NAS 8th Edition Annotated Solutions
Chapter 3


Updated 5/23/2017
p. 1 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

◦
Updated 5/23/2017
p. 2 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
k  44.18*106
Updated 5/23/2017
p. 3 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
1  V 
   
V  P T
AP 

V  V0 1 

 BP
 A  B  P   AP 
V0 AB
V

 V0 
2
  B  P
  B  P 2
P


V0 AB
1 V
 

V P
 B  P
2
AP 

V0 1 

 BP

AB
 B  P  AP  B  P 
dV
  dP
V
dV
c
0.125 cm3 g-1
  dP 
dP 
dP
V
V  P  b
V  P  2700 bar 
dV 
0.125 cm3 g-1
dP
 P  2700 bar 
dW   PdV  P
Updated 5/23/2017
0.125 cm3 g-1
dP
 P  2700 bar 
p. 4 of 182
NAS 8th Edition Annotated Solutions
500 bar
W

P
1 bar
Chapter 3
500 bar
0.125 cm3 g -1
dP  0.125 cm3 g -1  P   2700 bar  ln  P  2700 bar  
1 bar
 P  2700 bar 

 3200 bar  
3 -1
W  0.125 cm3 g -1  499 bar   2700 bar  ln 
   5.1591 bar cm g
2701
bar



W  0.51591 Pa m3 g -1  515.91 Pa m3 kg -1  515.91 J kg -1
1  V 
   
V  P T




1
dV   dP   3.9  106  0.1 109 P dP
V
V 
ln    3.9  106  Po  P   5  1011 P 2  Po2
 Vo 




V  Vo exp 3.9  106  Po  P   5  1011 P 2  Po2




 dP
dW   3.9  10  0.1  10 P  PV exp  3.9  10  P  P   5  10  P  P   dP


dW   PVdP   PVo exp 3.9  106  Po  P   5  1011 P 2  Po2
6
9
6
o
Updated 5/23/2017
11
o
2
2
o
p. 5 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

 
dW   3.9  10 P  0.1  10 P  exp  3.9  10 P  5  10
 
P  dP
dW  3.9  106  0.1  109 P P exp 3.9  106 1  P   5  1011 P 2  1 dP
6
9
6
2
11
2
3000
W
 3.9  10 P  0.1 10 P  exp  3.9  10 P  5  10 P  dP
6
9
6
2
11
2
1


dV
  dT  1.2  103 K -1 dT
V
V 
ln  2    T2  T1   1.2  103 K -1 T2  T1 
 V1 
V 
ln  2   1.2  103 K -1  20 K   0.024
 V1 


V2  V1 exp  0.024  6.29 10-4 m3 kg-1 *exp  0.024   6.44 10-4 m3 kg-1



Updated 5/23/2017
p. 6 of 182
NAS 8th Edition Annotated Solutions
Chapter 3








dV
  dT   dP
V
V 
ln    T  P
 Vo 
V 
ln    (104 K 1 ) *(5 K)  (4.8 105 bar -1 ) * 20 bar  5 104  9.6 104  0.0015
 Vo 
Updated 5/23/2017
p. 7 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
𝜕
𝜕
Updated 5/23/2017
p. 8 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
°

Updated 5/23/2017
°
°
1  V 
   
V  P T
1
dV   dP
V
p. 9 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
V 
ln      P  P1 
 V1 
V  A T  exp   P 
V  V1 exp    P  P1  
A T   V1 exp  P1 

dW   PVdP   PV1 exp    P  P1   dP   A T  P exp   P  dP
P

W   A T  P exp   P  dP 
A T 

  P1  1 exp   P1    P  1 exp   P  
P1

W

V V
1
 P1  1V1   P  1 V   PV

1 1  PV 


1



Updated 5/23/2017

p. 10 of 182
NAS 8th Edition Annotated Solutions
R
Chapter 3
2
T2  P2  C p  1  7
 
    0.5520
T1  P1 
8



8 bar
(a)
P
(b)
(c)
1 bar
0.00624
m3 mol-1
0.0275
m3 mol-1
0.0499
m3 mol-1
V
Updated 5/23/2017
p. 11 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 12 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 13 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 14 of 182
NAS 8th Edition Annotated Solutions
10 bar
Chapter 3
1
P
2
3 bar
2 bar
4
0.00499
m3 mol-1
3
0.0166
m3 mol-1
V
V1 
RT1 8.314 Pa m3mol-1K -1 *600 K

 0.00499 m3 /mol
P1
106 Pa
RT2 8.314 Pa m3mol-1K -1 *600 K
V2 

 0.0166 m3 /mol
6
P2
0.3*10 Pa
Updated 5/23/2017
p. 15 of 182
NAS 8th Edition Annotated Solutions
V4 
Chapter 3
RT4 8.314 Pa m3mol-1K -1 *400 K

P4
0.2*106 Pa


J mol-1K-1
0.0166 m3 /mol
0.00499m3 /mol



J mol-1K-1


J mol-1K-1






J mol-1K-1


J mol-1K-1



Updated 5/23/2017
p. 16 of 182
NAS 8th Edition Annotated Solutions
Chapter 3




J mol-1K-1


J mol-1K-1

V1 
RT1 8.314 Pa m3mol-1K -1 *300 K

P1
105 Pa
Updated 5/23/2017
p. 17 of 182
NAS 8th Edition Annotated Solutions
V1 
Chapter 3
RT1 8.314 Pa m3mol-1K -1 *300 K

P1
5*105 Pa

J mol-1K-1





J mol-1K-1


J mol-1K-1



J mol-1K-1



J mol-1K-1






Updated 5/23/2017
p. 18 of 182
NAS 8th Edition Annotated Solutions
Cv
Vi  P1 
 
V1  P2 
W (adiabatic) 
Cp
Chapter 3
3
1 5
    0.225
 12 
P2Vi  PV
(12 bar * 2.70 m3 )  (1 bar*12 m3 )
1 1

 30.6 bar m3  30.60 *105 J = 3060 kJ
 1
(5 / 3)  1

Cp
Pi  V1 
 
P1  V2 
Cv
 12  3  62.90
5
PV
(62.9bar *1 m3 )  (1 bar *12 m3 )
i 2  PV
1 1
W (adiabatic) 

 76.3 bar m3  76.30 *105J = 7630 kJ
 1
5 / 3 1
Updated 5/23/2017
p. 19 of 182
NAS 8th Edition Annotated Solutions
Chapter 3


(c)
12
bar
(b)
P
(d)
(a)
(e)
1 bar
1 m3
12 m3
V
∗𝑃
Updated 5/23/2017
−1
𝑑𝑃
+ 𝑃
𝑑𝑧
𝑑𝑇
=
𝑑𝑧
∗
𝑃
𝑃
𝑑𝑃
+ 𝑃
𝑑𝑧
𝑑𝑇
=0
𝑑𝑧
p. 20 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
𝑑𝑃 𝑑𝑇
+
𝑃
𝑑𝑧 𝑑𝑧
=0
𝑑𝑃 𝑑𝑇
+
=0
𝑑𝑧 𝑑𝑧
∗ −𝑀 𝑔 +
𝑀𝑔
(
𝑅
dt
d  mU 
dt
𝑑𝑇
𝑑𝑧
Inlet mass
flow rate = m '
enthalpy = H’
dm
 m'
dt
d  mU 
)=
𝑑𝑇
=0
𝑑𝑧
Inside tank,
m = 0 at t = 0
 m'H '

dm
H'
dt
 mU t   mU t 0   mH 't   mH 't 0
Updated 5/23/2017
p. 21 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 mU t   mH 't

Tsupply =318 K
Psupply =1500 kPa
T= 298 K
dn
n
dt
n
Updated 5/23/2017
p. 22 of 182
NAS 8th Edition Annotated Solutions
d  nU 
dt
Chapter 3
 nHsupply  Q
Q
n
dU
dn
dn
U
 n Usupply  RTsupply   Q 
Usupply  RTsupply   Q
dt
dt
dt
dn
dt
n
Q
Q U
dn dn
dn

Usupply  RTsupply   U  Usupply  RTsupply 

dt dt
dt
Q   Qdt   U  Usupply  RTsupply 
dn
dt  U  Usupply  RTsupply   n final  ninitial 
dt
dn
 n'
dt
Updated 5/23/2017
p. 23 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
n'
dm
 m'
dt
d  nU 
dt
d  nU 
dt
 n ' H ' Q

dn
H ' Q
dt
n2U 2  n1U1   n2  n1  H ' Q
n2 U 2  H '  n1 U1  H '  Q
U 2  H '  U 2  U ' P 'V '  U 2  U ' RT '  Cv T2  T '  RT '  CvT2  C pT '






n2 CvT2  C pT '  n1 CvT1  C pT '  Q
n2 CvT2  C pT '  Q
Updated 5/23/2017
p. 24 of 182
NAS 8th Edition Annotated Solutions

Chapter 3

n2 CvT2  C pT '  0
CvT2  C pT '
T2   T '

T2   T '
n
PV
300000 Pa  4 m3

 346 mol
RT 8.3145 Pa m3 mol-1 K  417.2 K



n2 CvT2  C pT '  mtank C p,tank T2  298 K 
n2  2.5T2  3.5  298 K   8.3145 J mol-1 K  400 kg  460 J kg -1 K -1 T2  298 K 
n2 
P2V2
300000 Pa  4 m3

RT2 8.3145 Pa m3 mol-1 K -1  T2
1200000
T2
 2.5T2  3.5  298 K   400  460 T2  298 K 
3000000T2  1251600000  184000T2 2  54832000T2
T2 2  281.70T2  6802.2  0
T2  304.1 K or -44.7 K (of course only the first answer is physically meaningful)
Updated 5/23/2017
p. 25 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
T = f(P) ?
dn
 n
dt
dU t
 nH
dt
Updated 5/23/2017
p. 26 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
dU
d
dU
dn
dn
dn
U
  n U  RT   U
 RT
Utank  nU   tank  n
dt
dt
dt
dt
dt
dt
or
dUtank
dU
dn
n
 RT
dt
dt
dt
n
dn
 RT
 C tank  nC  dT
dt
dt
v,
v
1 dT
R
dn

T dt  Cv ,tank  nCv  dt
 T  R  Cv ,tank  n2 Cv 
ln  2  
ln 


 T1  Cv  Cv ,tank  n1Cv 
PV


Cv ,tank  2 tank Cv 

T  R
RT2

ln  2  
ln 
PV
T
C


1 tank
 1
v
 Cv ,tank  RT Cv 

1

R
P2Vtank

 Cv
C

C
v
,
tank
v

T2 
RT2


PV
T1 

1 tank
 Cv ,tank  RT Cv 

1

Updated 5/23/2017
p. 27 of 182
NAS 8th Edition Annotated Solutions
V 
RT 8.314 Pa m3mol-1K -1 *303 K

P
100*103 Pa
V2 
RT2 8.314 Pa m3mol-1K -1 *480 K

P2
500*103 Pa
Updated 5/23/2017
Chapter 3
p. 28 of 182
NAS 8th Edition Annotated Solutions
Chapter 3


















Updated 5/23/2017



p. 29 of 182
NAS 8th Edition Annotated Solutions
Chapter 3













Updated 5/23/2017
p. 30 of 182
NAS 8th Edition Annotated Solutions
Chapter 3














Updated 5/23/2017
p. 31 of 182
NAS 8th Edition Annotated Solutions
Chapter 3









T2  T1  
T2  T1 
T2  423.15K 
Updated 5/23/2017
u22  u12
2C p
u22  u12
2C p
(50 m/s) 2  (2.5 m/s) 2
*0.028 kg/mol  421.95 K or 148.8 C
J
2*3.5*8.314
molK
p. 32 of 182
NAS 8th Edition Annotated Solutions
Chapter 3































