1
$\begingroup$

When studying entropy in terms of order and disorder, we can consider the total entropy, $S$ for a material at a temperature, $T$ to consist of two components:

  1. Thermal entropy: The entropy by virtue of the thermal energy supplied to the material. Given by the Clausius equation from the classical notion of entropy $$S = \int_{0}^{T} \frac{dq}{T}$$
  2. Configurational Entropy: The entropy by virtue of multiple possible distinct 'arrangements' of the constituent particles of the material. Given by the Boltzmann formula from the statistical notion of entropy: $$S = k \ln (W)$$

It is clear that the first component is meant to capture disorder due to thermal motion (such as vibration) of the constituent atoms (or other particles). Similarly, the second is meant to capture disorder due to multiple possible atomic permutations corresponding to the same thermodynamic state.


However, this appears to become a hairy distinction when we consider the physical changes which might occur in a real substance with temperature:

  1. An ionic solid like $\text{Na} \text{Cl}$ might start to see the breakage of an increasing number of bonds and breakdown of its strict order with higher temperature.
  2. A substance like ice might undergo a phase transformation to become water in the liquid phase.

etc


I have been told the configurational entropy should strictly depend on the composition of the system so that we have say $S = -R \cdot (X_{A} \ln(X_{A}) + X_{B} \ln(X_{B}))$ for a binary solid solution. This would imply that the configurational entropy should not change and the entropy generated in the above examples is entirely thermal.

However, I cannot help but think that changes of these kinds should increase the space of possible configurations leading to increased configurational entropy.

Thus, I am asking about the proper classification for entropy generated by changes of this kind. Maybe I getting confused by the "configurational" in the name and there is some method by which we can show that the supposedly increased configuration space is already accounted for in the thermal entropy.

[I understand that the Boltzmann and Clausius formulae are supposed to be equivalent and that the heat supplied to the system during a transformation increases the thermal entropy. So this should(?) be possible since we want to avoid double counting of entropy changes.]

Is there some way to deal with these cases given the above definitions? If not, is there some well-accepted convention for assigning the entropy? Or do these notions break down entirely when we consider, say, phase transformations?

Alternatively, is it correct to say that the configurational entropy (as defined above) has increased with temeprature in such cases?

$\endgroup$

1 Answer 1

1
$\begingroup$

There exist several definitions of entropy (see here for a bit more). Notably, the entropy as defined in phenomenological thermodynamics (i.e., the Clausius definition), the entropy as defined by Boltzmann (configuration entropy) and the Shannon's information entropy are not exactly the same, although in many cases they can be shown to be equivalent.

Secifically, within the equilibrium thermodynamics and statistical phsyics context the Clausius entropy and the configuration entropy can be shown to be equivalent - the former is defined phenomenologically, while the latter is its microscopic explanation.

When an isolated system is in thermodynamic equilibrium we assume that all its states with the same energy are equally accessible, i.e., have equal probability (incidentally, in this case the configurational entropy is equal to the information entropy). Adding energy of the system changes the microstates microstates that it can occupy, and hence changes its entropy.

If some physical change occurs due to an increase in temperature, is the resulting entropy thermal or configurational?

Clausius inequality deals with the change in entropy - it cannot assign a definite entropy to a state (it is defined up to a constant). On the other hand, in statistical physics the confugurational entropy can be associated with the number of microstates - also up to a constant, but the constant is the same for all systems, it appears in the definition of the entropy. In quantum statistical phsycis we further define the value of this constant.

Thus, the answer is: the change in entropy is thermal, but the entropy itself is configurational.

$\endgroup$
3
  • $\begingroup$ Per this, is it correct to conclude that when there is an entropy increase due to breaking bonds in a cryatal, we should assign the increase in entropy to the thermal head? $\endgroup$
    – user0
    Commented Feb 22, 2022 at 12:47
  • $\begingroup$ I understand that ther is an integration constant which appears from the Clausius formula. But there would be some fixed change in such a situation which will be indepndent of said constant. I am trying to understand the nature of this new entropy - is it thermal or configurational? Also, why is the configuratioanl entropy variable upto a constant? Aren't both k and the number of microstates fixed for a given system? $\endgroup$
    – user0
    Commented Feb 22, 2022 at 12:51
  • $\begingroup$ @user0 what I tried to day un my answer, is that it is the same entropy, but expressed in different languages. $\endgroup$
    – Roger V.
    Commented Feb 22, 2022 at 17:12

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.