Question

1.     The exhaust of an automobile is found to contain 2% CO, at a temperature of 80°C. Express the concentration of CO in the exhaust in mg m^-3.

2.     California’s air quality standard for nitrogen dioxide (NO₂) is 470mg m^-3 (at a temperature of 25°C and 1 atm of pressure). Express the concentration in ppmv.

Answer 1

since the exhaus is gas, volume %= Mole %

Basis : 1 mole of exhaust. Moles of CO= 2/100=0.02 , molar mass of CO= 12+16= 28 g/mole

moles= mass/molar mass , mass = moles* molar mass =0.02*28 =0.56 gm=0.56*1000mg= 560 mg

Given P= 1atm, T= 80deg.c= 80+273=353K, R= 0.0821 L.atm/mole.K

from gas law , V= nRT/P= 1*0.0821* 353/1 = 28.98 L, 1000L= 1m3, 28.98L= 28.88*10^-3 m3

concentration of CO= 560/(28.88*10^-3) mg/m3=19390.6 mg/m3

2. Basis : Basis :   1 m3 of mixture, V= 1000L, T= 25 deg.c= 25+273= 298K, P= 1 atm

n= no of moles of gas= PV/RT =1* 1000/(0.0821*298)= 40.87 moles of gas mixture

mass of NO2= 470mg =470*10^-3 gm of NO2, molar mass of NO2= 14+32= 46 g/mole

moles of NO2= mass/molar mass =470*10^-3/ 46 =0.0102 moles

moles fraction of of NO2 =0.0102/40.87 =0.00025

since at the temperatuture, volume fraction = mole%

hence ppm of NO2= 0.00025

or 0.00025*10⁶/106 =250/10⁶ = 250 ppmV