0
$\begingroup$

In Reif's Fundamentals of Statistical and Thermal Physics he outlines a "proof" (sections 3.8 and 3.9) of the equation $dS=\frac{\delta Q}{T}$ for any quasi-static, infinitesimal process (i.e. any changes of the parameters are infinitesimal). He defines a reversible process as a process where $\Omega_f=\Omega_i$, i.e. the number of states accessible to the system doesn't change. But when I look up the formula for $dS$, I see it given as $dS=\frac{\delta Q_{rev}}{T}$, which is different than what Reif has.

Im confused because by Reifs definition, any reversal process must have $dQ=0$ or else $dS>0$ (i.e. $\Omega_f \neq \Omega_i$), so the definition of $dS=\frac{\delta Q_{rev}}{T}$ doesn't really make sense. I know I'm not being crazy because Reif says "the performance of quasi static work changes the energy of a thermally isolated system but does not affect the number of states accessible to it... such a process is thus reversible" (page 116). His discussion seems to imply that reversible processes are a subset of processes in which $dS=\frac{\delta Q}{T}$ holds, in contrast with the definition $dS=\frac{\delta Q_{rev}}{T}$. Is Reif wrong? Is he just using different notation? What's the discrepancy?

Improve this question
$\endgroup$
6
  • $\begingroup$ A quasi static infinitesimal process is a reversible one. $\endgroup$
    – Matt Hanson
    Commented Jul 3 at 1:49
  • $\begingroup$ @MattHanson so, from Reif's definition of reversible, this would mean that the formula $dS=\frac{\delta Q}{T}$ only applies when $\delta Q=0$... i.e. $dS=0$. Do you mean that reversible=quasi static+infinitesmal is the standard definition of reversible, and Reif's is not? $\endgroup$
    – user62783
    Commented Jul 3 at 1:54
  • 1
    $\begingroup$ In reversible heat transfer, $\delta Q\neq 0$, and $\Delta\Omega$ for the system and heat source/sink are equal and opposite; thus, the total $\Omega_\mathrm{final}=\Omega_\mathrm{initial}$. $\endgroup$
    – Chemomechanics
    Commented Jul 3 at 2:14
  • 1
    $\begingroup$ @MattHanson I don't understand your comment. Two bodies of different temperature connected by a link with minuscule thermal conductivity can reasonably be modeled as undergoing a quasistatic process of temperature equilibration, but entropy is generated with every infinitesimal energy shift down the thermal gradient. That process is not reversible. $\endgroup$
    – Chemomechanics
    Commented Jul 3 at 5:42
  • $\begingroup$ @Chemomechanics I could be overlooking something blatant but if heat can be transferred between two systems, doesn't this nearly always mean $\Omega_f>\Omega_i$? If the two systems have energies $E_1$ , $E_2$ and $E_1+E_2=E$, now isn't $\Omega_f=\sum_{E_i}\Omega^{1} (E_i)\Omega^{2} (E-E_i)>\Omega^{1} (E_1)\Omega^{2} (E_2)=\Omega_i$ (where the summation is over all energies accessible to system 1, and we assume system 1 can assume more energies than just $E_1$). Or am I being stupid (which I now think I am) and this won't apply since the change isn't infinitesimal? $\endgroup$
    – user62783
    Commented Jul 3 at 5:45

1 Answer 1

Reset to default
0
$\begingroup$

Reif perhaps means that in a reversible process, entropy of the supersystem (the system + the reservoir that exchanges heat with it) remains constant. Together with Boltzmann's/Planck's observation that thermodynamic entropy is proportional to log of number of compatible microstates, this implies that the number of compatible microstates of the supersystem remains constant.

In a reversible process, thermodynamic entropy and thus the number of compatible microstates of the system can change. E.g. in a process where the system does no work, but accepts heat (isochoric heating), both increase.

Improve this answer
$\endgroup$

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.