PHYSICAL CHEMISTRY IN BRIEF

CHAP. 7: PHASE EQUILIBRIA [CONTENTS] 196

Specified Calculated From equation How

T, x 1 p p=x 1 ps 1 +x 2 ps 2 analytically

y 1 y 1 =x 1 ps 1 /p analytically

T, y 1 p p= (y 1 /ps 1 +y 2 /ps 2 )−^1 analytically

x 1 x 1 =p y 1 /ps 1 analytically

T, p x 1 x 1 = (p−ps 2 )/(ps 1 −ps 2 ) analytically

y 1 y 1 =x 1 ps 1 /p analytically

p, x 1 T p=x 1 ps 1 (T) +x 2 ps 2 (T) numerically

y 1 y 1 =x 1 ps 1 /p analytically

p, y 1 T p= (y 1 /ps 1 (T) +y 2 /ps 2 (T))−^1 numerically

x 1 x 1 =p y 1 /ps 1 analytically

x 1 , y 1 T y 1 /y 2 =ps 1 (T)x 1 /(ps 2 (T)x 2 ) numerically

p p=x 1 ps 1 +x 2 ps 2 analytically

Figure7.5shows the dependence of the total and partial pressures of individual components
on the composition of the liquid phase in a binary system. If Raoult’s law applies, the relations
p=f(x 1 ) andp 1 =pyi=f(x 1 ) are linear (p=ps 2 + (ps 1 −ps 2 )x 1 ). The values given by Raoult’s
law are represented by line segments in the diagram.

Note:For most liquid mixtures, the approximation using an ideal solution is rather rough.

In practice, therefore, more accurate relations presented in the following section are usually

used in place of Raoult’s law.

7.6.3 Liquid-vapour equilibrium with an ideal vapour and a real liquid phase

If the vapour phase behaves as an ideal gas but the liquid phase is not an ideal mixture, we
can write
p yi=γixipsi, i= 1. 2 ,... , k , (7.24)