Solutions to Chemical and Engineering Thermodynamics, 3e
4
4.1 Using the Mollier diagram
µ=∂∂
FH IK = FH IK =−( )°
× − ×
= × ° = °−
T
P
T
PH H
∆∆
510 490
1241 10 7 929 10
4 463 10 4 463
7 6
6
C
Pa
C Pa C MPa
. .
. .
c h
κSS S
T
P
T
P=
∂∂
FH IK ≈ FH IK =−( )°
× − ×
= × ° = °−
∆∆
510 490
1069 10 9 515 10
1702 10 1702
7 6
5
C
Pa
C Pa C MPa
. .
. .
c h
∂ ∂∂ ∂
= ∂∂
× ∂∂
= ∂∂
× ∂∂
=∂∂
×∂∂
= = ( )
H S
H S
H T
S T
S P
H P
H T
H P
S P
S T
T H
P H
P S
T S
T
P
S
a fa f
a fa f
a fa f
a fa f
a fa f
a fa f
a fa f
,
,
,
,
,
,
,
,
,
,
,
,.
µκ
0262 unitless
4.2 (a) Start from eqn. 4.4-27
H T P H T P RT Z TdP
dTP dV
VV
V
, ,( ) − ( ) = −( ) + FH IK −LNM
OQP=∞
zIG 1
PRT
V b
a T
V bV b=
−−
( )+ −2 22
; ∂∂
FH IK =−
−+ −
P
T
R
V b
da dt
V bV bV2 22
so
H T P H T P
RT Z TR
V b
da dt
V bV b
RT
V b
a T
V bV bdV
RT Z a Tda
dT
dV
V bV b
V
V
V
V
, ,( ) − ( )
= −( ) +−
−+ −
RSTUVW −
−+
( )+ −
LNM
OQP
= −( ) + +FH IK + −
=∞
=∞
zz
IG
12 2
12
2 2 2 2
2 2
From integral tables we have
dx
a x b x c b a c
a x b b a c
a x b b a c′ + ′ + ′=
′ − ′ ′
′ + ′ − ′ − ′ ′
′ + ′ + ′ − ′ ′z 2 2
2
2
1
4
2 4
2 4ln for 4 02′ ′ − ′ <a c b
In our case ′ =a 1 , ′ =b b2 , c b= − 2 ; so 4 4 1 2 82 2 2 2′ ′ − ′ = ⋅ ⋅ − − ( ) = −a c b b b bc h and
′( ) − ′ ′ = =b a c b b2 24 8 2 2 .
Solutions to Chemical and Engineering Thermodynamics, 3e
H T P H T P
RT Za Tda dT
b
V b b
V b b
V b b
V b b
RT Za Tda dT
b
V b
V b
V V
, ,
ln ln
ln
( ) − ( )
= −( ) + − + −+ +
− + −+ +
LNMM
OQPP
= −( ) +− + −
+ +
=∞
IG
12 2
2 2 2 2
2 2 2 2
2 2 2 2
2 2 2 2
12 2
1 2
1 2
a f
a f d id i
or finally
H T P H T P RT ZTda dT a
b
Z B
Z B, , ln( ) − ( ) = −( ) +
− + +
+ −
LNMM
OQPP
IG 12 2
1 2
1 2
a f d id i
(b) This part is similar except that we start from eqn. (4.4-28)
S T P S T P R ZdP
dT
R
Vdv
R ZR
V b
da dT
V bV b
R
VdV
R Z RV b
V
da dT
b
V b
V b
R Z Bda dT
b
Z B
Z B
VV
V
V
V
V
V
V
V
, , ln
ln
ln ln ln
ln ln
( ) − ( ) = + FH IK −LNM
OQP
= +−
−+ −
−LNM
OQP
= + − ++ +
+ −
= −( ) ++ +
+ −
FHG
IKJ
=∞
=∞
=∞ =∞
zz
IG
2 22
2 2
1 2
1 2
2 2
1 2
1 2
d id i
d id i
4.3 Start with eqn. (4.2-21): dU C dT TP
TP dV
V
= + FHG
IKJ −
LNMM
OQPPV
∂∂
. Thus ∂∂
U
TC
V
FHG
IKJ = V ;
∂∂
∂∂
∂∂
U
TC T
P
TP
V
TP V P
FHG
IKJ = + F
HGIKJ −
LNMM
OQPPFHG
IKJV and
∂∂
∂∂
∂∂
∂∂
U
T
U
TT
P
TP
V
TP V V P
FHG
IKJ − F
HGIKJ = F
HGIKJ −
LNMM
OQPPFHG
IKJ .
