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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!

BooleanDesigns
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1 Answers1

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  1. 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.

  2. 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.

Chemomechanics
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