The air pollutants are represented in either volume fraction/total volume or weight/volume. The former one may be of several types. For example, moles of pollutant/ total moles of air (mole fraction) in specific volume, number of molecules of pollutant/total number of molecules in the specific volume, etc. Overall, the concentration has quantity but no units. For example, 400 ppmv of CO2 is a representation by 400 moles of CO2 / 1000000 moles of air in a given volume. Therefore, it just a number. Another way of expressing the gas concentration is the weight of pollutant/volume (g/m3 or mg/m3 or μg/m3).

The conversion of ppmv, ppbv to mg/m3 or μg/m3 respectively will depend on the molecular weight of the pollutant, and atmospheric temperature and pressure as follows.

(PPM/1000000)*(1000mg/1g)*(MW g/1 mole)*(1 mole/22.4L)*(1000L/1m3) at STP

Canceling out the numerators and denominators will yield the following equation.

=> PPM*MW/22.4 mg/m3 at STP. This equation clearly shows that molecular weight is very important to convert. However, atmospheric properties change dramatically. Therefore, by using the equation of state, we can calculate the volume occupied by 1 mole of gas as follows.

PV= nRT, where P is atmospheric pressure in hPa, V in Liters, n is a number of moles, T temperature in Kelvin and the R is gas constant of 83.144 L hPa/K mole.

V= RT/P because n is 1 mole. Substituting this at 22.4 L in the above equation will give the full controlled equation of MW, P, T that can be used to convert the pollutants into weights/volume.

Concentration in mg/m3 = ppm*WM*P/83.144*T

In the same way,

(PPB/1,000,000,000)*(1000,000 μg/1g)*(MW g/1 mole)*(1 mole/22.4L)*(1000L/1m3) at STP

Cancelling out the numerators and denominators will yield the following equation.

Therefore, the concentration in μg/m3 = ppb*MW*P/83.144*T

Problem1:

What is the concentration of CO2 in mg/m3 when it has 400 ppmv and the atmospheric temperature and pressure are 298K and 1010 hPa, respectively?

Concentration of CO2 = 400*44*1010/83.144*298 mg/m3

= 717.44 mg/m3

Problem 2:

What is the concentration of Ozone in μg/m3 when it has 80 ppbv, and the atmospheric temperature and pressure are 298K and 1010 hPa, respectively? The molecular weight of O3 is 48.

Concentration of O3 = 80*48*1010/83.144*298 μg/m3

= 156.5 μg/m3

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## About Dr. Kishore Ragi

I am critical reviewer of literature on the science. Here, I write critical review on the published research, my research and research related business plans.