PHYSICAL CHEMISTRY IN BRIEF

CHAP. 2: STATE BEHAVIOUR [CONTENTS] 59

2.4.2 Amagat’s law

According to Amagat’s law, the volumeV, the molar volumeVmand the compressibility factor
zof a mixture are given by

V=

∑k

j=1

Vj•, Vm=

∑k

j=1

xjVm•,j, z=

∑k

j=1

xjzj•, (2.39)

whereVj•=Vj•(T, p, nj) is the volume,Vm•,j=Vm•,j(T, p) is the molar volume, andz•j=z•(T, p)
is the compressibility factor of a pure substance j at the temperature and pressure of the
mixture and in the same state of matter as the mixture.
If pure gases obey the equation of state of an ideal gas, we must have

Vi•=V xi, (2.40)

whereV is the volume of the mixture andxiis the mole fraction of substanceiin the mixture.

Example

At a temperature of 25◦C and a pressure of 101 325 Pa, the molar volume of water is

18.07 cm^3 mol−^1 , and the molar volume of methanol is 40.73 cm^3 mol−^1. Using Amagat’s law,

estimate the molar volume of a mixture containing 2 moles of water and 5 moles of methanol at

the same temperature and pressure.

Solution

The mole fractions of the components in the mixture are

xH 2 O=

2

2 + 5

=

2

7

, xmethanol=

5

7

.

We estimate the molar volume of the mixture using the second of equations (2.39).

Vm=

2

7

18 .07 +

5

7

40 .73 = 34. 26 cm^3 mol−^1.