A house has an air infiltration rate of 0.300 air exchanges per hour. During an episode...

A house has an air infiltration rate of 0.300 air exchanges per hour. During an episode of photochemical smog, the outdoor concentration of peroxyacetyl nitrate (PAN) is 79.6 ppb. If the family remains indoors, and the initial concentration of PAN inside is 12.1 ppb, how long (in hours) will it be before the PAN concentration inside rises to 39.0 ppb?

Solutions

Expert Solution

Here, since the pollutant concentration increases, the up-going curve can be used.

C = C_max [1-e^(-kt)]

C_max is the outdoor concentration = 79.6 ppb. Since the relationship between time and concentration is not linear, first we need to calculate the time required to reach 39.0 ppb from zero ppb. Then we need to calculate the time required to reach 12.1 ppb from zero ppb. The subtraction of these two will be our result.

39.0 ppb = 79.6 ppb x [1-e^(-0.3)]

= [ln (1-39.0/79.6)] / -0.3 hour^-1 = 2.2441 hour

12.1 ppb = 79.6 ppb x [1-e^(-0.3)]

= [ln (1-12.1/79.6)] / -0.3 hour^-1 = 0.5496 hour

2.2441 hour - 0.5496 hour = 1.694 hour

queen_honey_blossom answered 1 year ago

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