On 02/06/2017 22:43, jacobnavia wrote:
Le 02/06/2017 à 09:16, Poutnik a écrit :

Thermodynamics generally does not care,
what time it takes for a system
to get into the preferred final state.

"Final state" implies time.

No it doesn't apart from perhaps having to wait an eternity to get
there. Metastable system states may persist almost forever if the
activation energy to escape from a local minima is too large.

Atoms A and B in gas state are inserted into a container in proportion
1:1. The final state is an almost perfect distribution of a mixture of
both gases. We do not expect the gases to appear separated after some
time. That is the accepted final state of a smooth distribution for two
gases in a container at room temperature say.

Give or take a random fluctuations yes.

If there are N atoms in the volume and you choose to split the space
down the middle with an imaginary line then the probability of a split
N-n, n across that line is given by the terms in the binomial expansion
with 2^N states in total. The most common states being determined by
their degeneracy factor - essentially one derivation of entropy.

N!/((N-n)!n!)

Formally it is obviously maximised when n=N/2 although you would be very
surprised if you didn't see variation. If N is small enough then it
isn't so long to wait to catch all of them in one half by chance.

But if N is Avagadro's number - well you do the maths.

But is it the final state?

Surely not, since if not given any external energy, the final state of
the mixture could be a separated mixture of frozen A and B at almost
absolute zero. Let's suppose that when freezing, gases A and B do not
mix easily.

Perhaps more apposite to the original question one of the methods of
separating U235 from U238 relies on making UF6 gas and putting it
through a cascade of centrifuges to impose a large potential gradient on
the maximum entropy distribution in the spinning centrifuges.

Time is always there in all physics. The concept of "final state"
implies time, you see?

The concept of "final state" implies that ultimately it has a lower
energy than all other possible states. Although something may still be
long term metastable despite a lower energy state being available if
there is an activation energy needed to get there that isn't available.

Common window glass for example.

--
Regards,
Martin Brown