I'm not a chemist, but I need some insight into chemical thermodynamics.

In the book that I'm reading (Nevers, 2012), equation (4.28) relates the chemical potential of a pure species to it's Gibbs energy per mol:

$\qquad\mu_a^{(1)} = \bigg( \dfrac{\partial G}{\partial n_a} \bigg)_{T,P,n_b...} = g_a^{(1)}$

where the subscript stands for the species $a$, the superscript stands for it's phase, $\mu$ is the chemical potential and $g$ is the Gibbs energy per mol.

I understand the first equal sign. Though, I don't understand the second equal sign. What am I missing?

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    $\begingroup$ Partial derivative of the Gibbs free energy with respect to the amount of the species, all other species' concentrations in the mixture remaining constant, and at constant temperature. When pressure is constant, chemical potential is the partial molar Gibbs free energy. At chemical equilibrium or in phase equilibrium the total sum of chemical potentials is zero, as the free energy is at a minimum. Which is also defined in book in Example 4.1. It's physical properties of substances we can only observe them and understand rules, but why it is so, is question for particle physics. $\endgroup$ – sigrlami Jul 19 '13 at 12:40
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    $\begingroup$ @Sigrlami I disagree with your “why it is so, is question for particle physics” — sebastian is asking why, given the basic principles and axioms of thermodynamics, the second equality holds. There is a good reason for that, as explained in Max's answer… while one has to beware of why questions in science, they're not all bad! $\endgroup$ – F'x Jul 20 '13 at 10:23
  • $\begingroup$ @F'x I agree. I find Sigrlami's comment rather confusing. Max' answer below is what I was looking for. $\endgroup$ – seb Jul 20 '13 at 14:27

The key is that the Gibbs free energy is an extensive quantity. This means that for a single-component system it can be written as $$ G\left(T,P,N\right) = N\cdot g\left(T,P\right) $$ where N is the number of particles and g is the Gibbs free energy per particle. In other words, if you double the number of particles and double the volume (at fixed temperature and pressure), then G will also double.

From the above expression you can see that the chemical potential is just the Gibb's free energy per particle: $$\mu\left(T,P\right) = g\left(T,P\right)$$

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    $\begingroup$ Excellent and clear answer! You write it for pure phase, but the same reasoning extends to mixtures and thus answers sebastian's question. $\endgroup$ – F'x Jul 20 '13 at 10:23

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