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So, I have just derived that $dG\leq0$ for a closed system at constant temperature and volume. But, I am finding it hard to understand how this corresponds to the equilibrium state which has minimum Gibbs free energy. I mean, mathematically, it is a second derivative, so I am thinking like the change in Gibbs free energy can be either negative or zero, but how it can be minimum? It is simply the first derivative, for the function like $G$, for it to be minimum, both its first derivative should be zero and second derivative be greater than zero. How can I comprehend this relation regarding Gibbs free energy be negative?

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  1. The condition for $dG \le 0$ should be $T = const$ and $P=const$, instead of $V=const$.
  2. $dG\le0$ describes the free energy change of a process, not a state. For a reversible process, $dG=0$, while for a irreversible process $dG<0$.
  3. The second law of thermodynamics says: Isolated systems spontaneously evolve towards thermodynamic equilibrium, the state with maximum entropy. Minimum $G_{system}$ corresponds to maximum entropy $S_{system + environment}$. In this sense, minimum $G$ at equilibrium is true by definition, instead of being a consequence of $dG\le0$.
  4. As far as we know, there are no perpetual motion machines. So every system we know has a minimum energy $G_{min}$. We say states of system that has $G_{min}$ is in thermodynamic equilibrium. Since $dG\le0$, we know every system is heading towards equilibrium, however slowly it may go.
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