Check that equations 5.69 and 5.70 satisfy the identityG=NAμA+NBμB (equation 5.37)

Short Answer

Expert verified

Hence, the identity is satisfied.

G=NAμA+NBμB

Step by step solution

01

Given information

The identityG=NAμA+NBμB

02

Explanation

The chemical potentials of the solvent and solute are determined by the following equations:

μA=μ0-NBkTNAμB=f+klnNBNA

Where, μ0andfare functions of T and P.

We need to check that these equation satisfy equation 5.37, which is given by:

G=NAμA+NBμB

substitute with the above equations to get:

NAμA+NBμB=NAμ0-NBkT+NBf+NBklnNBNA

By using ln(A/B)=ln(A)-ln(B), we can write G as

NAμA+NBμB=NAμ0-NBkT+NBf+NBklnNB-NBkln(N)(1)

The Gibbs free energy for pure solvent is given by (from equation 5.68):

G=NAμ0+NBf-NBkTlnNA+NBkTlnNB-NBkT(2)

Comparing (1) and (2)

G=NAμA+NBμB

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