PHYSICAL CHEMISTRY IN BRIEF

CHAP. 2: STATE BEHAVIOUR [CONTENTS] 44

Molar and specific volumes and densities may be mutually converted using the relations

Vm=M v , c=ρ/M , (2.3)

whereM is the molar mass.

2.1.3 Compressibility factorz

Definition

z=

pV

nRT

, (2.4)

whereRis the universal gas constant,R= 8.314 J mol−^1 K−^1.
U Main unit:dimensionless quantity. The compressibility factor of a pure substance is a function
of temperature and pressure or temperature and molar volume. In mixtures, the compressibility
factor is also a function of composition.

Example

At a temperature of 100◦C and a pressure of 27.6 MPa, two moles of carbon dioxide occupy a

volume of 140.06 cm^3. Calculate the compressibility factor.

Solution

From equation (2.4) we get

z=

27. 6 × 106 × 140. 06 × 10 −^6

2 × 8. 314 × 373. 15

= 0. 6230.

2.1.4 Critical point

The term critical point relates to a group of three valuesTc, pc, Vc, where

Tcis thecritical temperature,

pcis thecritical pressure,

Vcis thecritical volume.

At the critical point the properties of liquid and gas (vapour) coincide. The critical temperature

is the highest temperature at which a pure substance may exist in the liquid state.