(a) Ideal gas PV RT=
TP
TP
U
T
U
TV P V
∂∂
∂∂
∂∂
FHG
IKJ − = ⇒ F
HGIKJ = F
HGIKJ0
(b) van der Waals gas
PRT
V b
a
V=
−− 2 ;
∂∂
P
T
R
V bV
FHG
IKJ =
−; ⇒
FHG
IKJ −
LNMM
OQPP =T
P
TP
a
VV
∂∂ 2
Also: dPRdT
V b
RT
V bdV
a
VdV=
−−
−( )+
2 3
2
Solutions to Chemical and Engineering Thermodynamics, 3e
4.6 d SC
TdT
V
TdP
P
= − FHGIKJ
P ∂∂
[eqn. (4.2-20)]
For the ideal gas d SC
TdT
R
PdPIG P= −
*
. Thus, at constant temperature
d S SR
P
V
TdP
P
− = − FHGIKJ
LNM
OQP
IGd i ∂∂
and
S T P S T P S T P S T PR
P
V
TdP
PP
P
, , , ,( ) − ( ) − =( ) − =( ) = − FHG
IKJ
RSTUVW=
zIG IG0 00
∂∂
However, S T P S T P, ,=( ) − =( ) =0 0 0IG , since all fluids are ideal at P = 0 . Also
PV Z T P RTr r= ,a f . Thus
∂∂
∂∂
V
T PRZ T P RT
Z T P
TPr r
r r
P
FHG
IKJ = + F
HGIKJ
RSTUVW
1,
,a f a f
and
R
P
V
T
R
PZ T P
RT
P
Z
T
S T P S T P RZ
P
T
P
Z
TdP
RZ T P
P
T
P
Z
TdP
Pr r
P
PT P
T P
r r
r
r
r r Pr
T P
T P
rr r
r r
−FHG
IKJ = − − −
FHG
IKJ
⇒ ( ) − ( ) = − − + FHG
IKJ
LNM
OQP
= −−
+FHG
IKJ
LNMM
OQPP
=
=
zz
∂∂
∂∂
∂∂
∂∂
,
, ,
,
,
,
,
,
a fk p
a f
1
1
1
0
0
IG
4.7 (a) Ideal gas
PV NRT
N
=
=( )
+( ) × × ⋅=−
50 bar 100 m
27315 150 8314 10142 12
3
3K bar m kmol K kmol
c h. .
.
Energy balance, closed nonflow system
∆U Q PdV Q W= − = +z .
However, for ideal gas ∆U = 0 since T is constant (isothermal). Thus
W Q PdVNRTV
dV NRTVV
NRTPP
Q
= − = − = − = − = −
= − × × +( ) × FH IK= × == −
z z ln ln
. . ln
. .
.
2
1
1
2
3
142 1 27315 15050
300
895 9 10 895 9
895 9
kmol 8.314 J mol K
kJ MJ
MJ
Also, by Ideal Gas Law at fixed T and N
Solutions to Chemical and Engineering Thermodynamics, 3e
PV PV V VP
P1 1 2 2 2 11
2
100 m50
3001667= ⇒ = = × =3 3 m.
(b) Corresponding states
T P ZH H
TS Sr r
C
IGIG
initial state
final state 1.391300
73.76cal
mol K
cal
mol K
− −
+ = =
=
150 27315
304 21391
50
73760679 094 07 0 4
4067 0765 45 2 4
.
..
.. . . .
. . . .
Number of moles of gas = = = =NPV
ZRT
142 1
094151 2
.
