Can you prove the interesting relation in equilibrium thermodynamics?
Problem and background
I noted the following interesting relation in a paper discussing the liquid-vapor phase change, in which it was given directly without any derivation and reference:
$$d\ln p^\text{eq}_\text{v}=\frac{1}{ \rho R T}dp+ \frac{H}{RT^2}dT, \quad (1)$$
where $p^\text{eq}_\text{v}$ is the equilibrium vapor pressure of a liquid, $\rho$ is the density of the liquid, $R$ is the gas constant, $T$ is the temperature, $H$ is the molar enthalpy. However, I really want to understand how the differential relation has been obtained.
I have some basic knowledge of equilibrium thermodynamics and thought that Eq.(1) has a close relationship with the Gibbs-Duhem relation
$$d \mu=-sdT +\nu dp, \quad (2)$$ in which $\mu$ is the chemical potential, $s$ and $\nu$ are the molar entropy and volume, respectively.
I also noted that Eq.(1) include the Clausius-Clapeyron (C-C) equation in this form:
$$\frac{d \ln p^\text{eq}_\text{v}}{dT}=\frac{H}{RT^2}. \quad (3)$$ That is, the 2nd term on the right hand side (RHS) of Eq.(1) appears to be from the C-C equation.
What I have tried
I have tried to derive Eq.(1) along the following two directions.
1. From Eq.(2), the chemical potential of vapor at equilibrium can be written as
$$\mu^\text{eq}\text{v}=\mu^0\text{v}+RT \ln p^\text{eq}_\text{v}, \quad (4)$$
because $dT=0$ at equilibrium. 749
Here $\mu^0_\text{v}$ is a function of $T$ arising from integration. Plug Eq.(4) in (2)
$$d \mu^\text{eq}\text{v}=\frac{d \mu^0\text{v}}{dT}dT+R(Td \ln p^\text{eq}\text{v}+\ln p^\text{eq}\text{v} d T)=-sdT +\nu dp.$$
Solving for $d \ln p^\text{eq}_\text{v}$ yields
$$d \ln p^\text{eq}\text{v}=\frac{1}{\rho R T}dp+(-\frac{s}{RT}-\frac{1}{RT}\frac{d \mu^0\text{v}}{dT}-\frac{\ln p^\text{eq}_\text{v}}{T})dT, $$ where $p=\rho RT$ has been used. As you can see, I don't know how to reduce the terms in the parentheses to $\frac{H}{RT^2}$.
2. On the other hand, I tried to integrate Eq.(2) for vapor phase from a reference value to equilibrium value,
$$\int_{\mu^\star} ^ {\mu_\text{v} ^\text{eq}} d \mu=-\int_{T^\star}^{T_\text{v}^\text{eq}}s_\text{v}dT +\int_{p^\star}^{p_\text{v}^\text{eq}}\nu dp. \quad (5)$$
Recalling that $\nu=\frac{RT}{p}$ for vapor. 374