13-1
Chapter 13
GAS MIXTURES
Composition of Gas Mixtures
13-1C It is the average or the equivalent gas constant of the gas mixture. No.
13-2C No. We can do this only when each gas has the same mole fraction.
13-3C It is the average or the equivalent molar mass of the gas mixture. No.
13-4C The mass fractions will be identical, but the mole fractions will not.
13-5C Yes.
13-6C The ratio of the mass of a component to the mass of the mixture is called the mass fraction (mf),
and the ratio of the mole number of a component to the mole number of the mixture is called the mole
fraction (y).
13-7C From the definition of mass fraction,
M
mi
N M
= i i = y i  i
mm N m M m
 Mm
mf i =




13-8C Yes, because both CO2 and N2O has the same molar mass, M = 44 kg/kmol.
13-9 A mixture consists of two gases. Relations for mole fractions when mass fractions are known are to
be obtained .
Analysis The mass fractions of A and B are expressed as
mf A =
MA
mA
N M
= A A = yA
Mm
mm N m M m
and
mf B = y B
MB
Mm
Where m is mass, M is the molar mass, N is the number of moles, and y is the mole fraction. The apparent
molar mass of the mixture is
Mm =
mm N A M A + N B M B
=
= y AM A + yB M B
Nm
Nm
Combining the two equation above and noting that y A + y B = 1 gives the following convenient relations
for converting mass fractions to mole fractions,
yA =
MB
M A (1 / mf A − 1) + M B
and
yB = 1 − y A
which are the desired relations.
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13-2
13-10 The molar fractions of the constituents of moist air are given. The mass fractions of the constituents
are to be determined.
Assumptions The small amounts of gases in air are ignored, and dry air is assumed to consist of N2 and O2
only.
Properties The molar masses of N2, O2, and H2O are 28.0, 32.0, and 18.0 kg/kmol, respectively (Table A1).
Analysis The molar mass of moist air is
M=
∑y M
i
i
= 0.78 × 28.0 + 0.20 × 32.0 + 0.02 × 18 = 28.6 kg / kmol
Then the mass fractions of constituent gases are determined to be
N2 :
mf N 2 = y N 2
O2 :
mfO 2 = yO 2
M N2
M
M O2
M
mf H 2 O = y H 2 O
H 2O :
= (0.78)
28.0
= 0.764
28.6
= (0.20)
32.0
= 0.224
28.6
M H 2O
M
= (0.02)
Moist air
78% N2
20% O2
2% H2 O
(Mole fractions)
18.0
= 0.013
28.6
Therefore, the mass fractions of N2, O2, and H2O in the air are 76.4%, 22.4%, and 1.3%, respectively.
13-11 The molar fractions of the constituents of a gas mixture are given. The gravimetric analysis of the
mixture, its molar mass, and gas constant are to be determined.
Properties The molar masses of N2, and CO2 are 28.0 and 44.0 kg/kmol, respectively (Table A-1)
Analysis Consider 100 kmol of mixture. Then the mass of each component and the total mass are
N N 2 = 60 kmol 
→ m N 2 = N N 2 M N 2 = (60 kmol)(28 kg/kmol) = 1680 kg
N CO 2 = 40 kmol 
→ m CO 2 = N CO 2 M CO 2 = (40 kmol)(44 kg/kmol) = 1760 kg
m m = m N 2 + m CO 2 = 1680 kg + 1760 kg = 3440 kg
Then the mass fraction of each component (gravimetric analysis) becomes
mf N 2 =
mf CO 2 =
m N2
mm
m CO 2
mm
=
1680 kg
= 0.488 or 48.8%
3440 kg
=
mole
60% N2
40% CO2
1760 kg
= 0.512 or 51.2%
3440 kg
The molar mass and the gas constant of the mixture are determined from their definitions,
Mm =
mm
3,440 kg
=
= 34.40 kg / kmol
N m 100 kmol
Rm =
Ru
8.314 kJ / kmol ⋅ K
=
= 0.242 kJ / kg ⋅ K
Mm
34.4 kg / kmol
and
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13-3
13-12 The molar fractions of the constituents of a gas mixture are given. The gravimetric analysis of the
mixture, its molar mass, and gas constant are to be determined.
Properties The molar masses of O2 and CO2 are 32.0 and 44.0 kg/kmol, respectively (Table A-1)
Analysis Consider 100 kmol of mixture. Then the mass of each component and the total mass are
N O 2 = 60 kmol 
→ m O 2 = N O 2 M O 2 = (60 kmol)(32 kg/kmol) = 1920 kg
N CO 2 = 40 kmol 
→ m CO 2 = N CO 2 M CO 2 = (40 kmol)(44 kg/kmol) = 1760 kg
m m = m O 2 + m CO 2 = 1920 kg + 1760 kg = 3680 kg
Then the mass fraction of each component (gravimetric analysis) becomes
mf O 2 =
mf CO 2 =
mO2
mm
m CO 2
mm
=
1920 kg
= 0.522 or 52.2%
3680 kg
=
mole
60% O2
40% CO2
1760 kg
= 0.478 or 47.8%
3680 kg
The molar mass and the gas constant of the mixture are determined from their definitions,
Mm =
mm
3680 kg
=
= 36.80 kg/kmol
N m 100 kmol
and
Rm =
Ru
8.314 kJ/kmol ⋅ K
=
= 0.226 kJ/kg ⋅ K
Mm
36.8 kg/kmol
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13-4
13-13 The masses of the constituents of a gas mixture are given. The mass fractions, the mole fractions, the
average molar mass, and gas constant are to be determined.
Properties The molar masses of O2, N2, and CO2 are 32.0, 28.0 and 44.0 kg/kmol, respectively (Table A-1)
Analysis (a) The total mass of the mixture is
m m = m O 2 + m N 2 + m CO 2 = 5 kg + 8 kg + 10 kg = 23 kg
Then the mass fraction of each component becomes
mf O 2 =
mf N 2 =
mf CO 2 =
mO2
mm
=
5 kg
= 0.217
23 kg
=
8 kg
= 0.348
23 kg
m N2
mm
m CO 2
=
mm
5 kg O2
8 kg N2
10 kg CO2
10 kg
= 0.435
23 kg
(b) To find the mole fractions, we need to determine the mole numbers of each component first,
N O2 =
N N2 =
N CO 2 =
mO2
M O2
m N2
M N2
=
5 kg
= 0.156 kmol
32 kg/kmol
=
8 kg
= 0.286 kmol
28 kg/kmol
m CO 2
=
M CO 2
10 kg
= 0.227 kmol
44 kg/kmol
Thus,
N m = N O 2 + N N 2 + N CO 2 = 0.156 kmol + 0.286 kmol + 0.227 kmol = 0.669 kmol
and
y O2 =
y N2 =
y CO 2 =
N O2
Nm
N N2
Nm
N CO 2
Nm
=
0.156 kmol
= 0.233
0.699 kmol
=
0.286 kmol
= 0.428
0.669 kmol
=
0.227 kmol
= 0.339
0.669 kmol
(c) The average molar mass and gas constant of the mixture are determined from their definitions:
Mm =
mm
23 kg
=
= 34.4 kg/kmol
N m 0.669 kmol
and
Rm =
Ru
8.314 kJ/kmol ⋅ K
=
= 0.242 kJ/kg ⋅ K
34.4 kg/kmol
Mm
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13-5
13-14 The mass fractions of the constituents of a gas mixture are given. The mole fractions of the gas and
gas constant are to be determined.
Properties The molar masses of CH4, and CO2 are 16.0 and 44.0 kg/kmol, respectively (Table A-1)
Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each component
and the total number of moles are
→ N CH 4 =
m CH 4 = 75 kg 
→ N CO 2 =
m CO 2 = 25 kg 
m CH 4
M CH 4
m CO 2
M CO 2
=
75 kg
= 4.688 kmol
16 kg/kmol
mass
=
25 kg
= 0.568 kmol
44 kg/kmol
75% CH4
25% CO2
N m = N CH 4 + N CO 2 = 4.688 kmol + 0.568 kmol = 5.256 kmol
Then the mole fraction of each component becomes
y CH 4 =
y CO 2 =
N CH 4
Nm
N CO 2
Nm
=
4.688 kmol
= 0.892 or 89.2%
5.256 kmol
=
0.568 kmol
= 0.108 or 10.8%
5.256 kmol
The molar mass and the gas constant of the mixture are determined from their definitions,
Mm =
mm
100 kg
=
= 19.03 kg/kmol
N m 5.256 kmol
and
Rm =
Ru
8.314 kJ/kmol ⋅ K
=
= 0.437 kJ/kg ⋅ K
Mm
19.03 kg/kmol
13-15 The mole numbers of the constituents of a gas mixture are given. The mass of each gas and the
apparent gas constant are to be determined.
Properties The molar masses of H2, and N2 are 2.0 and 28.0 kg/kmol, respectively (Table A-1)
Analysis The mass of each component is determined from
N H 2 = 8 kmol 
→ m H 2 = N H 2 M H 2 = (8 kmol)(2.0 kg/kmol) = 16 kg
N N 2 = 2 kmol 
→ m N 2 = N N 2 M N 2 = (2 kmol)(28 kg/kmol) = 56 kg
8 kmol H2
2 kmol N2
The total mass and the total number of moles are
m m = m H 2 + m N 2 = 16 kg + 56 kg = 72 kg
N m = N H 2 + N N 2 = 8 kmol + 2 kmol = 10 kmol
The molar mass and the gas constant of the mixture are determined from their definitions,
Mm =
mm
72 kg
=
= 7.2 kg/kmol
N m 10 kmol
and
Rm =
Ru
8.314 kJ/kmol ⋅ K
=
= 1.155 kJ/kg ⋅ K
Mm
7.2 kg/kmol
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13-6
13-16E The mole numbers of the constituents of a gas mixture are given. The mass of each gas and the
apparent gas constant are to be determined.
Properties The molar masses of H2, and N2 are 2.0 and 28.0 lbm/lbmol, respectively (Table A-1E).
Analysis The mass of each component is determined from
N H 2 = 5 lbmol 
→ m H 2 = N H 2 M H 2 = (5 lbmol)(2.0 lbm/lbmol) = 10 lbm
N N 2 = 4 lbmol 
→ m N 2 = N N 2 M N 2 = (4 lbmol)(28 lbm/lbmol) = 112 lbm
The total mass and the total number of moles are
m m = m H 2 + m N 2 = 10 lbm + 112 lbm = 122 lbm
5 lbmol H2
4 lbmol N2
N m = N H 2 + N N 2 = 5 lbmol + 4 lbmol = 9 lbmol
The molar mass and the gas constant of the mixture are determined from their definitions,
m
122 lbm
Mm = m =
= 13.56 lbm/lbmol
N m 9 lbmol
and
Rm =
Ru
1.986 Btu/lbmol ⋅ R
=
= 0.1465 Btu/lbm ⋅ R
Mm
13.56 lbm/lbmol
13-17 The mass fractions of the constituents of a gas mixture are given. The volumetric analysis of the
mixture and the apparent gas constant are to be determined.
Properties The molar masses of O2, N2 and CO2 are 32.0, 28, and 44.0 kg/kmol, respectively (Table A-1)
Analysis For convenience, consider 100 kg of the mixture. Then the number of moles of each component
and the total number of moles are
mO2
20 kg
m O 2 = 20 kg 
→ N O 2 =
=
= 0.625 kmol
M O 2 32 kg/kmol
m N 2 = 20 kg 
→ N N 2 =
mN2
M N2
→ N CO 2 =
m CO 2 = 50 kg 
30 kg
=
= 1.071 kmol
28 kg/kmol
m CO 2
M CO 2
=
50 kg
= 1.136 kmol
44 kg/kmol
mass
20% O2
30% N2
50% CO2
N m = N O 2 + N N 2 + N CO 2 = 0.625 + 1.071 + 1.136 = 2.832 kmol
Noting that the volume fractions are same as the mole fractions, the volume fraction of each component
becomes
N O 2 0.625 kmol
=
= 0.221 or 22.1%
y O2 =
2.832 kmol
Nm
y N2 =
N N2
Nm
y CO 2 =
=
N CO 2
Nm
1.071 kmol
= 0.378 or 37.8%
2.832 kmol
=
1.136 kmol
= 0.401 or 40.1%
2.832 kmol
The molar mass and the gas constant of the mixture are determined from their definitions,
m
100 kg
Mm = m =
= 35.31 kg/kmol
N m 2.832 kmol
and
Rm =
Ru
8.314 kJ/kmol ⋅ K
=
= 0.235 kJ/kg ⋅ K
Mm
35.31 kg/kmol
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13-7
P-v-T Behavior of Gas Mixtures
13-18C Normally yes. Air, for example, behaves as an ideal gas in the range of temperatures and pressures
at which oxygen and nitrogen behave as ideal gases.
13-19C The pressure of a gas mixture is equal to the sum of the pressures each gas would exert if existed
alone at the mixture temperature and volume. This law holds exactly for ideal gas mixtures, but only
approximately for real gas mixtures.
13-20C The volume of a gas mixture is equal to the sum of the volumes each gas would occupy if existed
alone at the mixture temperature and pressure. This law holds exactly for ideal gas mixtures, but only
approximately for real gas mixtures.
13-21C The P-v-T behavior of a component in an ideal gas mixture is expressed by the ideal gas equation
of state using the properties of the individual component instead of the mixture, Pivi = RiTi. The P-v-T
behavior of a component in a real gas mixture is expressed by more complex equations of state, or by Pivi
= ZiRiTi, where Zi is the compressibility factor.
13-22C Component pressure is the pressure a component would exert if existed alone at the mixture
temperature and volume. Partial pressure is the quantity yiPm, where yi is the mole fraction of component i.
These two are identical for ideal gases.
13-23C Component volume is the volume a component would occupy if existed alone at the mixture
temperature and pressure. Partial volume is the quantity yiVm, where yi is the mole fraction of component i.
These two are identical for ideal gases.
13-24C The one with the highest mole number.
13-25C The partial pressures will decrease but the pressure fractions will remain the same.
13-26C The partial pressures will increase but the pressure fractions will remain the same.
13-27C No. The correct expression is “the volume of a gas mixture is equal to the sum of the volumes
each gas would occupy if existed alone at the mixture temperature and pressure.”
13-28C No. The correct expression is “the temperature of a gas mixture is equal to the temperature of the
individual gas components.”
13-29C Yes, it is correct.
13-30C With Kay's rule, a real-gas mixture is treated as a pure substance whose critical pressure and
temperature are defined in terms of the critical pressures and temperatures of the mixture components as
Pcr′ , m =
∑y P
i cr ,i
and Tcr′ , m =
∑yT
i cr ,i
The compressibility factor of the mixture (Zm) is then easily determined using these pseudo-critical point
values.
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-8
13-31 A tank contains a mixture of two gases of known masses at a specified pressure and temperature.
The volume of the tank is to be determined.
Assumptions Under specified conditions both O2 and CO2 can be treated as ideal gases, and the mixture as
an ideal gas mixture.
Analysis The total number of moles is
8 kmol O2
10 kmol CO2
N m = N O 2 + N CO 2 = 8 kmol + 10 kmol = 18 kmol
Then
290 K
150 kPa
N RT
(18 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(290 K)
Vm = m u m =
= 289.3 m 3
Pm
150 kPa
13-32 A tank contains a mixture of two gases of known masses at a specified pressure and temperature.
The volume of the tank is to be determined.
Assumptions Under specified conditions both O2 and CO2 can be treated as ideal gases, and the mixture as
an ideal gas mixture.
Analysis The total number of moles is
8 kmol O2
10 kmol CO2
N m = N O 2 + N CO 2 = 8 kmol + 10 kmol = 18 kmol
Then
400 K
150 kPa
N RT
(18 kmol)(8.314 kPa ⋅ m 3/kmol ⋅ K)(400 K)
Vm = m u m =
= 399.1 m 3
Pm
150 kPa
13-33 A tank contains a mixture of two gases of known masses at a specified pressure and temperature.
The mixture is now heated to a specified temperature. The volume of the tank and the final pressure of the
mixture are to be determined.
Assumptions Under specified conditions both Ar and N2 can be treated as ideal gases, and the mixture as
an ideal gas mixture.
Analysis The total number of moles is
0.5 kmol Ar
2 kmol N2
N m = N Ar + N N 2 = 0.5 kmol + 2 kmol = 2.5 kmol
And
Vm
N R T
(2.5 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(280 K)
= m u m =
= 23.3 m 3
Pm
250 kPa
Q
280 K
250 kPa
Also,
T
P2V 2 P1V1
400 K
=

