Answer:
a) [tex]\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)][/tex]
b) [tex]x=4[/tex]
c) [tex]\Delta G_{max}=-2721.9 J/mol[/tex]
Explanation:
Gas mixture:
[tex]n_{Ar}= 4 mol[/tex]
[tex]n_{Ne}= x mol[/tex]
[tex]n_{Xe}= y mol[/tex]
[tex]n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}[/tex]
[tex] n_{Ne} + n_{Xe}=2*n_{Ar}[/tex]
[tex] x + y=8 mol[/tex]
[tex]y=8 mol- x[/tex]
Mol fractions:
[tex]x_{Ar}=\frac{4 mol}{12 mol}=1/3[/tex]
[tex]x_{Ne}=\frac{x mol}{12 mol}=x/12[/tex]
[tex]x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12[/tex]
Expression of [tex]\Delta G_{mixing}[/tex]
[tex]\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)[/tex]
[tex]\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)][/tex]
[tex]\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)][/tex]
Expression of [tex]\Delta G_{max}[/tex]
[tex]\frac{d \Delta G_{mixing}}{dx}=0[/tex]
[tex]\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12][/tex]
[tex]0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)[/tex]
[tex]0=[ln (x/12)-ln ((8-x)/12)[/tex]
[tex]ln (x/12)=ln ((8-x)/12)[/tex]
[tex]x=(8-x)[/tex]
[tex]x=4[/tex]
[tex]\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)][/tex]
[tex]\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)][/tex]
[tex]\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)][/tex]
[tex]\Delta G_{max}=-2721.9 J/mol[/tex]
