Question

proof that (x1 × d ln f1/d x1) + (x2 × d ln f2/d x1)=0 in chemical engineering thermodynamics

Answer 1

Answer:

One of the formulations of Gibbs-Duhem equation is:

$\Sigma{x_id\bar{Z_i}} = 0,$

where $x_i$ is the molar fraction of a component i and $\bar{Z}_i$ is the partial molar property of this component.

Fugacity is linked to the chemical potential $\mu_i$ in a following way:

$\mu_i = \mu_i^o + RT\text{ln}f_i$

Therefore, the change of the chemical potential is:

$d\mu_i = RT d\text{ln}f_i$

When we substitute chemical potential in the equation of Gibbs-Duhem, we get:

$\Sigma{x_id\mu_i} = 0$

$RT\Sigma{x_id\text{ln}f_i} = 0, \text{or } \Sigma{x_id\text{ln}f_i} = 0$

Let's use the obtained equation for a binary mixture:

$x_1d\text{ln}f_1 +x_2d\text{ln}f_2 = 0$

For an infinitesimal change of the composition of the mixture, we get:

$x_1d\text{ln}f_1/dx_1 +x_2d\text{ln}f_2/dx_1 = 0$ .