Chemical equilibrium conditions for mixture of ideal gases vs. ideal gas separated by partition

Question

I am not sure whether chemical equilibrium conditions are the same or different for the two scenarios below:

i. Closed rigid container of volume $V$ with uniform temperature $T$ and pressure $P$ filled with ideal gas of one kind with total particle count $N$. A partition divides the container into two components 1 and 2 with volumes $V_1,V_2$ and particle counts $N_1$,$N_2$. The components are allowed to exchange particles through this partition.

ii. Closed rigid container of volume $V$ that is isolated from the rest of the universe with uniform temperature $T$ and pressure $P$ filled and mixed with two species of ideal gases 1 and 2 with particle counts $N_1$,$N_2$. The total number of particles $N=N_1+N_2$ is fixed.

As discussed in the previous question, in scenario (i) the equilibrium conditions are

$$\tag{1} \mu_1=\mu_2$$ $$\tag{2} d\mu_1=d\mu_2=0$$

I still have doubts on why (2) must be true. Gibbs-Duhem equation for components read

$$d\mu_1=-sdT+vdP$$ $$d\mu_1=-sdT+vdP$$

where $s$ and $v$ are molar entropies and volumes. The answer in the linked question says that (2) is true because pressure and temperature are constant $dP=dT=0$ but I am still not convinced because it makes me to believe that (2) is true regardless of whether the container is in chemical equilibrium or not since pressure and temperature are constant all times. I understand that as particles get transferred from one component to another, chemical potentials must change $d\mu_1\ne0$ and $d\mu_2\ne0$. When particles stop transferring during chemical equilibrium, $d\mu_1=d\mu_2=0$.

In scenario (ii), the first equilibrium conditions is

$$\mu_1=\mu_2$$

Gibbs-Duhem equation for the ideal gas mixture is

$$N_1d\mu_1+N_2d\mu_2=-SdT+VdP=0$$

since pressure and temperature are constant, so the other equilibrium condition is

$$d\mu_1=-\frac{N_2}{N_1}d\mu_2$$

It implies that chemical potentials still change when the mixture is in chemical equilibrium.

Did I misunderstand anything?

Answer 1

In scenario 2, for an ideal gas$$\mu_1=\mu_1^0(T)+RT\ln{P}+RT\ln{\left(\frac{N_1}{N_1+N_2}\right)}$$$$\mu_2=\mu_1^0(T)+RT\ln{P}+RT\ln{\left(\frac{N_2}{N_1+N_2}\right)}$$where $\mu_j^0$ is the chemical potential of pure species j at temperature T and 1 bar and P is the pressure in bars. In addition, we have that $$G=\mu_1N_1+\mu_2N_2$$ and, from this, it follows that, at constant temperature and pressure, $$dG=\mu_1dN_1+\mu_2dN_2$$. (Substitute yourself and you can confirm this). So, from the previous two equations, you have that $$N_1d\mu_1+N_2d\mu_2=0$$All this Gibbs-Duhem equation tells you is how $\mu_1$ and $\mu_2$ must vary mutually as N1 and N2 very.

Answer 2

Chemical equilibrium is defined as:

In a chemical reaction, chemical equilibrium is the state in which both the reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system.1 This state results when the forward reaction proceeds at the same rate as the reverse reaction. The reaction rates of the forward and backward reactions are generally not zero, but they are equal. Thus, there are no net changes in the concentrations of the reactants and products.

Important here is that we are talking about transformation of particles from one type to another, e.g., when they cross the partition dividing the container in the two parts. However, the OP clarifies in the comments that no chemical transformation takes place.

Case (i)
In absence of a chemical transformation, setup (i) does not make much sense. Specifically, the condition

A partition divides the container into two components 1 and 2 with volumes $V_1$, $V_2$ and particle counts $N_1$, $N_2$.

contradicts the sentence that follows:

The components are allowed to exchange particles through this partition.

If the components are allowed to mix, then the state where the species of each gas are on one side of the partition is not an equilibrium state - the particles will filter through the partition, till the concentration for each species would be the same on each side of the partition. (This setup is very similar to the one used in the Gibbs paradox.)

