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Modern Engineering Thermodynamics

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12.2 Thermodynamic Properties of Gas Mixtures 411<br />

EXAMPLE 12.2<br />

The molar composition of air that is normally used to determine the thermodynamic properties of air at standard temperature<br />

and pressure is<br />

Component Molar %<br />

Nitrogen 78.09<br />

Oxygen 20.95<br />

Argon 0.930<br />

CO 2 and trace gases 0.0300<br />

Total 100.00%<br />

Determine (a) the equivalent molecular mass, (b) the gas constant for this mixture, and (c) the composition of air on a<br />

mass (or weight) basis.<br />

Solution<br />

a. Since we are given the molar composition for air, we can find its equivalent molecular weight from Eq. (12.11).<br />

Assuming that argon and carbon dioxide are the only minor components present, Table C.13 provides the necessary<br />

molecular masses as<br />

Then, Eq. (12.11) gives<br />

M air = ∑ 4<br />

b. Equation (12.15) gives the gas constant of this mixture as<br />

i=1<br />

M nitrogen = 28:02 kg=kgmole<br />

M oxygen = 32:00 kg=kgmole<br />

M argon = 39:94 kg=kgmole<br />

M carbon dioxide = 44:01 kg=kgmole<br />

χ i M i = χ N2<br />

M N2 + χ O2<br />

M O2 + χ Ar M Ar + χ CO2 M CO2<br />

= 0:7809ð28:02 kg=kgmoleÞ + 0:2095ð32:00 kg=kgmoleÞ<br />

+ 0:00930ð39:94 kg=kgmoleÞ + 0:000300ð44:10 kg=kgmoleÞ<br />

= 28:97 kg=kgmole<br />

R air =<br />

R = 8:3143 kJ=ðkgmole .KÞ<br />

= 0:287 kJ<br />

M air 28:97 kg=kgmole<br />

kg .K<br />

which agrees with the values given in Table C.13b in Thermodynamic Tables to accompany <strong>Modern</strong> <strong>Engineering</strong> <strong>Thermodynamics</strong>.<br />

c. Equation (12.13) can be used to determine the corresponding mass or weight fraction composition as<br />

<br />

w N2 = χ N2<br />

M N2<br />

M air<br />

<br />

w O2 = χ O2<br />

M O2<br />

M air<br />

<br />

= 0:7809<br />

<br />

= 0:2095<br />

<br />

<br />

28:02 kg=kgmole<br />

28:97 kg=kgmole<br />

<br />

<br />

32:00 kg=kgmole<br />

28:97 kg=kgmole<br />

= 0:7553 = 75:53% by mass<br />

= 0:2314 = 23:14% by mass<br />

<br />

<br />

M<br />

w Ar = χ Ar<br />

39:94 kg=kgmole<br />

Ar = 0:00930 = 0:0128 = 1:28% by mass<br />

M air<br />

28:97 kg=kgmole<br />

<br />

<br />

<br />

M CO2<br />

44:01 kg=kgmole<br />

w CO2 = χ CO2 = 0:000300 = 0:00046 = 0:0456% by mass<br />

M air<br />

28:97 kg=kgmole<br />

Note the difference between the mass and mole fraction composition values.<br />

Exercises<br />

1. Research has suddenly discovered that the composition given in Example 12.1 was wrong. It should have been 30.0%<br />

propane and 70.0% air on a mass basis. Determine the proper molar basis composition for this new mixture.<br />

Answer: χ propane = 0.220 and χ air = 0.780.<br />

(Continued )

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