The molar volume for a binary mixture solution of species 1 and species 2 at 298

Question

The molar volume for a binary mixture solution of species 1 and species 2 at 298.15 K and 1 bar is given as V = x_1x_2 (45x_1 + 25x_2) + 110x_1 + 90x_2 where x_1 and x_2 are mole fractions of species 1 and 2 in the solution, respectively. V is in the unit of cm^3/mol. Determine expressions for the partial molar volumes, bar V_1 and bar V_2; Show that the derived expressions in part (a) satisfy the Gibbs/Duhem equation; To prepare a 1,000 cm^3 solution consisting of 60 mol-% species 1 and 40 mol-% of species 2, calculate the required volumes of pure species 1 solution and of pure species 2 solution.

Explanation / Answer

V=x1x2*(45x1+25x2)+ 110x1+90x2

V=x1*(1-x1)*(45x1+25*(1-x1))+ 110x1+90*(1-x1)

= (x1-x12)*( 25+20x1)+ 90+20x1

= 25x1+20x12 – 25x12-20x13+90+20x1

= -20x13 -5x12 +45x1+90

(dV/dx1)= -60x12-10x1+45

V1- =partial molal volume of species 1= V+x2*dV/dx1

= -20x13 -5x12 +45x1+90 +(1-x1)*( -60x12-10x1+45)

= -20x13 -5x12 +45x1+90 -60x12-10x1+45 +60x13+10x12-45x1

= 40x13-55x12-10x1+135

V2- = partial molal property of species -2

= V-x1*(dV/dx)= -20x13 -5x12 +45x1+90 –x1*( -60x12-10x1+45)

=40x13+5x12+90

According to Gibbs duhem equation

, x1*dV1- +x2*dV2- =0

Or x1*dV1-/dx1+(1-x1)*dV2-/dx1=0

, x1*(120x12-110x1-10)+ (1-x1)* (120x12+10x1)

=120x13-110x12 +120x12 -10x1-120x13-10x12 +10x1= 0

So   Gibbs-Duhem theorem is satisfied

For x1=0.6 and x2=0.4

V= x1V1- +x2V2-

V= 0.6*(40*0.63-55*0.62-10*0.6+135) +0.4*(40*0.63+5*0.62+90)=110.88 cm3/mol

Number of moles of mixture in solution =1000/110.88=9.018 moles

Since x1= n1/n, n1=0.6*9.018=5.4, n2= 8.162-4.89=3.618

Where n1 and n2 are moles of species 1 and 2 in the mixture

For x1=0 the volume is pure component volume of species 2

V=x1x2*(45x1+25x2)+ 110x1+90x2

V2=90 cm3/mol

Similarly V1= 110 cm3/mole by x2=0

V1ttoal = 110*5.4 = 594 cm3

V2total = 90*3.618=326 cm3

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