EGR 334 Thermodynamics: Homework 09
Problem 3: 92
Determine the volume in m3 occupied by 20 kg of hydrogen at 1170 kPa, and -220 deg C.
----------------------------------------------------------------------------------------------------------------------------- -----------m= 20 kg
p = 1170 kPa = 11.70 bar
T = -220 deg C. = 53 K.
for hydrogen: R=4.1240 kJ/kg-K
Rbar = 8.314 kJ/kmol-K
M=2.016
pc = 13.0 bar
T c = 33.2
Calculate the reduced pressure and temperature:
pr 
p 11.7

 0.9
pc 13.0
Tr 
T
53

 1.59
Tc 33.2
from Chart A-1: the compressibility factor, Z is about 0.94.
which indicates that this is behaving very much like an ideal gas
Using:
Z
v
pv
RT
ZRT 0.94(4.1240kJ / kg  K )(53K ) kPa kN  m

 0.1765m3 / kg
2
p
1170kPa
kN / m
kJ
therefore:
V  mv  (20kg )(0.1765m3 / kg )  3.53m3
EGR 334 Thermodynamics: Homework 09
Problem 3: 93
Carbon Monoxide with mass of 150 lb occupies a volume at 500 deg R and 3500 psi. Determine the volume, in ft 3
---------------------------------------------------------------------------------------------------- ------------------------------------Carbon Monoxide:
m = 150 lbm
T = 500 deg R
p = 3500 psi= 238 atm
R = 0.07090 Btu/lbm-oR
for CO:
M = 28.01 lb/lbmol
T c = 239 R
pc= 34.5 atm
calculate the reduced temperature and pressure:
pr 
p 238

 6.90
pc 34.5
Tr 
T 500

 2.09
Tc 239
from Figure A-2: the compressibility factor, Z is about 1.05.
which indicates that this is behaving very much like an ideal gas
Using:
Z
v
pv
RT
2
ZRT 1.05(0.0709Btu / lbm  o R )(500 o R )  1 ft  778.17 ft  lb f

 0.05747 ft 3 / lbm


2
p
3500lbf / in
Btu
 12in 
therefore:
V  mv  (150lbm)(0.05747 ft 3 / lbm)  8.62 ft 3
EGR 334 Thermodynamics: Homework 09
Problem 3: 96
Butane (C4H10) in a piston cylinder assembly undergoes an isothermal compression at 173 deg C from p 1 = 1.9 MPa
to p2 = 2.5 MPa. Determine the work in kJ/kg.
----------------------------------------------------------------------------------------------------------------------------------------Butane:
State 1: T1 = 173 C = 446 K
State 2: T 2 = 173 C = 446 K
p1= 1.9 MPa = 19 bar
p 2 = 2.5 MPa = 25 bar
To solve this we would like to assume that this will behave as an ideal gas…let's check by finding the
compressibility factor:
For Butane: M = 58.12 kg/kmol
Tc = 425 K pc = 38.0 bar Rbar =8.314 kJ/kmol-K
Calculate the reduced temperature and pressure: based on p2
pr 
p 25

 0.658
pc 38
From figure A-1
pr 
Tr 
T 446

 1.05
Tc 426
Z2 = 0.78 which falls significantly below the behavior as an ideal gas.
p 19

 0.5
pc 38
Checking the Z value based on p1, the Z factor is still only about Z1= 0.84
Possible Options to determine work done:
a) treat as ideal gas even though Z is less than 1:
pv  ZRT
R = Rbar/M
then
p
V=vm
ZRT
v
Work:
W   pdV  
ZR mT V2
ZRT
1
dV  ZRmT  dV  bar
ln
V /m
V
M
V1
for isothermal:
pV
 mRT  constant
Z
p1V1 p2V2

Z1
Z2
V2 p1 Z 2

V1
p2 Z1
therefore:
W ZRbarT V2 ZRbarT  p1 Z 2 

ln 
ln 

m
M
V1
M
 p2 Z1 
0.8(8.314kJ / kmol  K )(446 K )  19 0.78 
W
ln 
  17.8 kJ / kg
58.12kg / kmol
 25 0.84 
Note: this uses an average of Z=0.8 instead even though Z varies from 0.78 to 0.84 for this range of states. This
approximation will give some error to the answer.
EGR 334 Thermodynamics: Homework 09
Problem 3: 99
For what ranges of pressure and temperature can air be considered an ideal gas? Explain your reasoning.
------------------------------------------------------------------------------------------------------- -----------------------------------In order for air to be considered an ideal gas, pv = mRT or when expressed as
Z
pv
RT
then the substance behaves as an ideal gas is Z is sufficiently close to 1. For the sake of argument, let's say within
10% error is sufficiently close…then values of 0.9 < Z < 1.1 is a valid range.
Looking at the Generalized compressibility figure will provide some guidance to appropriate temperatures and
pressures for Air with
M = 28.97 kg/kmol
Tc = 133 K
pc = 37.7 bar.
Picking off some selected points:
at pr = 0.5 the Tr should be at least 1.20 to be above Z = 0.9
This equates to
p  pr  pc  0.5(37.7bar )  18.85 bar
and
T  Tr  Tc  1.2(133K )  160K  113 oC
at pr = 1.00 the Tr should be at least 1.40 to be above Z =0.9
This equates to
p  pr  pc  1.0(37.7bar )  37.7 bar
and
T  Tr  Tc  1.4(133K )  186.2 K  86.8 oC
at pr = 5.00 the Tr should be at least 1.8 to be above Z =0.9
This equates to
p  pr  pc  5.0(37.7bar )  188.5bar
and
T  Tr  Tc  1.8(133K )  239.4 K  33.6 oC
at pr > 10 the Z rises above 1.1 for many temperatures….so it would be better to stay below p r > 10
p  pr  pc  10.0(37.7bar )  377bar
In general…Air can be treated as an ideal gas if you keep the pressure below about 400 bar and above temperatures
of -40 deg C.