.. kmol ( 142 1. =
PV
RT from above)
Final volume = =VZNRT
Pff
=× × × × +( )−0765 151 2 8 314 10 273 15 150
300
2. . . .=1356 m. 3
Energy balance on gas: ∆U Q W= +
Entropy balance on gas processes in gas are reversible: ∆SQ
TS S= + ⇒ =gen gen 0 . Therefore
∆SQ
T= or Q T S= ∆
∆
∆
∆
S S S N S S N S S S S S S
NP
P
N N
Q T S
W U Q N U U Q N H H N P V PV Q
f i f i f f f i i i
f
i
f i f i f f i i
= − = − = − + − − −
= − × − − − ×( )RSTUVW
= − −RSTUVW = −
= = +( ) × × −( ) = −
= − = − − = − − − −
c h c h c h c hm rIG IG IG IG
J K
kmol MJ
2 4 4184 8 314 04 4184
8 368 8314300
502326
27315 150 151 2 2326 14885
. . . ln . .
. . ln .
. . . .
=−
+LNMMN T
H H
TCf f
C
IGd iH Hf
IGiIG−c h
0Since process
is isothermal.
−−
− +OQPP −T
H H
TZ RT Z RT QC
i i
Cf f i i
IGd i
=− − −( )( ) × −
× +( ) × −( )LNM
OQP +
= × − + + = − +=
151 2304 2 45 07 4184 8314
273 15 150 0 765 09414885
151 2 10 48365 615 7 14885 638 2 14885
850 3
3
.. . . . .
. . ..
. . . . . .
.
kmol MJ
J MJ MJ
MJ
(c) Peng-Robinson E.O.S.Using the program PR1 with T = 27315. , P = 1 bar as the reference state, we obtainT = °150 C , P = 50 barZ = 0 9202. ; V = × −06475 10 3. m mol3 ; H = 470248 . J mol ; S = −1757. J mol K .
T = °150 C , P = 300 bar
Solutions to Chemical and Engineering Thermodynamics, 3e
Z = 0 7842. ; V = × −0 9197 10 4. m mol3 ; H = −6009. J mol ; S = −4124. J mol .
NV
V= =
×=−
100 m
06475 10154 443
3
3 m mol kmol
..
Q TN S= = +( ) × × − − −( )( ) = −∆ 273 15 150 154 44 4124 1757 1546 9. . . . . MJ
W U Q H PV H PV Q
N H PV H PV Q
f i
f i
= − = −( ) − −( ) −
= −( ) − −( ) −
=
− − × ×
× ⋅ −
+ × × ×
L
N
MMMM
O
Q
PPPP× + ×
=
−
−
∆
154 44
6009 300 0 9197 10
10 4702 48
50 0 6475 10 10
10 1546 9 10
885 25
4
5
3 5
3 6.
. .
.
.
.
.
J bar m
MJ
3
{Note that N, Q and W are close to values obtained from corresponding states.}
4.8 ∂∂
∂∂
∂∂
∂∂
∂ ∂∂ ∂
T
P
T S
P S
T S
P T
P T
P S
S T P T
S P T PS
FHG
IKJ =
( )( )
=( )( )
⋅( )
= −( ) ( )( ) ( )
,
,
,
,
,
,
, ,
, ,a f =
−= =
∂ ∂∂ ∂
∂ ∂ αS P
S T
V T
C T
V T
CT
P
Pa fa f
a fP P
and
κκ
∂
∂∂ ∂∂ ∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
S
T
S
T
V P
V V dP
V V dP
V S P S
V T P T
V S
V T
P T
P S
S V
T V
T P
S P
S
T
T
S
C
T
T
C
C
C
= =( ) ( )( ) ( )
=( )( )
⋅( )( )
=( )( )
⋅( )( )
= FHG
IKJ ⋅ FHG
IKJ = ⋅ =
1
1
a fa fa fa f
