0
$\begingroup$

As far as I know, for isolated system ∆U=q=work=0. And ∆S=q reversible/T So shouldn't ∆S also be equal to 0. But for an isolated system ∆S is positive. I know that entropy is the degree of randomness and disorderness of the system but here the formula doesn't seem to work.

$\endgroup$

1 Answer 1

1
$\begingroup$

Forget randomness and disorderliness. You have written quite correctly that $\Delta S = \frac{q_{rev}}{T}$ and also noted that in an isolated system $q_{rev}=0$. So when it is an isolated system AND the process is reversible indeed we have $\Delta S=0$ but otherwise in an isolated system when the process is irreversible, that is a spontaneous process, $\Delta S >0$.

$\endgroup$
4
  • $\begingroup$ But ΔS > 0, if some spontaneous reaction occurs in this isolated system. $\endgroup$
    – Maurice
    Commented Jan 24 at 10:01
  • $\begingroup$ Imagine an isolated system that is subdivided into subsystems, each of which is at thermal equilibrium. There is a constraint between the two subsystems that prevents them from interacting. Now you remove the constraint and allow them to re-equilibrate with one another spontaneously. Do you think that this is a reversible process? If not, do you think that the entropy change is zero? $\endgroup$ Commented Jan 24 at 11:47
  • $\begingroup$ @Maurice as I have written above otherwise, and do not forget that $q_{rev}$ denotes the externally supplied thermal energy between two equilibrium states, which by definition is zero for an isolated system irrespective of the process being reversible or irreversible. So when the process is reversible and isolated $\Delta S=0$ but when it is isolated and is not reversible, i.e., irreversible, $\Delta S \ne 0$. Experience and its formulation the 2nd Law says that it is then $\Delta S > 0$ $\endgroup$
    – hyportnex
    Commented Jan 24 at 12:20
  • $\begingroup$ And if you want to avoid any lingering confusion in your mind then always, always, and always think of thermal energy transfer as transport of entropy at some temperature. If you write $\Delta S \ge \Delta S_0$ where $\Delta S_0$ is the transported entropy into the system from the reservoir at temperature $T_0$ then the 2nd law states that the system's entropy change $\Delta S$ is never less than $\Delta S_0$ and $\Delta S- \Delta S_0\ge 0$ is the entropy generated by the process meanwhile the system's internal energy has increased by $T_0\Delta S_0$. Now there cannot be any confusion. $\endgroup$
    – hyportnex
    Commented Jan 24 at 12:27

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.