Chapter 13
Gas Mixtures
Study Guide in PowerPoint
to accompany
Thermodynamics: An Engineering Approach, 5th edition
by Yunus A. Çengel and Michael A. Boles
The discussions in this chapter are restricted to nonreactive ideal-gas mixtures.
Those interested in real-gas mixtures are encouraged to study carefully the material
presented in Chapter 12.
Many thermodynamic applications involve mixtures of ideal gases. That is, each of
the gases in the mixture individually behaves as an ideal gas. In this section, we
assume that the gases in the mixture do not react with one another to any significant
degree.
We restrict ourselves to a study of only ideal-gas mixtures. An ideal gas is one in
which the equation of state is given by
PV  mRT or
PV  NRu T
Air is an example of an ideal gas mixture and has the following approximate
composition.
Component
N2
O2
Argon
CO2 + trace elements
% by Volume
78.10
20.95
0.92
0.03
2
Definitions
Consider a container having a volume V that is filled with a mixture of k different
gases at a pressure P and a temperature T.
A mixture of two or more gases of fixed chemical composition is called a nonreacting
gas mixture. Consider k gases in a rigid container as shown here. The properties of
the mixture may be based on the mass of each component, called gravimetric
analysis, or on the moles of each component, called molar analysis.
k gases
T = Tm
P = Pm
V = Vm
m = mm
The total mass of the mixture mm and the total moles of mixture Nm are defined as
k
mm   mi
i 1
k
and
N m   Ni
i 1
3
The composition of a gas mixture is described by specifying either the mass fraction
mfi or the mole fraction yi of each component i.
mf i 
mi
mm
and
yi 
Ni
Nm
Note that
k
 mf
k
=1
i
and
i 1
y
i
1
i 1
The mass and mole number for a given component are related through the molar
mass (or molecular weight).
mi  N i M i
To find the average molar mass for the mixture Mm , note
k
k
i 1
i 1
mm   mi   N i M i  N m M m
Solving for the average or apparent molar mass Mm
k
k
mm
Ni
Mm 

Mi   yi Mi
N m i 1 N m
i 1
( kg / kmol )
4
The apparent (or average) gas constant of a mixture is expressed as
Rm 
Ru
Mm
Can you show that Rm is given as
( kJ / kg  K )
k
Rm   mf i Ri
i 1
To change from a mole fraction analysis to a mass fraction analysis, we can show
that
yM
mf i  k i i
 yi Mi
i 1
To change from a mass fraction analysis to a mole fraction analysis, we can show that
mf / M i
yi  k i
 mf i / Mi
i 1
5
Volume fraction (Amagat model)
Divide the container into k subcontainers, such that each subcontainer has only one
of the gases in the mixture at the original mixture temperature and pressure.
Amagat's law of additive volumes states that the volume of a gas mixture is equal to
the sum of the volumes each gas would occupy if it existed alone at the mixture
temperature and pressure.
k
Vm   Vi (Tm , Pm )
Amagat's law:
i 1
The volume fraction of the vfi of any component is
and
Vi (Tm , Pm )
vf i 
Vm
k
 vf
i 1
i
=1
6
For an ideal gas mixture
Vi 
N i Ru Tm
Pm
and Vm 
N m Ru Tm
Pm
Taking the ratio of these two equations gives
Vi
Ni
vf i 

 yi
Vm N m
The volume fraction and the mole fraction of a component in an ideal gas mixture are
the same.
Partial pressure (Dalton model)
The partial pressure of component i is defined as the product of the mole fraction and
the mixture pressure according to Dalton’s law. For the component i
Pi  yi Pm
k
Dalton’s law:
Pm   Pi (Tm ,Vm )
i 1
7
Now, consider placing each of the k gases in a separate container having the volume
of the mixture at the temperature of the mixture. The pressure that results is called
the component pressure, Pi' .
Pi ' 
N i Ru Tm
Vm
and
Pm 
N m Ru Tm
Vm
Note that the ratio of Pi' to Pm is
Pi ' Vi
N

 i  yi
Pm Vm N m
For ideal-gas mixtures, the partial pressure and the component pressure are the
same and are equal to the product of the mole fraction and the mixture pressure.
8
Other properties of ideal-gas mixtures
The extensive properties of a gas mixture, in general, can be determined by summing
the contributions of each component of the mixture. The evaluation of intensive
properties of a gas mixture, however, involves averaging in terms of mass or mole
fractions:
k
k
k
U m   U i   mi ui   N i ui
i 1
i 1
i 1
k
k
k
(kJ)
Hm   Hi   mi hi   N i hi
i 1
k
i 1
i 1
k
k
Sm   Si   mi si   N i si
i 1
and
i 1
k
um   mf i ui
and
i 1
um   yi ui
(kJ / kg or kJ / kmol)
i 1
k
k
and
i 1
hm   yi hi
(kJ / kg or kJ / kmol)
i 1
k
sm   mf i si
(kJ / K)
k
i 1
hm   mf i hi
(kJ)
k
and
i 1
sm   yi si
(kJ / kg  K or kJ / kmol  K)
i 1
k
Cv , m   mf i Cv , i
k
and
i 1
i 1
k
C p , m   mf i C p , i
i 1
Cv , m   yi Cv , i
k
and
C p , m   yi C p , i
i 1
9
These relations are applicable to both ideal- and real-gas mixtures. The properties or
property changes of individual components can be determined by using ideal-gas or
real-gas relations developed in earlier chapters.
Ratio of specific heats k is given as
km 
C p ,m
Cv ,m