→ P2 = 2 P1 =
(250 kPa ) = 357.1 kPa
T1
280 K
T2
T1
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13-9
13-34 The masses of the constituents of a gas mixture at a specified pressure and temperature are given.
The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined.
Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixture
as an ideal gas mixture.
Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 kg/kmol, respectively (Table A-1)
Analysis The mole numbers of the constituents are
mCO 2
1 kg
mCO 2 = 1 kg

→
N CO 2 =
=
= 0.0227 kmol
MCO 2 44 kg / kmol
1 kg CO2
mCH 4 = 3 kg

→
N CH 4 =
mCH 4
MCH 4
=
3 kg
= 0.1875 kmol
16 kg / kmol
N m = NCO 2 + NCH 4 = 0.0227 kmol + 0.1875 kmol = 0.2102 kmol
yCO 2 =
N CO 2
Nm
N CH 4
=
3 kg CH4
300 K
200 kPa
0.0227 kmol
= 0108
.
0.2102 kmol
0.1875 kmol
= 0.892
Nm
0.2102 kmol
Then the partial pressures become
PCO 2 = y CO 2 Pm = (0.108)(200 kPa ) = 21.6 kPa
yCH 4 =
=
PCH 4 = y CH 4 Pm = (0.892)(200 kPa ) = 178.4 kPa
The apparent molar mass of the mixture is
m
4 kg
Mm = m =
= 19.03 kg / kmol
N m 0.2102 kmol
13-35E The masses of the constituents of a gas mixture at a specified pressure and temperature are given.
The partial pressure of each gas and the apparent molar mass of the gas mixture are to be determined.
Assumptions Under specified conditions both CO2 and CH4 can be treated as ideal gases, and the mixture
as an ideal gas mixture.
Properties The molar masses of CO2 and CH4 are 44.0 and 16.0 lbm/lbmol, respectively (Table A-1E)
Analysis The mole numbers of gases are
m CO 2
1 lbm
m CO 2 = 1 lbm 
→ N CO 2 =
=
= 0.0227 lbmol
M CO 2 44 lbm/lbmol
1 lbm CO2
m CH 4 = 3 lbm 
→ N CH 4 =
m CH 4
M CH 4
=
3 lbm
= 0.1875 lbmol
16 lbm/lbmol
N m = N CO 2 + N CH 4 = 0.0227 lbmol + 0.1875 lbmol = 0.2102 lbmol
y CO 2 =
N CO 2
Nm
N CH 4
=
3 lbm CH4
600 R
20 psia
0.0227 lbmol
= 0.108
0.2102 lbmol
0.1875 lbmol
= 0.892
Nm
0.2102 lbmol
Then the partial pressures become
PCO 2 = y CO 2 Pm = (0.108)(20 psia ) = 2.16 psia
y CH 4 =
=
PCH 4 = y CH 4 Pm = (0.892)(20 psia ) = 17.84 psia
The apparent molar mass of the mixture is
m
4 lbm
Mm = m =
= 19.03 lbm/lbmol
N m 0.2102 lbmol
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
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13-10
13-36 The masses of the constituents of a gas mixture at a specified temperature are given. The partial
pressure of each gas and the total pressure of the mixture are to be determined.
Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture as
an ideal gas mixture.
Analysis The partial pressures of constituent gases are
PN 2
(0.6 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)(300 K)
 mRT 
= 178.1 kPa
=
 =
0.3 m 3
 V  N2
(0.4 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(300 K)
 mRT 
PO 2 = 
= 103.9 kPa
 =
0.3 m 3
 V  O2
0.3 m3
0.6 kg N2
0.4 kg O2
300 K
and
Pm = PN 2 + PO 2 = 178.1 kPa + 103.9 kPa = 282.0 kPa
13-37 The volumetric fractions of the constituents of a gas mixture at a specified pressure and temperature
are given. The mass fraction and partial pressure of each gas are to be determined.
Assumptions Under specified conditions all N2, O2 and CO2 can be treated as ideal gases, and the mixture
as an ideal gas mixture.
Properties The molar masses of N2, O2 and CO2 are 28.0, 32.0, and 44.0 kg/kmol, respectively (Table A-1)
Analysis For convenience, consider 100 kmol of mixture. Then the mass of each component and the total
mass are
N N 2 = 65 kmol 
→ m N 2 = N N 2 M N 2 = (65 kmol)(28 kg/kmol) = 1820 kg
N O 2 = 20 kmol 
→ m O 2 = N O 2 M O 2 = (20 kmol)(32 kg/kmol) = 640 kg
N CO 2 = 15 kmol 
→ m CO 2 = N CO 2 M CO 2 = (15 kmol)(44 kg/kmol) = 660 kg
m m = m N 2 + m O 2 + m CO 2 = 1820 kg + 640 kg + 660 kg = 3120 kg
65% N2
20% O2
15% CO2
350 K
300 kPa
Then the mass fraction of each component (gravimetric analysis) becomes
mf N 2 =
mfO 2 =
mfCO 2 =
mN 2
mm
mO 2
mm
mCO 2
mm
=
1820 kg
= 0.583 or 58.3%
3120 kg
=
640 kg
= 0.205 or 20.5%
3120 kg
=
660 kg
= 0.212 or 21.2%
3120 kg
For ideal gases, the partial pressure is proportional to the mole fraction, and is determined from
PN 2 = y N 2 Pm = (0.65)(300 kPa ) = 195 kPa
PO 2 = y O 2 Pm = (0.20)(300 kPa ) = 60 kPa
PCO 2 = y CO 2 Pm = (0.15)(300 kPa ) = 45 kPa
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13-11
13-38 The masses, temperatures, and pressures of two gases contained in two tanks connected to each other
are given. The valve connecting the tanks is opened and the final temperature is measured. The volume of
each tank and the final pressure are to be determined.
Assumptions Under specified conditions both N2 and O2 can be treated as ideal gases, and the mixture as
an ideal gas mixture
Properties The molar masses of N2 and O2 are 28.0 and 32.0 kg/kmol, respectively (Table A-1)
Analysis The volumes of the tanks are
(1 kg)(0.2968 kPa ⋅ m 3 /kg ⋅ K)(298 K)
 mRT 
= 0.295 m 3
 =
300
kPa
P