Updated 5/23/2017
p. 33 of 182
NAS 8th Edition Annotated Solutions
Chapter 3





















Updated 5/23/2017

p. 34 of 182
NAS 8th Edition Annotated Solutions
Chapter 3




Updated 5/23/2017
p. 35 of 182
NAS 8th Edition Annotated Solutions
R
Chapter 3
2
 P  CP
 10 bar  5
Tintermidiate = T1  2  = 298.15K × 
 = 335.947 K
 1 bar 
 P1 
U = W = CV T = 1.5 × 8.314 J·mol1·K 1 × (335.947 K - 298.15 K) = 471.366 J·mol1
H = U + W = 471.366 J·mol1 + 471.366 J·mol1 = 942.732 J·mol1
U = CV T = 1.5 × 8.314 J·mol1·K 1 × (573.15 K - 335.947 K) = 2958.16 J·mol1
H = CP T = 2.5 × 8.314 J·mol1·K1 × (573.15 K - 335.947 K) = 4930.26 J·mol1

W = -RT = -8.314 J·mol1·K1 × (573.15 K - 335.947 K) = -1972.11 J·mol1

P 
Tintermidiate = T1  2 
 P1 
R
CP
2
 10 bar  5
= 298.15K × 
 = 335.947 K
 1 bar 
U = W = CV T = 1.5 × 8.314 J·mol1·K 1 × (335.947 K - 298.15 K) = 471.366 J·mol1
Updated 5/23/2017
p. 36 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
H = U + W = 471.366 J·mol1 + 471.366 J·mol1 = 942.732 J·mol1
Q = U = CV T = 1.5 × 8.314 J·mol1·K 1 × (573.15 K - 335.947 K) = 2958.16 J·mol1
H = CP T = 2.5 × 8.314 J·mol1·K1 × (573.15 K - 335.947 K) = 4930.26 J·mol1

R
2
 P  CP
 10 bar  5
Tintermidiate = T1  2  = 298.15K × 
 = 335.947 K
 1 bar 
 P1 
U = W = CV T = 1.5 × 8.314 J·mol1·K 1 × (335.947 K - 298.15 K) = 471.366 J·mol1
H = U + W = 471.366 J·mol1 + 471.366 J·mol1 = 942.732 J·mol1


Updated 5/23/2017
p. 37 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
RTA RTB 2 RTo


p
p
po
R
2
TB  p  C p  p  7
 
 
To  po 
 po 
Updated 5/23/2017
p. 38 of 182
NAS 8th Edition Annotated Solutions
Chapter 3








2
TB  p  7
 
To  po 
TA TB 2 p
 
To To po
TA
2.5 319.75


 1.4342
300
1
300
2
TB  p  7
 
To  po 
TA TB 2 p
 
To To po
2
TA  p  7 2 p
  
To  po 
po
Updated 5/23/2017
p. 39 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
2
p 1   p  7 TA 
   
po 2   po 
To 


2
TB  p  7
 
To  po 
TA TB 2 p
 
To To po
TA 2 p TB 2*1.3233 325

 

 1.563
To po To
1
300
TA TB 2 p
 
To To po
744.3  TB TB 2 p


300
300 po
7
p  TB  2
 
po  To 
Updated 5/23/2017
2
2
 p 7
TB  To    300 1.2405 7  319.05 K
 po 
p. 40 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 

PV  RT   b 
P
RT 

V
RT 
 
 b 

P 
RT 
Updated 5/23/2017
p. 41 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

1  V 
   
V  P T
 RT
 V 

  2
P
 P T
RT

1
P2

RT 
 

P 
 
 b 
b
 P 1 


P 
RT 
RT 
RT  


 

P V   b 
  RT
RT  


RT
P

V b
RT
 
1 d  
 
d 

 
V b
R  RT  

V b
R






2
RT 
RT dT  
RT 
dT T
RT
 P 



  
2
2
 T V
 
 


V  b 

V  b 

RT 
RT 





RT
V b
RT

P
RT
d 
2
R

 P 
P
dT T   P    P     d 



  
 

2
 T V
 T   RT   T dT 
 RT 


 P 
Updated 5/23/2017
p. 42 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 43 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
Z
PV
B
C
D
 1   2  3  ...  1  B   C  2  D  3  ...
RT
V V
V
B
dZ
d   0
C
1 d2Z
2 d2
 0
P
a T 
RT

V  b V   b V   b 
Z
a T  V
PV
V


RT V  b RT V   b V   b 
1
Z
1 b

Z


a T  1

   1    b
RT 1   b
a T  
1

1  b  RT 1   b  1   b  
Updated 5/23/2017
p. 44 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
a T   1   b  1   b      b 1   b     b 1   b    
dZ
b




2
2
d  1  b  2
RT 

1


b

1


b








a T  
dZ
b
1   b 2  2




d  1  b  2
RT  1   b  2 1   b  2 


a T 
dZ
b
d   0
RT
B
a T 
dZ
b
d   0
RT



2
2
2
2
2
2 2

2 b 1   b  1   b    2 b 1   b   1   b   
a T   2 b  1   b   1   b    1   b 




4
4
RT 
d  2 1  b  3
1   b   1   b  



d2Z
2b 2
d2Z
2b 2
d
1  b  

2
d2Z
d
2

3
 2b 2 
a T 
RT
 0
1 d 2Z
C
2 d 2


2
2 2


a T   2 b  1   b  1   b    1   b   2 b 1   b    2 b 1   b    

RT 
1   b  3 1   b  3



 2     b   2  b2 

 b2 
a T      b 

RT

a T     b
 0
RT
PV
B C
1  2
RT
V V
12 bar  V
1
140 cm3 mol-1 7200 cm 6 mol-2

V
V2
83.145 bar cm3 mol-1 K -1  298 K
 140 cm3 mol-1 7200 cm6 mol-2 
V  2064.8 cm3 mol-1 1 



V
V2


Updated 5/23/2017
p. 45 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

B 0  0.083 
0.422
B1  0.139 
0.172
Tr1.6
Tr4.2
 0.3040
 0.00218
BPc
 B 0   B1  0.3040  0.087  0.00218  0.3038
RTc
 BP  P
0.2381
Z  1   c  r  1  0.3038
 0.9315
1.056
 RTc  Tr
P
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
1.05560.5  83.1452  282.32
a(T )  0.42748 r
 0.42748
 4548100 bar cm 6 mol-2
Pc
50.40
b  0.08664
Updated 5/23/2017
RTc
83.145  282.3
 0.08664
 40.35 cm3 mol-1
Pc
50.40
p. 46 of 182
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T  V  b
RT
b
P
P V V  b 

4548100 V  40.35  
3
-1
V   2064.8  40.35 
 cm mol

12
V
V

40.35

 


379008 V  40.35  
3
-1
V   2105.2 
 cm mol

V V  40.35  

1   0.480  1.574  0.176 1  T  R T
a(T )  0.42748
2
2
0.5
r
2 2
c
Pc
1  0.480  1.574  0.087  0.176 0.087   1  1.0556   83.145  282.3
a(T )  0.42748
2
2
2
2
50.40
a(T )  0.42748
0.9731  83.145  282.3
50.40
2
2
a(T )  4547150 bar cm6 mol-2
b  0.08664
V
RTc
83.145  282.3
 0.08664
 40.35 cm3 mol-1
Pc
50.40
a T  V  b
RT
b
P
P V V  b 

4547150 V  40.35  
3
-1
V   2064.8  40.35 
 cm mol

12
V
V

40.35





378929 V  40.35  
3
-1
V   2105.2 
 cm mol

V
V

40.35




Updated 5/23/2017
p. 47 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V  2.4142b V  0.4142b 
1   0.37464  1.54226  0.2699 1  T  R T
a(T )  0.45724
2
0.5
r
2 2
c
Pc
 
a(T )  0.45724
1  0.37464  1.54226  0.087  0.2699  0.087 
a(T )  0.45724
2
2

1  1.0556
  83.1452  282.32
2
50.40
0.9724  83.145  282.3
50.40
2
2
a(T )  4860171 bar cm6 mol-2
RT
83.145  282.3
b  0.07779 c  0.07779
 36.23 cm3 mol-1
Pc
50.40
V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


4860171V  36.23
3
-1
V   2064.8  36.23 
 cm mol

12
V

15.005
V

87.46





405014 V  36.23 
3
-1
V   2101.0 
 cm mol

V  15.005V  87.46  

PV
B C
1  2
RT
V V
15 bar  V
1
156.7 cm3 mol-1 9650 cm 6 mol-2

V
V2
83.145 bar cm3 mol-1 K -1  323.15 K
 156.7 cm3 mol-1 9650 cm6 mol-2 
V  1791.2 cm3 mol-1 1 



V
V2


Updated 5/23/2017
p. 48 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

B 0  0.083 
0.422
B1  0.139 
0.172
Tr1.6
Tr4.2
 0.3026
 0.003266
BPc
 B 0   B1  0.3026  0.100  0.003266  0.3023
RTc
 BP  P
0.3079
Z  1   c  r  1  0.3023
 0.9120
1.058
 RTc  Tr
P
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
1.0580.5  83.1452  305.32
a(T )  0.42748 r
 0.42748
 5496600 bar cm 6 mol-2
Pc
48.72
b  0.08664
V
RTc
83.145  305.3
 0.08664
 45.14 cm3 mol-1
Pc
48.72
a T  V  b
RT
b
P
P V V  b 

5496600 V  45.14  
3
-1
V  1791.2  45.14 
 cm mol

15V V  45.14  


366440 V  45.14  
3
-1
V  1836.3 
 cm mol

V
V

45.14




Updated 5/23/2017
p. 49 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
1   0.480  1.574  0.176 1  T  R T
a(T )  0.42748
2
0.5
r
2
2 2
c
Pc
1   0.480  1.574  0.100  0.176  0.100   1  1.058    83.145  305.3

a(T )  0.42748
2
2
2
2
48.72
a(T )  0.42748
0.96398  83.145  305.3
48.72
2
2
a(T )  5450090 bar cm6 mol-2
V
a T  V  b
RT
b
P
P V V  b 

5450090 V  45.14  
3
-1
V  1791.2  45.14 
 cm mol

15
V
V

45.14





363399 V  45.14  
3
-1
V  1836.3 
 cm mol

V V  45.14  

P
a T 
RT

V  b V  2.4142b V  0.4142b 
Updated 5/23/2017
p. 50 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
1   0.37464  1.54226  0.2699 1  T  R T
a(T )  0.45724
2
0.5
r
2
2
2
c
Pc
1  0.37464  1.54226  0.100  0.2699 0.100  1  1.058   83.145  305.3
a(T )  0.45724
2
2
2
48.72
a(T )  0.45724
0.97014  83.145  305.3
48.72
2
2
a(T )  5866378 bar cm6 mol-2
RT
83.145  305.3
b  0.07779 c  0.07779
 40.53 cm3 mol-1
Pc
48.72
V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 

5866378 V  40.53 
3
-1
V  1791.2  40.53 
 cm mol

15
V

16.78
V

97.85






391091 V  40.53 
3
-1
V  1831.7 
 cm mol

V

16.78
V

97.85





PV
B C
1  2
RT
V V
15 bar  V
1
194 cm3 mol-1 15300cm 6 mol-2

V
V2
83.145 bar cm3 mol-1 K -1  348 K
 194 cm3 mol-1 15300cm6 mol-2 
V  1929.0 cm3 mol-1 1 



