A home has volume = 400m^3 and an infiltration rate of 0.3 airchanges per hour.

Question

A home has volume = 400m^3 and an infiltration rate of 0.3 airchanges per hour. During an episode of smog, the outdoorconcentration of PAN is 75 ppb. How long will it take to rise to 40ppb inside the house if the initial concentration inside is 9ppb? The relevant equation is C(t) = Cmax *(e^(-kt)). I believe you have to assume steady state to find Cmax.At steady state F = CVk, where F is flow rate, C is theconcentration, and k is the first order rate constant. Please help, I am desperate for any suggestions. This is a question from Hites' Elements of EnvironmentalChemistry A home has volume = 400m^3 and an infiltration rate of 0.3 airchanges per hour. During an episode of smog, the outdoorconcentration of PAN is 75 ppb. How long will it take to rise to 40ppb inside the house if the initial concentration inside is 9ppb? The relevant equation is C(t) = Cmax *(e^(-kt)). I believe you have to assume steady state to find Cmax.At steady state F = CVk, where F is flow rate, C is theconcentration, and k is the first order rate constant. Please help, I am desperate for any suggestions. This is a question from Hites' Elements of EnvironmentalChemistry

Explanation / Answer

I wont give you the solution right away but Ill give you some hintsand then message me if you still need help. Dont worry about theVolume of the house it doesnt matter. The Cmax is 75ppb and youneed to use the other equation. C=Cmax [1-e^(-kt)]. This is theincreasing concentration equation, the one you have written is thedecreasing concentration and is by the way written incorrectlyanyways. Good luck

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