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which a single homogeneous phase cannot exist. Two-phase regions are shown shaded Section 12.6
Two-Component
in this chapter. Liquid–Vapor Equilibrium
A point in a two-phase region of a two-component phase diagram gives the sys-
tem’s overall composition, and the compositions of the two phases in equilibrium are
given by the points at the ends of the tie line through that point.

EXAMPLE 12.3 Phase compositions at a point on a tie line

For a liquid–vapor system whose state corresponds to point E in the Fig. 12.10
phase diagram, find the overall mole fraction of B in the system and find the
mole fraction of B in each phase present in the system. Assume the scales in the
phase diagram are linear.
The overall x [Eq. (12.34)] corresponds to the x value at point E. The length
B
B
from x 0to x 1in Fig. 12.10 is 5.98 cm. The distance from x 0tothe
B
B
B
intersection of the vertical line from E with the x axis is 3.59 cm. Therefore the
B
overall x for the system at E is x 3.59/5.98 0.60 . The system at E consists
B
0
B
of liquid and vapor phases in equilibrium. The liquid-phase composition is given
by point H at the left end of the tie line HEI as x . The distance from x 0to
B
B,3
l
x x B,3 is 2.79 cm, so x B,3 2.79/5.98 0.46 x . The vapor-phase compo-
7
B
B
v
sition is given by point I on the HEI tie line as x B,2 4.22/5.98 0.70 x .
6
B
Exercise
For a liquid–vapor system whose state corresponds to the point that is the inter-
l
v
section of the vertical line from H with line JF, find the overall x , x , and x .
B
B
B
v
l
(Answer: x 0.46 , x 0.33 , x 0.60 .)
2
B
B
7
0
B
For a two-phase, two-component system, the number of degrees of freedom is f
c ind p 2 2 2 2 2. In the Fig. 12.10 phase diagram, T is held fixed and
this reduces f to 1 in the two-phase region of Fig. 12.10. Hence, once P is fixed, f is 0
v
l
in this two-phase region. For a fixed P, both x and x are thus fixed. For example, at
B
B
l
v
pressure P in Fig. 12.10, x is fixed as x B,2 and x is fixed as x . The overall x de-
B
B
E
B
B,3
pends on the relative amounts of the liquid and vapor phases that are present in equi-
librium. Recall that the masses of the phases, which are extensive variables, are not
considered in calculating f.
Different relative amounts of liquid and vapor phases at the pressure P in Fig.
E
12.10 correspond to different points along the tie line HEI with different values of the
v
l
overall mole fraction x but the same value of x and the same value of x . We now
B
B
B
relate the location of point E on the tie line to the relative amounts of the two phases
l
v
present. For the two-phase, two-component system, let n , n , and n be the total num-
B
ber of moles of B, the total number of moles in the liquid phase, and the total number
of moles in the vapor phase, respectively. The overall mole fraction of B is x
B
v
l
v
l
l
v
l
v v
l
n /(n n ), so n x n x n . Also, n n n x n x n . Equating these
B
B
B
B
B
B
B
B
B
two expressions for n , we get
B
v
v v
l
l
l
x n x n x n x n
B
B
B
B
v
v
l
l
n 1x x 2 n 1x x 2 (12.39)
B
B
B
B
v
l
n EH n EI (12.40)
where EH and EI are the lengths of the lines from E to the liquid and vapor curves in
l
v
Fig. 12.10 and n and n are the total numbers of moles in the liquid and vapor phases,
respectively. Equation (12.40) is the lever rule. Note its resemblance to the lever law
of physics: m l m l , where m and m are masses that balance each other on a
1 1 2 2 1 2
seesaw with fulcrum a distance l from mass m and l from m . When point E in
1 1 2 2