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.

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 .m3 .mol-1 .K-1)

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

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One must be aware that when dealing with concentration of gases in air, the approximation of 1 ppm= 1 mg.L -1 is no longer valid as it is with dilute aqueous solutions.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

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

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).

Converting Micrograms per Cubic Meter to Parts per Million

Where MW is the molecular weight of the pollutant

in units of grams per mole.

1.414.22 molLMW

MV pp

Where Mp 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 Va= 1.00

m3

3

221

.1000

)/325.101()273/(.414.22)/(

mLV

PKPaKTmolLMWMppm

a

p

Example 2-24:

A 1m3 sample of air was found to contain 80 μg. M-3 of SO2. the temperature

and pressure were 25.0°C and 103.193 KPa when the air sample was taken.

What was the SO2 concentration in parts per million?

Solution:

First we must determine the MW of SO2 form the chart inside the cover of this

book, we find:

MW of SO2 = 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

333

1

.10

)193.103/325.101()273/298(1.414.22).06.64/80(

mLm

KPaLPaKKmolLmolgg

= 0.030 ppm of SO2