C p ,m
Cv ,m
The entropy of a mixture of ideal gases is equal to the sum of the entropies of the
component gases as they exist in the mixture. We employ the Gibbs-Dalton law that
says each gas behaves as if it alone occupies the volume of the system at the
mixture temperature. That is, the pressure of each component is the partial pressure.
For constant specific heats, the entropy change of any component is
10
The entropy change of the mixture per mass of mixture is
The entropy change of the mixture per mole of mixture is
11
In these last two equations, recall that
Pi , 1  yi , 1 Pm, 1
Pi , 2  yi , 2 Pm, 2
Example 13-1
An ideal-gas mixture has the following volumetric analysis
Component
N2
CO2
% by Volume
60
40
(a)Find the analysis on a mass basis.
For ideal-gas mixtures, the percent by volume is the volume fraction. Recall
yi  vf i
12
Comp.
yi
N2
CO2
0.60
0.40
Mi
yiMi
mfi = yiMi /Mm
kg/kmol kg/kmol kgi/kgm
28
16.8
0.488
44
17.6
0.512
Mm = yiMi = 34.4
(b) What is the mass of 1 m3 of this gas when P = 1.5 MPa and T = 30oC?
Rm 

mm 
Ru
Mm
( kJ / kg  K )
kJ
kmol  K  0.242 kJ
kg
kg  K
34.4
kmol
8.314
PmVm
RmTm
15
. MPa (1m3 )
103 kJ

(0.242 kJ / ( kg  K ))(30  273) K m3 MPa
 20.45 kg
13
(c) Find the specific heats at 300 K.
Using Table A-2, Cp N2 = 1.039 kJ/kgK and Cp CO2 = 0.846 kJ/kgK
2
C p , m   mf i C p ,i  (0.488)(1039
. )  (0.512)(0.846)
1
 0.940
Cv , m
kJ
kgm  K
kJ
 C p , m  Rm  (0.940  0.242)
kgm  K
 0.698
kJ
kgm  K
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(d) This gas is heated in a steady-flow process such that the temperature is
increased by 120oC. Find the required heat transfer. The conservation of mass and
energy for steady-flow are
m 1  m 2  m
m 1h1  Q in  m 2 h2
Q  m (h  h )
in
2
1
 p , m (T2  T1 )
 mC
The heat transfer per unit mass flow is
Q in
qin 
 C p , m (T2  T1 )
m
kJ
 0.940
(120 K )
kgm  K
kJ
 112.8
kgm
15
(e) This mixture undergoes an isentropic process from 0.1 MPa, 30oC, to 0.2 MPa.
Find T2.
The ratio of specific heats for the mixture is
k
Cp ,m
Cv , m

0.940
 1347
.
0.698
Assuming constant properties for the isentropic process
(f) Find Sm per kg of mixture when the mixture is compressed isothermally from 0.1
MPa to 0.2 MPa.
16
But, the compression process is isothermal, T2 = T1. The partial pressures are given
by
Pi  yi Pm
The entropy change becomes
For this problem the components are already mixed before the compression process.
So,
yi , 2  yi ,1
Then,
17
2
sm   mf i si
i 1
 (0.488
kg N 2
 0167
.
kJ
kgm  K
kgm
)( 0.206
kJ
kg N 2  K
)  (0.512
kgCO2
kgm
)( 0131
.
kJ
kgCO2  K
)
Why is sm negative for this problem? Find the entropy change using the average
specific heats of the mixture. Is your result the same as that above? Should it be?
(g) Both the N2 and CO2 are supplied in separate lines at 0.2 MPa and 300 K to a
mixing chamber and are mixed adiabatically. The resulting mixture has the
composition as given in part (a). Determine the entropy change due to the mixing
process per unit mass of mixture.
18
Take the time to apply the steady-flow conservation of energy and mass to show that
the temperature of the mixture at state 3 is 300 K.
But the mixing process is isothermal, T3 = T2 = T1. The partial pressures are given by
Pi  yi Pm
The entropy change becomes
19
But here the components are not mixed initially. So,
y N 2 ,1  1
yCO2 , 2  1
and in the mixture state 3,
y N 2 , 3  0.6
yCO2 , 3  0.4
Then,
20
Then,
2
sm   mf i si
i 1
 (0.488
 0163
.
kg N 2
kgm
)(0152
.
kJ
kg N 2  K
)  (0.512
kgCO2
kgm
)(0173
.
kJ
kgCO2  K
)
kJ
kgm  K
If the process is adiabatic, why did the entropy increase?
Extra Assignment
Nitrogen and carbon dioxide are to be mixed and allowed to flow through a
convergent nozzle. The exit velocity to the nozzle is to be the speed of sound for the
mixture and have a value of 500 m/s when the nozzle exit temperature of the mixture
is 500oC. Determine the required mole fractions of the nitrogen and carbon dioxide to
produce this mixture. From Chapter 17, the speed of sound is given by
C  kRT
Answer: yN2 = 0.589, yCO2 = 0.411
NOZZLE
Mixture
N2 and CO2
C = 500 m/s
T = 500oC
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