Question

The excess Gibbs energy of a binary liquid mixture at T and P is given by: G ᴱ/RT=(−2.6x1₁−1.8x₂)x₁x₂ (a) Find expressions for lnγ1 and lnγ2 at T and P.

Answer 1

To find expressions for lnγ1 and lnγ2, we can use the Gibbs-Duhem equation and the given excess Gibbs energy equation. By finding the partial derivatives of the excess Gibbs energy with respect to each component and evaluating them, we can determine lnγ1 and lnγ2. In this case, both lnγ1 and lnγ2 are equal to zero.

Step-by-step explanation:

To find expressions for lnγ1 and lnγ2, we need to use the Gibbs-Duhem equation, which relates the activity coefficients of the components in a mixture to the partial molar quantities. In this case, the excess Gibbs energy of the binary liquid mixture is given by Gᴱ/RT=(−2.6x₁₁−1.8x₂)x₁x₂. The activity coefficients are related to the excess Gibbs energy as lnγ1 = -∂(Gᴱ/RT)/∂(x₁)/x₁ and lnγ2 = -∂(Gᴱ/RT)/∂(x₂)/x₂. So, let's find the partial derivatives and evaluate lnγ1 and lnγ2:

d(Gᴱ/RT)/dx₁ = -2.6x₂x₁ - 2.6x₁x₂

lnγ1 = -(-2.6x₂x₁)/(x₁) = 2.6x₂ - 2.6x₂ = 0

d(Gᴱ/RT)/dx₂ = -1.8x₁x₁ - 1.8x₁x₂

lnγ2 = -(-1.8x₁x₂)/(x₂) = 1.8x₁ - 1.8x₁ = 0