 N2
V N2 = 
(3 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(298 K)
 mRT 
= 0.465 m 3
 =
500 kPa
 P  O2
V O2 = 
1 kg N2
3 kg O2
25°C
300 kPa
25°C
500 kPa
V total = V N 2 + V O 2 = 0.295 m 3 + 0.465 m 3 = 0.76 m 3
Also,
m N 2 = 1 kg 
→ N N 2 =
m O 2 = 3 kg 
→ N O 2 =
m N2
M N2
mO2
M O2
=
1 kg
= 0.03571 kmol
28 kg/kmol
=
3 kg
= 0.09375 kmol
32 kg/kmol
N m = N N 2 + N O 2 = 0.03571 kmol + 0.09375 kmol = 0.1295 kmol
Thus,
 NRu T
Pm = 
 V
(0.1295 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(298 K)

 =
= 422.2 kPa
0.76 m 3
m
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educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-12
13-39 The volumes, temperatures, and pressures of two gases forming a mixture are given. The volume of
the mixture is to be determined using three methods.
Analysis (a) Under specified conditions both O2 and N2 will considerably deviate from the ideal gas
behavior. Treating the mixture as an ideal gas,
 PV
N O 2 = 
 RuT

(8000 kPa)(0.3 m 3 )
 =
= 1.443 kmol
3

O 2 (8.314 kPa ⋅ m /kmol ⋅ K)(200 K)
 PV
N N 2 = 
 RuT

(8000 kPa)(0.5 m 3 )
 =
= 2.406 kmol
3

 N 2 (8.314 kPa ⋅ m /kmol ⋅ K)(200 K)
N m = N O 2 + N N 2 = 1.443 kmol + 2.406 kmol
= 3.849 kmol
Vm =
0.3 m3 O2
200 K
8 MPa
0.5 m3 N2
200 K
8 MPa
N2 + O2
200 K
8 MPa
N m RuTm (3.849 kmol)(8.314 kPa ⋅ m3 /kmol ⋅ K)(200 K)
=
= 0.8 m 3
8000 kPa
Pm
(b) To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of
the mixture using the critical point properties of O2 and N2 from Table A-1. But we first need to determine
the Z and the mole numbers of each component at the mixture temperature and pressure (Fig. A-15),



 Z O 2 = 0.77
8 MPa
=
= 1.575 

5.08 MPa

T R ,O 2 =
Tm
200 K
=
= 1.292
Tcr,O 2 154.8 K
PR ,O 2 =
Pm
Pcr,O 2
O 2:
TR, N 2 =
N 2:
PR , N 2 =



 Z N 2 = 0.863
8 MPa
=
= 2.360 

3.39 MPa

Tm
=
Tcr, N 2
Pm
Pcr, N 2
200K
= 1.585
126.2K
 PV
N O 2 = 
 ZRu T

(8000 kPa)(0.3 m 3 )
 =
= 1.874 kmol

3
 O 2 (0.77)(8.314 kPa ⋅ m /kmol ⋅ K)(200 K)
 PV
N N 2 = 
 ZRu T

(8000 kPa)(0.5 m 3 )
 =
= 2.787 kmol

3
 N 2 (0.863)(8.314 kPa ⋅ m /kmol ⋅ K)(200 K)
N m = N O 2 + N N 2 = 1.874 kmol + 2.787 kmol = 4.661 kmol
The mole fractions are
y O2 =
y N2 =
Tcr′ , m =
N O2
Nm
N N2
Nm
=
1.874kmol
= 0.402
4.661kmol
=
2.787kmol
= 0.598
4.661kmol
∑yT
i cr ,i
= y O 2 Tcr ,O 2 + y N 2 Tcr , N 2
= (0.402)(154.8K) + (0.598)(126.2K) = 137.7K
Pcr′ , m =
∑y P
i cr ,i
= y O 2 Pcr ,O 2 + y N 2 Pcr , N 2
= (0.402)(5.08MPa) + (0.598)(3.39MPa) = 4.07 MPa
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13-13
Then,
TR =
PR =
Thus,
Vm =
Tm



 Z m = 0.82
8 MPa
=
= 1.966 

4.07 MPa

=
'
Tcr,
O2
Pm
'
Pcr,
O2
200 K
= 1.452
137.7 K
(Fig. A-15)
Z m N m Ru Tm (0.82)(4.661 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(200 K)
=
= 0.79 m 3
Pm
8000 kPa
(c) To use the Amagat’s law for this real gas mixture, we first need the Z of each component at the mixture
temperature and pressure, which are determined in part (b). Then,
Thus,
Zm =
∑y Z
Vm =
Z m N m Ru Tm (0.83)(4.661 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(200 K)
=
= 0.80 m 3
Pm
8000 kPa
i
i
= y O 2 Z O 2 + y N 2 Z N 2 = (0.402 )(0.77 ) + (0.598)(0.863) = 0.83
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13-14
13-40 [Also solved by EES on enclosed CD] The mole numbers, temperatures, and pressures of two gases
forming a mixture are given. The final temperature is also given. The pressure of the mixture is to be
determined using two methods.
Analysis (a) Under specified conditions both Ar and N2 will considerably deviate from the ideal gas
behavior. Treating the mixture as an ideal gas,
Initial state : P1V1 = N1RuT1 
N 2T2
(4)(200 K)
P1 =
(5 MPa ) = 18.2 MPa
 P2 =
Final state : P2V 2 = N 2 RuT2 
N1T1
(1)(220 K)
(b) Initially,
TR =
PR =



 Z Ar = 0.90 (Fig. A-15)
5 MPa
=
= 1.0278 

4.86 MPa

T1
220 K
=
= 1.457
Tcr,Ar 151.0 K
P1
Pcr,Ar
1 kmol Ar
220 K
5 MPa
3 kmol N2
190 K
8 MPa
Then the volume of the tank is
V =
ZN Ar Ru T (0.90)(1 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(220 K)
=
= 0.33 m 3
P
5000 kPa
After mixing,




V m / N Ar
v Ar

=
=
 PR = 0.90
Ru Tcr,Ar / Pcr,Ar Ru Tcr,Ar / Pcr,Ar


3
(0.33 m )/(1 kmol)

1
.
278
=
=

(8.314 kPa ⋅ m 3 /kmol ⋅ K)(151.0 K)/(4860 kPa)

(Fig. A-15)


Tcr, N 2


v N2
V m / N N2

=
=
 PR = 3.75
Ru Tcr, N 2 / Pcr, N 2
Ru Tcr, N 2 / Pcr, N 2


3
(0.33 m )/(3 kmol)
=
= 0.355 

(8.314 kPa ⋅ m 3 /kmol ⋅ K)(126.2 K)/(3390 kPa)

