Solucionario fender cap8
8.1
a.
U (T ) = 25.96T + 0.02134T 2 J / mol
U (0 o C) = 0 J / mol
U (100 o C) = 2809 J / mol
Tref = 0 o C (since U(0 o C) = 0)
b. We can never know the true internal energy. U (100 o C) is just the change from U (0 o C) to
c.
U (100 o C) .
Q − W = ΔU + ΔE k + ΔE p
ΔE k = 0, ΔE p = 0, W = 0
Q = ΔU = (3.0 mol)[(2809 − 0) J / mol] = 8428 J ⇒ 8400 J
d.
Cv=
F ∂U I
GH ∂T JK
z
=
V
dU
= [25.96 + 0.04268T ] J / (mol⋅ o C)
dT
T2
ΔU =
z
F
GG
H
100
Cv (T )dT =
(25.96 + 0.04268T )dT = 25.96T + 0.04268
0
T1
T2
2
OP
QP
100
0
I
JJ J / mol
K
ΔU = (3.0 mol) ⋅ ΔU (J / mol)
= (3.0 mol) ⋅ [25.96(100 − 0) + 0.02134(100 2 − 0)] (J / mol) = 8428 J ⇒ 8400 J
8.2
a.
b
g
b
gb
Cv = Cp − R ⇒ Cv = 35.3 + 0.0291T [J / (mol⋅° C)] − 8.314 [J / (mol ⋅ K)] 1 K 1° C
g
⇒ Cv = 27.0 + 0.0291T [J / (mol⋅° C)]
100
b.
ˆ
ΔH =
∫ C p dT = 35.3T ]25 + 0.0291
100
25
z
100
c.
ΔU =
Cv dT =
25
z
100
z
100
T2 ⎤
⎥ = 2784 J mol
2 ⎦ 25
100
C p dT −
25
b
g
25
d.
8.3
gb
RdT = ΔH − RΔT = 2784 − 8.314 100 − 25 = 2160 Jmol
H is a state property
a.
Cv [ kJ / (mol⋅ o C)] = 0.0252 + 1547 × 10 −5 T − 3.012 × 10 −9 T 2
.
n=
PV
(2.00 atm)(3.00 L)
=
= 0.245 mol
RT (0.08206[atm ⋅ L / (mol ⋅ K)](298 K)
z
1000
Q1 = nΔU 1 = (0.245 mol) ⋅
z
z
0.0252 dT ( kJ / mol) = 6.02 kJ
25
1000
Q2 = nΔU 2 = (0.245) ⋅
[0.0252 + 1547 × 10 −5 T ] dT = 7.91 kJ
.
25
1000
Q3 = nΔU 3 =(0.245) ⋅
[0.0252 + 1547 × 10 −5 T − 3.012 × 10 −9 T 2 ] dT = 7.67 kJ
.
25
6.02 - 7.67
× 100% = −215%
.
7.67
7.91- 7.67
% error in Q2 =
× 100% = 313%
.
7.67
% error in Q1 =
8-1
8.3 (cont’d)
b.
C p = Cv + R
C p [ kJ / (mol⋅ o C)] = (0.0252 + 1547 × 10 −5 T − 3.012 × 10 −9 T 2 ) + 0.008314
.
= 0.0335 + 1547 × 10 −5 T − 3.012 × 10 −9 T 2
.
z
T2
Q = ΔH = n C P dTT1
z
1000
= (0.245 mol) ⋅
[0.0335 + 1547 × 10 −5 T − 3.012 × 10 −9 T 2 ] dT [kJ / (mol⋅ o C)] = 9.65 × 10 3 J
.
25
Piston moves upward (gas expands).
c.
a.
(C )
b.
8.4
The difference is the work done on the piston by the gas in the constant pressure process.
dC i
( 40 C ) = 0.1265 + 23.4 × 10 ( 40 ) = 0.1360 [kJ/(mol ⋅ K)]
−5
o
p C H (l )
66
bg b40° Cg = 0.07406 + 32.95 × 10
−5
p CH v
66
b40g − 25.20 × 10 b40g
−8
2
bg
+ 77.57 × 10 −12 40
3
= 0.08684 [kJ / (mol⋅ o C)]
c.
d.
e.
8.5
.
dC i b g b313 Kg = 0.01118 + 1095 × 10 b313g − 4.891 × 10 b313g
−5
2
p Cs
ΔHC6 H6 bv g
= 0.009615 [ kJ / (mol ⋅ K)]
32.95 × 10 −5 2 2520 × 10 −8 3 77.57 × 10 −12 4
.
= 0.07406T +
T−
T+
T
3
2
41095 × 10 −5 2
.
= 0.01118T +
T + 4.891 × 10 2 T −1
2
ΔH C b sg
−2
OP
PQ
OP
PQ
300
= 3171 kJ mol
.
40
573
= 3.459 kJ / mol
313
H 2 O (v, 100 o C, 1 atm) → H 2 O (v, 350 o C, 100 bar)
a.
H = 2926 kJ kg − 2676 kJ kg = 250 kJ kg
z
350
b.
H=
0.03346 + 0.6886 × 10 −5 T + 0.7604 × 10 −8 T 2 − 3593 × 10 −12 T 3 dT
.
100
= 8.845 kJ mol ⇒491.4 kJ kg
Difference results from assumption in (b) that H is independent of P. The numerical difference
is ΔH for H 2 O v, 350° C, 1 atm → H 2 O v, 350° C, 100 bar
b
8.6
b.
g
b
z
g
80
dC i
p n − C H (l)
6 14
= 0.2163 kJ / (mol⋅ o C) ⇒ ΔH = [0.2163] dT = 1190 kJ / mol
.
25
The specific enthalpy of liquid n-hexane at 80oC relative to liquid n-hexane at25oC is 11.90 kJ/mol
c.
dC i
p n − C H (v) [ kJ
6 14
z
/ (mol⋅ o C)] = 013744 + 40.85 × 10 −5 T − 23.92 × 10 −8 T 2 + 57.66 × 10 −12 T 3
.
0
ΔH =
[013744 + 40.85 × 10 −5 T − 23.92 × 10 −8 T 2 + 57.66 × 10 −12 T 3 ] dT = −110.7 kJ / mol
.
500
The specific enthalpy of hexane vapor at 500oC relative to hexane vapor at 0oC is 110.7 kJ/mol. The
specific enthalpy of hexane...
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