1. Fundamental of Gases Ideal Gas Law The behavior of chemicals in air with respect to temperature and pressure can be assumed to be ideal (in the chemical sense) because the concentration of these pollutants are usually sufficiently low. Thus, we can assume that at the same temperature and pressure, different kinds of gases have densities proportional to their molecular masses.

2. Ideal gas law ρ = P X M R X T Where ρ = density of gas (g. m -3 ) P = absolute pressure (Pa) M= molecular mass (g. mol -1 ) T= absolute temperature (K) R= ideal gas constant = 8.3143 J.K -1.mol -1 (or Pa.m 3.mol -1.K - 1 )

3. Because density is defined as mass per unit volume, or the number of moles per unit volume, n / V, the expression may be rewritten in the general form as PV= nRT (The ideal gas law) where V is the volume occupied by n moles of gas. At 273.15 k and 101.325 KPa, one mole of an ideal gas occupies 22.414 L

4. Concentration of Pollutants in Air One must be aware that when dealing with concentration of gases in air, the approximation of 1 ppm= 1 mg.L -1 is no longer valid as it is with dilute aqueous solutions. This is because the density of air is not 1 g.mL -1 and varies significantly with temperature. With air, concentrations are often reported in units of micrograms per cubic meter or parts per million. With, air the units of parts per million are reported on a volume - volume basis

5. 1.The units of parts per million have the advantage over micrograms per cubic meter in that changes in temperature and pressure do not change the ratio of the volume of pollutant to volume of air. 2.Thus, it is possible to compare concentration given in parts per million, without considering effects of pressure or temperature. The concentration of particulate matter may be reported only as micrograms per cubic meter. The micrometer unit is used to report particle size

6. Converting Micrograms per Cubic Meter to Parts per Million The conversion between micrograms per cubic meter and parts per million is based on the fact that at standard conditions (0°C and 101.325 KPa), one mole of an ideal gas occupies 22.414 L. Thus, we may write an equation that converts the mass of the pollutant, Mp, in grams to its equivalent volume, Vp, in liters at standard temperature and pressure (STP).

7. Converting Micrograms per Cubic Meter to Parts per Million Where MW is the molecular weight of the pollutant in units of grams per mole.

8. Where M p is the mass of the pollutant of interest in micrograms.The factors converting micrograms to grams and liters to million of liters cancel one another. Unless otherwise stated, it is assumed that V a = 1.00 m 3

9. Example 2-24: A 1m 3 sample of air was found to contain 80 μg. M -3 of SO 2. the temperature and pressure were 25.0°C and 103.193 KPa when the air sample was taken. What was the SO 2 concentration in parts per million? Solution: First we must determine the MW of SO 2 form the chart inside the cover of this book, we find: MW of SO 2 = 32.06 + 2(15.9994) = 64.06 g.mol -1 Next we must convert the temperature form Celsius to Kelvin. Thus 25°C+ 273 K= 298 K Now using Equation 2-97, we find Concentration = 0.030 ppm of SO 2