Solucionario fender cap8

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CHAPTER EIGHT
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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