Question

The excess gibbs free energy of a binary liquid mixture at t and p is given by: T = (−2.6 x1 − 1.8x2)x1x2 , a) find expression of ln 1 and ln 2 at t and p.

Answer 1

To find ln γ1 and ln γ2 for a binary mixture with the given excess Gibbs free energy, differentiating the excess Gibbs energy expression with respect to the mole fractions and applying the appropriate formulas yields ln γ1 and ln γ2 as functions of the composition.

Step-by-step explanation:

To find the expressions for ln γ1 and ln γ2, we can refer to the Gibbs-Duhem relationship and the fact that the excess Gibbs free energy (Gex) of a mixture is related to the chemical potential (and thus activity coefficients) of the components. The given expression for excess Gibbs free energy is Gex = −(2.6x1 + 1.8x2)x1x2,

To determine the activity coefficients γ1 and γ2, we use the relations: ln γ1 = (∂Gex)/(∂n1)T,P,n2 and ln γ2 = (∂Gex)/(∂n2)T,P,n1. Differentiating Gex with respect to n1 and n2 while holding temperature and pressure constant yields:

ln γ1 = −(2.6 − 2.6x2)x2

ln γ2 = −(1.8 − 1.8x1)x1

Remember that x1 + x2 = 1 for a binary mixture, so you can simplify the expressions further if necessary.