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. 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.
I then arrived
$$\mu_\text{v} ^\text{eq}-\mu^\star=-s_\text{v}\int_{T^\star}^{T_\text{v}^\text{eq}}dT+
RT\int_{p^\star}^{p_\text{v}^\text{eq}}\frac{dp}{p}=-s_\text{v}(T_\text{v}^\text{eq}-T^\star)+RT \ln \frac{p_\text{v}^\text{eq}}{p^\star},$$
in which I have assumed that the variations in $s_\text{v}$ and $T$ are small in the RHS-two terms, respectively.
My question is how to derive equation (1). Thank you very much.
 A: I get a slightly different expression from (1), but maybe looking over this derivation will be useful, as I may have made a mistake.
Let $$\Delta \mu^\circ=\mu_V^\circ-\mu_L^\circ\tag{1}$$ where $V$ is the vapor state and $L$ is the liquid state and $^\circ$ indicates a reference condition. At equilibrium, $$\mu_V=\mu_V^\circ+RT\ln a_V \tag{2}$$ must equal $$\mu_L=\mu_L^\circ+RT\ln a_L\tag{3}$$. Now assume that the activity of the vapor is equal to the partial pressure $p$ and that the activity of the liquid is 1: $$RT\ln p=-\Delta \mu^\circ=-\Delta h^\circ+T\Delta s^\circ\tag{4}$$ or $$\ln p=-\frac{\Delta h^\circ}{RT}+\frac{\Delta s^\circ}{R}\tag{5}$$ Take the differential: $$d(\ln p)=\frac{\Delta h^\circ}{RT^2}dT-\frac{1}{RT}\left[\left(\frac{\partial\Delta h^\circ}{\partial T}\right)_P dT+\left(\frac{\partial\Delta h^\circ}{\partial P}\right)_T dP\right]+\frac{1}{R}\left[\left(\frac{\partial\Delta s^\circ}{\partial T}\right)_P dT+\left(\frac{\partial\Delta s^\circ}{\partial P}\right)_T d\right]\tag{6}$$
But $$\frac{1}{RT}\left(\frac{\partial\Delta h^\circ}{\partial T}\right)_P=\frac{1}{R}\left(\frac{\partial\Delta s^\circ}{\partial T}\right)_P=\frac{\Delta c_P}{RT}\tag{7}$$ so some terms cancel out, leaving $$d(\ln p)=\frac{\Delta h^\circ}{RT^2}dT-\frac{1}{RT}\left(\frac{\partial\Delta h^\circ}{\partial P}\right)_T dP+\frac{1}{R}\left(\frac{\partial\Delta s^\circ}{\partial P}\right)_T d\tag{8}P$$
$$d(\ln p)=\frac{\Delta h^\circ}{RT^2}dT-\frac{1}{RT}\left(\frac{\partial\Delta \mu^\circ}{\partial P}\right)_T dP\tag{9}$$ From $\Delta \mu^\circ=-\Delta s^\circ\,dT+\Delta v^\circ\,dP$, we get $$d(\ln p)=\frac{\Delta h^\circ}{RT^2}dT-\frac{\Delta v^\circ}{RT}dP\tag{10}$$
For vaporization, $$\Delta v^\circ\approx\frac{RT}{P}\tag{11}$$ for a large expansion to the vapor phase; thus,
 $$d(\ln p)=\frac{\Delta h^\circ}{RT^2}dT-\frac{1}{\rho R T}dP\tag{12}$$ As I said, a little different (one change in sign and one difference in molar enthalpy rather than absolute molar enthalpy).
