1
$\begingroup$

In my textbook there is written

$$\Delta S = \int_R \frac{\delta Q}{T}$$

where the $R$ means calculated along a reversible transformation.

The variation of entropy only depends upon the initial and the final state, and thus has the same value regardless if the transformation was reversible or not.

[...]

We can then found the relation

$$\Delta S = \int_R \frac{\delta Q}{T}$$ $$\Delta S > \int_I \frac{\delta Q}{T}$$

I understood that the RHS in the second equation is just an integral and does not represent the variation of entropy whatsoever.

It is then added that

in a close system $\delta Q = 0$ implies

$$\Delta S \ge 0$$ The equal sign holds for a reversible process, the $>$ for an irreversible one.

Now, I don't understand this. $\Delta S = \int_R \frac{\delta Q}{T} = 0$.

Why the difference between reversible and irreversible ones, while above there is clearly stated that the the value should be the same?

Thanks in advance

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

The book is correctly saying that entropy (S) of the system only depends on the state of system (Pressure, Temperature, Volume; P,V,T). The entropy can change, but if it does there is a change in state (P,V and/or T).

However, if the change in entropy is being expressed as a function of heat, which is not a state fuction, then it matters whether the process is reversible.

If there is a reverible adiabatic process (no heat), than entropy does not change.

Improve this answer
$\endgroup$
4
  • $\begingroup$ yes there is written is a function of heat, but calculated along reversible process, regardless if the actual process is reversible or not. In an irreversible adiabatic process the entropy should be zero, too, because it would be calculated along the reversible one! I know it is not zero, I am trying to understand what's wrong $\endgroup$
    – Ant
    Commented Mar 15, 2014 at 17:53
  • $\begingroup$ If an adiabatic process is not reversible, dissipation must be involved. This is equivalent a heat input dQ. Therefore you end up at a different state with different entropy as if it was reversible. $\endgroup$
    – Mirc Breitschuh
    Commented Mar 15, 2014 at 18:21
  • $\begingroup$ @MircBreitschuh maybe I got it. Let me recap: entropy is calculated along a reversible process, but if the process I'm studying is irreversible, then the initial and final state will be different than if it was reversible, and this explains the difference. $\endgroup$
    – Ant
    Commented Mar 15, 2014 at 19:20
  • 1
    $\begingroup$ Yes in case the process is adiabatic. On the other hand you could have an arbitrary process (eg. involving heating/dissipation and cooling) between two different isentropic states. However you would find an increase in entropy if the system that "provided" the heat transfer is included in the analysis. So you can also have an irreversible process which has a deltaS =0, if your closed system only is considered. $\endgroup$
    – Mirc Breitschuh
    Commented Mar 15, 2014 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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