V
V2


Updated 5/23/2017
p. 51 of 182
2
NAS 8th Edition Annotated Solutions
Chapter 3

B 0  0.083 
0.422
B1  0.139 
0.172
Tr1.6
Tr4.2
 0.2836
 0.02011
BPc
 B 0   B1  0.2836  0.286  0.02011  0.2779
RTc
 BP  P
0.3989
Z  1   c  r  1  0.2779
 0.8985
1.0919
 RTc  Tr
P
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
1.09190.5  83.1452  318.7 2
a(T )  0.42748 r
 0.42748
 7639640 bar cm 6 mol-2
Pc
37.6
b  0.08664
V
RTc
83.145  318.7
 0.08664
 61.06 cm3 mol-1
Pc
37.6
a T  V  b
RT
b
P
P V V  b 

7639640 V  61.06  
3
-1
V  1929.0  61.06 
 cm mol

15V V  61.06  


509309 V  61.06  
3
-1
V  1990.1 
 cm mol

V
V

61.06




Updated 5/23/2017
p. 52 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
1   0.480  1.574  0.176 1  T  R T
a(T )  0.42748
2
2
0.5
r
2 2
c
Pc
1  0.480  1.574  0.286  0.176 0.286  1  1.0919   83.145  318.7
a(T )  0.42748
2
2
2
2
37.6
a(T )  0.42748
0.9193  83.145  318.7
36.7
2
2
a(T )  7519399 bar cm6 mol-2
V
a T  V  b
RT
b
P
P V V  b 

7519399 V  61.06  
3
-1
V  1929.0  61.06 
 cm mol

15
V
V

61.06





501293 V  61.06  
3
-1
V  1990.1 
 cm mol

V
V

61.06

 

P
a T 
RT

V  b V  2.4142b V  0.4142b 
1   0.37464  1.54226  0.2699 1  T  R T
a(T )  0.45724
2
0.5
r
2
2 2
c
Pc
1   0.37464  1.54226  0.286  0.2699  0.286   1  1.0919    83.145  318.7

a(T )  0.45724
2
2
2
37.6
a(T )  0.45724
0.92994  83.145  318.7
37.6
2
2
a(T )  7940485 bar cm6 mol-2
RT
83.145  318.7
b  0.07779 c  0.07779
 54.82 cm3 mol-1
Pc
37.6
Updated 5/23/2017
p. 53 of 182
2
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 

7940485 V  54.82  
3
-1
V  1929.0  54.82 
 cm mol

15
V

22.71
V

132.35





529365 V  54.82  
3
-1
V  1983.8 
 cm mol

V  22.71V  132.35  

Updated 5/23/2017
p. 54 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V V  b 
Updated 5/23/2017
p. 55 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 56 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

cm3
cm6 
724
93866

cm 
mol 
mol 2 
V  16629
1 
mol 
V
V2





3
Updated 5/23/2017
p. 57 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 BP  P
0.38750
Z  1   c  r 1  0.8010*
 0.9532
0.6637
 RTc  Tr
P
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
0.66370.5 *83.1452 *4522
a T   0.42748 r
 0.42748*
 19149726 bar cm6 mol 2
Pc
38.7
b  0.08664
RTc
83.145*452
 0.08664*
 84.14 cm3mol 1
Pc
38.7
V  16629  84.14 
Updated 5/23/2017
19149726 V  84.14 
1.5
V V  84.14 
p. 58 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
1   0.48  1.574  0.176 1  T  R T
a T   0.42748
2
0.5
r
2
2
2
c
Pc
1   0.48  1.574*0.086  0.176*0.086 1  0.6637  * 83.145 * 452
a T   0.42748*
2
0.5
2
2
2
38.7
a T   19350625.74 bar cm6 mol 2
b  0.08664
RTc
83.145*452
 0.08664*
 84.14 cm3mol 1
Pc
38.7
1   0.37464  1.54226*0.086  0.2699*0.086 1  0.6637  *83.145 * 452
a T   0.45724*
2
0.5
2
2
38.7
a T  19956222.093 bar cm6 mol 2
Updated 5/23/2017
p. 59 of 182
2
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 60 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 BP  P
2.13
Z  1   c  r 1  0.8010*
 0.6833
1.282
 RTc  Tr
P
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
1.2820.5 *83.1452 *2342
a T   0.42748 r
 0.42748*
 3204293 bar cm6 mol 2
Pc
44.6
b  0.08664
RTc
83.145*234
 0.08664*
 37.80 cm3mol 1
Pc
44.6
V  262.56  37.80 
Updated 5/23/2017
3204293 V  37.80 
95 V V  37.80 
p. 61 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
1   0.48  1.574  0.176 1  T  R T
a T   0.42748
2
0.5
r
2
2
2
c
Pc
1   0.48  1.574*0.126  0.176*0.126 1  1.282  * 83.145 * 234
a T   0.42748*
2
0.5
2
2
2
44.6
a T   3008454.99 bar cm6 mol 2
b  0.08664
RTc
83.145*234
 0.08664*
 37.80 cm3mol 1
Pc
44.6
V  262.56  37.80 
3008454.99 V  37.80 
95
V V  37.80 
1   0.37464  1.54226*0.126  0.2699*0.126 1  1.282  *83.145 * 234
a T   0.45724*
2
0.5
2
2
2
44.6
a T   3322337.2 bar cm6 mol 2
b  0.07779
RTc
 33.93 cm3mol 1
Pc
Updated 5/23/2017
p. 62 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
PV
B C
1  2
RT
V V
18 bar  V
1
152.5 cm3 mol-1 -5800 cm 6 mol-2

V
V2
83.145 bar cm3 mol-1 K -1  523 K
 152.5 cm3 mol-1 5800 cm6 mol-2 
V  2415.8 cm3 mol-1 1 



V
V2



Updated 5/23/2017
p. 63 of 182
NAS 8th Edition Annotated Solutions
B 0  0.083 
0.422
B1  0.139 
0.172
Tr1.6
Tr4.2
Chapter 3
 0.5103
 0.2816
BPc
 B 0   B1  0.5103  0.345  0.2816  0.6075
RTc
 BP  P
0.08161
Z  1   c  r  1  0.6075
 0.9387
0.8082
 RTc  Tr

Z  1  BP  CP2  DP3
 Z 
'
' 2
   B  2C P  3D P
 P T
Z  1  B   C  2  D 3
Updated 5/23/2017
p. 64 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

P (bar)
Zi
Updated 5/23/2017
Z1i
Z2i
p. 65 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
1
0.95
0.9
Zi
0.85
0.8
Z1i 0.75
Zi
0.7
Z1i
0.65
Z2i
Z2i
0.6
0.55
0.5
0
50
100
150
200
P (bar)
P
a T 
RT

V  b V V  b 
T
T 0.5 R 2Tc2
a(T )  0.42748 r
 0.42748
Pc
0.5
R 2Tc2.5
Pc
 0.42748
T 0.5  83.1452  369.82.5
42.48

1.8295  108
bar cm6 mol-2
T
RT
83.145  369.8
b  0.08664 c  0.08664
 62.71 cm3 mol-1
Pc
42.48

Updated 5/23/2017

p. 66 of 182
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T  V  b
RT
b
P
P V V  b 

10338000 V  62.71 
3
-1
V  1899.1  62.71 
 cm mol

13.71  V V  62.71 


754082 V  62.71 
3
-1
V  1961.8 
 cm mol

V
V

62.71




 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  313.15  62.17  13.71  13.71V 
V  62.71  V V  62.71 

10338000


 1963.4  V 
V  62.71  V V  62.71 

 754048 
0.2857
1T
V sat  Vc Zc r 
10.8468
V sat  200  0.276
0.2857
Z  1  B0
Updated 5/23/2017
Pr
P
  B1 r
Tr
Tr
p. 67 of 182
NAS 8th Edition Annotated Solutions
B 0  0.083 
0.422
B1  0.139 
0.172
Tr1.6
Tr4.2
B 0  0.083 
B1  0.139 
Chapter 3
0.422
0.84681.6
0.172
0.84684.2
 0.4676
 0.2068
Z  1   0.4676  0.152  0.2068
0.3227
 0.8098
0.8468


V
a T  V  b
RT
b
P
P V V  b 

10177000 V  62.71 
3
-1
V  1565.8  62.71 
 cm mol

17.16

V
V

62.71





593080 V  62.71 
3
-1
V  1628.5 
 cm mol

V
V

62.71




Updated 5/23/2017
p. 68 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  323.15  62.17  17.16  17.16V 
V  62.71  V V  62.71 

10177000


 1627.8  V 
V  62.71  V V  62.71 

 593065 
0.2857
1T
V sat  Vc Zc r 
10.8739
V sat  200  0.276
0.2857
Z  1  B0
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
B 0  0.083 
B1  0.139 
Updated 5/23/2017
Tr1.6
Tr4.2
0.422
0.87391.6
0.172
0.87394.2
 0.4406
 0.1640
p. 69 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
Z  1   0.4406  0.152  0.1640
0.4040
 0.8014
0.8739

V

a T  V  b
RT
b
P
P V V  b 

10023300 V  62.71 
3
-1
V  1305.4  62.71 
 cm mol

21.22

V
V

62.71





472351 V  62.71 
3
-1
V  1368.1 
 cm mol

V V  62.71 

 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  333.15  62.17  21.22  21.22V 
V  62.71  V V  62.71 

100233000


 1367.5  V 
V  62.71  V V  62.71 

 472351 
0.2857
1T
V sat  Vc Zc r 
V
sat
10.9009
 200  0.276
Updated 5/23/2017
0.2857
p. 70 of 182
NAS 8th Edition Annotated Solutions
Z  1  B0
Chapter 3
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
B 0  0.083 
B1  0.139 
Tr1.6
Tr4.2
0.422
0.90091.6
0.172
0.90094.2
 0.4157
 0.1276
Z  1   0.4157  0.152  0.1276
P
0.4995
 0.7588
0.9009
a T 
RT

V  b V V  b 
Updated 5/23/2017
p. 71 of 182
NAS 8th Edition Annotated Solutions
T
T 0.5 R 2Tc2
a(T )  0.42748 r
 0.42748
Pc
b  0.08664
Chapter 3
0.5
R 2Tc2.5
Pc
 0.42748
T 0.5  83.1452  369.82.5
42.48

1.8295  108
bar cm6 mol-2
T
RTc
83.145  369.8
 0.08664
 62.71 cm3 mol-1
Pc
42.48


V
a T  V  b
RT
b
P
P V V  b 

9876000 V  62.71 
3
-1
V  1099.9  62.71 
 cm mol

25.94

V
V

62.71





380725 V  62.71 
3
-1
V  1162.6 
 cm mol

V
V

62.71

 

 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  343.15  62.17  25.94  25.94V 
V  62.71  V V  62.71 

9876000


 1162.0  V 
V  62.71  V V  62.71 

 380725 
0.2857
1T
V sat  Vc Zc r 
10.9279
V sat  200  0.276
0.2857
Updated 5/23/2017
p. 72 of 182
NAS 8th Edition Annotated Solutions
Z  1  B0
Chapter 3
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
B 0  0.083 
B1  0.139 
Tr1.6
Tr4.2
0.422
0.92791.6
0.172
0.92794.2
 0.3927
 0.09652
Z  1   0.3927  0.152  0.0965
Updated 5/23/2017
0.6106
 0.7319
0.9279
p. 73 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

P
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
T 0.5 R 2Tc2.5
T 0.5  83.1452  425.12.5
2.9006  108
a(T )  0.42748 r
 0.42748
 0.42748

bar cm6 mol-2
Pc
Pc
37.96
T
RT
83.145  425.1
b  0.08664 c  0.08664
 80.67 cm3 mol-1
Pc
37.96


V
a T  V  b
RT
b
P
P V V  b 

15016000 V  80.67  
3
-1
V   2013.3  80.67 
 cm mol

15.41

V
V

80.67




974430 V  80.67  
3
-1
V   2094.0 
 cm mol

V V  80.67  

Updated 5/23/2017
p. 74 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  373.15  80.67  15.41  15.41V 
V  80.67  V V  80.67  

