We have learned molar $G$ change due to formation of a regular solution is

$G_m^{\text{m}\text{i}\text{x}}=G_m^{\text{m}\text{i}\text{x}\text{,}\text{i}\text{d}\text{e}\text{a}\text{l}}+G_m^e=RT(x_{A}ln{}_A+x_{B}ln{}_B),$ molar $G$ change due to formation of an

ideal solution is $G_m^{\text{m}\text{i}\text{x}\text{,}\text{i}\text{d}\text{e}\text{a}\text{l}}=RT(x_{A}ln{x}_A+x_{B}ln{x}_B),$ the excess molar $G$ change is

$G_m^e=G_m^{\text{m}\text{i}\text{x}}-G_m^{\text{m}\text{i}\text{x}\text{,}\text{i}\text{d}\text{e}\text{a}\text{l}}=RT[x_{A}ln(\frac{{}_A}{x_A})+x_{B}ln(\frac{{}_B}{x_B})]=RT(x_{A}ln{}_A+x_{B}ln{}_B).$

We also learned the partial molar $G$ change of component A due to the formation of a regular

solution $i{s}_{{?}_{?}}\frac{}{b}ar(G{)}_A^{\text{m}\text{i}\text{x}}=RTln{}_A=RTln{}_A+RTln{x}_A$

Think what is the partial excess molar $G$ change of component $A$ in a solution $\frac{}{b}ar(G{)}_A^e$

The liquid Fe$-$Mn alloy at $1863K$ is a regular solution. Its excess molar $G$ is $703J$ when $x_{Mn}=0\mathrm{.}2.$

a$)$ Calculate $\frac{}{b}ar(G{)}_{Fe}^e$ and $\frac{}{b}ar(G{)}_{Mn}^e$ at $x_{Mn}=0\mathrm{.}6$; b$)$ calculate molar $G$ change due to formation of $Fe-$

Mn solution at $x_{Mn}=0\mathrm{.}4$; Draw molar $G$ change $($due to mixing$)-$composition diagram using any

software or method that you are familiar with. c$)$ calculate the partial pressures of $Mn$ and $Fe$

exerted by the alloy at $x_{Mn}=0\mathrm{.}2.$ Find relevant information in the appendix of the textbook if needed.$|$