(Fig. A-15)
T R ,Ar =
Ar:
V R , Ar
TR, N 2 =
N 2:
V R, N 2
Tm
200K
=
= 1.325
Tcr,Ar 151.0K
Tm
=
200K
= 1.585
126.2K
Thus,
PAr = ( PR Pcr ) Ar = (0.90)(4.86 MPa) = 4.37 MPa
PN 2 = ( PR Pcr ) N 2 = (3.75)(3.39 MPa) = 12.7 MPa
and
Pm = PAr + PN 2 = 4.37 MPa + 12.7 MPa = 17.1 MPa
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13-15
13-41 EES Problem 13-40 is reconsidered. The effect of the moles of nitrogen supplied to the tank on the
final pressure of the mixture is to be studied using the ideal-gas equation of state and the compressibility
chart with Dalton's law.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data"
R_u = 8.314 [kJ/kmol-K] "universal Gas Constant"
T_Ar = 220 [K]
P_Ar = 5000 [kPa] "Pressure for only Argon in the tank initially."
N_Ar = 1 [kmol]
{N_N2 = 3 [kmol]}
T_mix = 200 [K]
T_cr_Ar=151.0 [K] "Critical Constants are found in Table A.1 of the text"
P_cr_Ar=4860 [kPa]
T_cr_N2=126.2 [K]
P_cr_N2=3390 [kPa]
"Ideal-gas Solution:"
P_Ar*V_Tank_IG = N_Ar*R_u*T_Ar "Apply the ideal gas law the gas in the tank."
P_mix_IG*V_Tank_IG = N_mix*R_u*T_mix "Ideal-gas mixture pressure"
N_mix=N_Ar + N_N2 "Moles of mixture"
"Real Gas Solution:"
P_Ar*V_Tank_RG = Z_Ar_1*N_Ar*R_u*T_Ar "Real gas volume of tank"
T_R=T_Ar/T_cr_Ar "Initial reduced Temp. of Ar"
P_R=P_Ar/P_cr_Ar "Initial reduced Press. of Ar"
Z_Ar_1=COMPRESS(T_R, P_R ) "Initial compressibility factor for Ar"
P_Ar_mix*V_Tank_RG = Z_Ar_mix*N_Ar*R_u*T_mix "Real gas Ar Pressure in mixture"
T_R_Ar_mix=T_mix/T_cr_Ar
"Reduced Temp. of Ar in mixture"
P_R_Ar_mix=P_Ar_mix/P_cr_Ar
"Reduced Press. of Ar in mixture"
Z_Ar_mix=COMPRESS(T_R_Ar_mix, P_R_Ar_mix ) "Compressibility factor for Ar in mixture"
P_N2_mix*V_Tank_RG = Z_N2_mix*N_N2*R_u*T_mix "Real gas N2 Pressure in mixture"
T_R_N2_mix=T_mix/T_cr_N2 "Reduced Temp. of N2 in mixture"
P_R_N2_mix=P_N2_mix/P_cr_N2 "Reduced Press. of N2 in mixture"
Z_N2_mix=COMPRESS(T_R_N2_mix, P_R_N2_mix ) "Compressibility factor for N2 in mixture"
P_mix=P_R_Ar_mix*P_cr_Ar +P_R_N2_mix*P_cr_N2 "Mixture pressure by Dalton's law. 23800"
225000
Pmix
[kPa]
9009
13276
17793
23254
30565
41067
56970
82372
126040
211047
Pmix,IG
[kPa]
9091
13636
18182
22727
27273
31818
36364
40909
45455
50000
180000
P m ix [kPa]
NN2
[kmol]
1
2
3
4
5
6
7
8
9
10
Solution Method
135000
Chart
Ideal Gas
90000
45000
0
1
2
3
4
5
6
7
8
9
N N2 [kmol]
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10
13-16
13-42E The mole numbers, temperatures, and pressures of two gases forming a mixture are given. For a
specified final temperature, the pressure of the mixture is to be determined using two methods.
Properties The critical properties of Ar are Tcr = 272 R and Pcr = 705 psia. The critical properties of N2 are
Tcr = 227.1 R and Pcr = 492 psia (Table A-1E).
Analysis (a) Under specified conditions both Ar and N2 will considerably deviate from the ideal gas
behavior. Treating the mixture as an ideal gas,
P1V1 = N1RuT1 
N 2T2
(4)(360 R)
P1 =
(750 psia ) = 2700 psia
 P2 =
P2V 2 = N 2 RuT2 
N1T1
(1)(400 R)
Initial state :
Final state :
(b) Initially,
TR =
PR =



 Z Ar = 0.90 (Fig. A-15)
750 psia
=
= 1.07 

705 psia

T1
400 R
=
= 1.47
Tcr,Ar 272 R
P1
Pcr,Ar
1 lbmol Ar
400 R
750 psia
3 lbmol N2
340 R
1200 psia
Then the volume of the tank is
V =
ZN Ar Ru T (0.90)(1 lbmol)(10.73 psia ⋅ ft 3 /lbmol ⋅ R)(400 R)
=
= 5.15 ft 3
750 psia
P
After mixing,




V m / N Ar
v Ar

=
=
 PR = 0.82
Ru Tcr,Ar / Pcr,Ar Ru Tcr,Ar / Pcr,Ar


3
(5.15 ft )/(1 lbmol)
=
= 1.244 

(10.73 psia ⋅ ft 3 /lbmol ⋅ R)(272 R)/(705 psia)

(Fig. A-15)


Tcr, N 2


v N2
V m / N N2

=
=
 PR = 3.85
Ru Tcr, N 2 / Pcr, N 2
Ru Tcr, N 2 / Pcr, N 2


3
(5.15 ft )/(3 lbmol)

0
.
347
=
=

(10.73 psia ⋅ ft 3 /lbmol ⋅ R)(227.1 R)/(492 psia)

(Fig. A-15)
T R , Ar =
Ar:
v R , Ar
TR, N 2 =
N 2:
v R, N 2
Tm
360 R
=
= 1.324
Tcr,Ar 272 R
Tm
=
360 R
= 1.585
227.1 R
Thus,
PAr = ( PR Pcr ) Ar = (0.82)(705 psia) = 578 psia
PN 2 = ( PR Pcr ) N 2 = (3.85)(492 psia) = 1894 psia
and
Pm = PAr + PN 2 = 578 psia + 1894 psia = 2472 psia
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13-17
Properties of Gas Mixtures
13-43C Yes. Yes (extensive property).
13-44C No (intensive property).
13-45C The answers are the same for entropy.
13-46C Yes. Yes (conservation of energy).
13-47C We have to use the partial pressure.
13-48C No, this is an approximate approach. It assumes a component behaves as if it existed alone at the
mixture temperature and pressure (i.e., it disregards the influence of dissimilar molecules on each other.)
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13-18
13-49 Oxygen, nitrogen, and argon gases are supplied from separate tanks at different temperatures to form
a mixture. The total entropy change for the mixing process is to be determined.
Assumptions Under specified conditions all N2, O2, and argon can be treated as ideal gases, and the
mixture as an ideal gas mixture.
Properties The molar masses of O2, N2, and Ar are 32.0, 28.0, and 40.0 kg/kmol, respectively (TableA-1).
The properties of Argon are R = 0.2081 kJ/kg.K and cp = 0.5203 kJ/kg.K (Table A-2).
Analysis Note that volume fractions are equal to mole fractions in
ideal gas mixtures. The partial pressures in the mixture are
O2
PO 2 , 2 = y O 2 Pm = (0.21)(200 kPa) = 42 kPa
10°C
PN 2 , 2 = y N 2 Pm = (0.78)(200 kPa) = 156 kPa
N2
60°C
PAr, 2 = y Ar Pm = (0.01)(200 kPa) = 2 kPa
The molar mass of the mixture is determined to be
M m = yO 2 M O 2 + y N 2 M N 2 + y Ar M Ar
= (0.21)(32 kg/kmol) + (0.78)(28) + (0.01)(40) = 28.96 kg/kmol
21% O2
78% N2
21% Ar
200 kPa
Ar
200°C
The mass fractions are
M O2
32 kg/kmol
mf O 2 = y O 2
= (0.21)
= 0.2320
28.96 kg/kmol
Mm
M N2
mf N 2 = y N 2
mf Ar = y Ar
Mm
= (0.78)
28 kg/kmol
= 0.7541
28.96 kg/kmol
M Ar
40 kg/kmol
= (0.01)
= 0.0138
28.96 kg/kmol
Mm
The final temperature of the mixture is needed. The conservation of energy on a unit mass basis for steady
flow mixing with no heat transfer or work allows calculation of mixture temperature. All components of
the exit mixture have the same common temperature, Tm. We obtain the properties of O2 and N2 from EES:
ein = eout
mf O 2 h@ 10°C + mf N 2 h@ 60°C + mf Ar c p , ArTAr,1 = mf O 2 h@ Tm + mf N 2 h@ Tm + mf Ar c p , ArTm
(0.2320)(−13.85) + (0.7541)(36.47) + (0.0138)(0.5203)(200) = (0.2320)h@ Tm + (0.7541)h@ Tm
+ (0.0138)(0.5203)Tm
Solving this equation with EES gives Tm = 50.4ºC. The entropies of O2 and N2 are obtained from EES to be
T = 10°C, P = 200 kPa 
→ s O 2 ,1 = 6.1797 kJ/kg.K
T = 50.4°C, P = 42 kPa 
→ s O 2 ,2 = 6.7082 kJ/kg.K
T = 60°C, P = 200 kPa 
→ s N 2 ,1 = 6.7461 kJ/kg.K
T = 50.4°C, P = 156 kPa 
→ s N 2 , 2 = 6.7893 kJ/kg.K
The entropy changes are
∆s O 2 = s O 2 , 2 − s O 2 ,1 = 6.7082 − 6.1797 = 0.5284 kg/kg.K
∆s N 2 = s N 2 , 2 − s N 2 ,1 = 6.7893 − 6.7461 = 0.04321 kg/kg.K
∆s Ar = c p ln
T2
P
 50.4 + 273 
 2 
− R ln 2 = (0.5203) ln
 − (0.2081) ln
 = 0.7605 kJ/kg.K
T1
P1
 200 + 273 
 200 
The total entropy change is
∆s total = mf O 2 ∆s O 2 + mf O 2 ∆s O 2 + mf Ar ∆s Ar
= (0.2320)(0.5284) + (0.7541)(0.04321) + (0.0138)(0.7605) = 0.1656 kJ/kg.K
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13-19
13-50 Volumetric fractions of the constituents of a mixture are given. The mixture undergoes an adiabatic
compression process. The makeup of the mixture on a mass basis and the internal energy change per unit
mass of mixture are to be determined.
Assumptions Under specified conditions all CO2, CO, O2, and N2 can be treated as ideal gases, and the
mixture as an ideal gas mixture.
Properties 1 The molar masses of CO2, CO, O2, and N2 are 44.0, 28.0, 32.0, and 28.0 kg/kmol, respectively
(Table A-1). 2 The process is reversible.
Analysis Noting that volume fractions are equal to mole fractions in ideal gas mixtures, the molar mass of
the mixture is determined to be
M m = y CO 2 M CO 2 + y CO M CO + y O 2 M O 2 + y N 2 M N 2
= (0.15)(44) + (0.05)(28) + (0.10)(32) + (0.70)(28) = 30.80 kg/kmol
The mass fractions are
M CO 2
44 kg/kmol
mf CO 2 = y CO 2
= (0.15)
= 0.2143
30.80 kg/kmol
Mm
mf CO = y CO
mf O 2 = y O 2
M CO
28 kg/kmol
= (0.05)
= 0.0454
30.80 kg/kmol
Mm
M O2
Mm
M N2
= (0.10)
32 kg/kmol
= 0.1039
30.80 kg/kmol
15% CO2
5% CO
10% O2
70% N2
300 K, 1 bar
28 kg/kmol
= 0.6364
30.80 kg/kmol
Mm
The final pressure of mixture is expressed from ideal gas relation to be
T
T
P2 = P1 r 2 = (100 kPa )(8) 2 = 2.667T2
(Eq. 1)
300 K
T1
since the final temperature is not known. We assume that the process is reversible as well being adiabatic
(i.e. isentropic). Using Dalton’s law to find partial pressures, the entropies at the initial state are determined
from EES to be:
→ s CO 2 ,1 = 5.2190 kJ/kg.K
T = 300 K, P = (0.2143 × 100) = 21.43 kPa 
mf N 2 = y N 2
= (0.70)
→ s CO,1 = 79483 kJ/kg.K
T = 300 K, P = (0.04545 × 100) = 4.55 kPa 
→ s N 2 ,1 = 6.9485 kJ/kg.K
T = 300 K, P = (0.1039 × 100) = 10.39 kPa 
→ s O 2 ,1 = 7.0115 kJ/kg.K
T = 300 K, P = (0.6364 × 100) = 63.64 kPa 
The final state entropies cannot be determined at this point since the final pressure and temperature are not
known. However, for an isentropic process, the entropy change is zero and the final temperature and the
final pressure may be determined from
∆s total = mf CO 2 ∆s CO 2 + mf CO ∆s CO + mf O 2 ∆s O 2 + mf N 2 ∆s N 2 = 0
and using Eq. (1). The solution may be obtained using EES to be T2 = 631.4 K, P2 = 1684 kPa
The initial and final internal energies are (from EES)
u CO 2 , 2 = −8734 kJ/kg
u CO 2 ,1 = −8997 kJ/kg
u CO, 2 = −3780 kJ/kg
u CC,1 = −4033 kJ/kg
→
→
T2 = 631.4 K 
T1 = 300 K 
u O 2 , 2 = 156.8 kJ/kg
u O 2 ,1 = −76.24 kJ/kg
u N 2 , 2 = 163.9 kJ/kg
u N 2 ,1 = −87.11 kJ/kg,
The internal energy change per unit mass of mixture is determined from
∆u mixture = mf CO 2 (u CO 2 , 2 − u CO 2 ,1 ) + mf CO (u CO, 2 − u CO,1 ) + mf O 2 (u O 2 , 2 − u O 2 ,1 ) + mf N 2 (u N 2 , 2 − u N 2 ,1 )
= 0.2143[(−8734) − (−8997)] + 0.0454[(−3780) − (−4033)]
+ 0.1039[156.8 − (−76.24)]6 + 0.6364[163.9 − (−87.11)]
= 251.8 kJ/kg
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13-20
13-51 Propane and air mixture is compressed isentropically in an internal combustion engine. The work
input is to be determined.
Assumptions Under specified conditions propane and air can be treated as ideal gases, and the mixture as
an ideal gas mixture.
Properties The molar masses of C3H8 and air are 44.0 and 28.97 kg/kmol, respectively (TableA-1).
Analysis Given the air-fuel ratio, the mass fractions are determined to be
mf air =
mf C3H8
AF
16
=
= 0.9412
AF + 1 17
1
1
=
=
= 0.05882
AF + 1 17
The molar mass of the mixture is determined to be
Mm =
mf air
M air
1
1
=
= 29.56 kg/kmol
mf C3H8
0.9412
0.05882
+
+
28.97 kg/kmol 44.0 kg/kmol
M C3H8
Propane
Air
95 kPa
30ºC
The mole fractions are
y air = mf air
Mm
29.56 kg/kmol
= (0.9412)
= 0.9606
28.97 kg/kmol
M air
y C3H8 = mf C3H8
Mm
29.56 kg/kmol
= (0.05882)
= 0.03944
44.0 kg/kmol
M C3H8
The final pressure is expressed from ideal gas relation to be
P2 = P1 r
T2
T2
= (95 kPa )(9.5)
= 2.977T2
T1
(30 + 273.15) K
(1)
since the final temperature is not known. Using Dalton’s law to find partial pressures, the entropies at the
initial state are determined from EES to be:
T = 30°C, P = (0.9606 × 95) = 91.26 kPa 
→ s air ,1 = 5.7417 kJ/kg.K
T = 30°C, P = (0.03944 × 95) = 3.75 kPa 
→ s C3H8 ,1 = 6.7697 kJ/kg.K
The final state entropies cannot be determined at this point since the final pressure and temperature are not
known. However, for an isentropic process, the entropy change is zero and the final temperature and the
final pressure may be determined from
∆s total = mf air ∆s air + mf C3H8 ∆s C3H 8 = 0
and using Eq. (1). The solution may be obtained using EES to be
T2 = 654.9 K, P2 = 1951 kPa
The initial and final internal energies are (from EES)
T1 = 30°C 
→
u air ,1 = 216.5 kJ/kg
u C3H8 ,1 = −2404 kJ/kg
T2 = 654.9 K 
→
u air , 2 = 477.1 kJ/kg
u C3H8 , 2 = −1607 kJ/kg
Noting that the heat transfer is zero, an energy balance on the system gives
q in + win = ∆u m 
→ win = ∆u m
where
∆u m = mf air (u air,2 − u air,1 ) + mf C3H8 (u C3H8 ,2 − u C3H 8 ,1 )
Substituting,
win = ∆u m = (0.9412)(477.1 − 216.5) + (0.05882)[(−1607) − (−2404)] = 292.2 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-21
13-52 The moles, temperatures, and pressures of two gases forming a mixture are given. The mixture
temperature and pressure are to be determined.
Assumptions 1 Under specified conditions both CO2 and H2 can be treated as ideal gases, and the mixture
as an ideal gas mixture. 2 The tank is insulated and thus there is no heat transfer. 3 There are no other
forms of work involved.
Properties The molar masses and specific heats of CO2 and H2 are 44.0 kg/kmol, 2.0 kg/kmol, 0.657
kJ/kg.°C, and 10.183 kJ/kg.°C, respectively. (Tables A-1 and A-2b).
Analysis (a) We take both gases as our system. No heat, work, or mass crosses the system boundary,
therefore this is a closed system with Q = 0 and W = 0. Then the energy balance for this closed system
reduces to
E in − E out = ∆E system
0 = ∆U = ∆U CO 2 + ∆U H 2
0 =[mcv (Tm − T1 )]CO + [mcv (Tm − T1 )]H
2
2
Using cv values at room temperature and noting that m = NM,
the final temperature of the mixture is determined to be
CO2
H2
2.5 kmol
7.5 kmol
200 kPa
27°C
400 kPa
40°C
(2.5 × 44 kg )(0.657 kJ/kg ⋅ °C)(Tm − 27°C) + (7.5 × 2 kg )(10.183 kJ/kg ⋅ °C)(Tm − 40°C ) = 0
Tm = 35.8°C (308.8 K )
(b) The volume of each tank is determined from
 NRu T1 
(2.5 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(300 K)