15016000


 2094.1  V 
V  80.67  V V  80.67  

 974430 
0.2857
1T
V sat  Vc Zc r 
10.8778
V sat  255  0.274
0.2857
Z  1  B0
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
B 0  0.083 
B1  0.139 
Updated 5/23/2017
Tr1.6
Tr4.2
0.422
0.87781.6
0.172
0.87784.2
 0.4369
 0.1583
p. 75 of 182
NAS 8th Edition Annotated Solutions
Z  1   0.4369  0.200  0.1583
P
Chapter 3
0.4059
 0.7833
0.8778
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
T 0.5 R 2Tc2.5
T 0.5  83.1452  425.12.5
2.9006  108
a(T )  0.42748 r
 0.42748
 0.42748

bar cm6 mol-2
Pc
Pc
37.96
T
RT
83.145  425.1
b  0.08664 c  0.08664
 80.67 cm3 mol-1
Pc
37.96

Updated 5/23/2017

p. 76 of 182
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T  V  b
RT
b
P
P V V  b 

14818000 V  80.67  
3
-1
V  1707.2  80.67 
 cm mol

18.66

V
V

80.67




794105 V  80.67  
3
-1
V  1787.9 
 cm mol

V V  80.67  

 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  383.15  80.67  18.66  18.66V 
V  80.67  V V  80.67  

14818000


 1787.9  V 
V  80.67  V V  80.67  

 794105 
0.2857
1T
V sat  Vc Zc r 
10.9013
V sat  255  0.274
0.2857
Z  1  B0
Updated 5/23/2017
Pr
P
  B1 r
Tr
Tr
p. 77 of 182
NAS 8th Edition Annotated Solutions
B 0  0.083 
0.422
B1  0.139 
0.172
B 0  0.083 
B1  0.139 
Chapter 3
Tr1.6
Tr4.2
0.422
0.90131.6
0.172
0.90134.2
 0.4153
 0.1271
Z  1   0.4153  0.200  0.1271
Updated 5/23/2017
0.4916
 0.7596
0.9013
p. 78 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

P
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
T 0.5 R 2Tc2.5
T 0.5  83.1452  425.12.5
2.9006  108
a(T )  0.42748 r
 0.42748
 0.42748

bar cm6 mol-2
Pc
Pc
37.96
T
RT
83.145  425.1
b  0.08664 c  0.08664
 80.67 cm3 mol-1
Pc
37.96


V
a T  V  b
RT
b
P
P V V  b 

14628800 V  80.67  
3
-1
V  1460.6  80.67 
 cm mol

22.38  V V  80.67  


653655 V  80.67  
3
-1
V  1541.3 
 cm mol

V
V

80.67




 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  393.15  80.67  22.38  22.38V 
V  80.67  V V  80.67  

14628800


 1541.3  V 
V  80.67  V V  80.67  

 653655 
Updated 5/23/2017
p. 79 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
0.2857
1T
V sat  Vc Zc r 
10.9248
V sat  255  0.274
0.2857

Updated 5/23/2017
p. 80 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
T 0.5 R 2Tc2.5
T 0.5  83.1452  425.12.5
2.9006  108
a(T )  0.42748 r
 0.42748
 0.42748

bar cm6 mol-2
Pc
Pc
37.96
T
RTc
83.145  425.1
3
-1
b  0.08664
 0.08664
 80.67 cm mol
Pc
37.96


V
a T  V  b
RT
b
P
P V V  b 

14446000 V  80.67  
3
-1
V  1260.6  80.67 
 cm mol

26.59

V
V

80.67





543287 V  80.67  
3
-1
V  1341.3 
 cm mol

V
V

80.67

 

 RT  bP  VP 
V  b  V V  b  


a T 


83.14
 403.15  80.67  26.59  26.59V 

V  80.67  V V  80.67  

14446000


 1341.3  V 
V  80.67  V V  80.67  

 543287 
Updated 5/23/2017
p. 81 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
0.2857
1T
V sat  Vc Zc r 
10.9484
V sat  255  0.274
0.2857
Updated 5/23/2017
p. 82 of 182
NAS 8th Edition Annotated Solutions
T (K)
403.15
P (bar)
26.59
Chapter 3
Tc (K)
425.1
Pc (bar)
37.96

Tr
0.93
0.93
0.95
0.95
Pr
0.40
0.60
0.40
0.60
Z0
0.8059
0.6635
0.8206
0.6967
0.2
Tr
0.9484
Pr
0.7005
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
Interpolated Values
Final Values
Z
V (cm3/mol)
P
0.6310
Z1
-0.0763
-0.1662
-0.0589
-0.1110
-0.1432
0.6023
759.4
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
T 0.5 R 2Tc2.5
T 0.5  83.1452  408.12.5
2.7255  108
a(T )  0.42748 r
 0.42748
 0.42748

bar cm6 mol-2
Pc
Pc
36.48
T
RT
83.145  408.1
b  0.08664 c  0.08664
 80.59 cm3 mol-1
Pc
36.48

Updated 5/23/2017

p. 83 of 182
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T  V  b
RT
b
P
P V V  b 

14302000 V  80.59  
3
-1
V  1825.5  80.59 
 cm mol

16.54

V
V

80.59




864692 V  80.59  
3
-1
V  1906.1 
 cm mol

V V  80.59  

 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  363.15  80.59  16.54  16.54V 
V  80.59  V V  80.59  

14302000


 1906.0  V 
V  80.59  V V  80.59  

 864692 
0.2857
1T
V sat  Vc Zc r 
10.8899
V sat  262.7  0.282
0.2857
Updated 5/23/2017
p. 84 of 182
NAS 8th Edition Annotated Solutions
Z  1  B0
Chapter 3
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
Tr1.6
Tr4.2
B 0  0.083 
B1  0.139 
0.422
0.88991.6
0.172
0.88994.2
 0.4256
 0.1417
Z  1   0.4256  0.302  0.1417 
0.4534
 0.7614
0.8899



V
a T  V  b
RT
b
P
P V V  b 

14109000 V  80.59  
3
-1
V  1549.0  80.59 
 cm mol

20.03

V
V

80.59




704406 V  80.59  
3
-1
V  1629.6 
 cm mol

V V  80.59  

Updated 5/23/2017
p. 85 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  373.15  80.59  20.03  20.03V 
V  80.59  V V  80.59  

14109000


 1629.6  V 
V  80.59  V V  80.59  

 704406 
0.2857
1T
V sat  Vc Zc r 
10.9144
V sat  262.7  0.282
0.2857
Z  1  B0
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
Updated 5/23/2017
Tr1.6
Tr4.2
p. 86 of 182
NAS 8th Edition Annotated Solutions
B 0  0.083 
B1  0.139 
0.422
0.91441.6
0.172
0.91444.2
Chapter 3
 0.4040
 0.1115
Z  1   0.4040  0.302  0.1115
Updated 5/23/2017
0.5491
 0.7372
0.9144
p. 87 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 88 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
T 0.5 R 2Tc2.5
T 0.5  83.1452  408.12.5
2.7255  108
a(T )  0.42748 r
 0.42748
 0.42748

bar cm 6 mol-2
Pc
Pc
36.48
T
RT
83.145  408.1
b  0.08664 c  0.08664
 80.59 cm3 mol-1
Pc
36.48


V
a T  V  b
RT
b
P
P V V  b 

13920000 V  80.59  
3
-1
V  1326.8  80.59 
 cm mol

24.01

V
V

80.59





579922 V  80.59  
3
-1
V  1407.4 
 cm mol

V V  80.59  

Updated 5/23/2017
p. 89 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  383.15  80.59  24.01  24.01V 
V  80.59  V V  80.59  

13920000


 1407.4  V 
V  80.59  V V  80.59  

 579922 
0.2857
1T
V sat  Vc Zc r 
10.9389
V sat  262.7  0.282
0.2857
Z  1  B0
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
Updated 5/23/2017
Tr1.6
Tr4.2
p. 90 of 182
NAS 8th Edition Annotated Solutions
B 0  0.083 
B1  0.139 
0.422
0.93891.6
0.172
0.93894.2
Chapter 3
 0.3838
 0.0851
Z  1   0.3838  0.181  0.0851
Updated 5/23/2017
0.6582
 0.7201
0.9389
p. 91 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

V
a T  V  b
RT
b
P
P V V  b 

13746000 V  80.59  
3
-1
V  1145.8  80.59 
 cm mol

28.53

V
V

80.59





481800 V  80.59  
3
-1
V  1226.3 
 cm mol

V
V

80.59

 

 RT  bP  VP 
V  b  V V  b  


a T 


83.14
 393.15  80.59  28.53  28.53V 

V  80.59  V V  80.59  

13746000


 1226.3  V 
V  80.59  V V  80.59  

 481800 
0.2857
1T
V sat  Vc Zc r 
V
sat
10.9632
 262.7  0.282
Updated 5/23/2017
0.2857
p. 92 of 182
NAS 8th Edition Annotated Solutions
Z  1  B0
Chapter 3
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
B 0  0.083 
B1  0.139 
Tr1.6
Tr4.2
0.422
0.96321.6
0.172
0.96324.2
 0.3651
 0.0623
Z  1   0.3651  0.181  0.0623
Updated 5/23/2017
0.7821
 0.6944
0.9632
p. 93 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
T 0.5 R 2Tc2.5
T 0.5  83.1452  417.22.5
1.362  108
a(T )  0.42748 r
 0.42748
 0.42748

bar cm 6 mol-2
Pc
Pc
77.10
T
b  0.08664
RTc
83.145  417.2
 0.08664
 38.98 cm3 mol-1
Pc
77.10


V
a T  V  b
RT
b
P
P V V  b 

7466000 V  38.98  
3
-1
V  1521.1  38.98 
 cm mol

18.21

V
V

38.98





409980 V  38.98  
3
-1
V  1560.1 
 cm mol

V
V

38.98




Updated 5/23/2017
p. 94 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  333.15  38.98  18.21  18.21V 
V  38.98  V V  38.98  

7466000


 1560.1  V 
V  38.98  V V  38.98  

 409980 
0.2857
1T
V sat  Vc Zc r 
10.7985
V sat  124  0.265
0.2857
Z  1  B0
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
Updated 5/23/2017
Tr1.6
Tr4.2
p. 95 of 182
NAS 8th Edition Annotated Solutions
B 0  0.083 
B1  0.139 
Chapter 3
0.422
0.79851.6
0.172
0.79854.2
 0.5219
 0.3036
Z  1   0.5210  0.069  0.3036 
T (K)
333.15
P (bar)
18.21
0.2362
 0.8397
0.7985
Tc (K)
417.2
Pc (bar)
77.1
Tr
0.80
0.80
0.85
0.85
Pr
0.10
0.20
0.10
0.20
ω
0.069
Tr
0.7985
Pr
0.2362
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
Interpolated Values
Final Values
Z
V (cm3/mol)
0
Z
0.9319
0.8539
0.9436
0.8810
0.8247
1
Z
-0.0487
-0.1160
-0.0319
-0.0715
-0.1419
0.8149
1239.7

Updated 5/23/2017
p. 96 of 182
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T  V  b
RT
b
P
P V V  b 

7353000 V  38.98  
3
-1
V  1268.6  38.98 
 cm mol

22.49

V
V

38.98




326900 V  38.98  
3
-1
V  1307.6 
 cm mol

V V  38.98  

 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  343.15  38.98  22.49  22.49V 
V  38.98  V V  38.98  

