MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
Problem Set 11 Solutions
1. In class, you were given:
⎛ J ⎞
∆Fmix = cX 1 X 2 ⎜
⎟.
⎝ mole ⎠
a. Find ∆ F1 and ∆ F2 .
b. Check your answers by substituting them into the relation:
∆Fmix = X 1∆ F1 + X 2 ∆ F2 .
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MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
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MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
2. Similar to problem 1, the Helmholtz free energy of mixing is given by:
⎛ J ⎞
∆Fmix = ao X 12 X 22 ⎜
⎟.
⎝ mole ⎠
a. Find ∆ F1 and ∆ F2 .
b. Check your answers by substituting them into the relation:
∆Fmix = X 1∆ F1 + X 2 ∆ F2 .
c. Graph ∆ F1 and ∆ F2 and ∆Fmix all on the same plot as a function of X2 over a
range of ao from -5000 to 5000 J/mol plotted at 500 J/mol increments.
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MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
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MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
Part c: Blue: ∆Fmix; Red: ∆F1; Green: ∆F2
Note that the largest positive values correspond to ao = 5000 J/mol
and the most negative values correspond to ao = -5000 J/mol.
∆Fmix, ∆F1, ∆F2 H J L
mol
ao : −5000 to 5000 plotted at 500Jêmol increments
2000
1500
1000
500
0
-500
-1000
-1500
0
0.2
0.4
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X2
0.6
0.8
1
MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
3. In the following binary system, the partial molar heat of mixing of component 1 can
be fitted by the expression: .
⎛ J ⎞
∆ H 1 = ao X 22 X 1 ⎜
⎟.
⎝ mole ⎠
a. Calculate the heat of mixing and the partial molar heat of mixing of component
2.
b. Graph the partial molar heat of mixing for both components and the heat of
mixing all on the same plot as a function of X2 over a range of ao from -10 to
10 J/mol plotted at 1 J/mol increments.
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MSE 308
Thermodynamics of Materials
Dept. of Materials Science & Engineering
Spring 2005/Bill Knowlton
Part c: Blue: ∆Hmix; Red: ∆H1; Green: ∆H2
Note that the largest positive values correspond to ao = 10 J/mol and
the most negative values correspond to ao = -10 J/mol.
The fact that ∆H2 does not go to 0 when X2 is not present suggests an
unrealistic solution.
ao : −10 to 10 plotted at 1 Jêmol increments
∆Hmix, ∆H1, ∆H2 H J L
mol
2
1
0
-1
-2
0
0.2
0.4
0.6
0.8
1
X2
4. In problem 6 of the last problems set, you proved the Gibb’s-Duhem equation strictly
using the following form of the Gibb’s-Duhem equation:
2
∑X
i =1
k
d ∆ Bk = 0
when given the change in volume of each component for a particular solution with
two components. Make up your own problem analogous to problem 6 and provide the
solution to the problem.
Open ended problem.
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