, ,
, ,
,
,
,
,
,
,
,
,V
P
V
P
4.9 (a)∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
H
V
H T
V T
H T
P T
P T
V T
H
P
P
VT T T
FHG
IKJ =
( )( )
=( )( )
⋅( )( )
= FHG
IKJ
FHG
IKJ
,
,
,
,
,
,
Since ∂∂
P
V T
FHG
IKJ ≠ 0 (except at the critical point)
∂∂
H
V T
FHG
IKJ = 0 if
∂∂
H
P T
FHG
IKJ = 0
(b)∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂ ∂ α∂∂
α
S
V
S P
V P
S P
T P
T P
V P
S
T
T
V
C
T VV
dT
dV
C TV
V V T
C
TV
S
V
P P P
P P P
FHG
IKJ =
( )( )
=( )( )
⋅( )( )
=FHG
IKJ ⋅
FHG
IKJ
= ⋅ ⋅ FHG
IKJ = = ⇒ F
HGIKJ
−
,
,
,
,
,
,
~P P P1
11
a fa f4.10 (a) We start by using the method of Jacobians to reduce the derivatives
Solutions to Chemical and Engineering Thermodynamics, 3e
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂ ∂∂ ∂
∂∂
∂ ∂∂ ∂
∂ ∂∂
TV
T HV H
T HT P
T PT V
T VV H
H T
P T
P T V T
H V T V
H
P
P V
H T
H V
d H T
H
T
T
V
T
V
FHG
IKJ =
( )( )
=( )( )
⋅( )( )
⋅( )
= −( )( )
( ) ( )( ) ( )
= −FHGIKJ
= −
,,
,,
,,
,,
,
,
, ,
, ,
a f
a fa f
a fa f
Now from Table 4.1 we have that
∂∂
∂∂
H
PV T
V
TT P
FHG
IKJ = − F
HGIKJ and
∂∂
∂∂
∂∂
HT
C V TVT
PTV P V
FHG
IKJ = + − F
HGIKJ
LNM
OQPFHG
IKJP
alternatively, since H U PV= +
∂∂
∂∂
∂∂
H
T
U
T
PV
TC V
dP
dTV V V V
FHG
IKJ = F
HGIKJ +
( )FHG
IKJ = + FH IKV
Thus
∂∂
∂ ∂ ∂ ∂∂ ∂
∂ ∂ ∂ ∂
∂ ∂T
V
P V V T V T
C V P T
V P V T P T
C V P TH
T P
V
T V
V
FHG
IKJ =
− −+
=− +
+a f a f
a fa f a f
a fV V
Note: I have used ∂∂
∂∂
∂∂
P
V
V
T
P
TT P V
FHG
IKJ
FHG
IKJ = −FHG
IKJ .
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
∂∂
T
V
T S
V S
T S
V T
V T
V S
S T
V T
T V
S V
S
V
T
S
T
C
P
T
S
T V V
FHG
IKJ =
( )( )
=( )( )
⋅( )( )
= −( )( )
⋅( )( )
=FHG
IKJFHG
IKJ = −
FHG
IKJ
,
,
,
,
,
,
,
,
,
,
V
(b) For the van der Waals fluid
∂∂
P
T
R
V bV
FHG
IKJ =
−,
∂∂
P
V
RT
V b
a
VT
FHG
IKJ = −
−( )+2 3
2
Thus
∂∂
T
V
RTV V b a V RT V b
C V R V bH
FHG
IKJ =
− − −( ) + + −( )
+ −
2 22n sV
after simplification we obtain
∂∂
T
V
a V b RTV b
C V b V R V b VH
FHG
IKJ =
− −( ) −
−( ) + −( )
2 2 2
2 2 3C
Solutions to Chemical and Engineering Thermodynamics, 3e
and
∂∂
T
V
RT
C V bS
FHG
IKJ = −
−( )V
4.11 There are a number of ways to solve this problem. The method I use is a little unusual, but thesimplest that I know of. At the critical point all three roots of V are equal, and equal to V C .
