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Sav [38]
3 years ago
14

In class we derived the Gibbs energy of mixing for a binary mixture of perfect gases. We also discussed that the same result is

obtained for liquids when the resulting solution is ideal. For real solutions we introduced activities and activity coefficients. Derive the molar Gibbs energy of mixing and the molar excess Gibbs energy of mixing in terms of activity coefficients.
Chemistry
1 answer:
GaryK [48]3 years ago
4 0

Answer:

Attached below

Explanation:

Free energy of mixing = ΔGmix = Gf - Gi

attached below is the required derivation of the

<u>a) Molar Gibbs energy of mixing</u>

ΔGmix = Gf - Gi

hence : ΔGmix = ∩RT ( X1 In X1 + X2 In X2 + X3 In X3 + ------- )

<u>b) molar excess Gibbs energy of mixing</u>

Ni = chemical potential of gas

fi = Fugacity

N°i = Chemical potential of gas when Fugacity = 1

ΔG = RT In ( a2 / a1 )  

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