Question

Dry air is a mixture of what 5 gases? Which of the 5 is the most abundant (about 79%), and which is second (about 21%)? As an aside, the molecular weight of dry air is 28.97 lb/lbmol or kg/kgmol etc

The Dalton model assumes that each component of an ideal gas mixture (recall PV=mRT) occupying a closed vessel (V) acts as if it were alone at the temperature (T) and volume (V). Furthermore, each component (i) is at a partial pressure Pi, with all of the partial pressures summing to a total pressure (P). An added touch is that the partial-to-total pressure ratio of each gas component is equal to the mole fraction.

How can you find specific internal energy, specific enthalpy, specific volume (or density), specific heat, etc. of a gas mixture? Answer: All of these specific properties are “per unit mass” therefore mass fraction weight the properties for specific mixture properties (hm-specific mixture enthalpy – dry air and water vapor) or simply multiply by mass for total properties (H – enthalpy).

Question. Calculate the constant pressure specific heat (Cp) of dry air if nitrogen and oxygen values are 0.178 and 0.158 Btu/lb-°F, respectively.

Answer 1

cpm= specific heat for the mixture=

Volume fraction given for Nitrogen=79%=mole fraction yN2=0.79

Volume fraction for Oxygen=21%= mole fraction yO2=.21

Gases	Mole fraction y	Molecular weight Mi(kg)	MiYi	Number of moles	Mass(mi) MiNi(kg)	Mass fraction mfi=mi/Mm	Cpi Btu/lb-°F	mfi.cpi
N2	0.79	28	22.12	79	2212	0.76	0.178	0.135
O2	0.21	32	6.72	21	672	0.23	0.158	0.036
	1	60	28.84 (given value=28.97kg/kmol)	100	2884 kg=Total mass of mixture mm	100		specific heat for mixture 0.171 Btu/lb-°F

specific heat for mixture=Cpm=CpN2.mf N2+CpO2.mf o2

=0.178 x0.135+0.158 x0.036

=0.171Btu/lb-°F