How does one show the equivalence (or difference) of Darcy's law written using Hubbert's potential of a real gas to that of Darcy's law written using the concept of pseudo-pressure (aka pseudo-potential)?
To further describe my question and my efforts so far, here are my explanations and sticking points:
If we take Hubbert's potential to be defined as $$\tag{1} \Phi^h=gz+\int_{p_b}^p \frac{dp}{\rho}$$
with the elevation coordinate ($z$) taken as positive upward,
then Darcy's law can be written as $$\tag{2} q=-\frac{kA\rho}{\mu} \frac{d\Phi^h}{ds}$$
where $s$ is the distance in the direction of flow, which is taken as positive.
Invoking Eqn 1 in Eqn 2, and distributing the density ($\rho$) term, we get, $$\tag{3} q=-\frac{kA}{\mu} \frac{d(\rho gz+\int_{p_b}^p dp)}{ds}$$
To obtain Eqn 3 in difference form, noting that dynamic viscosity ($\mu$) is a function of pressure, we separate variables and integrate from distance $0$ to $L$, where the pressures are $p_1$ and $p_2$, respectively $$\tag{4} q \int_{0}^L ds=-kA \int_{p_1}^{p_2} \frac{1}{\mu} d\left(\rho gz+\int_{p_b}^p dp\right)$$
I should state that the relationship between pressure ($p$) and gas density is
$$\tag{5} \rho=\frac{pM_w}{z_gRT}$$
If we take the gas molecular weight ($M_w$), universal gas constant ($R$), and temperature ($T$) as constants, then we can simplify the expression as
$$\tag{6} \rho=\frac{p}{z_g}$$
Also, another aspect of gas flow I should mention is that as the gas flows from high potential to low potential it expands, i.e. the volumetric rate ($q$) is not constant. However, the mass rate ($\dot m$) is a constant, and the relationship between mass rate and volumetric rate is $$\tag{7} q=\frac{\dot m}{\rho}$$
Continuing from Eqn 4, I think I can further separate variables as so: $$\tag{8} q\int_0^L ds=-kA\left[\int_{p_1}^{p_2}\frac{\rho g}{\mu} dz+\int_{p_1}^{p_2} \int_{p_b}^{p}\frac{dp}{\mu}\right]$$
Substituting the relationship between mass rate and volumetric rate (Eqn 7) and then multiplying through by density, $$\tag{9} \dot m \int_0^L ds=-kA\left[\int_{p_1}^{p_2}\frac{\rho^2 g}{\mu} dz+\int_{p_1}^{p_2} \int_{p_b}^{p}\frac{\rho}{\mu} dp\right]$$
I will note here that the generalize form for pseudo-pressure (denoted as $m(p)$) is written as $$\tag{10} m(p)=\int_{p_b}^p \frac{\rho}{\mu}dp$$
therefore,
$$\tag{11} \dot m \int_0^L ds=-kA\left[\int_{p_1}^{p_2}\frac{\rho^2 g}{\mu} dz+\int_{p_1}^{p_2} m(p)\right]$$
At this point I'm not sure how to handle the integral of the first term on the right hand side of Eqn 11. Performing the other integrals, I believe I am stuck here:
$$\tag{12} \dot m L=-kA\left[\int_{p_1}^{p_2}\frac{\rho^2 g}{\mu} dz+(m(p_2)-m(p_1))\right]$$
and then
$$\tag{13} \dot m L=-kA\left[\int_{p_1}^{p_2}\frac{\rho^2 g}{\mu} dz+\Delta m(p)\right]$$
