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Modern Engineering Thermodynamics

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12.9 Mixtures of Real Gases 435<br />

Then, Eq. (12.38) gives the Amagat compressibility factor for the mixture as<br />

Z Am = ∑ 3<br />

i=1<br />

<br />

w i M m<br />

M i<br />

Z Ai<br />

<br />

= w <br />

<br />

ammoniaM m<br />

ðZ A Þ<br />

M ammonia<br />

+<br />

w <br />

chlorineM m<br />

ammonia<br />

M chlorine<br />

=<br />

<br />

<br />

ð0:333Þð31:4 kg/kgmole mixtureÞ<br />

ð0:64Þ<br />

17:030 kg/kgmole ammonia<br />

<br />

<br />

ð0:333Þð31:4kg/kgmole mixtureÞ<br />

+ ð0:55Þ<br />

70:906 kg/kgmole chlorinee<br />

ðZ A Þ chlorine<br />

+<br />

<br />

<br />

ð0:333Þð31:4 kg/kgmole mixtureÞ<br />

+ ð0:86Þ = 0:68<br />

44:013 kg/kgmole nitrous oxide<br />

The mixture gas constant can now be calculated from<br />

<br />

<br />

w nitrous oxideM m<br />

M nitrous oxide<br />

ðZ A Þ nitrous oxide<br />

R m =<br />

R M m<br />

= 8:3143 kJ/ðkgmole .KÞ<br />

31:4 kg/kgmole<br />

= 0:265 kJ<br />

kg .K<br />

Then, Eq. (12.37) gives the total volume of the mixture as<br />

V m = Z Am m m R m T m<br />

p m<br />

= ð0:68Þð3:00 kgÞð0:265 kJ/ðkg .KÞÞð500: KÞ<br />

ð20:0 MPaÞð1000: kPa/MPaÞ<br />

= 0:0135 m 3<br />

Exercises<br />

39. Determine the volume in Example 12.15 if the composition is changed to 1.00 kg of ammonia and chlorine but 5.00 kg<br />

of nitrous oxide. Answer: V = 0:0297 m 3 .<br />

40. If the mixture temperature in Example 12.15 is increased from 500. K to 1000. K, what volume would then be required<br />

to hold the mixture if all the other variables remain unchanged? Answer: V = 0:0401 m 3 .<br />

41. If the nitrous oxide gas is eliminated from the mixture in Example 12.15 and the masses of ammonia and chlorine are<br />

each increased to 1.50 kg, what volume would then be required to hold this new mixture if all the other variables<br />

remain unchanged? Answer: V = 0:0141 m 3 .<br />

It should be clear from the preceding formulae that Z Di ≠ Z Ai ; therefore, the resultant Dalton and Amagat mixture<br />

compressibility factors also in general are not equal, or Z Dm ≠ Z Am .<br />

Dalton’s law for mixtures of real gases (Eq. (12.35)) is based on the premise that each gas in the mixture acts as<br />

though it alone occupies the entire volume of the mixture at the temperature of the mixture. Therefore, the<br />

gases are assumed not to interact in any manner. We do not find this to be true experimentally except at very<br />

low pressures or very high temperatures.<br />

Amagat’s law for mixtures of real gases (Eqs. (12.37)), on the other hand, incorporates the resultant mixture<br />

pressure and, therefore, automatically takes gas molecular interactions into account. Consequently, it tends to be<br />

more accurate than Dalton’s law at high pressures and low temperatures.<br />

A third method of incorporating the compressibility factor charts into predicting the behavior of real gas mixtures<br />

involves defining a pseudocritical pressure and a pseudocritical temperature for the mixture as<br />

and<br />

Kay’s law<br />

P cm = ∑ N<br />

x i P ci (12.39)<br />

i=1<br />

T cm = ∑ N<br />

i=1<br />

x i T ci (12.40)

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