7353000


 1307.6  V 
V  38.98  V V  38.98  

 326900 
0.2857
1T
V sat  Vc Zc r 
10.8225
V sat  124  0.265
0.2857
Z  1  B0
Updated 5/23/2017
Pr
P
  B1 r
Tr
Tr
p. 97 of 182
NAS 8th Edition Annotated Solutions
B 0  0.083 
0.422
B1  0.139 
0.172
B 0  0.083 
B1  0.139 
Chapter 3
Tr1.6
Tr4.2
0.422
0.82251.6
0.172
0.82254.2
 0.4939
 0.2518
Z  1   0.4939  0.069  0.2518
T (K)
343.15
P (bar)
22.49
0.2917
 0.8187
0.8225
Tc (K)
417.2
Pc (bar)
77.1
Tr
0.80
0.80
0.85
0.85
Pr
0.10
0.20
0.10
0.20
ω
0.069
Tr
0.8225
Pr
0.2917
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
Interpolated Values
Final Values
Z
V (cm3/mol)
Updated 5/23/2017
0
Z
0.9319
0.8539
0.9436
0.8810
0.8009
1
Z
-0.0487
-0.1160
-0.0319
-0.0715
-0.1462
0.7908
1003.4
p. 98 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
T 0.5 R 2Tc2.5
T 0.5  83.1452  417.22.5
1.362  108
a(T )  0.42748 r
 0.42748
 0.42748

bar cm 6 mol-2
Pc
Pc
77.10
T
b  0.08664
RTc
83.145  417.2
 0.08664
 38.98 cm3 mol-1
Pc
77.10


V
a T  V  b
RT
b
P
P V V  b 

7248000 V  38.98  
3
-1
V  1070.5  38.98 
 cm mol

27.43  V V  38.98  


264224 V  38.98  
3
-1
V  1109.4 
 cm mol

V
V

38.98




 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  353.15  38.98  27.43  27.43V 
V  38.98  V V  38.98  

7248000


 1109.4  V 
V  38.98  V V  38.98  

 264224 
Updated 5/23/2017
p. 99 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
0.2857
1T
V sat  Vc Zc r 
10.8465
V sat  124  0.265
0.2857
Z  1  B0
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
B 0  0.083 
B1  0.139 
Tr1.6
Tr4.2
0.422
0.84651.6
0.172
0.84654.2
 0.4680
 0.2073
Z  1   0.4680  0.069  0.2073
Updated 5/23/2017
0.3558
 0.7973
0.8465
p. 100 of 182
NAS 8th Edition Annotated Solutions
T (K)
353.15
P (bar)
27.43
Chapter 3
Tc (K)
417.2
Pc (bar)
77.1
Tr
0.80
0.80
0.85
0.85
Pr
0.10
0.20
0.10
0.20
ω
0.069
Tr
0.8465
Pr
0.3558
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
Interpolated Values
Final Values
Z
V (cm3/mol)
Z0
0.9319
0.8539
0.9436
0.8810
0.7799
Z1
-0.0487
-0.1160
-0.0319
-0.0715
-0.1394
0.7703
824.6

V
a T  V  b
RT
b
P
P V V  b 

7147000 V  38.98  
3
-1
V   912.8  38.98 
 cm mol

33.08

V
V

38.98





216057 V  38.98  
3
-1
V   951.7 
 cm mol

V
V

38.98

 

Updated 5/23/2017
p. 101 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  363.15  38.98  33.08  33.08V 
V  38.98  V V  38.98  

7147000


 951.7  V 
V  38.98  V V  38.98  

 216057 
0.2857
1T
V sat  Vc Zc r 
10.8704
V sat  124  0.265
0.2857
Z  1  B0
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
B 0  0.083 
B1  0.139 
Updated 5/23/2017
Tr1.6
Tr4.2
0.422
0.87041.6
0.172
0.87044.2
 0.4439
 0.1691
p. 102 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
Z  1   0.4439  0.069  0.1691
T (K)
363.15
P (bar)
33.08
0.4291
 0.7754
0.8704
Tc (K)
417.2
Pc (bar)
77.1
Tr
0.90
0.90
0.93
0.93
Pr
0.20
0.40
0.20
0.40
ω
0.069
Tr
0.8704
Pr
0.4291
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
Interpolated Values
Final Values
Z
V (cm3/mol)
T (K)
363.15
P (bar)
33.08
Z0
0.9015
0.7800
0.9115
0.8059
Z1
-0.0442
-0.1118
-0.0326
-0.0763
0.7346
-0.1600
ω
0.069
Tr
0.8704
0.7235
660.5
Tc (K)
417.2
Pc (bar)
77.1
Tr
0.85
0.85
0.90
0.90
Pr
0.10
0.20
0.10
0.20
Pr
0.4291
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
Interpolated Values
Final Values
Z
V (cm3/mol)
Updated 5/23/2017
Z0
0.9436
0.8810
0.9528
0.9015
0.7566
Z1
-0.0319
-0.0715
-0.0205
-0.0442
-0.1361
0.7472
682.1
p. 103 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V V  b 
T 0.5 R 2Tc2
T 0.5 R 2Tc2.5
T 0.5  83.1452  430.82.5
1.4439  108
a(T )  0.42748 r
 0.42748
 0.42748

bar cm6 mol-2
Pc
Pc
78.84
T
b  0.08664
RTc
83.145  430.8
 0.08664
 39.363 cm3 mol-1
Pc
78.84


V
a T  V  b
RT
b
P
P V V  b 

7683000 V  39.363 
3
-1
V  1573.6  39.363 
 cm mol

18.66  V V  39.363 


411736 V  39.363 
3
-1
V  1613.0 
 cm mol

V
V

39.363




 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  353.15  39.363  18.66  18.66V 
V  39.363  V V  39.363 

7683000


 1613.0  V 
V  39.363  V V  39.363 

 411736 
Updated 5/23/2017
p. 104 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
0.2857
1T
V sat  Vc Zc r 
10.8198
V sat  122  0.269
0.2857
Z  1  B0
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
B 0  0.083 
B1  0.139 
Tr1.6
Tr4.2
0.422
0.81981.6
0.172
0.81984.2
 0.4969
 0.2572
Z  1   0.4969  0.245  0.2572 
Updated 5/23/2017
0.2367
 0.8282
0.8198
p. 105 of 182
NAS 8th Edition Annotated Solutions
T (K)
353.15
P (bar)
18.66
Chapter 3
Tc (K)
430.8
Pc (bar)
78.84
Tr
0.80
0.80
0.85
0.85
Pr
0.10
0.20
0.10
0.20
ω
0.245
Tr
0.8198
Pr
0.2367
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
Interpolated Values
Final Values
Z
V (cm 3/mol)
Z0
0.9319
0.8539
0.9436
0.8810
Z1
-0.0319
-0.0715
-0.0205
-0.0442
0.8382
-0.0729
0.8204
1291.0

V
a T  V  b
RT
b
P
P V V  b 

7577000 V  39.363 
3
-1
V  1295.3  39.363 
 cm mol

23.31  V V  39.363 


325053 V  39.363 
3
-1
V  1334.7 
 cm mol

V
V

39.363




Updated 5/23/2017
p. 106 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  363.15  39.363  23.31  23.31V 
V  39.363  V V  39.363 

7577000


 1334.7  V 
V  39.363  V V  39.363 

 325053 
0.2857
1T
V sat  Vc Zc r 
10.8430
V sat  122  0.269
0.2857
Z  1  B0
Pr
P
  B1 r
Tr
Tr
B 0  0.083 
0.422
B1  0.139 
0.172
Updated 5/23/2017
Tr1.6
Tr4.2
p. 107 of 182
NAS 8th Edition Annotated Solutions
B 0  0.083 
B1  0.139 
0.422
0.84301.6
0.172
0.84304.2
Chapter 3
 0.4726
 0.2134
Z  1   0.4726  0.245  0.2134 
T (K)
363.15
P (bar)
23.11
0.2957
 0.8159
0.8430
Tc (K)
430.8
Pc (bar)
78.84
Tr
0.80
0.80
0.85
0.85
Pr
0.10
0.20
0.10
0.20
ω
0.245
Tr
0.8430
Pr
0.2931
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
Interpolated Values
Final Values
Z
V (cm 3/mol)
Z0
0.9319
0.8539
0.9436
0.8810
0.8169
Z1
-0.0319
-0.0715
-0.0205
-0.0442
-0.0722
0.7992
1044.3

Updated 5/23/2017
p. 108 of 182
NAS 8th Edition Annotated Solutions
P
a T 
RT

V  b V V  b 
a(T )  0.42748
b  0.08664
Chapter 3
 Tr  R 2Tc2
 Tr   83.1452  369.82
 0.42748
 9.5134  106 Tr  bar cm 6 mol-2
Pc
42.48
RTc
83.145  369.8
 0.08664
 62.71 cm3 mol-1
Pc
42.48



 
 T   1   0.480  1.574 * 0.152  0.176 * 0.152 1  0.9279    1.0532
2
 Tr   1  0.480  1.574  0.176 2 1  Tr0.5
2
0.5
2
r

V
a T  V  b
RT
b
P
P V V  b 

10020000 V  62.71 
3
-1
V  1099.9  62.71 
 cm mol

25.94

V
V

62.71





386259 V  62.71 
3
-1
V  1162.6 
 cm mol

V
V

62.71

 

 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  343.15  62.17  25.94  25.94V 
V  62.71  V V  62.71 

10020000


 1162.0  V 
V  62.71  V V  62.71 

 386259 
Updated 5/23/2017
p. 109 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

P
a T 
RT

V  b V V  b 
a(T )  0.42748
b  0.08664
 Tr  R 2Tc2
 Tr   83.1452  425.12
 0.42748
 1.4068  107 Tr  bar cm6 mol-2
Pc
37.96
RTc
83.145  425.1
 0.08664
 80.67 cm3 mol-1
Pc
37.96



 
 T   1   0.480  1.574 * 0.200  0.176 * 0.200 1  0.8778    1.1019
2
 Tr   1  0.480  1.574  0.176 2 1  Tr0.5
2
0.5
2
r

V
a T  V  b
RT
b
P
P V V  b 

15502000 V  80.67  
3
-1
V   2013.3  80.67 
 cm mol

15.41

V
V

80.67




1005940 V  80.67  
3
-1
V   2094.0 
 cm mol

V V  80.67 


Updated 5/23/2017
p. 110 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  373.15  80.67  15.41  15.41V 
V  80.67  V V  80.67  

15502000


 2094.1  V 
V  80.67  V V  80.67  

 1005940 
P
a T 
RT

V  b V V  b 
a(T )  0.42748
b  0.08664
 Tr  R 2Tc2
 Tr   83.1452  425.12
 0.42748
 1.4068  107 Tr  bar cm6 mol-2
Pc
37.96
RTc
83.145  425.1
 0.08664
 80.67 cm3 mol-1
Pc
37.96



 
 T   1   0.480  1.574 * 0.200  0.176 * 0.200 1  0.9013    1.0814
2
 Tr   1  0.480  1.574  0.176 2 1  Tr0.5
2
0.5
2
r

V
a T  V  b
RT
b
P
P V V  b 

15212000 V  80.67  
3
-1
V  1707.2  80.67 
 cm mol

18.66

V
V

80.67





815252 V  80.67  
3
-1
V  1787.9 
 cm mol

V
V

80.67




Updated 5/23/2017
p. 111 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  383.15  80.67  18.66  18.66V 
V  80.67  V V  80.67  

15212000


 1787.9  V 
V  80.67  V V  80.67  

 815252 
P
a T 
RT

V  b V V  b 
a(T )  0.42748
b  0.08664
 Tr  R 2Tc2
 Tr   83.1452  425.12
 0.42748
 1.4068  107 Tr  bar cm6 mol-2
Pc
37.96
RTc
83.145  425.1
 0.08664
 80.67 cm3 mol-1
Pc
37.96



 
 T   1   0.480  1.574 * 0.200  0.176 * 0.200 1  0.9248    1.0613
2
 Tr   1  0.480  1.574  0.176 2 1  Tr0.5
2
0.5
2
r

