0
$\begingroup$

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.

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ Thanks. To be clear, equation 2 says that the total Gibbs energy of the two-component system is the average of chemical potentials of each component (assuming that each component is filled with particles of one type)? I’m not sure how equation 3 is derived. Can you give more details? $\endgroup$
    – Jimmy Yang
    Mar 18, 2023 at 19:05
  • $\begingroup$ @JimmyYang (i) Yes, your reading of Eq 2 is correct. In fact, the first equation in your post is the differential of my Eq 2. (ii) Eq 3 is a special result for ideal mixtures, namely mixtures whose enthalpy is $H=n_1 h_1 + n_2 h_2$, and their entropy is $S = N_1 s_1+N_2 s_2 - R(N_1\ln\frac{N_1}{N_1+N_2} + N_2 \ln\frac{N_2}{N_1+N_2})$, where $h_i$ and $s_i$ are the properties of the pure components. Ideal gas mixtures and ideal solutions (solutions of chemically similar molecules) satisfy these equations. Set $G = H-T S$ and you will get to Eq (3). $\endgroup$
    – Themis
    Mar 18, 2023 at 19:27
  • $\begingroup$ I see. If each component is filled with a different particle type, then the total Gibbs energy is $G=N_1\mu_1+N_2\mu_2$? I still have doubts about the $N_1+N_2$ factor. Also, the differential for equation 2 is $\mu_1 dN_1+\mu_2 dN_2=GdN_1+N_1dG+GdN_2+N_2dG$ which reduces to $(\mu_1-\mu_2)dN_1=(N_1+N_2)dG=NdG$. I don't see how it results in my equation $dG=(\mu_1-\mu_2)dN_1$. $\endgroup$
    – Jimmy Yang
    Mar 18, 2023 at 20:08
  • 1
    $\begingroup$ (i) You are right (I corrected it), I should have said $(N_1+N_2)g = G$. (ii) According to the Gibbs-Duhem equation (which is easy to prove), $N_1 d\mu_1+N_2 d\mu_2 = 0$. $\endgroup$
    – Themis
    Mar 18, 2023 at 20:22
  • $\begingroup$ Is equation 3 intensive or extensive? I understand that $g_i$ is intensive but not sure about $RT \ln \frac{N_i}{N_1+N_2}$ term. I don't think that equation is intensive since it changes as $N_i$ changes? $\endgroup$
    – Jimmy Yang
    Mar 18, 2023 at 21:54
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ You're right, I forgot to write constant $N_2$. $\endgroup$
    – Themis
    Mar 18, 2023 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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