PHYSICAL CHEMISTRY IN BRIEF

CHAP. 6: THERMODYNAMICS OF HOMOGENEOUS MIXTURES [CONTENTS] 151

b

b

Vm

V 2

Vm,• 2

V 1

Vm,• 1

id.mix

Vm=f(x 1 )

x 1

A

A

A

AAU

b

@

@

@@R

Obr.6.1:Graphical determination of the partial molar volume using the method of intercepts. The
thick curve describes the dependence of the molar volumeVmon the mole fractionx 1 of component

1. The intercepts cut out by the tangent to this curve on the vertical lines forx 1 = 0 andx 1 = 1
   determine the partial molar volumes of componentsV 2 andV 1 .Vm•, 1 andVm•, 2 are the volumes of pure
   substances 1 and 2. The line segment connecting them describes the dependence of the molar volume
   of an ideal mixture on the mole fraction.

Solution

Forx 1 = 0.4 we have

Vm = 0. 4 × 73 .936 + 0. 6 × 89. 412 − 0. 4 × 0. 6 × 0 .272 = 83. 156 cm^3 mol−^1 ,

(

∂Vm

∂x 1

)

= 73. 936 − 89. 412 − 0. 272 ×(1− 2 × 0 .4) =− 15. 530 cm^3 mol−^1

We substitute into (6.52)V 1 = 83.156 + 0. 6 ×(− 15 .530) = 73. 838 cm^3 mol−^1 ,

V 2 = 89. 368 cm^3 mol−^1.