Engineering Materials Engineering M12
Moorpark College
Chapter 4 Homework 4.3 Calculate the energy for vacancy formation in silver, given that the equilibrium number of vacancies at 800C (1073 K) is 3.6X1023m-3. The atomic weight and density (at 800C) for Silver are respectively, 107.9 g/mol, and 9.5 g/cm3 This problem calls for the computation of the energy for vacancy formation in silver. Upon examination of Equation (4.1), all parameters besides Q are given except N, the total number of atomic sites. However, N is v related to the density, (ρ), Avogadro's number (N ( N ), and the atomic weight (A ( A) according to Equation (4.2) as A
N =
N A ρ Ag A
Ag
(6.023 x 10 = = 5.30 x 10
22
23
)(
atoms / mol 9.5 g / cm3
)
107.9 g / mol
3 28 3 atoms/cm = 5.30 x 10 atoms/m
Now, taking natural logarithms of both sides of Equation (4.1), and, after some algebraic manipulation
N − RT ln V N
QV =
(
= − 8.62 x 10
-5
3.60 x 1023 m−3 eV/atom - K )(1073 K) ln 28 −3 5.30 x 10 m = 1.10 eV/atom
4.9 Calculate the composition, in weight percent, of an alloy that consists 105 kg of iron, 0.2 kg of Carbon, and 1.0 kg of chromium The concentration, in weight percent, of an element in an alloy may be computed using a modification of Equation (4.3). For this alloy, the concentration of iron (C (C ) is just Fe
CFe =
=
mFe m
Fe
105 kg 105 kg + 0.2 kg
+ mC + mCr
+ 1.0 kg
x 100
x 100 = 98.87 98.87 wt%
Similarly, for carbon
1
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Engineering Materials Engineering M12
Moorpark College
CC =
0.2 kg 105 kg + 0.2 kg
+ 1.0 kg
CCr =
1.0 kg 105 kg + 0.2 kg
+ 1.0 1.0 kg
x 100 = 0.19 0.19 wt%
And for chromium
x 100 = 0.94 wt%
4.11 What is the composition, in atom percent, of an alloy that contains 44.5 lbm of silver, 83.7 lbm of gold, and 5.3 lbm of Cu In this problem we are asked to determine the concentrations, concentrations, in atom percent, of the Ag-Au-Cu Ag-Au-Cu alloy. It is first necessary to convert the amounts of Ag, Au, and Cu into grams.
' m Ag = (44. (44.5 5 lbm )(453.6 g/lb m ) = 20,185 20,185 g
' m Au = (83.7 (83.7 lbm )(453.6 g/lb m ) = 37,96 37,966 6 g
' mCu = (5.3 (5.3 lbm)(453.6 g/lbm ) = 2,404 g
These masses must next be converted into moles [Equation (4.4)], as
nm
' m Ag
=
A
Ag
Ag
nm
Au
Cu
20,185 g = 187.1 187.1 mol 107. 107.87 87 g / mol mol
37,966 g = 192.8 192.8 mol 196.97 g/mol
=
nm
=
=
2,404 ,404 g = 37.8 mol 63.5 63.55 5 g /mol
Now, employment of a modified form of Equation (4.5)
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Engineering Materials Engineering M12
Moorpark College
' C Au =
192.8 mol 187.1 mol + 192.8 192.8 mol mol
+ 37.8 mol
37.8 37.8 mol 187.1 mol + 192. 192.8 8 mol
+ 37.8 mol
' CCu =
x 100 = 46.2 46.2 at%
x 100 = 9.0 at%
4.21 Niobium forms a substitutional solid solution with vanadium. Compute the number of niobium atoms per cubic centimeter for a niobium-vanadium alloy that contains 24 wt% Nb and 76 wt% V. The densities of pure niobium and vanadium are 8.57 and 6.10 g/cm3 respectively This problem asks us to determine the number of niobium atoms per cubic centimeter for a 24 wt% Nb-76 wt% V solid solution. To solve this problem, employment of Equation (4.18) is necessary, using the following values: C = C = 24 wt% 1 Nb
ρ1 = ρNb = 8.57 g/cm 3 ρ2 = ρV = 6.10 g/cm 3 A = A = 92.91 g/mol 1 Nb Thus
NNb =
N ACNb CNb ANb
ρNb
(
ANb
+
ρV
23
6.023 x 10
=
(24)(92.91 (24)(92.91 g/ mol) (8.5 (8.57 7 g /cm 3 )
+
= 1.02 x 10
(100 − C ) Nb
)
atom atoms s / mol mol (24)
92.91 g/mol 6.10 g/cm3 22
atoms/cm
3
(100 −
24)