=
= 31.18 m 3
200 kPa
 P1  CO 2
V CO 2 = 
 NRu T1 
(7.5 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(313 K)
 =
= 48.79 m 3
400
kPa
P
1
 H2

V H 2 = 
Thus,
V m = V CO2 + V H 2 = 31.18 m 3 + 48.79 m 3 = 79.97 m 3
N m = N CO 2 + N H 2 = 2.5 kmol + 7.5 kmol = 10.0 kmol
and
Pm =
N m Ru Tm (10.0 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(308.8 K)
=
= 321 kPa
Vm
79.97 m 3
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-22
13-53 The temperatures and pressures of two gases forming a mixture are given. The final mixture
temperature and pressure are to be determined.
Assumptions 1 Under specified conditions both Ne and Ar can be treated as ideal gases, and the mixture as
an ideal gas mixture. 2 There are no other forms of work involved.
Properties The molar masses and specific heats of Ne and Ar are 20.18 kg/kmol, 39.95 kg/kmol, 0.6179
kJ/kg.°C, and 0.3122 kJ/kg.°C, respectively. (Tables A-1 and A-2).
Analysis The mole number of each gas is
 PV 
(100 kPa)(0.45 m3 )
= 0.0185 kmol
N Ne =  1 1  =
3
 RuT1  Ne (8.314 kPa ⋅ m /kmol ⋅ K)(293 K)
 PV 
(200 kPa)(0.45 m3 )
= 0.0335 kmol
N Ar =  1 1  =
3
 RuT1  Ar (8.314 kPa ⋅ m /kmol ⋅ K)(323 K)
Ne
Ar
100 kPa
20°C
200 kPa
50°C
Thus,
15 kJ
N m = N Ne + N Ar = 0.0185 kmol + 0.0335 kmol = 0.0520 kmol
(a) We take both gases as the system. No work or mass crosses the system boundary, therefore this is a
closed system with W = 0. Then the conservation of energy equation for this closed system reduces to
E in − E out = ∆E system
− Qout = ∆U = ∆U Ne + ∆U Ar 
→ − Qout = [mcv (Tm − T1 )]Ne + [mcv (Tm − T1 )]Ar
Using cv values at room temperature and noting that m = NM, the final temperature of the mixture is
determined to be
−15 kJ = (0.0185 × 20.18 kg )(0.6179 kJ/kg ⋅ °C )(Tm − 20°C )
+ (0.0335 × 39.95 kg )(0.3122 kJ/kg ⋅ °C )(Tm − 50°C )
Tm = 16.2°C (289.2 K )
(b) The final pressure in the tank is determined from
Pm =
N m Ru Tm (0.052 kmol)(8.314 kPa ⋅ m 3 / kmol ⋅ K)(289.2 K)
=
= 138.9 kPa
Vm
0.9 m 3
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-23
13-54 The temperatures and pressures of two gases forming a mixture are given. The final mixture
temperature and pressure are to be determined.
Assumptions 1 Under specified conditions both Ne and Ar can be treated as ideal gases, and the mixture as
an ideal gas mixture. 2 There are no other forms of work involved.
Properties The molar masses and specific heats of Ne and Ar are 20.18 kg/kmol, 39.95 kg/kmol, 0.6179
kJ/kg.°C, and 0.3122 kJ/kg.°C, respectively. (Tables A-1 and A-2b).
Analysis The mole number of each gas is
 PV 
(100 kPa)(0.45 m3 )
= 0.0185 kmol
N Ne =  1 1  =
3
 RuT1  Ne (8.314 kPa ⋅ m /kmol ⋅ K)(293 K)
 PV 
(200 kPa)(0.45 m3 )
= 0.0335 kmol
N Ar =  1 1  =
3
 RuT1  Ar (8.314 kPa ⋅ m /kmol ⋅ K)(323 K)
Ne
Ar
100 kPa
20°C
200 kPa
50°C
Thus,
8 kJ
N m = N Ne + N Ar = 0.0185 kmol + 0.0335 kmol = 0.0520 kmol
(a) We take both gases as the system. No work or mass crosses the system boundary, therefore this is a
closed system with W = 0. Then the conservation of energy equation for this closed system reduces to
E in − E out = ∆E system
− Qout = ∆U = ∆U Ne + ∆U Ar
− Qout = [mcv (Tm − T1 )]Ne + [mcv (Tm − T1 )]Ar
Using cv values at room temperature and noting that m = NM, the final temperature of the mixture is
determined to be
−8 kJ = (0.0185 × 20.18 kg )(0.6179 kJ/kg ⋅ °C )(Tm − 20°C )
+ (0.0335 × 39.95 kg )(0.3122 kJ/kg ⋅ °C )(Tm − 50°C )
Tm = 27.0°C (300.0 K )
(b) The final pressure in the tank is determined from
Pm =
N m RuTm
Vm
=
(0.052 kmol)(8.314 kPa ⋅ m3 /kmol ⋅ K)(300.0 K)
0.9 m3
= 144.1 kPa
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-24
13-55 [Also solved by EES on enclosed CD] The temperatures and pressures of two gases forming a
mixture in a mixing chamber are given. The mixture temperature and the rate of entropy generation are to
be determined.
Assumptions 1 Under specified conditions both C2H6 and CH4 can be treated as ideal gases, and the
mixture as an ideal gas mixture. 2 The mixing chamber is insulated and thus there is no heat transfer. 3
There are no other forms of work involved. 3 This is a steady-flow process. 4 The kinetic and potential
energy changes are negligible.
Properties The specific heats of C2H6 and CH4 are 1.7662
kJ/kg.°C and 2.2537 kJ/kg.°C, respectively. (Table A-2b).
20°C
Analysis (a) The enthalpy of ideal gases is independent of
9 kg/s C2H6
pressure, and thus the two gases can be treated independently
even after mixing. Noting that W& = Q& = 0 , the steady-flow
200 kPa
energy balance equation reduces to
E& in − E& out = ∆E& system ©0 (steady) = 0
45°C
4.5 kg/s
E& in = E& out
∑ m& h = ∑ m& h
0 = ∑ m& h − ∑ m& h = m&
0 = [m& c (T − T )]
+ [m& c
i i
CH4
e e
e e
p
e
i i
i
(he − hi )C H
p (Te − Ti )]CH
C2H6
C2H6
2
6
+ m& CH 4 (he − hi )CH
4
4
Using cp values at room temperature and substituting, the exit temperature of the mixture becomes
0 = (9 kg/s )(1.7662 kJ/kg ⋅ °C )(Tm − 20°C ) + (4.5 kg/s )(2.2537 kJ/kg ⋅ °C )(Tm − 45°C )
Tm = 29.7°C (302.7 K )
(b) The rate of entropy change associated with this process is determined from an entropy balance on the
mixing chamber,
S&in − S&out + S&gen = ∆S&system Ê0 = 0
& ( s1 − s2 )]C2 H6 + [m
& ( s1 − s2 )]CH4 + S&gen = 0 → S&gen = [m
& ( s2 − s1 )]C2 H 6 + [m
& ( s2 − s1 )]CH 4
[m
The molar flow rate of the two gases in the mixture is
4.5 kg/s
 m& 
=
= 0.2813 kmol/s
N& CH 4 =  
 M  CH 4 16 kg/kmol
9 kg/s
 m& 
=
= 0.3 kmol/s
N& C 2 H 6 =  
 M  C 2 H 6 30 kg/kmol
Then the mole fraction of each gas becomes
0.3
0.2813
yC 2 H 6 =
= 0.516
yCH 4 =
= 0.484
0.3 + 0.2813
0.3 + 0.2813
Thus,
y Pm,2 



