How can we say that a system is reversible or irreversible if not stated?

I encountered a book example wherein a piston cylinder device encounters an isobaric process and experiences a transfer of heat to the system. And the change of entropy is to be determined.

In this problem, I first thought that the the entropy balance would be Q/T +(Sgenerated) = Delta S. In other words, the total change in entropy between the system boundary and the entropy generated by the system is equal to the change in entropy. But I was wrong, in the book it stated that the entropy generated within the system will not be accounted for the equation since no irreversibilities occured in the system boundary during the process.

This is where I am confused, how can we know if irreversibilities occured within the system boundary? Is it because the system underwent an isobaric process or is it a different factor? While it was not stated that no irreversibilities occured making the process reversible.

And how would we know if the system undergoes an irreversible or irreversible process if not stated? Is it by intuition itself?

I know that the entropy generated by the system would be zero if it undergoes a reversible process. I just find it hard to know when it would occur.

  • $\begingroup$ Irreversibility means that energy is moving down a gradient somewhere (e.g., a temperature gradient), which generates entropy. If your system is being heated by a reservoir at the same temperature as the system, for example, then entropy is transferred, but no entropy is generated. Note that this process is an idealized one that can't occur in reality. $\endgroup$ Commented Oct 2, 2017 at 18:33
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    $\begingroup$ Maybe we could help you better if you provided the exact statement of the problem. Any chance? I agree with you that if the process is reversible, then there is no entropy generated •within• the system. I think this would have been better wording than "by the system." $\endgroup$ Commented Oct 3, 2017 at 1:22
  • $\begingroup$ See the ADDENDUM to my answer. $\endgroup$ Commented Oct 3, 2017 at 13:17

2 Answers 2


If you apply a constant external pressure to the gas, equal to the initial gas pressure (throughout the isobaric process), while the gas is being heated by a fixed amount of heat Q, the final state of the gas will be fully determined, since, in this case, $\Delta H=nC_p\Delta T=Q$. I'm guessing that you were able to determine the final temperature and volume of the gas in the final state. With this being the case, then it doesn't matter whether the actual process was reversible or irreversible. The entropy change will be the same for both, since entropy is a function only of state, and the final state is established.

So, how do you determine the change in entropy for a system that has experienced an irreversible process? Well, what you do first is completely disregard the actual irreversible process path that gave rise to the change. Instead, you focus exclusively on the initial and final states of the system, and devise a reversible process path to take the system from the initial state to the final state. The entropy change for this reversible path will also be the entropy change for the actual irreversible path.


So, for the process you descried, $$\Delta S=nC_p\ln{(T_f/T_i)}\tag{1}$$
But, how do you know whether the actual process is reversible or not? A mathematical way of determining this is to apply the Clausius inequality: For an irreversible process on a closed system, $$\Delta S>\int{\frac{dQ}{T_B}}\tag{2}$$ where $T_B$ is the temperature at the boundary through which the heat dq flows. Usually this boundary is taken to be an ideal reservoir (or reservoirs). It is a little-emphasized fact in thermodynamics textbooks that in applying the Clausius inequality, one must, in the integration, use the boundary temperature through which the heat dq flows. If the process is reversible, relationship (2) will turn out to be an equality, but if the process is irreversible, the inequality in relationship (2) will apply.


It is not a correct statement that "the entropy generated by the system would be zero if it undergoes a reversible process". As you already pointed out, if the system undergoes a reversible process, the following equality is valid. You can see, entropy can change if there is heat flow.
$$\triangle S = \frac {\triangle Q}T$$ For your problem, the process is isobaric and there is heat transfer. The temperature in the system will vary. Thus there will be temperature difference between the system and its surrounding temperature. This will direct the heat flow in one direction (from high temperature to low temperature). Thus the process is not reversible. For this process, the following inequality will be valid. $$\triangle S > \frac {\triangle Q}T$$


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