Modern Engineering Thermodynamics

12.3 Mixtures of Ideal Gases 415
Consequently, for a mixture of ideal gases, we have
p i
= π i = V i
= ψ
p m V i = n
i
M
= χ
m n i = w m
i
m M i
or
M
π i = ψ i = χ i = w m
i
M i
(12.23)
which relates all four composition measures. Thus, the pressure fraction, volume fraction, and mole fraction are
all equal and differ from the mass fraction by only a molecular mass ratio.
EXAMPLE 12.3
An analysis of the exhaust gas from an experimental engine produces the following results on a molar basis:
Carbon dioxide = 9.51%
Water = 19.01%
Nitrogen = 71.48%
Assuming ideal gas behavior, (a) determine the volume fraction, pressure fraction, and mass fraction composition of the
mixture, and (b) if the total pressure of the mixture is 14.7 psia, determine the partial pressure of the water vapor in the
exhaust gas mixture.
Solution
a. For ideal gas behavior, Eq. (12.23) tells us that the mole fractions, volume fractions, and the pressure fractions are all
the same, or
χ CO2 = ψ CO2 = π CO2 = 9:51%
χ H2O = ψ H2O = π H2O = 19:01%
χ N2
= ψ N2
= π N2 = 71:48%
The equivalent molecular mass of this ideal gas mixture is given by Eq. (12.11) as
M m = ∑ N
i=1
χ i M i = χ CO2 M CO2 + χ H2OM H2O + χ N2
M N2
= 0:0951ð44:01 kg/kgmoleÞ + 0:1901ð18:02 kg/kgmoleÞ + 0:7148ð28:02 kg/kgmoleÞ
= 27:64 kg/kgmole
and the corresponding mass fractions are given by Eq. (12.23) as
M CO2
44:01 kg/kgmole
w CO2 = ψ CO2 = ð9:51%Þ = 15:14%
M m
27:64 kg/kgmole
M H2O
18:02 kg/kgmole
w H2O = ψ H2O = ð19:01%Þ = 12:39%
M m
27:64 kg/kgmole
M N2
28:02 kg/kgmole
w N2 = ψ N2
= ð71:48%Þ = 72:46%
M m
27:64 kg/kgmole
b. If the total pressure is 14.7 psia, then the partial pressure of the water vapor in the exhaust gas mixture is given by
Eq. (12.23) as
p H2O = p m χ H2O = ð14:7 psiaÞð0:1901Þ = 2:79 psia
Exercises
4. Find the partial volume of the nitrogen in the exhaust gas in Example 12.3 if the total volume of the gas mixture is
8.00 ft 3 . Answer: V N2 = 2:14 ft 3 :
5. If there are 20.0 moles of the exhaust gas mixture in Example 12.3, how many moles of carbon dioxide would be in the
mixture? Answer: n CO2 = 14:3 moles:
6. If there is 0.650 lbm of the exhaust gas mixture in Example 12.3, how many lbm of water vapor would be in the
mixture? Answer: m water vapor = 0.081 lbm.