T
T

( s 2 − s1 ) C 2 H 6 =  c p ln 2 − R ln
=  c p ln 2 − R ln y 

T
P
T
1
1 C H
1
 C2H6


2 6
= (1.7662 kJ/kg ⋅ K) ln
302.7 K
− (0.2765 kJ/kg ⋅ K) ln(0.516) = 0.240 kJ/kg ⋅ K
293 K
y Pm,2

T
( s 2 − s1 ) CH 4 =  c p ln 2 − R ln
T1
P1

= (2.2537 kJ/kg ⋅ K) ln



T

=  c p ln 2 − R ln y 

T1
 CH 4
 CH 4 
302.7 K
− (0.5182 kJ/kg ⋅ K) ln(0.484) = 0.265 kJ/kg ⋅ K
318 K
Noting that Pm, 2 = Pi, 1 = 200 kPa and substituting,
S&gen = (9 kg/s )(0.240 kJ/kg ⋅ K ) + (4.5 kg/s )(0.265 kJ/kg ⋅ K ) = 3.353 kW/K
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-25
13-56 EES Problem 13-55 is reconsidered. The effect of the mass fraction of methane in the mixture on
the mixture temperature and the rate of exergy destruction is to be investigated.
Analysis The problem is solved using EES, and the solution is given below.
"Input from the Diagram Window"
{Fluid1$='C2H6'
Fluid2$='CH4'
m_dot_F1=9 [kg/s]
m_dot_F2=m_dot_F1/2
T1=20 [C]
T2=45 [C]
P=200 [kPa]}
{mf_F2=0.1}
{m_dot_total =13.5 [kg/s]
m_dot_F2 =mf_F2*m_dot_total}
m_dot_total = m_dot_F1 + m_dot_F2
T_o = 25 [C]
"Conservation of energy for this steady-state, steady-flow control volume is"
E_dot_in=E_dot_out
E_dot_in=m_dot_F1*enthalpy(Fluid1$,T=T1) +m_dot_F2 *enthalpy(Fluid2$,T=T2)
E_dot_out=m_dot_F1*enthalpy(Fluid1$,T=T3) +m_dot_F2 *enthalpy(Fluid2$,T=T3)
"For entropy calculations the partial pressures are used."
Mwt_F1=MOLARMASS(Fluid1$)
N_dot_F1=m_dot_F1/Mwt_F1
Mwt_F2=MOLARMASS(Fluid2$)
N_dot_F2=m_dot_F2 /Mwt_F2
N_dot_tot=N_dot_F1+N_dot_F2
y_F1=IF(fluid1$,Fluid2$,N_dot_F1/N_dot_tot,1,N_dot_F1/N_dot_tot)
y_F2=IF(fluid1$,Fluid2$,N_dot_F2/N_dot_tot,1,N_dot_F2/N_dot_tot)
"If the two fluids are the same, the mole fractions are both 1 ."
"The entropy change of each fluid is:"
DELTAs_F1=entropy(Fluid1$, T=T3, P=y_F1*P)-entropy(Fluid1$, T=T1, P=P)
DELTAs_F2=entropy(Fluid2$, T=T3, P=y_F2*P)-entropy(Fluid2$, T=T2, P=P)
"And the entropy generation is:"
S_dot_gen=m_dot_F1*DELTAs_F1+m_dot_F2*DELTAs_F2
"Then the exergy destroyed is:"
X_dot_destroyed = (T_o+273)*S_dot_gen
mfF2
0.01
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0.99
T3
[C]
95.93
502.5
761.4
948.5
1096
1219
1324
1415
1497
1570
1631
Xdestroyed
[kW]
20.48
24.08
27
29.2
30.92
32.3
33.43
34.38
35.18
35.87
36.41
38
2000
36
]
C
[
3
T
34
1600
32
]
W
1200k[
30
28
F1 = C2H6
26
F2 = CH4
X
mtotal = 13.5 kg/s
24
800
400
22
20
0
0,2
0,4
mfF2
0,6
d
e
y
or
t
s
e
d
0,8
0
1
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-26
13-57 An equimolar mixture of helium and argon gases expands in a turbine. The isentropic work output of
the turbine is to be determined.
Assumptions 1 Under specified conditions both He and Ar can be treated as ideal gases, and the mixture as
an ideal gas mixture. 2 The turbine is insulated and thus there is no heat transfer. 3 This is a steady-flow
process. 4 The kinetic and potential energy changes are negligible.
Properties The molar masses and specific heats of He and Ar are 4.0 kg/kmol, 40.0 kg/kmol, 5.1926
kJ/kg.°C, and 0.5203 kJ/kg.°C, respectively. (Table A-1 and Table A-2).
Analysis The Cp and k values of this equimolar mixture are determined from
M m = ∑ y i M i = y He M He + y Ar M Ar = 0.5 × 4 + 0.5 × 40 = 22 kg/kmol
mf i =
mi
mm
=
2.5 MPa
1300 K
Ni M i
y M
= i i
NmM m
Mm
He-Ar
turbine
y M
y M
c p,m = ∑ mf i c p,i = He He c p,He + Ar Ar c p,Ar
Mm
Mm
0.5 × 4 kg/kmol
(5.1926 kJ/kg ⋅ K ) + 0.5 × 40 kg/kmol (0.5203 kJ/kg ⋅ K )
=
22 kg/kmol
22 kg/kmol
= 0.945 kJ/kg ⋅ K
200 kPa
and
km = 1.667
since k = 1.667 for both gases.
Therefore, the He-Ar mixture can be treated as a single ideal gas with the properties above. For isentropic
processes,
P
T2 = T1  2
 P1



(k −1) / k
 200 kPa 

= (1300 K )
 2500 kPa 
0.667/1.667
= 473.2 K
From an energy balance on the turbine,
E& in − E& out = ∆E& system ©0 (steady) = 0
E& in = E& out
h1 = h2 + wout
wout = h1 − h2
wout = c p (T1 − T2 ) = (0.945 kJ/kg ⋅ K )(1300 − 473.2 )K = 781.3 kJ/kg
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-27
13-58E [Also solved by EES on enclosed CD] A gas mixture with known mass fractions is accelerated
through a nozzle from a specified state to a specified pressure. For a specified isentropic efficiency, the
exit temperature and the exit velocity of the mixture are to be determined.
Assumptions 1 Under specified conditions both N2 and CO2 can be treated as ideal gases, and the mixture
as an ideal gas mixture. 2 The nozzle is adiabatic and thus heat transfer is negligible. 3 This is a steadyflow process. 4 Potential energy changes are negligible.
Properties The specific heats of N2 and CO2 are cp,N2 = 0.248 Btu/lbm.R, cv,N2 = 0.177 Btu/lbm.R, cp,CO2 =
0.203 Btu/lbm.R, and cv,CO2 = 0.158 Btu/lbm.R. (Table A-2E).
Analysis (a) Under specified conditions both N2 and CO2 can be treated as ideal gases, and the mixture as
an ideal gas mixture. The cp, cv, and k values of this mixture are determined from
c p,m =
∑ mf c
cv ,m =
∑ mf cv
i p ,i
= mf N 2 c p , N 2 + mf CO 2 c p ,CO 2
= (0.8)(0.248) + (0.2)(0.203) = 0.239 Btu/lbm ⋅ R
i
,i
= mf N 2 cv , N 2 + mf CO 2 cv ,CO 2
= (0.8)(0.177 ) + (0.2)(0.158) = 0.173 Btu/lbm ⋅ R
km =
c p,m
cv , m
=
90 psia
1800 R
80% N2
20% CO2
12 psia
0.239 Btu/lbm ⋅ R
= 1.382
0.173 Btu/lbm ⋅ R
Therefore, the N2-CO2 mixture can be treated as a single ideal gas with above properties. Then the
isentropic exit temperature can be determined from
P
T2 s = T1  2
 P1



