Question

bases oo and ibbs emuy et of

Answer 1

7.

Given information: α and β are two liquid phases. γ^αβ is the interfacial tension between these two liquid phases.

Let, the area of the interface is 1 m², so if we separate these two phases apart we will get 1 m² interface between pure α and vapour with interfacial tension of γ^αv and similarly another 1 m² interface between pure β and vapour with interfacial tension of γ^βv. Therefore, the change in Gibbs energy for this transformation is,

ΔG_A = w_A^αβ = γ^αv + γ^βv - γ^αβ                  …      … (1)

This increase in Gibbs energy is known as work of adhesion, w_A^αβ, between the phases α and β.

Now, if we separate a pure column of α, 2 m² of α-vapour interface will form and then,

ΔG_C = w_C^α = 2γ^αv

This Gibbs energy change, w_C^α, is known as work of cohesion of α. So, for β it will be, w_C^β = 2γ^βv.

Therefore, we can write,

w_A^αβ = (1/2) w_C^α + (1/2) w_C^β - γ^αβ

or,     γ^αβ = (1/2)(w_C^α + w_C^β) - w_A^αβ        …      … (2)

or, γ^αβ = (1/2)(w_C^α + w_C^β) - ΔG_A          …      … (3)

So, from eqn. (3) we can conclude that γ^αβ decreases with the increase in Gibbs energy of adhesion (ΔG_A).

Now, from eqn. (2) when γ^αβ = 0; which means spontaneous mixing of two liquids; that is, there is no resistance to the extension of the two interfaces between α and β, we get,

w_A^αβ = (1/2)(w_C^α + w_C^β)

So, in this condition (when γ^αβ = 0) the work of adhesion becomes the average of work of cohesion of these two liquid phases α and β.