Question

16) The concentration of carbon monoxide is 5.5 g/m3. Convert this to ppmy at temperature of 30 °C and pressure of 700 mm Hg. (7 points) 17)Determine the density of air at 75 °F and 14.7 psia. The molecular weight of air is 29 g/gmole? (6 points)

Answer 1

16. Temperature, T = 30+273.15 =303.15 K and [CO]=5.5 g/m3= 5500 mg/m3

Mass of CO per m3 volume = 5.5 g

Moles of CO per m3 volume = Mass/molar mass

=5.5 g/(28.0104 g/mol) = 0.196355 mol

Ideal gas equation PV =nRT

   Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Given, pressure, P = 760 mmHg = 1.00 atm

            1.00 atm x V = 0.196355 mol x (0.0821 atm L mol^-1K^-1) x 303.15 K

            Or, V = 4.8870199 atm L / 1.00 atm

            Hence, V = 4.8870199 L

# We have, volume of solute CO = 4.8870199 L = 4887019.9 uL

            Specified volume of air containing that CO = 1 m³ = 1000.0 L

Now,

            [CO], ppm = 48870199 uL / 1000 L = 4887.0199 ppm

17. Given T= 75 F=75+459.67 =534.67 R where R is Rankine scale

P= 14.7 psia , Molecular Weight of air (MW)= 29 g/gmole

                   PV = MRT

                   P=M/V * RT = ᵨ * RT

  ᵨ =P/RT

For R value R= R/Mw =10.731/29 =0.37

Density (ᵨ )= 14.7 /(0.37*534.67) = 0.0743 lb/ft3