Discussion
Outdoor air contains \( 0.04 \ \mathrm{mg/m^3} \) of a contaminant. With a source of \( 30 \ \mathrm{mg/h} \) and ventilation of \( 150 \ \mathrm{m^3/h} \), the indoor steady concentration is
Answer: C
Correct answer: \( 0.24 \ \mathrm{mg/m^3} \)
Ventilation dilutes towards the outdoor concentration, not towards zero, so the indoor level is the outdoor value plus the contribution of the indoor source:
\[ C = C_{out} + \frac{G}{Q} = 0.04 + \frac{30}{150} = 0.04 + 0.20 = 0.24\ \mathrm{mg/m^3} \]
The distractor 0.20 omits the outdoor term. That term is what limits how much ventilation can achieve: no amount of outdoor air can bring the indoor concentration below what the outdoor air already contains.
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