Mathematically this can be expressed as V V C− =a f3 0 which, on expansion, becomes
V V V V V VC C C3 2 2 33 3 0− + − = (1)
compare this with
PRT
V b
a
V V b b V b
RT
V b
a
V bV b=
−−
+( ) + −( )=
−−
+ −2 22
which multiplying through by the denominators can be written as
V V bRT
Pb
bRT
P
a
PV b
RTb
P
ab
P3 2 2 3
2
32
0+ −FH IK + − − +FH IK + + −FHG
IKJ = (2)
Comparing the coefficients of V in Eqns. (1) and (2) gives TC , PC
V bRT
PVC
CC
2 3: − = − (3)
V bbRT
P
a
PVC
C CC: − − + =3
232 2 (4)
V bRT
Pb
ab
PVC
C CC
0 3 2 3: + − = − (5)
From Eqn. (3)
P b
RT
P V
RTZC
C
C C
CC− = − = −1
33 or
P b
RTZC
CC= −1 3 (6)
For convenience, let y ZC= −1 3 or Zy
C =−1
3. Then
P b
RTyC
C
=
From Eqn. (4)
Solutions to Chemical and Engineering Thermodynamics, 3e
−FHG
IKJ − F
HGIKJ + =
⇒ − + + =−( )
3 2 3
3 23 1
9
2
22
22
2
P b
RT
P b
RT
aP
RTZ
y yaP
RT
y
C
C
C
C
C
CC
C
C
a f
a for expanding and rearranging
aP
RTy yC
Ca f c h221
310 4 1= + + (7)
Finally from eqn. (5)
P b
RT
bP
RT
P b
RT
aP
RTZC
C
C
C
C
C
C
CC
FHG
IKJ + F
HGIKJ − F
HGIKJFHG
IKJ = −
3 2
23
a fy y y y y y3 2 2 31
310 4 1
1
271+ − ⋅ + + = − −( )c h
or
64 6 12 1 03 2y y y+ + − = (8)
This equation has the solution y = 0077796074.
⇒ =bRT
PC
C
0077796074. (from Eqn. (6))
aRT
PC
C
= 04572355292
.a f
(from Eqn. (7))
Also Zy
C =−
=1
30307401309. .
Note that we have equated a and b to TC and PC only at the critical point. Therefore these functions
could have other values away from the critical point. However, as we have equated functions of V ,
we have assumed a and b would only be functions of T. Therefore, to be completely general wecould have
aRT
P
T
T
bRT
P
T
T
C
C C
C
C C
=FHG
IKJ
=FHG
IKJ
0457235529
0077796074
2
.
.
a fα
β
with αT
TC
FHG
IKJ → 1 as
T
TC
→ 1 and βT
TC
FHG
IKJ → 1 as
T
TC
→ 1 .
In fact, Peng and Robinson (and others) have set β= 1 at all temperatures and adjusted α a s a
function of temperature to give the correct vapor pressure (see chapter 5).
4.12 (also available as a Mathcad worksheet)
Solutions to Chemical and Engineering Thermodynamics, 3e
dN
dtN N N N
dUdt
N H N H QQN
H H
= + = ⇒ = −
= + + = ⇒ = −
& & & &
& & &&
1 2 2 1
1 1 2 21
2 1
0
0
Also, now using the program PR1 with T = 27315. , P = 1 bar reference state we obtain
T = °100 C T = °150 CP = 30 bar P = 20 barZ = 0 9032. Z = 0 9583.
V = × −09340 10 3. m mol3 V = × −01686 10 2. m mol3
H = 3609 72. J mol H = 679606 . J mol
S = −1584. J mol K S = −4 68 . J mol
&
& . . .Q
N= − =679606 3609 72 318634 J mol
4.13 (also available as a Mathcad worksheet)
Since process is adiabatic and reversible ∆S = 0 or S Si f= , i.e.,
S S T310 K, 14 bar 345 bar( ) = =( )?, . Using the program PR1 with the T = 27315. K and P = 1
bar reference state we obtain T = 310 K , P = 14 bar , Z = 0 9733. , V = × −01792 10 2. m mol3 ,
H = 109083. J mol and S = 1575. J mol K .
By trial and error (knowing P and S , guessing T) we obtain T = 34191. K , P = 345 bar ,
Z = 0 9717. , V = × −08007 10 4. m mol3 , H = 188609. J mol , S = 1575. J mol K
⇒ =Tf 34191. K .