V
a T  V  b
RT
b
P
P V V  b 

14930000 V  80.67  
3
-1
V  1460.6  80.67 
 cm mol

22.38

V
V

80.67





667113 V  80.67  
3
-1
V  1541.3 
 cm mol

V
V

80.67

 

Updated 5/23/2017
p. 112 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  393.15  80.67  22.38  22.38V 
V  80.67  V V  80.67  

14930000


 1541.3  V 
V  80.67  V V  80.67  

 667113 



 
 T   1   0.480  1.574 * 0.200  0.176 * 0.200 1  0.9484    1.0416
2
 Tr   1  0.480  1.574  0.176 2 1  Tr0.5
2
0.5
2
r
V
a T  V  b
RT
b
P
P V V  b 

14654000 V  80.67  
3
-1
V  1260.6  80.67 
 cm mol

26.59

V
V

80.67





551101V  80.67  
3
-1
V  1341.3 
 cm mol

V
V

80.67

 


Updated 5/23/2017
p. 113 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  403.15  80.67  26.59  26.59V 
V  80.67  V V  80.67  

14654000


 1341.3  V 
V  80.67  V V  80.67  

 551101 
P
a T 
RT

V  b V V  b 
a(T )  0.42748
b  0.08664
 Tr  R 2Tc2
 Tr   83.1452  430.82
 0.42748
 6.9565  106 Tr  bar cm6 mol-2
Pc
78.84
RTc
83.145  430.8
 0.08664
 39.363 cm3 mol-1
Pc
78.84



 
 T   1   0.480  1.574 * 0.245  0.176 * 0.245 1  0.8662    1.1220
2
 Tr   1  0.480  1.574  0.176 2 1  Tr0.5
2
0.5
2
r

V
a T  V  b
RT
b
P
P V V  b 

7805500 V  39.36  
3
-1
V  1079.5  39.36 
 cm mol

28.74

V
V

39.36





271591 V  39.36  
3
-1
V  1118.9 
 cm mol

V
V

39.36




Updated 5/23/2017
p. 114 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  373.15  39.36  28.74  28.74V 
V  39.36  V V  39.36  

7805500


 1118.9  V 
V  39.36  V V  39.36  

 271591 



 
 T   1   0.480  1.574 * 0.245  0.176 * 0.245 1  0.8894    1.0997
2
 Tr   1  0.480  1.574  0.176 2 1  Tr0.5
2
0.5
2
r

V
a T  V  b
RT
b
P
P V V  b 

7650000 V  39.36  
3
-1
V   909.9  39.36 
 cm mol

35.01  V V  39.36  


218515 V  39.36  
3
-1
V   949.3 
 cm mol

V
V

39.36




Updated 5/23/2017
p. 115 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V V  b  


a T 


 83.14  383.15  39.36  35.01  35.01V 
V  39.36  V V  39.36  

7650000


 949.3  V 
V  39.36  V V  39.36  

 218515 

P
a T 
RT

V  b V  2.4142b V  0.4142b 
a(T )  0.45724
a(T )  0.45724
 Tr  R 2Tc2
Pc
 Tr   83.1452  369.82
42.48
a(T )   Tr   1.0176  107 bar cm 6 mol-2
b  0.07779
RTc
83.145  369.8
 0.07779
 56.3 cm3 mol-1
Pc
42.48



 
 T   1   0.37464  1.54226 * 0.152  0.2699 * 0.152 1  0.9279    1.0448
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

Updated 5/23/2017
p. 116 of 182
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


10632000 V  56.3
3
-1
V  1099.9  56.3 
 cm mol

25.94
V

23.32
V

135.92






409851 V  56.3
3
-1
V  1156.2 
 cm mol

V  23.32 V  135.92  

 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  343.15  56.3  25.94  25.94V 
V  56.3  V  23.32 V  135.92  

10632000


 1156.2  V 
V  56.3  V  23.32 V  135.92  

 409851 

P
a T 
RT

V  b V  2.4142b V  0.4142b 
a(T )  0.45724
a(T )  0.45724
 Tr  R 2Tc2
Pc
 Tr   83.1452  425.12
37.96
a(T )   Tr   1.5048  107 bar cm 6 mol-2
b  0.07779
RTc
83.145  425.1
 0.07779
 72.43 cm3 mol-1
Pc
37.96

Updated 5/23/2017
p. 117 of 182
NAS 8th Edition Annotated Solutions
Chapter 3


 
 T   1   0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.8778    1.0866
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


16352000 V  72.43
3
-1
V   2013.3  72.43 
 cm mol

15.41 V  30.00 V  174.86  


1061129 V  72.43 
3
-1
V   2085.7 
 cm mol

V  30.00 V  174.86  

 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  373.15  72.43  15.41  15.41V 
V  72.43  V  30.00 V  174.96  

16352000


 2085.7  V 
V  72.43  V  30.00 V  174.96  

 1061129 
P
a T 
RT

V  b V  2.4142b V  0.4142b 
Updated 5/23/2017
p. 118 of 182
NAS 8th Edition Annotated Solutions
a(T )  0.45724
a(T )  0.45724
Chapter 3
 Tr  R 2Tc2
Pc
 Tr   83.1452  425.12
37.96
a(T )   Tr   1.5048  107 bar cm 6 mol-2
b  0.07779
RTc
83.145  425.1
 0.07779
 72.43 cm3 mol-1
Pc
37.96



 
 T   1   0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.9013    1.0692
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


16090000 V  72.43
3
-1
V  1707.2  72.43 
 cm mol

18.66
V

30.00
V

174.86






862226 V  72.43 
3
-1
V  1779.6 
 cm mol

V

30.00
V

174.86




 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  383.15  72.43  18.66  18.66V 
V  72.43  V  30.00 V  174.96  

16090000


 1779.6  V 
V  72.43  V  30.00 V  174.96  

 862226 
Updated 5/23/2017
p. 119 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V  2.4142b V  0.4142b 
a(T )  0.45724
a(T )  0.45724
 Tr  R 2Tc2
Pc
 Tr   83.1452  425.12
37.96
a(T )   Tr   1.5048  107 bar cm 6 mol-2
b  0.07779
RTc
83.145  425.1
 0.07779
 72.43 cm3 mol-1
Pc
37.96



 
 T   1   0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.9248    1.0522
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


15833000 V  72.43
3
-1
V  1460.6  72.43 
 cm mol

22.38
V

30.00
V

174.86






707471V  72.43 
3
-1
V  1533.0 
 cm mol

V

30.00
V

174.86




Updated 5/23/2017
p. 120 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  393.15  72.43  22.38  22.38V 
V  72.43  V  30.00 V  174.96  

15833000


 1533.0  V 
V  72.43  V  30.00 V  174.96  

 707041 



 
 T   1   0.37464  1.54226 * 0.200  0.2699 * 0.200 1  0.9484    1.0355
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


15582000 V  72.43
3
-1
V  1260.6  72.43 
 cm mol

26.59 V  30.00 V  174.86  


586008 V  72.43 
3
-1
V  1333.0 
 cm mol

V

30.00
V

174.86





 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  393.15  72.43  26.59  26.59V 
V  72.43  V  30.00 V  174.96  

15582000


 1333.0  V 
V  72.43  V  30.00 V  174.96  

 586008 
Updated 5/23/2017
p. 121 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V  2.4142b V  0.4142b 
a(T )  0.45724
a(T )  0.45724
 Tr  R 2Tc2
Pc
 Tr   83.1452  408.12
36.48
a(T )   Tr   1.4431  107 bar cm 6 mol-2
b  0.07779
RTc
83.145  408.1
 0.07779
 72.36 cm3 mol-1
Pc
36.48



 
 T   1   0.37464  1.54226 * 0.302  0.2699 * 0.302 1  0.8899    1.0946
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


15796000 V  72.36 
3
-1
V  1825.5  72.36 
 cm mol

16.54
V

29.97
V

174.69





955018 V  72.36  
3
-1
V  1897.9 
 cm mol

V  29.97 V  174.69  

Updated 5/23/2017
p. 122 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  363.15  72.36  16.54  16.54V 
V  72.36  V  29.97 V  174.69  

15796000


 1897.9  V 
V  72.36  V  29.97 V  174.69  

 955018 


 
 T   1   0.37464  1.54226 * 0.302  0.2699 * 0.302 1  0.9144    1.0727
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


15480000 V  72.36 
3
-1
V  1549.0  72.36 
 cm mol

20.03
V

29.97
V

174.69






772841 V  72.36  
3
-1
V  1621.3 
 cm mol

V

29.97
V

174.69




Updated 5/23/2017
p. 123 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  373.15  72.36  20.03  20.03V 
V  72.36  V  29.97 V  174.69  

15480000


 1621.3  V 
V  72.36  V  29.97 V  174.69  

 772841 
P
a T 
RT

V  b V  2.4142b V  0.4142b 
a(T )  0.45724
a(T )  0.45724
 Tr  R 2Tc2
Pc
 Tr   83.1452  408.12
36.48
a(T )   Tr   1.4431  107 bar cm 6 mol-2
b  0.07779
RTc
83.145  408.1
 0.07779
 72.36 cm3 mol-1
Pc
36.48



 
 T   1   0.37464  1.54226 * 0.181  0.2699 * 0.181 1  0.9389    1.0405
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

Updated 5/23/2017
p. 124 of 182
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


15015000 V  72.36 
3
-1
V  1326.8  72.36 
 cm mol

24.01
V

29.97
V

174.69





625364 V  72.36  
3
-1
V  1399.2 
 cm mol

V  29.97 V  174.69  

 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  383.15  72.36  24.01  24.01V 
V  72.36  V  29.97 V  174.69  

15015000


 1399.2  V 
V  72.36  V  29.97 V  174.69  

 625364 


 
 T   1   0.37464  1.54226 * 0.181  0.2699 * 0.181 1  0.9634    1.0240
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

Updated 5/23/2017
p. 125 of 182
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


14777000 V  72.36 
3
-1
V  1145.8  72.36 
 cm mol

28.53
V

29.97
V

174.69





517946 V  72.36  
3
-1
V  1218.1 
 cm mol

V  29.97 V  174.69  

 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  393.15  72.36  28.53  28.53V 
V  72.36  V  29.97 V  174.69  

14777000


 1218.1  V 
V  72.36  V  29.97 V  174.69  

 517946 
P
a T 
RT

V  b V  2.4142b V  0.4142b 
Updated 5/23/2017
p. 126 of 182
NAS 8th Edition Annotated Solutions
a(T )  0.45724
a(T )  0.45724
Chapter 3
 Tr  R 2Tc2
Pc
 Tr   83.1452  417.22
77.10
a(T )   Tr   7.136  10 bar cm 6 mol-2
6
b  0.07779
RTc
83.145  417.2
 0.07779
 35.00 cm3 mol-1
Pc
77.10



 
 T   1   0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.7985    1.1047
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


7883000 V  35.0 
3
-1
V  1521.1  35.00 
 cm mol

18.21
V

14.50
V

84.49






432903 V  35.00  
3
-1
V  1556.1 
 cm mol

V

14.50
V

84.49




 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  333.15  35.00  18.21  18.21V 
V  35.00  V  14.50 V  84.49  

7883000


 1556.1  V 
V  35.00  V  14.50 V  84.49  

 432903 
Updated 5/23/2017
p. 127 of 182
NAS 8th Edition Annotated Solutions
Chapter 3


 
 T   1   0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.8225    1.091
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


7788000 V  35.0 
3
-1
V  1268.6  35.00 
 cm mol

22.49
V

14.50
V

84.49





346265 V  35.00  
3
-1
V  1303.6 
 cm mol

V  14.50 V  84.49  

 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  343.15  35.00  22.49  22.49V 
V  35.00  V  14.50 V  84.49  

