### Solution to Class Test #2 - Bangladesh University of Engineering

```Department of Materials and Metallurgical Engineering
Bangladesh University of Engineering and Technology, Dhaka
MME6701
Thermodynamics and Kinetics of Materials
Solutions to Problems of Class Test 2
3. The activity coefficient of zinc in liquid brass has been expressed for the
temperature range 1000 to 1500 K by the equation
RT ln Zn = –4600(XCu)2
where R in J/mol-K and T in K. Calculate the activity of copper of a solution
containing 60 atom per cent copper at 1500 K.
Solution
ln
= −4600
ln
= −
= −
4600
4600
8.314 . 1773
= −0.312
Now,
.
ln
= −
ln
.
.
ln
= −
.
.
(−0.312 . 2
= −
.
.
= +0.624
.
.
= −0.624
.
= −0.624
ln
2
= −0.05
.
=
= 0.95
Then, the activity of copper
=
.
= 0.95 (0.60) = 0.57
)
4. A steel solution contains 0.6 % C, 1.0 % Mn, 0.3 % Si, and 0.8 % Cr (by weight).
Calculate the activity of carbon at 1000 °C, when the standard state of carbon is
1.0 % carbon. At 1000 °C, CC = 8.7, CMn = –4.3, eCSi= 11.2, eCCr= –12. The atomic
weights of C, Mn, Si and Cr are 12, 55, 28 and 52, respectively. What will be the
activity of carbon if the standard state of carbon is graphite. Use C0 = 7.9.
Solution
The Henrian activity of carbon
ℎ =
.
%
=
.
log ℎ = log
.
.
+ log
=
.
+ log
% +
%
%
+ log
+
+ log
%
+
%
%
+ log
+ log
%
%
Now since
0.2425
=
0.2425
=
=
0.2425
0.2425
(8.7) = 0.18
12
=
=
0.2425
(−4.3) = −0.02
55
Then
log ℎ =
% +
%
+
%
+
%
= (0.18)(0.6) + (−0.02)(1.0) + (11.2)(0.3) + (−12)(0.8) + log(0.6) = −6.37
ℎ = 4.23
10
Now since
ℎ
=
= ℎ
= (4.23 10 )(7.9) = 3.34
10
```