Duhem–Margules equation

The Duhem–Margules equation, named for Pierre Duhem and Max Margules, is a thermodynamic statement of the relationship between the two components of a single liquid where the vapour mixture is regarded as an ideal gas:

${\displaystyle \left({\frac {\mathrm {d} \,\ln \,P_{A}}{\mathrm {d} \,\ln \,x_{A}}}\right)_{T,P}=\left({\frac {\mathrm {d} \,\ln \,P_{B}}{\mathrm {d} \,\ln \,x_{B}}}\right)_{T,P}}$

where P_A and P_B are the partial vapour pressures of the two constituents and x_A and x_B are the mole fractions of the liquid.

Derivation

Duhem - Margulus equation give the relation between change of mole fraction with partial pressure of a component in a liquid mixture.

Let consider a binary liquid mixture of two component in equilibrium with their vapour at constant temperature and pressure. Then from Gibbs - Duhem equation is

${\displaystyle n_{A}\mathrm {d} \mu _{A}+n_{B}\mathrm {d} \mu _{B}=0\qquad [1]}$

Where n_A and n_B are number of moles of the component A and B while μ_A and μ_B is their chemical potential.

Dividing equation (1) by n_A + n_B , then

${\displaystyle {\frac {n_{A}}{n_{A}+n_{B}}}\mathrm {d} \mu _{A}+{\frac {n_{B}}{n_{A}+n_{B}}}\mathrm {d} \mu _{B}=0}$

Or

${\displaystyle x_{A}\mathrm {d} \mu _{A}+x_{B}\mathrm {d} \mu _{B}=0\qquad [2]}$

Now the chemical potential of any component in mixture is depend upon temperature, pressure and composition of mixture. Hence if temperature and pressure taking constant then chemical potential

${\displaystyle \mathrm {d} \mu _{A}=\left({\frac {\mathrm {d} \mu _{A}}{\mathrm {d} x_{A}}}\right)_{T,P}\mathrm {d} x_{A}\qquad [3]}$

${\displaystyle \mathrm {d} \mu _{B}=\left({\frac {\mathrm {d} \mu _{B}}{\mathrm {d} x_{B}}}\right)_{T,P}\mathrm {d} x_{B}\qquad [4]}$

Putting these values in equation (2), then

${\displaystyle x_{A}\left({\frac {\mathrm {d} \mu _{A}}{\mathrm {d} x_{A}}}\right)_{T,P}\mathrm {d} x_{A}+x_{B}\left({\frac {\mathrm {d} \mu _{B}}{\mathrm {d} x_{B}}}\right)_{T,P}\mathrm {d} x_{B}=0\qquad [5]}$

Because the sum of mole fraction of all component in the mixture is unity i.e.,

${\displaystyle x_{1}+x_{2}=1}$

Hence

${\displaystyle \mathrm {d} x_{1}+\mathrm {d} x_{2}=0}$

so equation (5) can be re-written:

${\displaystyle x_{A}\left({\frac {\mathrm {d} \mu _{A}}{\mathrm {d} x_{A}}}\right)_{T,P}=x_{B}\left({\frac {\mathrm {d} \mu _{B}}{\mathrm {d} x_{B}}}\right)_{T,P}\qquad [6]}$

Now the chemical potential of any component in mixture is such that

${\displaystyle \mu =\mu _{0}+RT\ln P}$

where P is partial pressure of component. By differentiating this equation with respect to the mole fraction of a component:

${\displaystyle {\frac {\mathrm {\mathrm {d} } \mu }{\mathrm {\mathrm {d} } x}}=RT{\frac {\mathrm {\mathrm {d} } \ln P}{\mathrm {d} x}}}$

So we have for components A and B

${\displaystyle {\frac {\mathrm {d} \mu _{A}}{\mathrm {d} x_{A}}}=RT{\frac {\mathrm {d} \ln P_{A}}{\mathrm {d} x_{A}}}\qquad [7]}$

${\displaystyle {\frac {\mathrm {d} \mu _{B}}{\mathrm {d} x_{B}}}=RT{\frac {\mathrm {d} \ln P_{B}}{\mathrm {d} x_{B}}}\qquad [8]}$

Substituting these value in equation (6), then

${\displaystyle x_{A}{\frac {\mathrm {d} \ln P_{A}}{\mathrm {d} x_{A}}}=x_{B}{\frac {\mathrm {d} \ln P_{B}}{\mathrm {d} x_{B}}}}$

or

${\displaystyle \left({\frac {\mathrm {d} \,\ln \,P_{A}}{\mathrm {d} \,\ln \,x_{A}}}\right)_{T,P}=\left({\frac {\mathrm {d} \,\ln \,P_{B}}{\mathrm {d} \,\ln \,x_{B}}}\right)_{T,P}}$

this is the final equation of Duhem- Margules equation.