Modern Engineering Thermodynamics

410 CHAPTER 12: Mixtures of Gases and Vapors
and
χ i = n
i M
n
= w m
i
m M i
(12.14)
Last, if the mixture is a gas or a vapor, we can determine its equivalent gas constant R m from the universal gas
constant R = 8:313 kJ/ ðkgmole.KÞ = 1545:35 ft .lbf/ ðlbm.RÞ and Eq. (12.11) or (12.12) as
R m = R M m
(12.15)
At this point, we have developed general formulae for determining the values of thermodynamic properties and
other important characteristics of mixtures of substances in their solid, liquid, vapor, or gaseous states. Before we
can continue further, we need to know how these mixture thermodynamic properties are related to each other.
Since the number of possible mixture compositions is infinite, the construction of thermodynamic tables or charts
for all possible mixtures is impractical except for very common mixtures, such as pure air and air–water vapor
mixtures.
EXAMPLE 12.1
A new gas furnace requires a mixture of 50.0% propane and 50.0% air on a mass basis. Determine (a) the equivalent molecular
mass of the mixture, (b) the mixture composition on a molar basis, and (c) the equivalent gas constant of the mixture.
Solution
a. The mixture composition on a mass basis is w propane = 0.500, and w air = 0.500. The molecular masses of the components
are found in Table C.13 in Thermodynamic Tables to accompany Modern Engineering Thermodynamics as
M propane = 44:09 kg/kgmole and M air = 28:97 kg/kgmole
Then, the equivalent molecular mass of the mixture is given by Eq. (12.12) as
M m =
1
1
kg
=
0:500
44:09 + 0:500 = 35:0
kgmole
28:97
M i
∑ N
w i
i=1
b. The equivalent molar values are given by Eq. (12.14) as
χ i = n
i M
= w m
i
n m M i
and this is used as follows:
M
χ propane = w m
35:0 kg/kgmole
propane
= 0:500 = 0:397
M propane
44:1 kg/kgmole
and the remaining component is
M
χ air = w m
35:0 kg/kgmole
air = 0:500 = 0:603
M air
28:97 kg/kgmole
so that the composition is 50.0% propane and 50.0% air on a mass basis, but it is 39.7% propane and 60.3% air
on a molar basis. (Note that, once we knew the propane molar concentration, we could have determined the air
molar concentration through Eq. (12.10), since w propane + w air = 1.0 and χ propane + χ air = 1.0, so χ air = 1.0 − χ propane =
1.0 − 0.397 = 0.603.)
c. The equivalent gas constant for this mixture is given by Eq. (12.15) as
R m = R M m
= 8:3143 kJ/kgmole .K
34:97 kg/kgmole
= 0:238 kJ
kg .K
The following example is slightly more complicated than Example 12.1, because it deals with a mixture containing
four components. The solution method, however, is the same as in Example 12.1.