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Why is this also the case in an isolated system. The internal energy is in this case constant. Even when the entropy is not at its maximum. How does this affect the amount of internal energy despite the fact that it does distribute more. But the internal energy is still constant and a "minimum" implies that the amount of energy changes

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    $\begingroup$ Re check your source. I think you’ll find the principle applies to a closed system, not an isolated system $\endgroup$
    – Bob D
    Apr 22 '20 at 9:16
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Start with the 1st and 2nd Laws written as $dU=\delta Q + \delta W$ and $\delta Q \le TdS$, where $\delta W$ is the work done by external forces on the system whose internal energy is $U$, and $\delta Q$ is the heat transferred from the environment to the system.

Then we have for any process that $$dU \le TdS + \delta W \tag{1}\label{1}$$ where in $\eqref{1}$ equality holds iff the process is reversible.

For now forget the words isolated, closed, adiabatic, etc., and assume that the process is such that the external work done is zero, then you have $dU\vert_{\delta W=0} \le TdS$, and if the process is also isentropic, that is $S=const,\; dS=0$, then you must have $$dU \le 0 \tag{2}\label{2}$$

What does $\eqref{2}$ mean? In an isentropic transformation there is entropy (heat) exchange but is such that the entropy of the system is unchanging while it absorbs and rejects entropy (heat); if the process is reversible then whatever goes in it also goes out, but for an irreversible process the internally generated entropy must also be rejected along with the absorbed one if the process to be isentropic internally.

Now assume that be the case and then we have that the internal energy cannot increase, and can only decrease for an irreversible process. When approaching equilibrium the internal energy of any system being bounded from below must reach a minimum at which point it stops changing. That is equilibrium.

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I am not sure if I totally understand your question but here is something that may help. The total internal energy of an isolated system remains constant. Any system tries to achieve an ideal state of minimum energy and maximum entropy to be more stable, as the energy is now more evenly distributed between the system and the surroundings. For example, a heated object tends to cool by losing heat to its surroundings by which the heat is distributed and not concentrated in one part of the environment. In the case of isolated objects since there are no surroundings to lose energy to, the energy, heat as in this example, will tend to redistribute/transfer all over the object in order to achieve an equilibrium. And the entropy will increase since the chaos/ movement of particles has increased. I hope this was helpful. let me know if you need more clarification over something or some part of the question I left unanswered.

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