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:

${\left(\frac{\mathrm{d}\ln P_A}{\mathrm{d}\ln x_A}\right)}_{T,P}={\left(\frac{\mathrm{d}\ln P_B}{\mathrm{d}\ln x_B}\right)}_{T,P}$

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 vapor at constant temperature and pressure. Then from Gibbs–Duhem equation is

$n_A\mathrm{d}{\mu }_A+n_B\mathrm{d}{\mu }_B=0$

(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

$\frac{n_A}{n_A+n_B}\mathrm{d}{\mu }_A+\frac{n_B}{n_A+n_B}\mathrm{d}{\mu }_B=0$

Or

$x_A\mathrm{d}{\mu }_A+x_B\mathrm{d}{\mu }_B=0$

(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

$\mathrm{d}{\mu }_A={\left(\frac{\mathrm{d}{\mu }_A}{\mathrm{d}{x}_A}\right)}_{T,P}\mathrm{d}{x}_A$

(3)

$\mathrm{d}{\mu }_B={\left(\frac{\mathrm{d}{\mu }_B}{\mathrm{d}{x}_B}\right)}_{T,P}\mathrm{d}{x}_B$

(4)

Putting these values in equation (2), then

$x_A{\left(\frac{\mathrm{d}{\mu }_A}{\mathrm{d}{x}_A}\right)}_{T,P}\mathrm{d}{x}_A+x_B{\left(\frac{\mathrm{d}{\mu }_B}{\mathrm{d}{x}_B}\right)}_{T,P}\mathrm{d}{x}_B=0$

(5)

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

$x_1+x_2=1$

Hence

$\mathrm{d}{x}_1+\mathrm{d}{x}_2=0$

so equation (5) can be re-written:

$x_A{\left(\frac{\mathrm{d}{\mu }_A}{\mathrm{d}{x}_A}\right)}_{T,P}=x_B{\left(\frac{\mathrm{d}{\mu }_B}{\mathrm{d}{x}_B}\right)}_{T,P}$

(6)

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

$\mu ={\mu }_0+RT\ln P$

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

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

So we have for components A and B

$\frac{\mathrm{d}{\mu }_A}{\mathrm{d}{x}_A}=RT\frac{\mathrm{d}\ln P_A}{\mathrm{d}{x}_A}$

(7)

$\frac{\mathrm{d}{\mu }_B}{\mathrm{d}{x}_B}=RT\frac{\mathrm{d}\ln P_B}{\mathrm{d}{x}_B}$

(8)

Substituting these value in equation (6), then

$x_A\frac{\mathrm{d}\ln P_A}{\mathrm{d}{x}_A}=x_B\frac{\mathrm{d}\ln P_B}{\mathrm{d}{x}_B}$

or

${\left(\frac{\mathrm{d}\ln P_A}{\mathrm{d}\ln x_A}\right)}_{T,P}={\left(\frac{\mathrm{d}\ln P_B}{\mathrm{d}\ln x_B}\right)}_{T,P}$

this is the final equation of Duhem–Margules equation.

Sources

• Atkins, Peter and Julio de Paula. 2002. Physical Chemistry, 7th ed. New York: W. H. Freeman and Co.
• Carter, Ashley H. 2001. Classical and Statistical Thermodynamics. Upper Saddle River: Prentice Hall.