Question

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

ofXe is prepared at a pressure of 1 bar and a temperature of
298K.The total number of moles in the mixture is three times thatof
Ar.Write an expression for the deltaGmixing in termsof
x. At what value of x does the magnitude ofdeltaGmixing
have its maximum value? CalculatedeltaGmixing for this
value of x.

Answer 1

a)
`\Delta G_(mixing)=(R*T)/(12)*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]`

b)
`x=4`

c)
`\Delta G_(max)=-2721.9 J/mol`

Step-by-step explanation:

Gas mixture:

`n_(Ar)= 4 mol`

`n_(Ne)= x mol`

`n_(Xe)= y mol`

`n_(tot)= n_(Ar) + n_(Ne) + n_(Xe)=3*n_(Ar)`

`n_(Ne) + n_(Xe)=2*n_(Ar)`

`x + y=8 mol`

`y=8 mol- x`

Mol fractions:

`x_(Ar)=(4 mol)/(12 mol)=1/3`

`x_(Ne)=(x mol)/(12 mol)=x/12`

`x_(Xe)=(8 - x mol)/(12 mol)=(8-x)/12`

Expression of
`\Delta G_(mixing)`

`\Delta G_(mixing)=R*T*\sum_{i]*x_i*ln (x_i)`

`\Delta G_(mixing)=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]`

`\Delta G_(mixing)=(R*T)/(12)*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]`

Expression of
`\Delta G_(max)`

`(d \Delta G_(mixing))/(dx)=0`

`(d \Delta G_(mixing))/(dx)=(R*T)/(12)*[ln (x/12)+12-ln ((8-x)/12)-12]`

`0=(R*T)/(12)*[ln (x/12)-ln ((8-x)/12)`

`0=[ln (x/12)-ln ((8-x)/12)`

`ln (x/12)=ln ((8-x)/12)`

`x=(8-x)`

`x=4`

`\Delta G_(max)=(8.314*298)/(12)*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]`

`\Delta G_(max)=(8.314*298)/(12)*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]`

`\Delta G_(max)=(8.314*298)/(12)*[12*ln (1/3)]`

`\Delta G_(max)=-2721.9 J/mol`