7788000


 1303.6  V 
V  35.00  V  14.50 V  84.49  

 317293 
Updated 5/23/2017
p. 128 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V  2.4142b V  0.4142b 
a(T )  0.45724
a(T )  0.45724
 Tr  R 2Tc2
Pc
 Tr   83.1452  417.22
77.10
a(T )   Tr   7.136  10 bar cm 6 mol-2
6
b  0.07779
RTc
83.145  417.2
 0.07779
 35.00 cm3 mol-1
Pc
77.10



 
 T   1   0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.8465    1.078
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


7694000 V  35.0 
3
-1
V  1070.5  35.00 
 cm mol

27.43
V

14.50
V

84.49






280490 V  35.00  
3
-1
V  1105.5 
 cm mol

V  14.50 V  84.49  

Updated 5/23/2017
p. 129 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  353.15  35.00  27.43  27.43V 
V  35.00  V  14.50 V  84.49  

7694000


 1105.5  V 
V  35.00  V  14.50 V  84.49  

 280490 


 
 T   1   0.37464  1.54226 * 0.069  0.2699 * 0.069 1  0.8705    1.065
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


7602000 V  35.0 
3
-1
V   912.8  35.00 
 cm mol

33.08
V

14.50
V

84.49





229807 V  35.00  
3
-1
V   947.8 
 cm mol

V  14.50 V  84.49  

Updated 5/23/2017
p. 130 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  363.15  35.00  33.08  33.08V 
V  35.00  V  14.50 V  84.49  

7602000


 047.8  V 
V  35.00  V  14.50 V  84.49  

 229807 
P
a T 
RT

V  b V  2.4142b V  0.4142b 
a(T )  0.45724
a(T )  0.45724
 Tr  R 2Tc2
Pc
 Tr   83.1452  430.82
78.84
a(T )   Tr   7.441  10 bar cm 6 mol-2
6
b  0.07779
RTc
83.145  430.8
 0.07779
 35.34 cm3 mol-1
Pc
78.84



 
 T   1   0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8198    1.1441
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

Updated 5/23/2017
p. 131 of 182
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


8513000 V  35.34 
3
-1
V  1573.6  35.34 
 cm mol

18.66
V

14.64
V

85.32





456217 V  35.34  
3
-1
V  1608.9 
 cm mol

V  14.64 V  85.32  

 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  353.15  35.34  18.66  18.66V 
V  35.34  V  14.64 V  85.32  

8513000


 1608.9  V 
V  35.34  V  14.64 V  85.32  

 456217 


 
 T   1   0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8430    1.1242
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

Updated 5/23/2017
p. 132 of 182
NAS 8th Edition Annotated Solutions
V
Chapter 3
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


8365000 V  35.34 
3
-1
V  1306.5  35.34 
 cm mol

23.11
V

14.64
V

85.32





361969 V  35.34  
3
-1
V  1341.9 
 cm mol

V  14.64 V  85.32  

 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  363.15  35.34  23.11  23.11V 
V  35.34  V  14.64 V  85.32  

8365000


 1341.9  V 
V  35.34  V  14.64 V  85.32  

 361969 
P
a T 
RT

V  b V  2.4142b V  0.4142b 
Updated 5/23/2017
p. 133 of 182
NAS 8th Edition Annotated Solutions
a(T )  0.45724
a(T )  0.45724
Chapter 3
 Tr  R 2Tc2
Pc
 Tr   83.1452  430.82
78.84
a(T )   Tr   7.441  10 bar cm 6 mol-2
6
b  0.07779
RTc
83.145  430.8
 0.07779
 35.34 cm3 mol-1
Pc
78.84



 
 T   1   0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8662    1.1047
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


8220000 V  35.34 
3
-1
V  1079.5  35.34 
 cm mol

28.74
V

14.64
V

85.32






285997 V  35.34  
3
-1
V  1114.9 
 cm mol

V

14.64
V

85.32




 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  373.15  35.34  28.74  28.74V 
V  35.34  V  14.64 V  85.32  

8220000


 1114.9  V 
V  35.34  V  14.64 V  85.32  

 285997 
Updated 5/23/2017
p. 134 of 182
NAS 8th Edition Annotated Solutions
Chapter 3


 
 T   1   0.37464  1.54226 * 0.245  0.2699 * 0.245 1  0.8894    1.0856
2
 Tr   1  0.37464  1.54226  0.2699 2 1  Tr0.5
2
0.5
2
r

V
a T 
RT
V b
b
P
P V  0.4142b V  2.4142b 


8078000 V  35.34 
3
-1
V   909.9  35.34 
 cm mol

35.01
V

14.64
V

85.32





230772 V  35.34  
3
-1
V   945.2 
 cm mol

V  14.64 V  85.32  

 RT  bP  VP 
V  b  V  0.4142b V  2.4142b  


a T 


 83.14  383.15  35.34  35.01  35.01V 
V  35.34  V  14.64 V  85.32  

8078000


 945.2  V 
V  35.34  V  14.64 V  85.32  

 230772 
Updated 5/23/2017
p. 135 of 182
NAS 8th Edition Annotated Solutions
Chapter 3




Updated 5/23/2017
p. 136 of 182
NAS 8th Edition Annotated Solutions
Chapter 3


B 0  0.083 
B1  0.139 
Updated 5/23/2017
0.422
1.46561.6
0.172
1.46564.2

 0.1459
 0.1045
p. 137 of 182
NAS 8th Edition Annotated Solutions
Z  1   0.1459  0.645  0.1045
Chapter 3
0.9759
 0.9477
1.4656
0.2857
0.2857
1T
V sat  Vc Zc r 
 200.0*0.276(1320 / 369.8)
 96.8 cm3/mol
Updated 5/23/2017
p. 138 of 182
NAS 8th Edition Annotated Solutions
Chapter 3


Updated 5/23/2017
p. 139 of 182
NAS 8th Edition Annotated Solutions
T (K)
333.15
Chapter 3
P (bar)
140
Tc (K)
305.3
Pc (bar)
48.72

Tr
1.05
1.05
1.10
1.10
Pr
2.00
3.00
2.00
3.00
Z0
0.3452
0.4604
0.3953
0.4770
0.1
Tr
1.0912
Pr
2.8736
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
Interpolated Values
Final Values
Z
V (cm3/mol)
Updated 5/23/2017
0.4630
Z1
-0.0432
-0.0838
0.0698
-0.0373
-0.0334
0.4597
91.0
p. 140 of 182
NAS 8th Edition Annotated Solutions
T (K)
388.8
Chapter 3
P (bar)
200
Tc (K)
305.3
Pc (bar)
48.72

Tr
1.40
1.40
1.50
1.50
Pr
3.00
5.00
3.00
5.00
Z0
0.7202
0.7761
0.7887
0.8200
0.1
Tr
1.2735
Pr
4.1051
Table Points
Tr (1)
Tr(1)
Tr(2)
Tr(2)
Pr(1)
Pr(2)
Pr(1)
Pr(2)
Interpolated Values
Final Values
Z
V (cm3/mol)
0.6816
Z1
0.2397
0.1737
0.2433
0.2309
0.1612
0.6978
112.8
454g
33.9975g/mol
P
a T 
RT

V  b V V  b 
.4278∗ Tr −0.5 ∗R2 ∗Tc2
Pc
.4278∗ 0.9021 −0.5 ∗83.1452 ∗324.82
65.4
R∗Tc
83.145∗324.8
0.08664 ∗
Pc
65.4
P
a T 
RT

V  b V V  b 
5022832.543
179.7(179.7+35.776)
Updated 5/23/2017
p. 141 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
454g
67.825g/mol
P
a T 
RT

V  b V V  b 
.4278∗ Tr −0.5 ∗R2 ∗Tc2
Pc
.4278∗ 1.1230 −0.5 ∗83.1452 ∗260.92
49.9
R∗Tc
83.145∗260.9
0.08664 ∗
Pc
49.9
P
a T 
RT

V  b V V  b 
358.5(358.5+37.6641)
454g
32.117 g/mol
P
a T 
RT

V  b V V  b 
.4278∗ Tr −0.5 ∗R2 ∗Tc2
Pc
.4278∗ 1.0864 −0.5 ∗83.1452 ∗269.72
48.4
Updated 5/23/2017
p. 142 of 182
NAS 8th Edition Annotated Solutions
R∗Tc
Pc
P
0.08664 ∗
Chapter 3
83.145∗269.7
48.4
a T 
RT

V  b V V  b 
169.78(169.78+40.14)
454g
76.62 g/mol
P
a T 
RT

V  b V V  b 
.4278∗ Tr −0.5 ∗R2 ∗Tc2
Pc
.4278∗ 0.9385 −0.5 ∗83.1452 ∗312.22
49.5
R∗Tc
83.145∗312.2
0.08664 ∗
Pc
49.5
P
a T 
RT

V  b V V  b 
405.06(405.06+45.434)
454g
77.945 g/mol
Updated 5/23/2017
p. 143 of 182
NAS 8th Edition Annotated Solutions
P
Chapter 3
a T 
RT

V  b V V  b 
.4278∗ Tr −0.5 ∗R2 ∗Tc2
Pc
.4278∗ 0.7855 −0.5 ∗83.1452 ∗3732
65.5
R∗Tc
83.145∗373
0.08664 ∗
Pc
65.5
P
a T 
RT

V  b V V  b 
412.044(412.044+41.022)
454g
71 g/mol
P
a T 
RT

V  b V V  b 
.4278∗ Tr −0.5 ∗R2 ∗Tc2
Pc
.4278∗ 1.2521 −0.5 ∗83.1452 ∗2342
44.6
R∗Tc
83.145∗234
0.08664 ∗
Pc
44.6
P
a T 
RT

