7
$\begingroup$

I understand why increasing the temperature of a fluid would increase its entropy, as the particles are free to move, and therefore an increase in kinetic energy would allow the particles to move more, thus increasing the fluid's entropy.

However, this doesn't make sense to me in the case of a solid - increasing the solid's temperature would result in an increase in K.E., but this would only result in the particles vibrating more. As entropy is defined as the 'disorderliness' of a system, and the molecules in a solid always remain in the same 'order' regardless of temperature, how is temperature proportional to entropy in a solid?

|improve this question
$\endgroup$
9
$\begingroup$

All right, someone bearing the standard of thermodynamics will give you the equations shortly... From a layman to another, here goes my attempt at a simpler explanation.

Entropy may be seen as the "disorderlyness" in some settings, but that is not a very useful way of seeing it. The metaphor is often used, but creates the wrong conclusions when looking closer at it. To me, is best to look at it as a quantity closely related to the number of possible (different) states available to a certain system. It is then obvious that if you increase the temperature, more modes of vibration are available, thus more states, ergo more entropy. The more stuff is in your room, the less probable is the state where everything is ordered - if all states are equal in probability ;)

|improve this answer
$\endgroup$
  • 2
    $\begingroup$ This pretty much summarises what happens, I would add that the entropy increases due to the fact that the number of energy levels available is increased (due to temperature increase) and so the number of ways of arranging the energy into these levels clearly increases and this is what the entropy measures so this also increases. Don't think of energy as 'disorder', which is rather a vague thing, but more correctly as the number of ways of placing the energy into all the available energy levels. Thus it applies to everything in the same way. $\endgroup$ – porphyrin Jun 22 '17 at 18:24

Your Answer

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

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