In a rigid container, gaseous octane is reacted completely with 100% excess air. The initial temperature and pressure are 25c and 1 atm. The molar specific entropy for octane at 25c and 1 atm is 467.2 kj/kmol-k.
a. If the container is well insulated, determine the temperature and pressure at the end of t
Answers
Answer:
Explanation:
Balanced chemical equation for combustion of liquid octane
2
C
8
H
8
+
25
O
2
→
18
H
2
O
+
16
C
O
2
+
e
n
e
r
g
y
From standard value chart of
Δ
G
0
f
values , we have
Δ
G
f
(
C
O
2
)
=
−
394.359
k
g
/
m
o
l
e
Δ
G
f
(
H
2
O
)
=
−
228.572
k
g
/
m
o
l
e
Δ
G
f
(
C
8
H
8
)
=
6.41
k
g
/
m
o
l
e
Δ
G
f
(
O
2
)
=
0
k
g
/
m
o
l
e
for 1 mole octane combustion the equation will be
C
8
H
8
+
25
2
O
2
→
9
H
2
O
+
8
C
O
2
+
e
n
e
r
g
y
Δ
G
=
(
9
×
(
−
228.572
)
+
8
×
(
−
394.359
)
)
−
(
25
2
×
(
0
)
+
1
×
6.4
)
=
5218.42
k
J
Specific heat of octane=255.68 kJ/K
We know
Q
=
m
S
Δ
T
where
m= mass of octane =
8
×
12
+
8
×
1
=
96
+
8
=
104
g
Q=energy=
5218.42
k
J
=
5218.42
×
10
3
J
/
K
S=Specific heat of octane = 255.68 J/K
Δ
T
=
Q
m
S
=
5218.42
×
10
3
104
×
255.68
=
0.196
×
10
3
K
Δ
T
=
196.2
K
For 104 g octane =
196.2
×
104
=
20404.8
K
=
20404
−
273
(a) Raise in temperature =
20131.8
0
C
(b) If air is increased to 400% , there will be no change because the mass of octane will remain the same
So, the temperature increasing the air to 400 % will remain as
20131.8
0
C
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