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 number of degrees of freedom for each alloy is 3.

Step-by-step explanation:

The number of degrees of freedom in an alloy can be determined using the Gibbs phase rule formula:

F = C - P + 2

Where F is the number of degrees of freedom, C is the number of components, and P is the number of phases.

For each alloy mentioned:

(a) 95 wt% Ag–5 wt% Cu at 780°C:

There are two components (Ag and Cu) and one phase, so the number of degrees of freedom is F = 2 - 1 + 2 = 3.

(b) 80 wt% Ni–20 wt% Cu at 1400°C:

There are two components (Ni and Cu) and one phase, so the number of degrees of freedom is F = 2 - 1 + 2 = 3.

(c) 44.9 wt% Ti–55.1 wt% Ni at 1310°C:

There are two components (Ti and Ni) and one phase, so the number of degrees of freedom is F = 2 - 1 + 2 = 3.

(d) 61.9 wt% Sn–38.1 wt% Pb at 183°C:

There are two components (Sn and Pb) and one phase, so the number of degrees of freedom is F = 2 - 1 + 2 = 3.

(e) 2.5 wt% C–97.5 wt% Fe at 1000°C:

There are two components (C and Fe) and one phase, so the number of degrees of freedom is F = 2 - 1 + 2 = 3.

Answer 2

The number of degrees of freedom for each alloy can be determined using the phase rule equation: F = C - P + 2. The alloys listed have 3 degrees of freedom.

Step-by-step explanation:

For alloys, the number of degrees of freedom can be determined using the phase rule equation:

F = C - P + 2

where F represents the degrees of freedom, C represents the number of components, and P represents the number of phases.

Using this equation, we can determine the number of degrees of freedom for each alloy:

(a) 95 wt% Ag–5 wt% Cu: C = 2 components (Ag and Cu), P = 1 phase (solid solution), F = 2 - 1 + 2 = 3 degrees of freedom.

(b) 80 wt% Ni–20 wt% Cu: C = 2 components (Ni and Cu), P = 1 phase (solid solution), F = 2 - 1 + 2 = 3 degrees of freedom.

(c) 44.9 wt% Ti–55.1 wt% Ni: C = 2 components (Ti and Ni), P = 1 phase (solid solution), F = 2 - 1 + 2 = 3 degrees of freedom.

(d) 61.9 wt% Sn–38.1 wt% Pb: C = 2 components (Sn and Pb), P = 1 phase (solid solution), F = 2 - 1 + 2 = 3 degrees of freedom.

(e) 2.5 wt% C–97.5 wt% Fe: C = 2 components (C and Fe), P = 1 phase (solid solution), F = 2 - 1 + 2 = 3 degrees of freedom.