The relation between Chemical potential and Gibbs free energy ($nmu = G$) is global?

Question

The change of Gibbs free energy (for a single phase or constitute system) is
$$dG = -SdT + VdP +{mu}dn$$
By using the fact that $T$ and $P$ are intensive properties and $n$(mole) is an extensive property,
$G = nmu$ is derived in my lecture note.
But I am not sure about that, because, with the definition of $dG$, $G$ is the function of $(T,P,n)$.
Where are the dependency of $T$ and $P$?
This interpretation is only valid for constant temperature and pressure situations?
like, $G = G_{0}(T,P) + nmu$?

J. Murray · Accepted Answer

$G$ depends on $T$ and $P$ through the chemical potential:
$$G(T,P,n) = mu(T,P,n) cdot n$$
Recall that $mu = left(frac{partial G}{partial n}right)_{T,P}$.  If $G$ is a function of some set of variables, then its derivatives (and in particular, $mu$) are functions of the same set of variables. Note also in this case that $left(frac{partial mu}{partial n}right)_{T,P} = 0$, in accordance with the assumptions used to derive $G = mu n$ (though these assumptions do not universally hold for all thermodynamical systems).

The relation between Chemical potential and Gibbs free energy ($nmu = G$) is global?

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