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Is it always true, for any cycle followed by a gas, that, if the cycle is "travelled" clockwise in the $P-V$ plane then the work exchanged by the gas is positive, and viceversa for the clockwise direction?

I came up with this situation, which I represented in the picture. The irreversible cycle is made of an isothermal compression, and two adiabatics. One of the adiabatics must be irreversible for the cycle to exists. enter image description here The cycle seems possible to me because since $$\Delta S_{universe}=\Delta S_{environment}=\Delta S_{environment, A->B}>0$$

This cycle is clockwise, but $$Q_{cycle}=W_{cycle}=Q_{A->B}<0$$

So this appears in contrast with the rule $$\mathrm{clockwise} \implies W>0$$ $$\mathrm{anticlockwise} \implies W<0$$

Is that possible or am I missing something?

Sign convention:

enter image description here

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    $\begingroup$ You question is equivalent to asking if the area under the left-going curves is necessarily less that the area under the right-going curves when the right-going curves are greater than or equal to the left-going ones over the whole considered domain. Right? So, what's the answer? $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jul 10, 2016 at 22:33
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    $\begingroup$ You shouldn't draw an irreversible cycle as a continuous line. I'm also not sure what sign convention you are using. Those have changed since my time at uni, but I thought nowadays, W and Q are positive when they go into the system (when internal energy increases). For a cycle, internal energy stays the same, so W = -Q $\endgroup$
    – Previous
    Commented Jul 10, 2016 at 23:42

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You have to be careful when drawing irreversible transformations on the PV plane. Apart from the fact that drawing them is meaningless because they are not a set of equilibrium points, you cannot just assign some arbitrary property to them.

You drew the blue curve and called it an "irreversible adiabatic", giving it some arbitrary slope. Fine, since we don't know what the "slope" of an irreversible adiabatic should be. But what if I drew another "irreversible adiabatic" connecting A and B?

We would then have

$$\Delta U = 0 \to Q = W$$

But $Q=0$ since every process is adiabatic, so that we would obtain

$$W=0$$

which is obviously wrong.

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  • $\begingroup$ Thanks for the reply and the example you made! I'm still a bit confused on what is wrong in what I asked. Is it the fact that irreversible processes cannot be represented, and so in a case like the one I proposed I cannot talk about "clockwise" or "anticlockwise" direction? (Although it seems to me that I can define a direction of travelling the cycle even if only one of the curves is not represented, for istance in my example - without B->C - the cycle would still be anticlockwise with no ambiguity) Or is it the fact that this "rule" ("clockwise" or "anticlockwise") does not hold always? $\endgroup$
    – Sørën
    Commented Jul 11, 2016 at 7:41
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    $\begingroup$ @valerio92 other than convention, the choice of direction is arbitrary. What matters is how you define directionality relative to the direction of energy flow, and maintaining consistency in your definition $\endgroup$
    – docscience
    Commented Jul 11, 2016 at 11:02
  • $\begingroup$ @Sørën The problem is that I suspect that the cycle you described is not possible, even if I haven't been able to come up with an explanation of why it should be impossible. Maybe we should run an experiment and see if it is possible to close the CAB cycle with an irreversible adiabatic process...My message anyway is that you have to be careful when using the PV plane: irreversible processes cannot be represented there, because curves in the PV plane are by definition made of equilibrium states. $\endgroup$
    – valerio
    Commented Jul 11, 2016 at 14:39
  • $\begingroup$ Note that it is perfectly fine to draw an irreversible process in the system's $p$-$V$ plane as long as the system undergoes only quasi-static (or quasi-equilibrium) processes, in which case the irreversibility is usually a consequence of heat transfer between system and environment while they are at different temperatures. On the other hand, I'm pretty sure that if the process is irreversible and adiabatic, it must also be non-quasi-static, which means that that irreversible adiabatic process should not be drawn in the $p$-$V$ plane. $\endgroup$
    – march
    Commented Jul 11, 2016 at 16:51
  • $\begingroup$ @march You are right, for some reason I had a non quasi-static process in mind when I wrote the post. Nevertheless, as you remarked, there are difficulties when drawing an irreversible adiabatic process in the PV plane. $\endgroup$
    – valerio
    Commented Jul 11, 2016 at 17:46
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The area enclosed within the loop is the net work and for the complete cycle you've illustrated, work (energy) flows both into and out of the system at different points within the cycle. But there is a net loss.

From $A$ to $B$ and $B$ to $C$, the direction is positive (by right hand convention) and so energy flows into the system. From $C$ to $A$ the direction is negative, and so energy flows out of the system.

Since integration along $V$ is negative along the direct path from $C$ to $A$ and greater in magnitude than the integration under the path from $A$ to $B$ then $B$ to $C$

(which is positive), then the net energy flow and work is negative.

But this can be directly determined by just integrating the area within the path and observing the path moves in a clockwise direction.

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  • $\begingroup$ May I know what is the right hand convention being mentioned in the answer? @docscience $\endgroup$
    – S Roy
    Commented Jan 16, 2023 at 18:21

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