PHYSICAL CHEMISTRY IN BRIEF

CHAP. 13: PHYSICAL CHEMISTRY OF SURFACES [CONTENTS] 449

Obr.13.5:Adsorption isobars

At higher adsorption values the dependence ofpsurbecomes more complicated, due to the
occurrence of phenomena corresponding to condensation etc.

Example

Calculate the pressure created by a surface film containing C 18 H 37 NH 2 ·HCl and spread on an area

of 14 × 35 cm^2 = 0. 049 m^2 in an amount of 1. 1 × 10 −^8 mol at a temperature of 22◦C. Assume

that the given substance practically does not dissolve in water, and that equation (13.31) applies.

Solution

Since the substance does not dissolve, all of it is absorbed in the surface phase andΓ 2 , 1 = Γ 2 =

n 2 /A. By substituting into (13.31) we obtain

psur=

1. 1 × 10 −^8

0. 049

8. 314 · 295 .15 = 0. 551 × 10 −^3 N m−^1 = 0. 551 mN m−^1.