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
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