System = contents of compressor
M.B.: dN
dtN N N N= = + ⇒ = −0 1 2 2 1& & & &
E.B.: dU
dtN H N H= = + +0 1 1 2 2& & &Q
0
adiabatic
+ −&W Ps
dV
dt
0
volume ofcompressor
constant
& & &W N H N HS = − +1 1 2 2 or &
& . . .W
NH HS = − = − =2 1 188609 109083 7952 6 J mol
4.14 (a) Pa
VV b RT
PV
RT
V
V b
a
RTV+
FHG
IKJ −( ) = ⇒ =
−−2
i) lim limPV
V
PV
RT
V
V b
a
RTV→→∞
→∞=
−−F
HGIKJ =
01
ii) B VPV
RTV
V
V b
a
RTVPV
V= −FH IK =
−− −RST
UVW→→∞
→∞lim lim
01 1
=− −( )
−( )−RST
UVW =−( )
−RSTUVW = −
→∞ →∞lim lim
V VV
V V b
V b
a
RTV
bV
V b
a
RTb
a
RT
Q.
N1
.N2
.M.B.
E.B.
Solutions to Chemical and Engineering Thermodynamics, 3e
iii) C VPVRT
BV
VbV b V b
V bb V
V bb
PV
V V= − −F
HGIKJ = − −( )
−RST
UVW =−
=→→∞
→∞ →∞lim lim lim
0
22
21
⇒ =C b2
(b) At the Boyle temperature: limP
VPV
RTB
→−FH IK = ⇒ =
01 0 0
0 = −ba
RTB
, Ta
RBb
= but aV RTc c=
9
8, b
V c=3
(Eqns. 4.6-3a)
TV RT
RVT TB
c c
cc c= = =
9 8
3
27
83375.
4.15 (a) From table: TB ~ 320 K , i.e., B T BBa f = ( ) =320 K 0
The inversion temperature is the temperature at which
∂∂
∂∂
∂∂
∂∂
∂∂
T
P CV T
V
T
TP T
RTP
BRP
dBdT
V TVT
RTP
BRTP
TdBdT
B TdBdT
H P
P P
P
FHG
IKJ = = − − F
HGIKJ
LNM
OQP
FHG
IKJ = +L
NMOQP = +
− FHG
IKJ = + − − = −
01
P
Thus, T inv is the temperature at which B TdB
dt− = 0 .
Plot up B vs. T, obtain dB dT either graphically, or numerically from the tabular data. I find
T inv ~ 600 K . Also, dB dT decreases with increasing temperature (i.e.,
dB dT ~ .456 cm mol K3 at 87.5 K and 0.027 cm mol K3 at 650 K. Presumably it is negative
at even higher temperature!)(b) Generally
µ ∂∂
∂∂
= FHG
IKJ = − − F
HGIKJ
RSTUVW = − −RST
UVWTP C
V TVT C
B TdBdTH P
1 1
P P
Using the data in the table it is easy to show that for T T< inv , B TdB
dT− < ⇒ >0 0µ , while for
T T> inv , B TdB
dT− > ⇒ <0 0µ .
(c) Since Fig. 2.4-3 for nitrogen is an H-P plot is easiest to proceed as follows
∂∂ ∂ ∂ ∂ ∂
∂ ∂∂ ∂
T
P H T P H
H P
H TH P T
T
P
FHG
IKJ =
−=
−1
a f a fa fa f
Since ∂ ∂H T CPa f = P is > 0 and less than ∞ [Except at a phase transition—see Chap. 5 and
Problem 5.1—however, µ has no meaning in the two-phase region], if dT dP Ha f is to be zero,
then d H dP Ta f must equal zero. That is, an inversion point occurs when isotherms are parallel
to lines of constant H (vertical line). This occurs at low pressures (ideal gas region) and at high
Solutions to Chemical and Engineering Thermodynamics, 3e
Pa
VV b RT
V V
T T
T
PRT
V b
a
V
ii
i
i f
f i
f
f
f
f f
+FHGG
IKJJ − =
⇒ = × = ×
− =− ⋅
× × ×= −
⇒ = − = = − °
=−
− = ×× − ×
−×
− −
−
− −
d id i
a f c h
d i c h
2
5 4
5
2 4 5
6 678 10 1336 10
02283
2 2725 6678 1062 73
29315 62 73 230 42 42 73
8314 230 42
1336 10 4 269 10
02283
1336
. ; . .
.
. ..
. . . .
. .
. .
.
