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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 of T , P,
A (the surface area), and n, so that
dU TdS − PdV + γdA + µdn
(5.5-1)
where
γ
∂U
∂A
S,V ,n
(5.5-2)
We can also write
dG dU + PdV + VdP − SdT − TdS
−SdT + VdP + γdA + µdn
(5.5-3)
so that
γ
∂G
∂A
T ,P,n
(5.5-4)
The quantity γ can be interpreted as the energy per unit area at constant S and V or as
the Gibbs energy per unit area at constant T and P. Work to create surface area is an
example of net work as discussed in Section 4.1. From Eq. (4.1-31) we recognize γ as
equal to the reversible work per unit area required to produce new surface at constant
T and P. 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
to L and a height above the liquid surface equal to x. There is a film of liquid within
the frame. We reversibly move the frame upward by a distance dx, increasing the area
inside the frame by Ldx and the energy by
dU TdS + γdA
(5.5-5)
We assume that the volume of the system does not change, so there is no PdV term in
this equation. The system is closed so there is no µdn term. Since there are two sides
to the liquid layer, the area of the surface increases by 2Ldx, and
dU TdS + 2Lγdx
(5.5-6)
Since dq rev TdS
dw rev dU − dq rev dU − TdS γdA 2Lγdx F rev dx
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 of T , P,
A (the surface area), and n, so that
dU TdS − PdV + γdA + µdn
(5.5-1)
where
γ
∂U
∂A
S,V ,n
(5.5-2)
We can also write
dG dU + PdV + VdP − SdT − TdS
−SdT + VdP + γdA + µdn
(5.5-3)
so that
γ
∂G
∂A
T ,P,n
(5.5-4)
The quantity γ can be interpreted as the energy per unit area at constant S and V or as
the Gibbs energy per unit area at constant T and P. Work to create surface area is an
example of net work as discussed in Section 4.1. From Eq. (4.1-31) we recognize γ as
equal to the reversible work per unit area required to produce new surface at constant
T and P. 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
to L and a height above the liquid surface equal to x. There is a film of liquid within
the frame. We reversibly move the frame upward by a distance dx, increasing the area
inside the frame by Ldx and the energy by
dU TdS + γdA
(5.5-5)
We assume that the volume of the system does not change, so there is no PdV term in
this equation. The system is closed so there is no µdn term. Since there are two sides
to the liquid layer, the area of the surface increases by 2Ldx, and
dU TdS + 2Lγdx
(5.5-6)
Since dq rev TdS
dw rev dU − dq rev dU − TdS γdA 2Lγdx F rev dx