# How could chemical potential be interpreted as the molar Gibbs free energy?

As known, the Gibbs free energy for a closed system at constant temperature $T$ and constant pressure $p$ (and thus constant volume $V$), $G=U+pV-TS$, will be minimized, where $U$ is the total internal energy of the system.

Could you explain in simple and understandable way why the chemical potential can be interpreted as the molar Gibbs free energy, that is, $G_\text{mol}=\frac{G}{n}=\mu$, where $n$ is the amount of substance in mole.

• The Wikipedia write up is fairly clear. Start there and refine your question if necessary. – Jon Custer Apr 11 '17 at 14:46

Let's first note that some of the ways we can add energy to a system are by heating it, doing mechanical work on it, or by adding mass; thus, $$dU=T\,dS-p\,dV+\mu\,dn$$ The Gibbs free energy potential $G=U-TS+PV$, which constitutes a Legendre transform, is of interest because (by differentiating) $$dG=dU-T\,dS-S\,dT+p\,dV+V\,dp=-S\,dT+V\,dp+\mu\,dn$$ which is particularly convenient to work with because many familiar processes occur at constant temperature and pressure. Under these conditions, $dG=\mu\,dn$ (or $\sum\mu_i\,n_i$ for composite systems).

Another approach is to simply start from $U=TS-PV+\mu n$, from which we can directly obtain $G=\mu n$.

A thermodynamic potential $P$ satisfies $$P(\lambda X, Y) = \lambda P(X,Y)$$ where $X$ are extensive and $Y$ are intensive parameters of state. As an example $P$ could be the internal energy of a system, it then doubles if we double all its extensive properties (volume, particle number, etc).

If we take $\lambda = \tfrac{1}{N}$, we get $$P(\tfrac{X}{N}, Y) = \tfrac{1}{N} P(X,Y)$$ Let us denote the intensive versions (such as energy per particle, volume per particle) of all extensive quantities with lower case letters: $x=\tfrac{X}{N}$, the same also applies for the potential: $p(X,Y)=\tfrac{1}{N}P(X,Y)$. We then get: $$P(x,Y) = \tfrac{1}{N} P(X,Y) = p(X,Y)$$.

Let us write this down for the Gibbs free energy $G=G(T,p,N)$. Note that its only extensive dependency is $N$, the other two, $p$ and $T$ are intensive. $$G(T,p,\tfrac{N}{N}) = G(T,p,1) = \frac{1}{N}G(T,p,N) = g(T,p,N)$$

Obviously, since the l.h.s does not depend on $N$, also the r.h.s. doesn't, so: $g=g(T,p)$. So we can rewrite the last step of the above equation as: $$G(T,p,N) = Ng(T,p)$$ Diffferentiating with respect to $N$ gives:

$$\left(\frac{\partial G}{\partial N}\right)_{T,p} = g(p,T)$$

On the other hand, from the total differential of $G$ that is given as $dG=-SdT+Vdp+\mu dN$ we know that $$\left(\frac{\partial G}{\partial N}\right)_{T,p} = \mu$$

Therefore, in total, we get: $$G(T,p,N) = Ng(T,p) = N\mu$$

Note that from this derivation you can follow Euler's relation. Indeed, since $G$ is defined as Legendre transform of $U$, $$G=U-TS+pV$$ we have, by plugging in $G=\mu N$ from above, $$N\mu = U-TS+pV$$ or $$U=TS-pV+N\mu$$

We know that the Gibbs Free energy is the Legendre Transform of the internal energy, So:

G = U - TS + PV

Using Euler's Relation(Callen, H. B. (n.d.). Thermodynamics and an introduction to thermostatistics) for single component(one kind of particle)

$U = TS - PV + \mu N$

On Substitution we get:

$G = \mu N$

Thus,

$\mu = \frac{G}{N}$

This is the Gibbs free energy per mole.