Saturation Relations with Surface Tension Effects

6. Saturation Relations with Surface Tension Effects

We have presented a physical interpretation of surface tension in Chapter 5. We now discuss a more rigorous explanation of the phenomena based on the energy minimum princi- ple. We first write an expression for energy change for a bubble (called the embryo phase β)

that has just appeared in a single component liquid (called the mother phase α). Surface ten- sion forces exist between the vapor and liquid phases within a thin layer γ that lies adjacent to bubble and has a thickness δ, volume V γ , pressure P γ , and entropy S γ . As δ→0, then the volume

V γ , appears like a surface. Now, consider an isolated system containing three subsystems α, β and γ. Assume that the state of the combined system (α+β+γ) lies infinitesimally away from equilibrium. Consequently, infinitesimal changes in volume occur in the α and β phases, which also change the volume of the subsystem γ.

For the open subsystem α, the infinitesimal energy changed U α =T α dS α –P α dV α + µ α dN α , where dN α denotes the moles transferred from it. Similarly, for the β phase,

dU β =T β dS β –P β dV β + µ β dN β .

We assume that the surface tension region “ γ” has uniform properties that are distinct from the α and β phases. Consequently,

dU γ =T γ dS γ –P γ dV γ – (– σ´) dV γ + µ γ dN γ .

For sake of illustration, consider a section of a cylindrical bubble of length L that has an arbi- trary arc length ds and a surface tension layer thickness δ. Its volume dV γ ≈ δL ds. Let σ´ de-

note the tensile stress and P σ the compressive stress exerted on the bubble. As δ→0, σ´dV σ = σ´ δL ds ≈ σ ds L = σ dA where σ= σ´ δ, and

dU γ =T γ dS γ –P γ dV γ + σ dA + µ γ dN γ .

For the combined system dU = dU α + dU β + dU γ =T α dS α –P α dV α + µ α dN α +T β dS β –P β dV β + µ β dN β

γ dS γ –P γ dV γ + µ +T γ dN γ + σ dA.

(140) (If the change in the properties of the region γ are neglected, dU = T α dS α –P α dV α + µ α dN α +

T β dS β – P β dV β + µ β dN β + σdA.) The entropy of the isolated system cannot change, and for constant S, V and M, dS = dS α + dS β + dS γ =0, i.e., dS α = –(dS β + dS γ )

(141) Likewise, (141) Likewise,

+ dN β ( β – α ) + dN γ ( γ – α µ µ µ µ )+ σ dA = 0. (143) Arbitrary changes in dS β and dS γ must satisfy Eq. (143) so that T β = T α = T γ = T,

which is the thermal equilibrium condition, and the same argument regarding dN β and dN γ implies that α

µ γ = µ = µ = µ, which is the phase equilibrium condition. Therefore,

dV β (P β –P α ) +dV γ (P γ –P α )– σ dA = 0. The compressive pressure within the thin region on which surface tension forces oc-

cur varies from P β to P α . If we include dV γ with the mother phase volume, then P α = P γ = P (which is the condition related to vaporization from the mother phase α, which is a liquid). Omitting the subscript for the mother phase α,

(144) where

dV β (P β – P) + 0 –

σ dA, = 0, i.e., (P β σ dA/dV .

β – P) =

β is the embryo phase. Similarly, for a condensation process from β (which is mother now) to α (embryo) and now combining γ with β , P β = P, and dV β = – dV α , so that Eq. (144)

can be written in the form (P α – P) =

σ dA/dV α . Generalizing, if P E is the pressure of embryo phase (say, vapor for a boiling process or a liquid

drop in a condensation process) and P is the pressure of mother phase (say, liquid in a vapori- zation process or vapor in condensation process), then

E E (P E – P) = σ dA /d V . Pressure of embryo- Pressure of mother) = Surface tension × surface area change with volume

change of the embryo phase. Considering a spherical embryo of radius a,

A E /d V E = d(4 2 )/d(4/3 πa 3 πa ) = 2/a, and (P E – P) =

E /d V σ dA E =2 σ/a

a. Remarks The pressure P denotes the mother phase pressure in case of both evaporation and condensation. In general, for spherical drops, the difference between the pressures in- side curved surface and in the bulk phase equals 2 σ/a.

The vapor bubbles may be generated sometimes at T > T sat (P) (cf. Chapter 10). The liquid at this state is called superheated liquid. However, tables of properties are not available at this condition. Hence, we have to generate properties of superheated liq- uid in terms of saturation properties. Thus, a general derivation is presented below for both phases. If the surface tension σ ≠ 0, the pressure in vapor and liquid phases is not

the same, even though the system is isothermal. Supposing that β is the vapor phase and α is the liquid mother phase, at phase equilibrium,

µ β (P β , T) = µ α (P α , T) or f β (P β , T) =f α (P α , T) (145) Using the Poynting correction,

β P fPT β ( β ,) v β ln

( ,) T P sat ∫ () T RT

dP

fP sat

Since f(P β , T) = φ(P β , T)P β and f(P sat (T), T) = φ(P sat (T), T)P sat (T), the above equation assumes the form

f sat β = v φ β ( PTP β ,) β = φ ( P ,) T P ( )( exp [ T dP ] (147)

sat

P sat ∫ () T RT

Similarly

= v φ ( P ,) TP = φ ( P ,) T P ( )( exp [ T dP ]

sat

sat