A 30 m³ room has ventilation500 m³/hour in/out and initially negligible VOC. A stove burns.25 kg wood/hour, emitting.2 gVOC/kg. VOC reacts first-order at .4/hour and entering air is clean. Find concentration after two hours in µg/m³,to one decimal.
VOC source is $.25(.2)=.05$ g/hour=50000 µg/hour. The mixed-room balance is $VdC/dt=E-(Q+kV)C$. Therefore $C(t)=E/(Q+kV)[1-e^{-(Q/V+k)t}]$. With Q+kV=512 m³/hour and t=2,the exponential is negligible and $C=50000/512=\mathbf{97.7}$ µg/m³ after rounding.
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