Question

(b) Convert the O3 mixing ratio of 100 ppbv (parts-per-billion by volume) to an O3 concentration (mass/volume) value using the ideal gas equation. Assume standard temperature and pressure (STP). The molecular weight of O3 is 48 g/mol.

Answer 1

Parts-per-billion by volume : 1ppb = 1 microliter / cubic meter

O₃ 100 ppbv = 100 microliter / m3 = 10^-7   m3 . = 0.0001 L

Using the ideal gas equation. PV = nRT (at STP)

From STP volume of one mole = 22.4 L

No. of moles of Ozone = 0.0001 L / 22.4 L = 4.46*10^-6 moles

Molecular weight of O₃ =  48 g/mol

Mass of ozone =  4.46*10^-6 mole * 48 g/mol = 2.14* 10^-4 gm.

So, concentration of ozone (mass/volume) = 2.14* 10^-4 gm. / cubic meter

=    2.14* 10^-4 gm. / 1000 L (1 c-m = 1000 L)

=  2.14* 10^-7 gm. /  L = 0.214 microgram / L