Question

The analysis on a mass basis of an ideal gas mixture at 30°F, 15 lbf/in.² is 40% CO₂, 10% CO, and 50% O₂.


Determine:

(a) the analysis in terms of mole fractions.

(b) the apparent molecular weight of the mixture.

(c) the partial pressure of each component, in lbf/in.²

(d) the volume occupied by 10 lb of the mixture, in ft³.

Answer 1

a.

Calculate the number of moles of ideal gas by

= mass/Molecular weight

Moles of CO2(n1) = 40%/44 = 0.90%

Moles of CO(n2) = 10%/28 =0.357%

Moles of O2(n3) = 50%/32 = 1.565%

Mole Fraction of CO2 = n1/(n1+n2+n3) = 32%

Mole Fraction of CO = n2/(n1+n2+n3) = 12.7%

Mole Fraction of O2 = n3/(n1+n2+n3) = 55.5%

b) Mixture Molecular weight = (n1*MWco2 + n2*MWco+ n3*MWo2)/100

Mixture Molecular weight = (32*44 + 12.7*28 + 55.5*32)/100 = 35.42

c) Pco2 = {n1/(n1+n2+n3)}*P = (n1/100)*P

Pco2 = 0.32*15 = 4.8 lbf/in^2

Pco = 0.127*15= 1.905 lbf/in^2

PO2= 0.555*15. = 8.325 bf/in^2

d) PV = nRT

n = Mass/MW

Mass = 10 lbb

MW = 35.43

T = 30 F= 489.67R

PV = nRT

15*V = (10/35.43) *1545.349*489.67

V = 14238.52 i.n^2*ft

V = 14238*0.0833^2 ft^3

V = 98.44 ft^3