The concentration of [tex] SO_{2} [/tex] in the stack gas = 12 ppmv
That means 12 L of [tex] SO_{2} [/tex] is present per [tex] 10^{6} L gas [/tex]
The given temperature is 273 K (0 C) and pressure is 1 atm. At these conditions, 1 mol of gas would occupy,
[tex] PV = nRT [/tex]
[tex] (1 atm) (V) = (1 mol)(0.08206\frac{L.atm}{mol.K}) (273 K) [/tex]
V = 22.4 L
1 mol [tex] SO_{2}[/tex] occupies 22.4 L
Moles of [tex] SO_{2} [/tex] = [tex] 12 L *\frac{1 mol}{22.4 L} = 0.5357 mol SO_{2} [/tex]
Mass of [tex] SO_{2} [/tex] =[tex] 0.5357 mol *\frac{64.06 g}{1 mol} = 34.32 g SO_{2} *\frac{10^{6} microgram}{1 g} [/tex] =[tex] 3.432 *10^{7} [/tex]μg
Converting [tex] 10^{6} L to m^{3} [/tex]:
[tex] 10^{6} L *\frac{1 m^{3}}{1000 L} [/tex] = [tex] 10^{3} m^{3} [/tex]
Calculating the concentration in μg/[tex] m^{3} [/tex]:
[tex] \frac{3.432 * 10^{7} microgram}{10^{3} L} = 3.432 * 10^{4} microgram/m^{3} [/tex]