(k −1) / k
 12 psia 

= (1800 R )
 90 psia 
0.382/1.382
= 1031.3 R
From the definition of adiabatic efficiency,
ηN =
c p (T1 − T2 )
h1 − h2
1,800 − T2
=

→ 0.92 =

→ T2 = 1092.8 R
h1 − h2 s c p (T1 − T2 s )
1,800 − 1031.3
(b) Noting that, q = w = 0, from the steady-flow energy balance relation,
E& in − E& out = ∆E& system ©0 (steady) = 0
E& in = E& out
h1 + V12 / 2 = h2 + V 22 / 2
0 = c p (T2 − T1 ) +
V 22 − V12
2
©0
 25,037 ft 2 /s 2
V 2 = 2c p (T1 − T2 ) = 2(0.239 Btu/lbm ⋅ R )(1800 − 1092.8) R 
 1 Btu/lbm


 = 2,909 ft/s


PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-28
13-59E EES Problem 13-58E is reconsidered. The problem is first to be solved and then, for all other
conditions being the same, the problem is to be resolved to determine the composition of the nitrogen and
carbon dioxide that is required to have an exit velocity of 2000 ft/s at the nozzle exit.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data"
mf_N2 = 0.8 "Mass fraction for the nitrogen, lbm_N2/lbm_mix"
mf_CO2 = 0.2 "Mass fraction for the carbon dioxide, lbm_CO2/lbm_mix"
T[1] = 1800 [R]
P[1] = 90 [psia]
Vel[1] = 0 [ft/s]
P[2] = 12 [psia]
Eta_N =0.92 "Nozzle adiabatic efficiency"
"Enthalpy property data per unit mass of mixture:"
" Note: EES calculates the enthalpy of ideal gases referenced to
the enthalpy of formation as h = h_f + (h_T - h_537) where h_f is the
enthalpy of formation such that the enthalpy of the elements or their
stable compounds is zero at 77 F or 537 R, see Chapter 14. The enthalpy
of formation is often negative; thus, the enthalpy of ideal gases can
be negative at a given temperature. This is true for CO2 in this problem."
h[1]= mf_N2* enthalpy(N2, T=T[1]) + mf_CO2* enthalpy(CO2, T=T[1])
h[2]= mf_N2* enthalpy(N2, T=T[2]) + mf_CO2* enthalpy(CO2, T=T[2])
"Conservation of Energy for a unit mass flow of mixture:"
"E_in - E_out = DELTAE_cv Where DELTAE_cv = 0 for SSSF"
h[1]+Vel[1]^2/2*convert(ft^2/s^2,Btu/lbm) - h[2] - Vel[2]^2/2*convert(ft^2/s^2,Btu/lbm) =0 "SSSF
energy balance"
"Nozzle Efficiency Calculation:"
Eta_N=(h[1]-h[2])/(h[1]-h_s2)
h_s2= mf_N2* enthalpy(N2, T=T_s2) + mf_CO2* enthalpy(CO2, T=T_s2)
"The mixture isentropic exit temperature, T_s2, is calculated from setting the
entropy change per unit mass of mixture equal to zero."
DELTAs_mix=mf_N2 * DELTAs_N2 + mf_CO2 * DELTAs_CO2
DELTAs_N2 = entropy(N2, T=T_s2, P=P_2_N2) - entropy(N2, T=T[1], P=P_1_N2)
DELTAs_CO2 = entropy(CO2, T=T_s2, P=P_2_CO2) - entropy(CO2, T=T[1], P=P_1_CO2)
DELTAs_mix=0
"By Dalton's Law the partial pressures are:"
P_1_N2 = y_N2 * P[1]; P_1_CO2 = y_CO2 * P[1]
P_2_N2 = y_N2 * P[2]; P_2_CO2 = y_CO2 * P[2]
"mass fractions, mf, and mole fractions, y, are related by:"
M_N2 = molarmass(N2)
M_CO2=molarmass(CO2)
y_N2=mf_N2/M_N2/(mf_N2/M_N2 + mf_CO2/M_CO2)
y_CO2=mf_CO2/M_CO2/(mf_N2/M_N2 + mf_CO2/M_CO2)
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13-29
SOLUTION of the stated problem
DELTAs_CO2=-0.04486 [Btu/lbm-R]
DELTAs_N2=0.01122 [Btu/lbm-R]
h[1]=-439.7 [Btu/lbm]
h_s2=-628.8 [Btu/lbm]
mf_N2=0.8 [lbm_N2/lbm_mix]
M_N2=28.01 [lbm/lbmol]
P[2]=12 [psia]
P_1_N2=77.64 [psia]
P_2_N2=10.35 [psia]
T[2]=1160 [R]
Vel[1]=0 [ft/s]
y_CO2=0.1373 [ft/s]
DELTAs_mix=0 [Btu/lbm-R]
Eta_N=0.92
h[2]=-613.7 [Btu/lbm]
mf_CO2=0.2 [lbm_CO2/lbm_mix]
M_CO2=44.01 [lbm/lbmol]
P[1]=90 [psia]
P_1_CO2=12.36 [psia]
P_2_CO2=1.647 [psia]
T[1]=1800 [R]
T_s2=1102 [R]
Vel[2]=2952 [ft/s]
y_N2=0.8627 [lbmol_N2/lbmol_mix]
SOLUTION of the problem with exit velocity of 2600 ft/s
DELTAs_CO2=-0.005444 [Btu/lbm-R]
DELTAs_N2=0.05015 [Btu/lbm-R]
h[1]=-3142 [Btu/lbm]
h_s2=-3288 [Btu/lbm]
mf_N2=0.09793 [lbm_N2/lbm_mix]
M_N2=28.01 [lbm/lbmol]
P[2]=12 [psia]
P_1_N2=13.11 [psia]
P_2_N2=1.748 [psia]
T[2]=1323 [R]
Vel[1]=0 [ft/s]
y_CO2=0.8543 [ft/s]
DELTAs_mix=0 [Btu/lbm-R]
Eta_N=0.92
h[2]=-3277 [Btu/lbm]
mf_CO2=0.9021 [lbm_CO2/lbm_mix]
M_CO2=44.01 [lbm/lbmol]
P[1]=90 [psia]
P_1_CO2=76.89 [psia]
P_2_CO2=10.25 [psia]
T[1]=1800 [R]
T_s2=1279 [R]
Vel[2]=2600 [ft/s]
y_N2=0.1457 [lbmol_N2/lbmol_mix]
PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
13-30
13-60 A piston-cylinder device contains a gas mixture at a given state. Heat is transferred to the mixture.
The amount of heat transfer and the entropy change of the mixture are to be determined.
Assumptions 1 Under specified conditions both H2 and N2 can be treated as ideal gases, and the mixture as
an ideal gas mixture. 2 Kinetic and potential energy changes are negligible.
Properties The constant pressure specific heats of H2 and N2 at 450 K are 14.501 kJ/kg.K and 1.049
kJ/kg.K, respectively. (Table A-2b).
Analysis (a) Noting that P2 = P1 and V2 = 2V1,
P2V 2 P1V1
2V
=

→ T2 = 1 T1 = 2T1 = (2 )(300 K ) = 600 K
V1
T2
T1
0.5 kg H2
1.6 kg N2
100 kPa
300 K
Also P = constant. Then from the closed system energy balance relation,
Ein − Eout = ∆Esystem
Qin − Wb,out = ∆U →
Qin = ∆H
Q
since Wb and ∆U combine into ∆H for quasi-equilibrium constant pressure processes.
[
]
[
]
Qin = ∆H = ∆H H 2 + ∆H N 2 = mc p ,avg (T2 − T1 ) H + mc p ,avg (T2 − T1 ) N
2
2
= (0.5 kg )(14.501 kJ/kg ⋅ K )(600 − 300)K + (1.6 kg )(1.049 kJ/kg ⋅ K )(600 − 300)K
= 2679 kJ
(b) Noting that the total mixture pressure, and thus the partial pressure of each gas, remains constant, the
entropy change of the mixture during this process is


T
P ©0 
T 
∆S H 2 = [m(s 2 − s1 )]H 2 = m H 2  c p ln 2 − R ln 2  = m H 2  c p ln 2 


T1
P1 
T1  H


H2
2
= (0.5 kg )(14.501 kJ/kg ⋅ K )ln
= 5.026 kJ/K
600 K
300 K


T
P ©0 
T
∆S N 2 = [m(s 2 − s1 )]N 2 = m N 2  c p ln 2 − R ln 2  = m N 2  c p ln 2


T1
P
T
1
1

 N2

= (1.6 kg )(1.049 kJ/kg ⋅ K )ln
= 1.163 kJ/K


 N2
600 K
300 K
∆S total = ∆S H 2 + ∆S N 2 = 5.026 kJ/K + 1.163 kJ/K = 6.19 kJ/K
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13-31
13-61 The states of two gases contained in two tanks are given. The gases are allowed to mix to form a
homogeneous mixture. The final pressure, the heat transfer, and the entropy generated are to be
determined.
Assumptions 1 Under specified conditions both O2 and N2 can be treated as ideal gases, and the mixture as
an ideal gas mixture. 2 The tank containing oxygen is insulated. 3 There are no other forms of work
involved.
Properties The constant volume specific heats of O2 and N2 are 0.658 kJ/kg.°C and 0.743 kJ/kg.°C,
respectively. (Table A-2).
Analysis (a) The volume of the O2 tank and mass of the nitrogen are
 mRT 
1
 =
V1,O 2 = 
P
 1 O
2
m N2
(1 kg)(0.2598 kPa ⋅ m 3 /kg ⋅ K)(288 K)
= 0.25 m 3
300 kPa
 PV 
(500 kPa)(2 m 3 )
=  1 1  =
= 10.43 kg
3
 RT1  N 2 (0.2968 kPa ⋅ m /kg ⋅ K)(323 K)
V total = V1,O 2 + V 1, N 2 = 0.25 m 3 + 2.0 m 3 = 2.25 m 3
O2
N2
1 kg
15°C
300 kPa
2 m3
50°C
500 kPa
Also,
m O 2 = 1 kg 
→ N O 2 =
mO 2
M O2
m N 2 = 10.43 kg 
→ N N 2 =
=
Q
1 kg
= 0.03125 kmol
32 kg/kmol
m N2
M N2
=
10.43 kg
= 0.3725 kmol
28 kg/kmol
N m = N N 2 + N O 2 = 0.3725 kmol + 0.03125 kmol = 0.40375 kmol
Thus,
 NRu T
Pm = 
 V
(0.40375 kmol)(8.314 kPa ⋅ m 3 /kmol ⋅ K)(298 K)

