Question

If expression 5.68 is correct, it must be extensive: Increasing both N_A and N_B by a common factor while holding all intensive variables fixed should increase G by the same factor. Show that expression 5.68 has this property. Show that it would not have this property had we not added the term proportional to In N_A!.

Short answer

Therefore,

$$\begin{gathered} G^{\text{'}}=xG \\ {G}^{\text{'}}\ne xG \end{gathered}$$

Step-by-step solution

Given information

Increasing both N_A and N_B by a common factor while holding all intensive variables fixed should increase G by the same factor.

Explanation

The Gibbs free energy for a pure solvent is calculated as follows:

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

We can show that G is an extensive quantity by replacing $N_A\text{with}x{N}_A\text{and}{N}_B\text{with}x{N}_B$while keeping the intensive quantities constant:

$G\text{'}=x{N}_A{\mu }_0+x{N}_{B}f-x{N}_{B}kT\ln \left(x{N}_A\right)+x{N}_{B}kT\ln \left(x{N}_B\right)-x{N}_{B}kT\left(2\right)$

By using $\ln \left(AB\right)=\ln \left(A\right)+\ln \left(B\right)$, we have

$$\begin{gathered} x{N}_{B}kT\ln \left(x{N}_A\right)=x{N}_{B}kT\left[\ln \left(N_A\right)+\ln \left(x\right)\right] \\ x{N}_{B}kT\ln \left(x{N}_B\right)=x{N}_{B}kT\left[\ln \left(N_B\right)+\ln \left(x\right)\right] \end{gathered}$$

Equation (2) will become

$$\begin{gathered} G^{\text{'}}=x{N}_A{\mu }_0+x{N}_{B}f-x{N}_{B}kT\ln \left(N_A\right)+x{N}_{B}kT\ln \left(N_B\right)-x{N}_{B}kT \\ {G}^{\text{'}}=x\left[N_A{\mu }_0+N_{B}f-N_{B}kT\ln \left(N_A\right)+N_{B}kT\ln \left(N_B\right)-N_{B}kT\right] \\ {G}^{\text{'}}=xG \end{gathered}$$

This means Gibbs energy is extensive quantity

Explanation

The Gibbs free energy will be: if the term $\ln \left(N_B!\right)$is not included to equation (1).

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

Replace $N_A\text{with}x{N}_A\text{and}{N}_B\text{with}x{N}_B$

$G^{\text{'}}=x{N}_A{\mu }_0+x{N}_{B}f-N_{B}kT\ln \left(x{N}_A\right)$

By using $\ln \left(AB\right)=\ln \left(A\right)+\ln \left(B\right)$

$$\begin{gathered} G=x{N}_A{\mu }_0+x{N}_{B}f-N_{B}kT\ln \left(N_A\right)-N_{B}kT\ln \left(x\right)\ne xG \\ {G}^{\text{'}}\ne xG \end{gathered}$$

Hence, Gibbs free energy will not be extensive.