to th
Solutions Chemical and Engineering Thermodynamics, 4 ed
1023.5 = 27.93 (T 420 ) +
So that the equation to be solved is
(
0.093 2
T 4202
2
)
The solution to this equation is T = 404.56 K so that the temperature change is -15.44 K
6.67
P( T) := R
T
V
+ R T
B( T)
Cv( T) := 27.93 + 0.093 T 8.314
2
V
d
B( T)
R
B( T)
dT
d
P( T)
+ R
+ R T
V
2
2
dT
V
V
5
R := 8.314 10
d
B( T)
B( T)
dT
d
P( T) dV R ln( V) R
R T
dT
V
V
B( T) := 297.6
256100
T
DB( T) := B( T) + T
6
10
d
B( T)
dT
d
B( T)
dT
2561
10000 T
2
4
DB( = 420) 2.976 10
This combination is a constant
4
B( 420) = 3.122 10
Solving for volumes
RTP1 := R
420
(
2
RTP1 + RTP1 4 RTP1 B( 420)
V1 :=
10
)0.5
3
V1 = 3.78 10
2
RTP2 := R
404.56
(
2
RTP2 + RTP2 4 RTP2 B( 404.56)
V2 :=
0.1
)0.5
2
Using Eq. (6.4-17a)
V2
V1
Term1 := 8.314 ln
Term2 :=
Term3 :=
Term1 = 37.325
DB( 420) 8.314
Term2 = 0.655
V1
DB( 404.56 8.314
)
3
Term3 = 7.349 10
V2
Term4 :=
404.56
Cv( T)
T
Term4 = 2.171
dT
420
DelS := Term1 + Term2 Term3 + Term4
J
DelS = 35.801
mol K
V2 = 0.337