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The question is simple:

In thermodynamics, how do you determine whether a process in reversible or not ?

For example, consider an a balloon containing some $He$, which is expanding in a constant pressure while changing its temperature, how can one determine whether the process that the system (balloon + He) undergoes a reversible or an irreversible process ?

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  • $\begingroup$ Possible duplicate, definitely related: physics.stackexchange.com/questions/243766/… $\endgroup$
    – jacob1729
    Commented May 30, 2019 at 16:28
  • $\begingroup$ Reversibility is an ideality but it's useful in practice, and often a decent approximation, if the entropy generation is small compared to entropy transfers/flows in and out of the system. $\endgroup$
    – Thermodynamix
    Commented May 30, 2019 at 16:53
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    $\begingroup$ All heat transfer is done through a zero temperature difference. There is no friction. Any heat transferred in one direction could just as easily be transferred in the reverse direction. Entropy changes are zero. Obviously, such a process does not exist in the real world. $\endgroup$
    – David White
    Commented May 30, 2019 at 16:57
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    $\begingroup$ You apply the first law of thermodynamics to determine the final states of the system and surroundings. Then, you calculate the change in entropy of the system and the change in entropy of the surroundings. If the sum of the changes in entropy of system and surroundings is zero, the process is reversible. $\endgroup$
    – Chet Miller
    Commented May 30, 2019 at 22:05

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For the expansion of your balloon to be reversible it has to be carried out quasistatically (very slowly) and without friction so that the pressure of the gas is always in equilibrium with the external pressure and the temperature of the gas in equilibrium with the surroundings . That means the heat transfer from the surroundings to the gas has to occur with an infinitely small temperature difference at each equilibrium state between the initial and final states.

Since helium can be considered an ideal gas, this means the ratio of the gas volume to its temperature must be constant at during the entire process, or $\frac{V}{T}$ = Constant.

Hope this helps.

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Well, I can't get into much technicality. The book from where I used to study thermodynamics and solved problems simply stated that

It is easy to identify irreversible process in a question. Wherever constant external pressure is present, the process is irreversible.

Hope that helps.

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    $\begingroup$ That’s not correct. You can have a reversible isobaric process, as long as the gas pressure is always in equilibrium with the external pressure. It requires the process be carried out quasistatically (very slowly) $\endgroup$
    – Bob D
    Commented May 30, 2019 at 17:31
  • $\begingroup$ That's what my concern was. Even i didnt agree with it. But i stated what was written in that paragraph... $\endgroup$
    – Aman Kumar
    Commented May 31, 2019 at 17:50
  • $\begingroup$ Do you remember the title and author of the book? $\endgroup$
    – Bob D
    Commented May 31, 2019 at 17:54
  • $\begingroup$ @BobD dear sir, it was not some specific book. It was in a coaching class booklet which i used to attend for college entrances... $\endgroup$
    – Aman Kumar
    Commented Jun 2, 2019 at 10:44
  • $\begingroup$ OK I thought perhaps if I could see the statement in context it might make sense, $\endgroup$
    – Bob D
    Commented Jun 2, 2019 at 19:38

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