Invariant Concentration of Ideal Gas Mixture with Altitude Consider the following scenario, We have two gasses, $A$ and $B$, both approximately ideal mixed together in a gravitational field of constant magnitude $g$. Let them have masses per particle of $m_a$ and $m_b$ respectively. At elevation $z = 0$, the concetration $\frac{N_a}{N_b+N_a} = c_0$, and the total pressure is $P_0$. We wish to find the concentration of the two gasses in the mixture at arbitrary elevation $z$.
The following is my very problematic analysis. Let's begin with pressures. Because they are both ideal gasses, they will both individually satisfy the barometric equation
\begin{equation}\frac{\partial P}{\partial z} = -\rho g\end{equation}
or in this particular case, the two equations
\begin{equation} \frac{\partial P}{\partial z} = \frac{-Pmgz}{kT}\end{equation}
Which leads to the obvious solutions
\begin{equation}
P_a(z) = P_{a0}e^{-m_agz/kT}
\end{equation}
\begin{equation}
P_b(z) = P_{b0}e^{-m_bgz/kT}
\end{equation}
Now, let's consider concentration. The chemical potential of a species $A$ in a mixture is the pure chemical potential $\mu_A^o(P,T)$, with an additional term for the entropy of mixing. Given that the entropy of mixing is
\begin{equation}
-\Delta S_{mix}/k = N_a \ln(1/c) + N_b\ln\left(\frac{1}{1-c}\right)
\end{equation}
Given that $c$ is a function of $z$. The chemical potential should be
\begin{equation}
\mu_{mix,A} = \mu_{A}^o + kT\frac{\partial \Delta S}{\partial N_a} = \mu_A^o + kT\ln(1/c)
\end{equation}
If we now utilize the fact that gas $A$ must be in equilibrium, we have that
\begin{equation}
\mu_{A}^o(P(z),T,0) + kT\ln(1/c) + m_Agz = \mu_{A}^o(P_0,T,0) + kT\ln(1/c_0)
\end{equation}
Equivalently,
\begin{equation}
kT\ln(c/c_0) = \mu_{A}^o(P(z),T,0) - \mu_A^o(P_0,T,0) + m_agz
\end{equation}
If we now utilize that the chemical potential of an ideal gas is just $kT\ln(P) - \chi(T)$, this becomes
\begin{equation}
kT\ln(c/c_0) = kT\ln(P_a/P_{0a}) + m_agz
\end{equation}
And
\begin{equation}
c = c_0\frac{P_a(z)}{P_{0a}}e^{m_agz/kT} = c_0e^{-m_agz/kT}e^{m_agz/kT} = c_0
\end{equation}
Thus, one of two things is true. Either I've blundered somewhere and I can't find the mistake, or this is correct and I need to be given an intuition why this is possible. We clearly expect something like a logistic curve where the lighter of the two species becomes more prevalent at higher altitudes. I can derive that result without mixing, but I cannot seem to get it with the mixing factored in. As such, please relieve me of my ignorance!
 A: *

*Your definition for the entropy of mixing has a sign error. The resulting chemical potential should be $\mu_{\text{mixture,}A}=\mu_A^\circ+kT\ln c$. That is, the chemical potential of a solute tends to be lower where it is more dilute, which promotes diffusional mixing because matter moves to areas of lower chemical potential.


*After writing "If we now utilize that the chemical potential of an ideal gas is just...," you conflate the total pressure $P$ with the partial pressure of A,
$P_A$. They aren't equal or linearly proportional. It's the total pressure that brings in information about B that's needed to explain the concentration dependence on height.
Here's another approach to compare:
The chemical potential of A must satisfy
$$\frac{\partial (\mu_{\text{mixture,}A}+m_Agz)}{\partial z}=0.$$
(This corresponds to your equation setting the chemical potentials equal at different heights, now expressed more powerfully as a derivative equaling zero.)
Since $\mu_{\text{mixture,}A}=\mu_{\text{mixture,}A}(T,P,c)$, we have
$$\frac{\partial \mu_{\text{mixture,}A}}{\partial T}\frac{\partial T}{\partial z}+\frac{\partial \mu_{\text{mixture,}A}}{\partial P}\frac{\partial P}{\partial z}+\frac{\partial \mu_{\text{mixture,}A}}{\partial c}\frac{\partial c}{\partial z}+m_Ag=0,$$
where we seek $\frac{\partial c}{\partial z}$. Now,

*

*For height-independent temperature, $\frac{\partial T}{\partial z}=0$, so the first term disappears;

*We always have $\frac{\partial \mu_i}{\partial P}=V_i$ (or $\frac{kT}{P}$ for an ideal gas);

*$\frac{\partial P}{\partial z}=-\rho g=-\frac{m_\text{mixture}}{V}g$ (hydrostatic equilibrium, where $m_\text{mixture}=cm_A+(1-c)m_B$ is the total mass); and

*$\frac{\partial \mu_{\text{mixture,}A}}{\partial c}=\frac{kT}{c}$ (from the equation in (1)), giving

$$0-m_\text{mixture}g+\frac{kT}{c}\frac{\partial c}{\partial z}+m_Ag=0;$$
$$kT\frac{\partial \ln c}{\partial z}=m_\text{mixture}g-m_Ag;$$
$$\frac{c}{c_0}=\exp\left(\frac{(m_\text{mixture}-m_A)gz}{kT}\right),$$
Do these steps make sense?
You can get to this result from your (corrected) equation
$$\mu_{A}^\circ(P,T) + kT\ln(c) + m_Agz = \mu_{A}^\circ(P_0,T) + kT\ln(c_0)$$
as long as we integrate $\frac{\partial\mu_A^\circ}{\partial P}=\frac{kT}{P}$ to obtain $\mu_A^\circ(P)=\mu_A^\circ(P_0)+kT\ln\left(\frac{P}{P_0}\right)=\mu_A^\circ(P_0)-m_\text{mixture}gz$. This again gives $$-m_\text{mixture}gz + kT\ln\left(\frac{c}{c_0}\right) + m_Agz = 0$$ and thus
$$\frac{c}{c_0}=\exp\left(\frac{(m_\text{mixture}-m_A)gz}{kT}\right).$$
As noted in a comment, the result from this derivation strategy matches that from simply using partial pressures as a surrogate for concentrations:
$$\frac{c}{c_0}=\frac{\frac{P_{A}}{P}}{\frac{P_{A,0}}{P_0}}=\frac{P_{A}P_0}{P_{A,0}P}=\exp\left(\frac{(m_\text{mixture}-m_A)gz}{kT}\right).$$
Since we’re assuming ideal gases, this a perfectly valid solution strategy, albeit less extensible to other materials than the chemical potential approach outlined above.
