# What is meant with internal energy tends to be at minimum?

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

• Re check your source. I think you’ll find the principle applies to a closed system, not an isolated system Apr 22 '20 at 9:16

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.