\subsection{Kempers: Evaluation of the steady state assumptions}
\label{subsec:method_steady_state}
To compare the predictions by the Kempers model to Kinetic gas theory, and investigate how the two relate to the Helmholtz energy of the two-bulb system, Kempers target function of minimisation,
\begin{equation}
    \tau \equiv \frac{A^{\alpha}}{T^{\alpha}} + \frac{A^{\beta}}{T^{\beta}}
    \label{eq:kempers_target_method}
\end{equation}
is computed for a two-container system with a small, finite temperature difference. The system average temperature is taken to be
\begin{equation}
    T = \frac{\Ta + \Tb}{2}.
\end{equation}
Note that this choice is made for simplicity, using a mole-average temperature may be reasonable, but makes the mathematics of the problem far more involved. It is assumed that when dealing with temperature differences on the order of \SI{0.1}{\kelvin}, the difference between the bulb average used here, and the mole average is negligible. The two bulb temperatures are related to the temperature difference, $\Delta T$ as

\begin{equation}
    T^{\alpha} = T - \frac{1}{2}\Delta T, \hspace{1cm} T^{\beta} = T + \frac{1}{2}\Delta T.
\end{equation}

The bulbs are considered as two separate, closed volumes held at $T^{\alpha}$ and $T^{\beta}$, with a fixed total number of moles $\vn = \vna + \vnb$. The volume of the two bulbs is held constant, and for simplicity is set to be equal. Given a distribution of component 1, $\Delta n_1 = \nb_1 - \na_1$ and component 2, $\Delta n_2 = \nb_2 - \na_2$, one can compute
\begin{equation}
    \na_1 = \frac{n_1 - \Delta n_1}{2}, \hspace{.5cm} \nb_1 = \frac{n_1 + \Delta n_1}{2}.
\end{equation}
Such that instead of considering the four mole numbers, we may instead consider the two component separations. These will be our free variables. Given that we have two degrees of freedom, imposing any constraint on the system will uniquely define a curve in the $(\Delta n_1, \Delta n_2)$-plane. Thus, the consequence of constraining the minimisation of $\tau$ can be investigated. The first curve we consider is that of equal bulb mass, $\Delta m = 0$, given by
\begin{equation}
    \begin{split}
        \Delta m = m_1 \Delta n_1 + m_2 \Delta n_2 &= 0\\
        \Delta n_2 &= - \frac{m_1}{m_2} \Delta n_1.
    \end{split}
    \label{eq:dm_0_curve}
\end{equation}
According to the Kempers-CoM model, the steady state is the minima in $\tau$ along this curve. The second to consider is the curve corresponding to no pressure difference, $\Delta p = 0$. From the total differential of pressure at constant volume,
\begin{equation}
    \begin{split}
        \d p &= \ppder{p}{T}_{\vn, V} \dT + \sum_i \ppder{p}{n_i}_{T, V, n_k} \dn_i = 0\\
        \Dn_2 &\approx - \frac{\ppder{p}{T}_{\vn, V} \DT + \ppder{p}{n_1}_{T, V, n_2} \Dn_1}{\ppder{p}{n_2}_{T, V, n_1}}.\\
    \end{split}
    \label{eq:dp_0_curve}
\end{equation}
The minima in $\tau$ along this curve is the steady state predicted by the Kempers-CoV model.

Finally, to investigate the steady state predicted by the Enskog solutions, we find the curve of fixed compositional difference $\xb_1 - \xa_1 = \Dx$, given by
\begin{equation}
    \begin{split}
        \Dx_1 &= \frac{\nb_1}{\nb_1 + \nb_2} - \frac{\na_1}{\na_1 + \na_2}\\
        &= \frac{n_1 + \Dn_1}{n + \Dn} - \frac{n_1 - \Dn_1}{n - \Dn}\\
        \Dx_1 \lrp n^2 - \lrp\Dn\rrp^2 \rrp &= \lrp n_1 + \Dn_1 \rrp \lrp n - \Dn \rrp - \lrp n_1 - \Dn_1 \rrp \lrp n + \Dn \rrp\\
        &= 2 \lrp n \Dn_1 - n_1 \Dn \rrp\\
        - \Dx_1 \lrp \Dn \rrp^2 + 2 n_1 \Dn + \Dx_1 n^2 - 2 n \Dn_1 &= 0.
    \end{split}
    \label{eq:dx_0_curve}
\end{equation}
Closer investigation reveals that only one of the roots of this equation will be a physical solution, the other will have $\abs{\Delta n} > n$ or $\abs{\Delta n_1} > n_1$.

In addition to computing these curves, it is of interest to investigate the steady state pressure gradient as predicted by the Enskog solutions. Recall Equation \eqref{eq:d12_kingas_driving_force}, 
\begin{equation}
    \bvu_1 - \bvu_2 = - \frac{1}{3}\rho \lsp \lcp \vD, \vD\rcp \Vec{d}_{12} + \lcp \vD, \vA \rcp \nabla \ln T\rsp.
\end{equation}
In the steady state, the left hand side vanishes. Expanding $\Vec{d}_{12}$ and inserting the transport coefficients for the bracket integrals yields
\begin{equation}
    \begin{split}
        0 = - \frac{1}{3} \rho \lsp \frac{3 \rho}{\rho_1 \rho_2}D_{12} \lrp \nabla x_1 + \frac{\rho_1 \rho_2 (m_2 - m_1)}{\rho \rho_m} \nabla \ln p\rrp + \frac{3 \rho}{\rho_1 \rho_2}D_T \nabla \ln T\rsp\\
        \nabla x_1 = -\frac{D_T}{T D_{12}}\nabla T - \frac{\rho_1 \rho_2 (m_2 - m_1)}{p \rho \rho_m} \nabla p.
    \end{split}
    \label{eq:dx_const_soret_varying_dp}
\end{equation}
Inserting for Equation \eqref{eq:dx_0_curve}, and taking finite differences as an approximation to the gradients yields
\begin{equation}
    \frac{n_1 + \Dn_1}{n + \Dn} - \frac{n_1 - \Dn_1}{n - \Dn} \approx -\frac{D_T}{T D_{12}} \DT - \frac{\rho_1 \rho_2 (m_2 - m_1)}{p \rho \rho_m} \Delta p.
    \label{eq:soret_constraint}
\end{equation}
From the total differential of pressure,
\begin{equation}
    \Delta p \approx \ppder{p}{T}_{\vn, V} \DT + \sum_i \ppder{p}{n_i}_{T, V, n_k} \Dn_i.
    \label{eq:appr_delta_p}
\end{equation}
Given a thermal diffusion ratio, and a separation of component 1, Equation \eqref{eq:soret_constraint} and \eqref{eq:appr_delta_p} can be solved numerically for $\Delta n_2$. The resulting curve describes the steady state relation between $\Dn_1$ and $\Dn_2$ for a given thermal diffusion ratio. Along this curve, both $\Delta x_1$ and $\Delta p$ may vary while Equation \eqref{eq:dx_const_soret_varying_dp} is obeyed. That is, if a thermal diffusion ratio is supplied from the Enskog solutions, this curve describes the steady states permitted by the Enskog solutions.

To summarise, the relation between the molar separation of the components of a binary system, $\Delta n_1$ and $\Delta n_2$, subjected to a temperature difference $\Delta T$ have been derived under three constraints. The first is that of equal container mass, the second is equal container pressure, the third is that the Enskog solutions are obeyed. Kempers assumes that the steady state is characterised by a minimum in the target function $\tau$, given in Equation \eqref{eq:kempers_target_method}, along one of the first two curves.