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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 substem, 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.

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.

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 substem, 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.

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 substem, 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.