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.

  • 1
    $\begingroup$ The Wikipedia write up is fairly clear. Start there and refine your question if necessary. $\endgroup$ – 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 $G=\mu n$.

| cite | improve this answer | |

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 $$U=TS-pV+N\mu$$

| cite | improve this answer | |

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:

$G = \mu N$


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

This is the Gibbs free energy per mole.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.