Question

Check that equations 5.69 and 5.70 satisfy the identity$G=N_A{\mu }_A+N_B{\mu }_B$ (equation 5.37)

Short answer

Hence, the identity is satisfied.

$G=N_A{\mu }_A+N_B{\mu }_B$

Step-by-step solution

Given information

The identity$G=N_A{\mu }_A+N_B{\mu }_B$

Explanation

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

$$\begin{gathered} {\mu }_A={\mu }_0-\frac{N_{B}kT}{N_A} \\ {\mu }_B=f+k\ln \left(\frac{N_B}{N_A}\right) \end{gathered}$$

Where, ${\mu }_0\text{and}f$are functions of T and P.

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

$G=N_A{\mu }_A+N_B{\mu }_B$

substitute with the above equations to get:

$N_A{\mu }_A+N_B{\mu }_B=N_A{\mu }_0-N_{B}kT+N_{B}f+N_{B}k\ln \left(\frac{N_B}{N_A}\right)$

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

$N_A{\mu }_A+N_B{\mu }_B=N_A{\mu }_0-N_{B}kT+N_{B}f+N_{B}k\ln \left(N_B\right)-N_{B}k\ln \left(N\right)\left(1\right)$

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

$G=N_A{\mu }_0+N_{B}f-N_{B}kT\ln \left(N_A\right)+N_{B}kT\ln \left(N_B\right)-N_{B}kT\left(2\right)$

Comparing (1) and (2)

$G=N_A{\mu }_A+N_B{\mu }_B$