Why chemical potential changes as a system approaches chemical equilibrium?

Question

Consider a closed system of constant volume $V$, constant pressure $P$, temperature $T$, and Gibbs energy $G$ that is in thermal and mechanical equilibrium with surroundings. It is filled with $N$ particles of one type.

We divide the system into two components: component 1 which has volume $V_1$, pressure $P$, temperature $T$, particle count $N_1$, chemical potential $\mu_1$, Gibbs energy $G_1$, and component 2 which has volume $V-V_1=V_2$, pressure $P$, temperature $T$, particle count $N-N_1=N_2$, chemical potential $\mu_2$, Gibbs energy $G-G_1=G_2$.

The total change in Gibbs energy for this two-component closed system is

$$dG=dG_1+dG_2=\mu_1 dN_1+\mu_2 dN_2=\mu_1 dN_1-\mu_2 dN_1=(\mu_1-\mu_2)dN_1$$

Suppose the system is initially not in chemical equilibrium $\mu_1>\mu_2$, it will decrease the particle count in component 1 $dN_1<0$ until $dG=0$.

However, I am not sure how $G_1$ and $G_2$ change as the system approaches chemical equilibrium. In terms of Gibbs energy, chemical potential is defined as

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

Since $G$ and $N$ are extensive quantities, it follows that $\mu$ is an intensive quantity. This means the value of $\mu$ is independent of particle count. For example, even if the particle count is doubled, $\mu$ stays the same.

So I am not sure how it is possible for $\mu_1$ to decrease as $N_1$ decreases if chemical potential is an intensive quantity.

Answer 1

The chemical potential of component $1$ in a mixture with component 2 is $$\tag{1} \mu_1 = \left(\frac{\partial G}{\partial N_1}\right)_{T,P,N_2} \color{red}\neq \frac{G}{N} $$ Notice that equality you wrote applies only in a one-component system. The relationship between the chemical potentials, the Gibbs energy of the mixture and the number of moles is $$\tag{2} N_1 \mu_1 + N_2 \mu_2 = (N_1+N_2) g = G $$ A simple example can be given for the case of an ideal mixture. Then $$\tag{3} \mu_i = g_i + RT \ln \frac{N_i}{N_1+N_2} $$ where $g_i=G_i/N_i$ is the molar Gibbs energy of pure $i$ at the same temperature and pressure as the mixture. This shows that the chemical potential changes with composition.

Answer 2

Your definition of the chemical potential should be $$\mu_1=\left(\frac{\partial G}{\partial N_1}\right)_{T,P,N_2}$$