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.


1 Answer 1


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).

  • $\begingroup$ Thanks for your efforts! A small error may took place just at the 1st equality of the 1st display eqn. At equilibrium (const $T$, $p_\text{V}=p_\text{L}=p_\text{V}^\text{eq}$): $\mu_\text{L}=\mu_\text{V}=\mu_\text{V}^0+RT \ln p_\text{V}^\text{eq}$. Note the index of my 1st RHS term is $\text{V}$ and the "activity" $a_\text{L}$ in the 2nd RHS term is taken as $p_\text{V}^\text{eq}$. I used subscript $\text{V}$ to distinguish the partial pressure of vapor. I think we should calculate $\mu_\text{L}$ via $\mu_\text{V}$ at equilibrium because liquid can not use ideal gas equation. $\endgroup$
    – jsxs
    Apr 18, 2017 at 1:35
  • $\begingroup$ I'm sorry, I don't follow you. Surely the activity of the liquid is 1, just as with any condensed phase? I am equating $\mu_L$ to $\mu_V$ at equilibrium. The only time the ideal gas law enters is when approximating the change in volume from the condensed phase to the vapor phase as $\Delta v^\circ\approx\frac{RT}{P}$. I numbered my equations with \tag to aid in referencing. $\endgroup$ Apr 18, 2017 at 2:45
  • $\begingroup$ It may be due to my lack in the knowledge of thermodynamics. I have two questions for discussion. 1. I don't remember the concept ''activity'', so I am not sure what are the activities of vapor and liquid. Could you pls explain in thermodynamics language? 2. In Eq.(7) I think you used the definition of molar heat capacity at constant $P$, which reads $c_P\equiv T(\partial s/\partial T)_P=\frac{1}{N}(\bar{d} Q/d T)_P$, where $\bar{d} Q$ denotes the quasi-static heat in an infinitesimal quasi-static process at constant mole no. $N$. $\endgroup$
    – jsxs
    Apr 18, 2017 at 8:02
  • $\begingroup$ @ Chemomechanics to continue...But I'm not sure whether $\bar{d} Q/N=d \Delta h$ in our present case. I only found a typo in the identity between Eqs. (9) and (10), on the LHS of which it should be $d \Delta \mu^0$. Furthermore, I guess there could be a sign mistake in the definition of $\Delta \mu^0$ or $\Delta h^0$. Especially, you use the difference in Eq.(7), if $\Delta c_P=c_{P,V}-c_{P,L}$ then $\Delta c_P<0$. I derived the expression twice using your method, but I have not found a mistake. $\endgroup$
    – jsxs
    Apr 18, 2017 at 8:20
  • $\begingroup$ @ Chemomechanics Is it the case in your derivation: $\Delta s^0=s_V^0-s_L^0>0$? $\endgroup$
    – jsxs
    Apr 18, 2017 at 9:18

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