Figure 1: Two-phase region on a schematic phase diagram of a binary system.
Let the point ( C 0, T 0) on the phase diagram belong to a region where two phases, 1 and 2, are present. Let W 1 and W 2 denote the weight fractions of the phases 1 and 2, respectively:
Wk = m k / m , where the index k enumerates the phases ( k = 1, 2), m k is the mass of the phase k , and m = m 1Â Â + m 2Â Â is the total mass of the mixture of the phases 1 and 2.
(a) Derive the lever rule: W 1 ? C 1 = W 2 ? C 2 ,  (*)
where ? C k = | C k - C 0| , C k being the borderline composition of phase k ( k = 1, 2) at the temperature T 0 , as shown in Figure 1 above.
1(b) Using the lever rule (*), express the weight fraction of each phase W k ( k = 1, 2) through their compositions C k .
Hint for Problem 1(a):
Start with the fact that the mass of the component B in the system equals m (B) = m 1(B) + m 2(B) ,
where m k (B) is the mass of the component B in the phase k ( k = 1, 2), and demonstrate that
W 1 C 1 + W 2 C 2 = C 0 .  (**) Remember that, by definition, the composition of phase k is
Ck = m k (B)/ m k , and the weight fraction of the component B in the system is
C 0Â Â = m (B) / m . Then use the obvious equality
W 1 + W 2 =1 together with Eq. (**) to prove the lever rule (*).
Solution
Mass balance of component B:
Mass of a component that is present in both phases equal to the mass of the component in one phase + mass of the component in the second phase => W 1 C 1 + W 2 C 2 = C 0
Total Mass balance
Sum of mass fractions =1 => W 1 + W 2 =1
Solving the above two equations we get W1 = (C2-C0)/(C2-C1)
and W2 = (C1-C0)/(C1-C2)
.
Measures of Central Tendency: Mean, Median and Mode
Figure 1- Two-phase region on a schematic phase diagram of a binary sy.docx
1. Figure 1: Two-phase region on a schematic phase diagram of a binary system.
Let the point ( C 0, T 0) on the phase diagram belong to a region where two phases, 1 and 2,
are present. Let W 1 and W 2 denote the weight fractions of the phases 1 and 2,
respectively:
Wk = m k / m , where the index k enumerates the phases ( k = 1, 2), m k is the mass of the
phase k , and m = m 1Â Â + m 2Â Â is the total mass of the mixture of the phases 1 and 2.
(a) Derive the lever rule: W 1 ? C 1 = W 2 ? C 2 ,  (*)
where ? C k = | C k - C 0| , C k being the borderline composition of phase k ( k = 1, 2) at the
temperature T 0 , as shown in Figure 1 above.
1(b) Using the lever rule (*), express the weight fraction of each phase W k ( k = 1, 2)
through their compositions C k .
Hint for Problem 1(a):
Start with the fact that the mass of the component B in the system equals m (B) = m 1(B) +
m 2(B) ,
where m k (B) is the mass of the component B in the phase k ( k = 1, 2), and demonstrate
that
W 1 C 1 + W 2 C 2 = C 0 .  (**) Remember that, by definition, the composition of phase
k is
Ck = m k (B)/ m k , and the weight fraction of the component B in the system is
C 0Â Â = m (B) / m . Then use the obvious equality
W 1 + W 2 =1 together with Eq. (**) to prove the lever rule (*).
Solution
2. Mass balance of component B:
Mass of a component that is present in both phases equal to the mass of the component in one
phase + mass of the component in the second phase => W 1 C 1 + W 2 C 2 = C 0
Total Mass balance
Sum of mass fractions =1 => W 1 + W 2 =1
Solving the above two equations we get W1 = (C2-C0)/(C2-C1)
and W2 = (C1-C0)/(C1-C2)
