## Q. 10.6

Gibbs Rule for an Isomorphous Phase Diagram
Determine the degrees of freedom in a Cu-40% Ni alloy at (a) 1300 °C, (b) 1250 °C, and (c) 1200 °C. Use Figure 10-8(a). ## Verified Solution

This is a binary system (C = 2). The two components are Cu and Ni. We will assume constant pressure. Therefore, Equation 10-2 (1 + C = F + P) can be used as follows.
(a) At 1300 °C, P = 1, since only one phase (liquid) is present; C = 2, since both copper and nickel atoms are present. Thus,
1 + C = F + P   ∴ 1 + 2 = F + 1 or F = 2
We must fix both the temperature and the composition of the liquid phase to completely describe the state of the copper-nickel alloy in the liquid region.
(b) At 1250 °C, P = 2, since both liquid and solid are present; C = 2, since copper and nickel atoms are present. Now,
1 + C = F + P   ∴ 1 + 2 = F + 2 or F = 1
If we fix the temperature in the two-phase region, the compositions of the two phases are also fixed. Alternately, if the composition of one phase is fixed, the temperature and composition of the second phase are automatically fixed.
(c) At 1200 °C, P = 1, since only one phase (solid) is present; C = 2, since both copper and nickel atoms are present. Again,
1 + C = F + P    ∴ 1 + 2 = F + 1 or F = 2
and we must fix both temperature and composition to completely describe the state of the solid.