V  b V V  b 
375.35(375.35+37.795)
Updated 5/23/2017
p. 144 of 182
NAS 8th Edition Annotated Solutions
B 0 = 0.083 
0.422
=  0.415
Tr1.6
B1 = 0.139 
0.172
=  0.126
Tr4.2
Chapter 3
BPc
= B 0 +  B1 =  0.415 + 0.045 ×  0.126  =  0.42067
RTc
 BP  P
0.3822
Z =1 +  c  r =1  0.42067 ×
= 0.8218
0.9021
 RTc  Tr
MWphosphine = 33.997 g·mol1
V = 2.4 L = 2400 cm3
Using this,
PV
25 bar × 2400 cm3
1
mass = MWPhosphine ×
= 33.997 g·mol ×
= 101.89 g
ZRT
0.8218 × 83.14 bar·cm3·mol1·K 1 × 293 K
Updated 5/23/2017
p. 145 of 182
NAS 8th Edition Annotated Solutions
B 0 = 0.083 
0.422
=  0.2675
Tr1.6
B1 = 0.139 
0.172
= 0.03333
Tr4.2
Chapter 3
BPc
= B 0 +  B1 =  0.2675 + 0.434 ×  0.03333 =  0.2530
RTc
 BP  P
0.5010
Z =1 +  c  r =1  0.2530 ×
= 0.887
1.1230
 RTc  Tr
MWBF3 = 67.82 g·mol1
V = 2.4 L = 2400 cm3
Using this,
PV
25 bar × 2400 cm3
1
mass = MWBF3 ×
= 67.82 g·mol ×
= 188.32 g
ZRT
0.887 × 83.14 bar·cm3·mol1·K 1 × 293 K
Updated 5/23/2017
p. 146 of 182
NAS 8th Edition Annotated Solutions
B 0 = 0.083 
0.422
=  0.2866
Tr1.6
B1 = 0.139 
0.172
= 0.01756
Tr4.2
Chapter 3
BPc
= B 0 +  B1 =  0.2866 + 0.094 ×  0.01756  =  0.2849
RTc
 BP  P
0.51652
Z =1 +  c  r =1  0.2849 ×
= 0.8645
1.0864
 RTc  Tr
MWSiH4 = 32.12 g·mol1
V = 2.4 L = 2400 cm3
Using this,
PV
25 bar × 2400 cm3
1
mass = MWSiH4 ×
= 32.12 g·mol ×
= 91.499 g
ZRT
0.8645 × 83.14 bar·cm3·mol1·K 1 × 293 K
Updated 5/23/2017
p. 147 of 182
NAS 8th Edition Annotated Solutions
B 0 = 0.083 
0.422
=  0.384
Tr1.6
B1 = 0.139 
0.172
=  0.0855
Tr4.2
Chapter 3
BPc
= B 0 +  B1 =  0.384 + 0.151 ×  0.0855  =  0.3969
RTc
 BP  P
0.5051
Z =1 +  c  r =1  0.3969 ×
= 0.7863
0.9385
 RTc  Tr
MWGeH4 = 76.62 g·mol1
V = 2.4 L = 2400 cm3
Using this,
PV
25 bar × 2400 cm3
1
mass = MWGeH4 ×
= 76.62 g·mol ×
= 239.99 g
ZRT
0.7863 × 83.14 bar·cm3·mol1·K 1 × 293 K
Updated 5/23/2017
p. 148 of 182
NAS 8th Edition Annotated Solutions
B 0 = 0.083 
0.422
=  0.5379
Tr1.6
B1 = 0.139 
0.172
=  0.3351
Tr4.2
Chapter 3
BPc
= B 0 +  B1 =  0.5379 + 0.011 ×  0.3351 =  0.5416
RTc
 BP  P
0.38168
Z =1 +  c  r =1  0.5416 ×
= 0.7368
0.7855
 RTc  Tr
MWAsH3 = 77.945 g·mol1
V = 2.4 L = 2400 cm3
Using this,
mass = MWAsH3 ×
Updated 5/23/2017
PV
25 bar × 2400 cm3
= 77.945 g·mol1 ×
= 260.54 g
ZRT
0.7368 × 83.14 bar·cm3·mol1·K 1 × 293 K
p. 149 of 182
NAS 8th Edition Annotated Solutions
B 0 = 0.083 
0.422
=  0.2115
Tr1.6
B1 = 0.139 
0.172
= 0.07209
Tr4.2
Chapter 3
BPc
= B 0 +  B1 =  0.2115 + 0.120 × (0.07209) = -0.2028
RTc
 BP  P
0.5605
Z =1 +  c  r =1  0.2028 ×
= 0.9092
1.2521
 RTc  Tr
MWNF3 = 71.00 g·mol1
V = 2.4 L = 2400 cm3
Using this,
mass = MWNF3 ×
Updated 5/23/2017
PV
25 bar × 2400 cm3
= 71 g·mol1 ×
= 192.33 g
ZRT
0.9092 × 83.14 bar·cm3·mol1·K 1 × 293 K
p. 150 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
P  Patmosphere   gh  1.01325bar  1027kg / m3 *9.8m / s 2 *300m


P
a T 
RT

V  b V V  b 
Updated 5/23/2017
p. 151 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
0.5
1.0561
T 0.5 R 2Tc2
a(T )  0.42748 r
 0.42748
Pc
b  0.08664
P
 83.1452  282.32
50.4
 4547019 bar cm 6 mol-2
RTc
83.145  282.3
 0.08664
 40.35 cm3 mol-1
Pc
50.4
83.145 * 298.15
4547019

 85.3 bar
105.2  40.35 105.2 105.2  40.35 
Updated 5/23/2017
p. 152 of 182
NAS 8th Edition Annotated Solutions
Chapter 3




Updated 5/23/2017

p. 153 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

Updated 5/23/2017
p. 154 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 155 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
BP
Bˆ  c  B 0   B1
RTc
B 0  0.083 
0.422
B1  0.139 
0.172
Tr1.6
Tr4.2

Updated 5/23/2017
p. 156 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

P

a T 
RT

V  b V V  b 
T
T 0.5 R 2Tc2
a(T )  0.42748 r
 0.42748
Pc
0.5
R 2Tc2.5
Pc
 0.42748
T 0.5  83.1452  126.22.5
34.00

1.5551  107
bar cm6 mol-2
T
RT
83.145  126.2
b  0.08664 c  0.08664
 26.74 cm3 mol-1
Pc
34.00
P
RT
900619
83.145 * 298.15
900619



 450.0 bar
V  26.74 V V  26.74  69.03  26.74 69.03  69.03  26.74 

Updated 5/23/2017
p. 157 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
Z  Z0   Z 1 = 0.322
Updated 5/23/2017
p. 158 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 159 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 160 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
For Z 0 9 and for approximately equal error in the five variables, a 1% maximum
error in B requires errors in the variables of less than about 0.02%. This is because the
divisor Z 1 0 1. In the limit as Z 1, the error in B approaches infinity.
Updated 5/23/2017
p. 161 of 182
NAS 8th Edition Annotated Solutions
Chapter 3

Updated 5/23/2017
p. 162 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 163 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
 (Tr ) 1/2 (Tr ) 3/2 0.42748 × (0.821) 1.5
q=
=
=
= 6.6326
Tr

0.08664
P
0.294
 =  r = 0.08664 ×
= 0.03102
Tr
0.821
Z =1+ -q × ×
Z -
Z Z +  
Z = 1 + 0.03102 - 6.6326 × 0.03102 ×
Updated 5/23/2017
Z - 0.03102
= 0.7904
Z  Z + 0.03102 
p. 164 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
n  1.5 108 / 0.8366  1.79 108 mol/day = 7470 kmol/hr
Updated 5/23/2017
p. 165 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 166 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 167 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 168 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
180
160
140
120
Zc
100
80
y = -0.4578x + 92.712
60
40
20
0
-20
0
50
100
150
200
250
ω
Updated 5/23/2017
p. 169 of 182
NAS 8th Edition Annotated Solutions
Chapter 3






















Updated 5/23/2017

p. 170 of 182
NAS 8th Edition Annotated Solutions
Chapter 3









Updated 5/23/2017

p. 171 of 182
NAS 8th Edition Annotated Solutions
Z-1/ρ
P Kpa
V (cm^3/g)
-118
-119
-120
-121
-122
-123
-124
-125
-126
-127
Chapter 3
/cm^3)
Z-1/
Z
2.63188E-05 0.996686
-125.9023437
3.1602E-05 0.996072
-124.2881683
3.6891E-05 0.995477
-122.6047371
4.21856E-05 0.994901
-120.8796292
4.74895E-05 0.994258
-120.9085917
5.28002E-05 0.993616
-120.9160244
5.81137E-05 0.993044
-119.697015
6.34358E-05 0.992433
-119.2916276
y = 166681x - 129.29
0
0.00002
0.00004
0.00006
0.00008
ρ (g/cm^3)
P Kpa
V (cm^3/g)
Updated 5/23/2017
/cm^3)
Z
Z-1/
2.41874E-05
0.997494
-103.6016566
2.90371E-05
0.997077
-100.6644948
3.38917E-05
0.996634
-99.32289962
p. 172 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
3.87526E-05
0.996138
-99.64691421
4.36138E-05
0.995747
-97.50696627
4.84847E-05
0.995235
-98.28634491
5.33532E-05
0.994861
-96.32176525
5.82313E-05
0.994386
-96.4013883
-95
-96
y = 189330x - 106.77
-97
Z-1/ρ
-98
-99
-100
-101
-102
-103
-104
0
0.00001
0.00002
0.00003
0.00004
0.00005
0.00006
0.00007
ρ (g/cm^3)
P Kpa
V (cm^3/g)
Updated 5/23/2017
/cm^3)
Z
Z-1/
2.23782E-05
0.998061
-86.63762145
2.68638E-05
0.997691
-85.94703438
3.13504E-05
0.997393
-83.14166567
3.58409E-05
0.997064
-81.92792085
4.03347E-05
0.996726
-81.17728694
4.48322E-05
0.996372
-80.92926009
4.93334E-05
0.99601
-80.88365086
5.38343E-05
0.995712
-79.65046047
p. 173 of 182
Z-1/ρ
NAS 8th Edition Annotated Solutions
-78
-79
-80
-81
-82
-83
-84
-85
-86
-87
-88
Chapter 3
y = 216165x - 90.772
0
0.00001
0.00002
0.00003
0.00004
0.00005
0.00006
ρ (g/cm^3)
1 + (1 - 0 -1) =3Zc
Updated 5/23/2017
p. 174 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
0 × 1 ×  2 -  0 +1 ( + 1) +  = 3(0.333) 2
or
1
( + 1) +  = 0.333
and
0 × 1 ×  2   + 1 +  = (0.333)3
 = 0.03693
(2)
solving equations 1 and  2  , we get
 = 0.086640
 = 0.427480
1 + (1 - (1 - 2) - (1 + 2)) = 3Z c
1 + (1 - 1 + 2 - 1 - 2)) = 3Z c
1 -  = 3Z c 1




(1 - 2) 1 + 2  2 - 1 - 2 + 1 + 2 ( + 1) +  = 3Z c2
(1 - 2) 2 - 1 + 1  ( + 1) +  = 3Z c2
3 2 - 2 +  = 3Z c2
1 -  
3 - 2 +  = 3 

 3 
9 2 - 6 + 3 =1 +  2 - 2
2
2
10 2 - 4 + 3 =1
Updated 5/23/2017
p. 175 of 182
NAS 8th Edition Annotated Solutions
=
1 + 102 + 4
3

Chapter 3
(2)

(1 - 2) 1 + 2 2   + 1 +  = Z c3
 1 + 102 + 4 
 1 
3 -  + 
 = 

3
 3 


3
(3)
By solving equation (3), we get
 = 0.33202
Also, from equation (1), we get
1 - 0.33202 = 3Z c
Z c =0.2227
and from equation (2), we get
=
1 + 10(0.33202) 2 + 4  0.33202 
3
 = 1.1435
Updated 5/23/2017
p. 176 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
-0.2
(Z-1)*Z*Tr/ Pr
-0.22
y = 0.0364x - 0.333
-0.24
-0.26
-0.28
-0.3
-0.32
-0.34
0
0.5
1
1.5
2
2.5
3
3.5
Pr/Z*Tr
Updated 5/23/2017
p. 177 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
ˆ  0.339  0* 0.033  0.339
B
Updated 5/23/2017
p. 178 of 182
NAS 8th Edition Annotated Solutions
Updated 5/23/2017
Chapter 3
p. 179 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
Z
1
a
b
a

 1

1  b RT
1  b RT
Z rep 
b
a
and Z atr 
1  b
RT
First, let’s put a and b equal to zero and check if it reduces to the ideal gas equation or not. If
equation reduces to the ideal gas equation, then the modification is reasonable.
a)
P=
RT
a
V -b V
when a = 0 and b = 0, then
P=
RT
V
Updated 5/23/2017
p. 180 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
The equation reduces to the ideal gas equation. Therefore, the modification is reasonable.
b) P =
RT
a
(V - b)2 V
when a = 0 and b = 0, then
P=
RT
V2
The equation does not reduce to the ideal gas equation. Therefore, the modification is not
reasonable.
c) P =
RT
a
- 2
V V - b  V
when a = 0 and b = 0, then
P=
RT
V2
The equation does not reduce to the ideal gas equation. Therefore, the modification is not
reasonable.
d) P =
RT
a
- 2
V V
when a = 0 and b = 0, then
P=
RT
V
The equation reduces to the ideal gas equation. Therefore, the modification is reasonable.
Updated 5/23/2017
p. 181 of 182
NAS 8th Edition Annotated Solutions
Chapter 3
dV
  dT   dP
V
V 
ln    T  P
 Vo 
 





Updated 5/23/2017
p. 182 of 182