Case (ii)

Closed rigid container of volume V that is isolated from the rest of the universe with uniform temperature $T$ and pressure $P$ filled and mixed with two species of ideal gases 1 and 2 with particle counts $N_1, N_2$. The total number of particles $N=N_1+N_2$ is fixed.

Again, in absence of a chemical transformation (or another process replacing particles of one kind with the other one), fixing the total number of particles also means fixing the number for each species. Since the particle number is fixed, we are working in canonical rather than grand canonical ensemble, and speaking of equality of chemical potential is meaningless.

Answer 3

The question is how to perform an experiment where a system is partitioned into two parts with the same $T$ and $P$ but different chemical potentials. If we use the same ideal gas the experiment is trivial because the ideal gas at fixed $T$ and $P$ has a single stable state with a single chemical potential. We can use a mixture, where now we can play with composition to change the chemical potential at fixed $T$ and $P$.

If we want to stay with a single component we need to move away from ideal gas and look for conditions under which a pure component can exist in more than one stable states at same $T$ and $P$. That's what happens at the phase boundary:

The stable isotherm is $ALVD$ but we also have the metastable branches $LB$ (supersaturated liquid) and $VC$ (supersaturated vapor). If we draw a random tie line $L'V'$, points $L'$ and $V'$ are at the same $T$ and $P$ but they will generally gave different chemical potentials. The chemical potentials are equal only on the stable tie line $LV$. This is how we determine the phase boundary of the van der Waals equation: we calculate $\mu_L$ on the liquid branch and $\mu_V$ and vapor branch and identify $P^\text{sat}$ as the pressure where the two are equal: $$\mu_L = \mu_V \qquad\text{at $P=P^\text{sat}$}$$ Above the saturation pressure we have $\mu_{L'}<\mu_{V'}$, which says that the liquid phase is the stable phase; below the saturation pressure the opposite is true, we find $\mu_{L'}>\mu_{V'}$, which makes vapor the stable phase. These relationships can be confirmed by calculations using the van der Waals equation or other cubic equations.

In summary The chemical potential is useful when we have multiple components, multiple phases, or both. The case of one phase/one component is trivial.

Answer 4

Since you are considering both a division in the system, along with having different components of gases, I guess adding a new label would fix one of the confusions: one label for the subsystem, and one for the component. Thus, $N_A^1$ would mean the number of particles of the $A$ component in the first compartment, or $N_B^2$ means the number of $B$-particles in the second compartment. You could define separate thermodynamic potentials for each subsystem and for the whole system as well. In this case, entropy is the best choice, since the whole system is isolated. So, you could say that the total entropy is conservative upon reaching equilibrium, i.e. \begin{equation} \delta S (N) = \delta S_1 (N_A^1,N_B^1) + \delta S_2 (N_A^2,N_B^2) = 0 \end{equation} ($S_i$ would depend also on other extensive parameters, like energy or volume, which I assume the existence of equilibrium with respect to them). From here it will be easy to show that for a system of several components, each having its own chemical potential, AND in absence of chemical reactions (conservation of $N^\alpha$) the chemical equilibrium is achieved if and only if \begin{equation} \mu_A^1=\mu_A^2 \ \ \ and \ \ \ \mu_B^1=\mu_B^2 \end{equation} This being said you could say that a biased interface, for instance, would only cause equilibrating w.r.t one component while the other would never equilibrate (of course it only means that the two open subsystems are open to diffuse $A$-particles but closed w.r.t exchanging $B$-particles, hence they are already in internal equilibrium w.r.t the $B$-component).

Now, the Gibbs-Duhem relation is merely a (powerful) statement that says: of all intensive parameters that define the equilibrium of the "subsystem" (which is open) in terms of an appropriate thermodynamic potential (Gibbs function which is a function of intensive variables), not all of them are independent. It does not indicate or imply equilibrium by itself. In the case of your problem, there are 4 intensive variables that indicate the equilibrium conditions, but they are dependent on each other via the Gibbs-Duhem relation, hence there are truly 3 independent thermodynamic variables.