A particulate matter sampler was operated for 24 hours at a flowrate of 1.4 m3 min-1. The starting weight of the filter paper before sampling was 2.05 grams. At the end of the sampling period, the filter's weight was 3.84 grams, and the gain in weight was due completely to the particulate matter that had collected on the surface of the filter paper during air sampling. Determine the particulate matter concentration, in micrograms per cubic meter (expressed to one decimal place), for the given sampling time.

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valetta valetta · Answer 1 · 2019-10-10T08:21:32+02:00

Answer:

887.89 μg/m³

Explanation:

Given:

Duration of mass flow = 24 hours = 24 × 60 = 1440 minutes

Flowrate = 1.4 m³/min

starting weight of the filter paper before sampling = 2.05 grams

Filter's weight = 3.84 grams

Now,

Mass of the particular matter collected

= Mass of filter after 24 hours - Initial weight of the filter

= 3.84 - 2.05

= 1.79

Now,

The total volume passing through filter = Flowerate × Time

= 1.4 × 1440

= 2016 m³

Particulate matter concentration = [tex]\frac{\textup{Mass of particulare matter collected}}{\textup{Volume of the sample}}[/tex]

or

Particulate matter concentration = [tex]\frac{\textup{1.79}}{\textup{2016}}[/tex]

or

Particulate matter concentration = 0.0008878 g/m³

also,

1 gram = 10⁶ micrograms

thus,

Particulate matter concentration = 0.0008878 × 10⁶ μg/m³

= 887.89 μg/m³