.
m mol m mol
Pa m mol
J mol K m mol K
K C
3 3
6 2
3
10
8286 10 8286 bar
4 2
6
−
= × =
c hPf . . Pa
(d) Here we will use the program PR1. Using the 273.15 K and 1 bar reference state we find that atthe initial conditions Z = 11005. , Vi = × −05365 104. m mol3 , Hi = −328185. J mol and
Si = −5912. J mol K. Therefore
U H P Vi i i i= − = − − × × × = −−328185 500 05365 10 10 5964354 5. . . J mol
Now since U Uf i= = −596435. J mol and
V Vf i= = × −2 1073 104. m mol3 .
We must, by trial-and-error, find the temperature and pressure of the state having theseproperties. I find the following as the solution Tf = 230 9. K ; P~ 99 bar (for which
V = 01075. m mol3 and U = −5968 4. J mol ). To summarize, we have the following answers
for the different parts of the problem:
Pf T f
Ideal gas 250 bar 293.15 KCorresponding states 101.1 bar 237 Kvan der Waals 82.86 bar 230.42 KPeng-Robinson 99 bar 230.9 K
Once again, the ideal gas solution is seriously in error.
4.20 Mass balance (system = both tanks): N N Ni f f1 1 2= +
energy balance (system = both tanks): N U N U N Ui i f f f f1 1 1 1 2 2= +
entropy balance (system = portion of initial contents of tank 1, also in there finally): S Si f1 1=
Also, P P Pf f f1 2= = ; N
V
Vi
i11
1
= ; NV
Vf
f11
1
= and NV
Vf
f22
2
=
(a) Ideal gas solution: obtain P
T
P
T
P
T
i
i
f
f
f
f1
1
1
1
2
2
= + from mass balance and
P P P P Pi f f f f1 1 2
72 250 bar 25 10= + = ⇒ = = ×. Pa from energy balance
Solutions to Chemical and Engineering Thermodynamics, 3e
T TP
PTf i
f
i
R Cf
1 11
11
8 314 35 565
20 273 151
2249 3
239
=FHG
IKJ ⇒ = +( )FH
IK =
= − °
P
K
C from entropy balance
. .
.
. .
and
1 2 1355 9 82 7
2 1 12
T T TT
f i ff= − ⇒ = = °. . K C
Also
N
N
P V
RT
RT
P V
f
i
f
f
i
i1
1
1 1
1
1
1 1
7
7
25 10
249 3
293 15
500 100588= ⋅ =
×⋅
×=
.
.
.
..
and
N
N
N
N
f
i
f
i2
1
1
1
1 0412= −FHG
IKJ = .
(b) Corresponding States Solution:
Initial conditions Tr = =29315
190 71538
.
.. ; Pr = ×
×=5 10
4 64 101077
7
6.. ; ⇒ =Z 122. ;
H H
TC
IG
J mol K−
= 180 . ; S SIG J mol K− = 96 . .
Mass balance:
P
Z TP
Z T Z T
i
i if
f f f f1
1 1 1 1 2 2
751 1 50 10
122 293 151398 10= +
RSTUVW
=×
×= ×
.
. .. (1)
Entropy balance:
S S S S S S S Sf i f f i i
1 1 1 1 1 1 1 10− = = − + − − −IG IG, IG, IGd i d i d i
or
S S CT
RPf
f f
1 11
729315 50 109 6− + −
×= −IG
Pd i ln.
ln.
. (2)
Energy balance:
N U N U N Ui i f f f f1 1 1 1 2 2= + but N N N N U U N U Ui f f f f i f f i
1 1 2 1 1 1 2 2 2 0= + ⇒ − + − =d i d i
or
Solutions to Chemical and Engineering Thermodynamics, 3e
P V
Z RTH P V H P V
P V
Z RTH P V H P V
Z TH H H H H H Z RT Z RT
Z TH H H H H H
f
f ff f f i i i
f
f ff f f i i i
f ff f f i i i f f i i
f ff f f i i i
1 1
1 11 1 1 1 1 1
2 2
2 22 2 2 1 1 1
1 11 1 1 1 1 1 1 1 1 1
2 22 2 2 1 1 1
0
1
1
− − − + − − − =
− + − − − − +
+ − + − − −
d i d in s d i d in s
d i d i d in s
d i d i
, IG , IG , IG , IG
, IG , IG , IG , IGd in s− + =Z RT Z RTf f i i2 2 1 1 0
Substituting in the known values gives
135565 29315 8314 6 406 5
135565 29315 8 314 6 406 5 0
1 11 1 1 1 1
2 22 2 2 2 2
Z TH H T Z T
Z TH H T Z T
f ff f f f f
f ff f f f f
− + − − +
+ − + − − + =
, IG
, IG
d i c hn s
d i c hn s
. . . , .
