other constraint is quite arbitrary, I find it quite confusing to try to compare it with a definite reversible path like $pV^{\gamma}=\text {cost}$, without any other information.

$\endgroup$
  • $\begingroup$ I disagree. All that the diagram suggests is that there is a continuous sequence of states in the $p$-$v$ space that the system passes through. This is just as possible for irreversible processes as it is for reversible ones. There is no implication of equilibrium states. Note also that the concepts of "equilibrium process" and "isentropic process" are indeed orthogonal. $\endgroup$ – Pirx Nov 23 '16 at 17:57
  • $\begingroup$ Hi @Pirx, this might be just a graphical issue, but the points in the $\text{P-V}$ diagram correspond to states in which the system has some definite pressure and volume, which is generally false (and meaningless) for a nonequilibrium state. Also, why do you say that "the concepts... are orthogonal"? And why do you say that the slope should be lower in the adiabatic case? $\endgroup$ – pppqqq Nov 23 '16 at 18:18
  • $\begingroup$ I realize I am going against (undergraduate) textbook wisdom here, but what I was trying to say is that we can most certainly draw a curve in a, say, $p$-$v$ diagram for any system, equilibrium or not, as long as we define what those $p$ and $v$ values mean. Thus, if we agree to simply show the volume-averaged pressure as a function of the specific volume, then there is no issue. I realize that this is not adhering to standard textbook rules, so that your answer may indeed be the one that was expected from the OP in his/her homework... ;-) $\endgroup$ – Pirx Nov 23 '16 at 18:24
  • $\begingroup$ Continued: As for my comment on "orthogonal concepts": It is possible to have equilibrium processes that are isentropic or non-isentropic, and it is possible to have isentropic non-equilibrium processes. $\endgroup$ – Pirx Nov 23 '16 at 18:26
  • $\begingroup$ I didn't mean to imply any categorical rule for drawing diagrams, however I see your point. However I find it quite strange to ask to compare an adiabatic reversible path with a non-reversible one, which might in principle be quite arbitrary. For example, the fact that you say, that the slope should be lower in the reversible path (which I guess comes from the fact $\text d s>0$) must surely break down at some point, if the irreversible path has to end in the same point as the reversible one $\endgroup$ – pppqqq Nov 23 '16 at 18:47
0
$\begingroup$

The reversible adiabatic process 4-1 would be isentropic, $\mbox{d}s=0$. For an ideal gas such a process is described by a curve given by $p\,v^\gamma=\mbox{const.}$, with $\gamma$ the ratio of specific heats. The curve would look somewhat similar to the one for 2-3, and somewhat "flatter" (lower absolute value of slope) compared to what you have in your diagram now.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.