Physical Chemistry Third Edition

224 5 Phase Equilibrium

In a system consisting of water and diethyl ether near room temperature, there is a

liquid phase that is mostly water, a liquid phase that is mostly diethyl ether, and a vapor

phase. There is one surface between the two liquid phases and one surface between

the upper liquid phase and the vapor phase. In addition, each phase has a surface with

the container. The energy, Gibbs energy, and other energy-related functions depend

separately on the area of each of these surfaces. The most important type of surface

in this system is the liquid–vapor surface. In many cases a system will have only one

significant surface. A liquid droplet suspended in a gas has only one surface. If a system

has only one component and one significant surface the energy is a function ofT,P,

A(the surface area), andn, so that

dUTdS−PdV+γdA+μdn (5.5-1)

where

γ

(

∂U

∂A

)

S,V,n

(5.5-2)

We can also write

dGdU+PdV+VdP−SdT−TdS

−SdT+VdP+γdA+μdn (5.5-3)

so that

γ

(

∂G

∂A

)

T,P,n

(5.5-4)

The quantityγcan be interpreted as the energy per unit area at constantSandVor as

the Gibbs energy per unit area at constantTandP. Work to create surface area is an

example ofnet workas discussed in Section 4.1. From Eq. (4.1-31) we recognizeγas

equal to the reversible work per unit area required to produce new surface at constant

TandP. In the case of a liquid–vapor surface, we identify it with the surface energy

calculated in Example 5.10 or Exercise 5.13.

Surface Tension

We now show that the surface energy corresponds to a force per unit length. The system

depicted in Figure 5.16 has a liquid phase in contact with a vapor phase. There is a

rectangular wire frame that protrudes from the surface of the liquid with a length equal

toLand a height above the liquid surface equal tox. There is a film of liquid within

the frame. We reversibly move the frame upward by a distancedx, increasing the area

inside the frame byLdxand the energy by

dUTdS+γdA (5.5-5)

We assume that the volume of the system does not change, so there is noPdVterm in

this equation. The system is closed so there is noμdnterm. Since there are two sides

to the liquid layer, the area of the surface increases by 2Ldx, and

dUTdS+ 2 Lγdx (5.5-6)

SincedqrevTdS

dwrevdU−dqrevdU−TdSγdA 2 LγdxFrevdx