. . . , . (3)
Eqns. (1-3) now must be solved. One possible procedure isi) Guess Pf
ii) Use Eqn. (2) to find T f1
iii) Use Eqn. (1) to find T f2
iv) Use Eqn. (3), together with T f1 and T f
2 to see if guessed Pf is correct. If not, go back to
step i.After many iterations, I found the following solution Pf = 9787. bar; T f
1 2216 K= . ;
T f2 259 4= . K ; N Nf i
1 1 0 645= . ; N Nf i2 1 0 355= . .
(c) Peng-Robinson equation of stateHere we use the equations
N N Ni f f1 1 2= + (4)
N U N U N Ui i f f f f1 1 1 1 2 2= + with U H PV= − (5)
S Si f1 = (6)
P P Pf f f1 2= =
and N V Vi i1 1 1= ; N V Vf f
1 1 1= ; N V V V Vf f f2 2 2 1 2= = since V V1 2= (value of V1 cancels out
of problem, so any convenient value may be used). Procedure I used to solve problem was as
follows. From PR1 we know V Ni i1 1⇒c h and Si
1 given initial conditions. Then
1. Guess value of T f1 , find P Pf f
1 = that satisfies S Sf i1 1=
2. Use T f1 , Pf and V f
1 to get N f1 ; then N N Nf i f
2 1 1= − so V f2 is known.
3. From Pf and V f2 find (trial-and-error with PR1) T f
2
4. See if eqn. (5) energy balance is satisfied; if not go back to step 1. After a number of
iterations I find Pf =103 6 bar. ; T f1 222 3= . K ; T f
2 2555= . K ; N Nf i1 1 0 619= . ;
N Nf i2 1 0381= . .
Solutions to Chemical and Engineering Thermodynamics, 3e
Summaryideal gas(part a)
Correspondingstates (part b)
P-R E.O.S.(part c)
Pf 250 bar 97.87 103.6
T f1
249.3 K 221.6 K 222.3 K
T f2
355.9 K 259.4 K 255.5 K
N Nf i1 1
0.588 0.645 0.619
N Nf i2 1
0.412 0.355 0.381
Clearly, the ideal gas assumption is seriously in error!
4.21 System = contents of compressor. This is a steady-state, open constant volume system.
mass balance: dN
dtN N= = +0 1 2& &
energy balance: dU
dtN H N H Q W Ps= = + + + −0 1 1 2 2& & & & dV
dt
0
⇒ = − + +0 1 1 2& & &N H H Q Wsa f
entropy balance: dS
dtN S N S
Q
T= = + + +0 1 1 2 2
& &&
S gen
0
.= − +&
&N S S
Q
T1 1 2a f
Thus,
& &Q TN S S= − −1 1 2a f&
&Q
NQ T S S
12 1= = −a f
and
& &
&W Q
NQ H HS +
= + = −1
2 1W
(a) Corresponding states solution
Q T S S T S S S S S S
T S S S S RP
P
= − = − + − − −
= − − − −RSTUVW
2 1 2 2 2 1 1 1
2 2 1 12
1
a f d i d i d in s
d i d i
IG IG IG IG
IG IG ln
Now Tr = =37315
405 6092
.
.. ; Pr , .
~ . 1 =1
112 80 009 ; Pr , .
. 2 = =50
112 80 443 . Thus
Q RT T S S S SPT
PT
r
r
r
r
= − + − − −RST
UVW= − × × + − −( ) = −
==
==
ln
. . ln . . .
, ,..
..
50
1
8314 37315 50 37315 523 0 14,088 1
2 2 0 4440 92
1 1 0 0090 92
2 1
IG IG
J mol
d i d i
and
W + = − = − + − − −
=
Q H H H H H H H H
T
2 1 2 2 20
1 1 1( ) ( ) ( )IG IG
sinceconstant
IG IG
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