Duhem–Margules equation

Derivation of Duhem–Margules Equation

Duhem–Margules Equation

Duhem–Margules Equation

Duhem–Margules equation 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.
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
nAA + nBB = 0 -----(equation-1)
Where nA and nB are number of moles of the component A and B while μA and μB is their chemical potential.
dividing equation-1 by nA + nB we get-
nA/(nA + nA) dμA + nB/(nA + nB) dμB = 0
or, XAA + XAB = 0 -----(equation-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-
A = (dμA/dXA)T,P . dXA -----(equation-3)
and
B = (dμB/dXB)T,P . dXB -----(equation-4)
Putting these values in (equation-2), then-
XA(dμA/dXA)T,P . dXA + XB(dμB/dXB)T,P . dXB = 0 -----(equation-5)
we know that sum of mole fraction of all component in the mixture is unity i.e., XA + XB = 1
Hence, dXA + dXB =0
So, equation -5 becomes
XA(dμA/dXA)T,P = XB(dμB/dXB)T,P -----(equation-6)
We know that the chemical potential of any component in mixture is
μ = μo + RTlnP
where P is partial pressure of component. By differentiating this equation with respect to the mole fraction of a component we get-
dμ/dX = RT(dlnP/dX)
so for two component A and B-
A/dXA = RT(dlnPA/dXA)
B/dXB = RT(dlnPB/dXB)
Now substituting these values in equation-6 we get-
XA(dlnPA/dXA) = XB(dlnPB/dXB) -----(equation-7)

or, (dlnPA/dlnXA)T,P = (dlnPB/dlnXB)T,P -----(equation-8)
Equation-7 and 8 are two forms of Duhem–Margules equation. and these equations are used in the derivation of Konowaloff's rule which is useful in the binary liquid mixtures by distillation.

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