 =
= 444.6 kPa
2.25 m 3
m
(b) We take both gases as the system. No work or mass crosses the system boundary, and thus this is a
closed system with W = 0. Taking the direction of heat transfer to be from the system (will be verified), the
energy balance for this closed system reduces to
E in − E out = ∆E system
− Qout = ∆U = ∆U O 2 + ∆U N 2 
→ Qout = [mcv (T1 − Tm )]O + [mcv (T1 − Tm )]N
2
2
Using cv values at room temperature (Table A-2), the heat transfer is determined to be
Qout = (1 kg )(0.658 kJ/kg ⋅ °C )(15 − 25)°C + (10.43 kg )(0.743 kJ/kg ⋅ °C )(50 − 25)°C
= 187.2 kJ
(from the system)
(c) For and extended system that involves the tanks and their immediate surroundings such that the
boundary temperature is the surroundings temperature, the entropy balance can be expressed as
S in − S out + S gen = ∆S system
−
Qout
+ S gen = m( s 2 − s1 )
Tb,surr
S gen = m( s 2 − s1 ) +
Qout
Tsurr
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13-32
The mole fraction of each gas is
yO 2 =
yN 2 =
NO2
Nm
NN2
Nm
=
0.03125
= 0.077
0.40375
=
0.3725
= 0.923
0.40375
Thus,
y Pm,2

T
( s 2 − s1 ) O 2 =  c p ln 2 − Rln
T1
P1

= (0.918 kJ/kg ⋅ K) ln



 O2
298 K
(0.077)(444.6 kPa)
− (0.2598 kJ/kg ⋅ K) ln
288 K
300 kPa
= 0.5952 kJ/kg ⋅ K
y Pm,2

T
( s 2 − s1 ) N 2 =  c p ln 2 − Rln
T1
P1

= (1.039 kJ/kg ⋅ K) ln



 N2
298 K
(0.923)(444.6 kPa)
− (0.2968 kJ/kg ⋅ K) ln
323 K
500 kPa
= −0.0251 kJ/kg ⋅ K
Substituting,
S gen = (1 kg )(0.5952 kJ/kg ⋅ K ) + (10.43 kg )(− 0.0251 kJ/kg ⋅ K ) +
187.2 kJ
= 0.962 kJ/K
298 K
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13-33
13-62 EES Problem 13-61 is reconsidered. The results obtained assuming ideal gas behavior with constant
specific heats at the average temperature, and using real gas data obtained from EES by assuming variable
specific heats over the temperature range are to be compared.
Analysis The problem is solved using EES, and the solution is given below.
"Input Data:"
T_O2[1] =15 [C];
T_N2[1] =50 [C]
T[2] =25 [C];
T_o = 25 [C]
m_O2 = 1 [kg];
P_O2[1]=300 [kPa]
V_N2[1]=2 [m^3];
P_N2[1]=500 [kPa]
R_u=8.314 [kJ/kmol-K];
MM_O2=molarmass(O2)
MM_N2=molarmass(N2);
P_O2[1]*V_O2[1]=m_O2*R_u/MM_O2*(T_O2[1]+273)
P_N2[1]*V_N2[1]=m_N2*R_u/MM_N2*(T_N2[1]+273)
V_total=V_O2[1]+V_N2[1];
N_O2=m_O2/MM_O2
N_N2=m_N2/MM_N2;
N_total=N_O2+N_N2
P[2]*V_total=N_total*R_u*(T[2]+273); P_Final =P[2]
"Conservation of energy for the combined system:"
E_in - E_out = DELTAE_sys
E_in = 0 [kJ]
E_out = Q
DELTAE_sys=m_O2*(intenergy(O2,T=T[2]) - intenergy(O2,T=T_O2[1])) +
m_N2*(intenergy(N2,T=T[2]) - intenergy(N2,T=T_N2[1]))
P_O2[2]=P[2]*N_O2/N_total
P_N2[2]=P[2]*N_N2/N_total
"Entropy generation:"
- Q/(T_o+273) + S_gen = DELTAS_O2 + DELTAS_N2
DELTAS_O2 = m_O2*(entropy(O2,T=T[2],P=P_O2[2]) - entropy(O2,T=T_O2[1],P=P_O2[1]))
DELTAS_N2 = m_N2*(entropy(N2,T=T[2],P=P_N2[2]) - entropy(N2,T=T_N2[1],P=P_N2[1]))
"Constant Property (ConstP) Solution:"
-Q_ConstP=m_O2*Cv_O2*(T[2]-T_O2[1])+m_N2*Cv_N2*(T[2]-T_N2[1])
Tav_O2 =(T[2]+T_O2[1])/2
Cv_O2 = SPECHEAT(O2,T=Tav_O2) - R_u/MM_O2
Tav_N2 =(T[2]+T_N2[1])/2
Cv_N2 = SPECHEAT(N2,T=Tav_N2) - R_u/MM_N2
- Q_ConstP/(T_o+273) + S_gen_ConstP = DELTAS_O2_ConstP + DELTAS_N2_ConstP
DELTAS_O2_ConstP = m_O2*( SPECHEAT(O2,T=Tav_O2)*LN((T[2]+273)/(T_O2[1]+273))R_u/MM_O2*LN(P_O2[2]/P_O2[1]))
DELTAS_N2_ConstP = m_N2*( SPECHEAT(N2,T=Tav_N2)*LN((T[2]+273)/(T_N2[1]+273))R_u/MM_N2*LN(P_N2[2]/P_N2[1]))
SOLUTION
Cv_N2=0.7454 [kJ/kg-K]
[kJ]DELTAS_N2=-0.262 [kJ/K]
DELTAS_O2=0.594 [kJ/K]
E_in=0 [kJ]
MM_N2=28.01 [kg/kmol]
m_O2=1 [kg]
[kmol]N_total=0.4036 [kmol]
P_N2[1]=500 [kPa]
[kPa]P_O2[2]=34.42 [kPa]
R_u=8.314 [kJ/kmol-K]
[kJ]Tav_N2=37.5 [C]
T_N2[1]=50 [C]
V_N2[1]=2 [m^3]
Cv_O2=0.6627 [kJ/kg-K]
DELTAE_sys=-187.7
DELTAS_N2_ConstP=-0.2625 [kJ/K]
DELTAS_O2_ConstP=0.594 [kJ/K]
E_out=187.7 [kJ]
MM_O2=32 [kg/kmol]
m_N2=10.43 [kg]
N_N2=0.3724 [kmol]
N_O2=0.03125
P[2]=444.6 [kPa]
P_Final=444.6 [kPa]
P_N2[2]=410.1 [kPa]
P_O2[1]=300
Q=187.7 [kJ]
Q_ConstP=187.8 [kJ]
S_gen=0.962 [kJ]
S_gen_ConstP=0.9616
Tav_O2=20 [C]
T[2]=25 [C]
T_o=25 [C]
T_O2[1]=15 [C]
V_O2[1]=0.2494 [m^3]
V_total=2.249 [m^3/kg]
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13-34
13-63 Heat is transferred to a gas mixture contained in a piston cylinder device. The initial state and the
final temperature are given. The heat transfer is to be determined for the ideal gas and non-ideal gas cases.
Properties The molar masses of H2 and N2 are 2.0, and 28.0 kg/kmol. (Table A-1).
Analysis From the energy balance relation,
E in − E out = ∆E
Qin − Wb,out = ∆U
(
Qin = ∆H = ∆H H 2 + ∆H N 2 = N H 2 h2 − h1
)H
2
(
+ N N 2 h2 − h1
)N
2
since Wb and ∆U combine into ∆H for quasi-equilibrium constant pressure processes
mH 2
6 kg
NH2 =
=
= 3 kmol
M H 2 2 kg / kmol
NN2 =
mN 2
MN 2
=
6 kg H2
21 kg N2
5 MPa
160 K
Q
21 kg
= 0.75 kmol
28 kg / kmol
(a) Assuming ideal gas behavior, the inlet and exit enthalpies of H2 and N2 are determined from the ideal
gas tables to be
H2 : h1 = h@160 K = 4,535.4 kJ / kmol,
h2 = h@ 200 K = 5,669.2 kJ / kmol
N2 :
h1 = h@160 K = 4,648 kJ / kmol,
h2 = h@ 200 K = 5,810 kJ / kmol
Thus, Qideal = 3 × (5,669.2 − 4,535.4 ) + 0.75 × (5,810 − 4,648) = 4273 kJ
(b) Using Amagat's law and the generalized enthalpy departure chart, the enthalpy change of each gas is
determined to be

Tm,1
160

=
= 4.805
T R1 ,H 2 =
Tcr,H 2 33.3

 Z h1 ≅ 0
Pm
5

=
= 3.846 
PR1 ,H 2 = PR2 ,H 2 =
(Fig. A-29)
H2:
Pcr,H 2 1.30
Zh ≅ 0
 2
Tm , 2
200

=
= 6.006
T R2 , H 2 =

Tcr,H 2 33.3

Thus H2 can be treated as an ideal gas during this process.

Tm,1
160

=
= 1.27
T R1 , N 2 =
Tcr, N 2 126.2

 Z h1 = 1.3
Pm
5

=
= 1.47 
PR1 , N 2 = PR2 , N 2 =
N2:
Pcr, N 2 3.39
 Z h = 0.7
 2
Tm, 2
200

=
= 1.58
T R2 , N 2 =

Tcr, N 2 126.2

Therefore,
h2 − h1
(
(Fig. A-29)
)H = (h2 − h1 )H ,ideal = 5,669.2 − 4,535.4 = 1,133.8kJ/kmol
(h2 − h1 )N
2
2
2
(
) (
)
= RuTcr Z h1 − Z h2 + h2 − h1 ideal
= (8.314kPa ⋅ m3/kmol ⋅ K)(126.2K)(1.3 − 0.7) + (5,810 − 4,648)kJ/kmol
= 1,791.5kJ/kmol
Substituting,
Qin = (3 kmol)(1,133.8 kJ/kmol) + (0.75 kmol)(1,791.5 kJ/kmol) = 4745 kJ
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