0
$\begingroup$

In my university exam recently, we had a problem in which we had to prove that Entropy is a state function. Now I tried doing that by using the Boltzmann definition of entropy (S=kln(W)), and arguing that since the change in entropy for a given set of initial and final points depends only upon the ratio of the number of microstates of the system at the initial and final points only, and not upon any intermediate (path dependant) information, we get that entropy is a state function.

Now, my instructor seems to think that this is a circular reasoning. According to him, assuming that S is given by the Boltzmann definition already encodes the fact that it is a state function, we didn't prove anything. He probably wanted us to show that using the definition of a state function, we can show that S is an exact differential.

Now, my question is, is my method tautological? Also, is there any equivalence between the statistical method and the fact that entropy can be written as an exact differential, to show that entropy is a state function.

$\endgroup$
2
  • 1
    $\begingroup$ I would agree with the instructor that given Boltzmann's formula it is trivial that entropy is a state function, (you literally have its value as a function of the state of the system) whether this is circular reasoning depends on what you take as a starting point. In any case it is probably fair to say the only way that the question makes sense is if you start from the thermodynamic definition $dS = \frac{\delta Q_\mathrm{rev}}{T}$ $\endgroup$ – By Symmetry Mar 13 '18 at 14:05
  • $\begingroup$ "[W]e had a problem in which we had to prove that Entropy is a state function." If that was all it asked, stick with the classical thermodynamic definition, as $By\ Symmetry$ says. The problem with using statistical thermods is that there are several approaches including those that start off with the equation $S=\sum_{i} P_{i} ln P_{i}$ and call it 'entropy' without (at first) showing that it is connected with $dS=\frac{\delta _{rev}}{T}$. $\endgroup$ – Philip Wood Mar 13 '18 at 15:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.