Question

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

Answer 1

$\phi=\sum\ x_i ln\ \hat{\phi_i}$ (As per standard notations).

Here $\phi$ is a molar property with $ln\ \hat{\phi_i}$ be the property of a component in solution.

By using Gibbs-Duhem equation,

At constant $p,$

$x_1\ (d\ ln\hat{\phi_1})+x_2\ (d\ ln\hat{\phi_2})=0$

And,$\hat{f_i} =x_i \hat{\phi_i}{p}\implies \hat{\phi_i}=\frac{\hat{f_i}}{x_i \ p}$

$\implies x_1\ (d\ ln\frac{\hat{f_1}}{p\ x_1})+x_2\ (d\ ln\frac{\hat{f_2}}{p \ x_2})=0$

$\implies x_1\ ( d(\ ln{\hat{f_1}})-d(ln{\ x_1 p}))$ $+x_2(\ (d\ ln{\hat{f_2}})-d(ln \ { x_2p\ }))=0$

As pressure is constant,$d(ln \ { x_1p\ })=d(ln \ { x_2p\ })=0$

The equation reduces to $x_1(d\ ln{\hat{f_1}})+x_2(d\ ln{\hat{f_2}})=0$

By using Lewis-Randall Rule,$\hat{f_1}=x_1f_1$ ,

We get, $x_1\times d\ ln{(\ x_1{f_1)}}+x_2\times d\ ln{{\ (x_2 f_2)}}=0$