Question

A mixture at 50°F and 35 psia has the following volumetric analysis; O2, 60 percent; CO2, 40 percent. Compute (a) the mass analysis, (b) the partial pressure of O2 in psia, and (c) the apparent molar mass of the mixture, in lbm/lbmol

Answer 1

Apply ideal gas law

PV = nRT

Assume a basis of v = 10 L

n = PV/(RT) = 35*10*/(10.73*(460+50)) = 0.06395 lbmol

a) mass analysis

solve for O2 and CO2

0.06395 * 0.6 = 0.03837 lbmolO2

0.06395 *0.4 = 0.02558 lbmolCO2

mass = mol*MW = 0.03837*32 = 1.22784 lb O2

mass = mol*MW = 0.02558 *44 = 1.12552 lb CO2

total mass = 1.22784 +1.12552 = 2.35336 lb

mass mass O2 = mass O2 / total mass = 1.22784 /2.35336 = 0.52

mass frac CO2 = mass CO2 / total mass = 1.12552 /2.35336 = 0.48

b)

Partial pressures are based on mol fraction

volume frac = mol fraction so

0.6*P1 = 35*0.6 = 21 psia of O2

0.4*P1 = 35*0.4 = 14 psia of CO2

c)

molar mass of mix

MW = mass/mol= 2.35336 /(0.06395 ) = 36.8 lbm/lbmol