Question

Specify the number of degrees of freedom for the following alloys:

(a) 95 wt% Ag–5 wt% Cu at 780°C

(b) 80 wt% Ni–20 wt% Cu at 1400°C

(c) 44.9 wt% Ti–55.1 wt% Ni at 1310°C

(d) 61.9 wt% Sn–38.1 wt% Pb at 183°C

(e) 2.5 wt% C–97.5 wt% Fe at 1000°C

Answer 1

The general form Gibbs phase rule is

P + F = C + N

For all the systems given in the question, the number of components, C, is 2, whereas N the number of noncompositional variables is 1, i.e. temperature. Hence, the phase rule becomes

P + F = 2 + 1 = 3

Therefore, F = 3 - P

where F is the number of degrees of freedom and P is the number of phases present at equilibrium.

For each of the alloys mentioned in the question, we need to determine the number of phases at the specified temperature to find F.

(a) From the Cu-Ag phase diagram, it can be found that an alloy of 95 wt% Ag–5 wt% Cu at 780°C will consist of only phase. Thus, P=1.

Therefore, F = 3 - 1 = 2

(b) From the Cu-Ni phase diagram, it can be found that an alloy of 80 wt% Ni–20 wt% Cu at 1400°C will be in the region, thus consisting of two phases. Hence, P = 2.

Therefore, F = 3 - 2 = 1

(c) From the Ti-Ni phase diagram, it can be found that an alloy of 44.9 wt% Ti–55.1 wt% Ni at 1310°C will be entirely liquid. Thus, P = 1.

Therefore, F = 3 - 1 = 2

(d) From the Sn-Pb phase diagram, it can be found that an alloy of 61.9 wt% Sn–38.1 wt% Pb at 183°C will be at the eutectic, which means that three phases are in equilibrium and thus P = 3.

Therefore, F = 3 - 3 = 0

(e) From the C-Fe phase diagram, it can be found that an alloy of 2.5 wt% C–97.5 wt% Fe at 1000°C will be in the two-phase region of Austenite + Cementite. Thus, P = 2.

Therefore, F = 3 - 2 = 1