Question

Your house is located in a polluted city with an ambient ozone concentration of 80 ppb by mole (exceeding the NAAQS!). The air intake into your house has a flow of 1500 L/min of air, which is the same as the outflow rate. Ozone is very reactive, and will oxidize chemicals (mostly at the surfaces of your house) such as the oils on the walls or on people’s skin. At steady state, if the indoor concentration of ozone in your house is 15 ppb, what is the rate in mg/min that ozone is being lost inside your house. Assume there are no sources of ozone inside the house (such as “air purification” machines, for example). One mole of air occupies 24.4 liters at room temperature and pressure.

Answer 1

of air. Colution & Given data is, * The ambient ozone concentration of soppb by mole. * The aur intake into your house has a flow of 1500 Ilmin * At Steady state, if the indoor concentration of ozone in your house is 15 ppb. * One mole of air occupies 24.4 liters at room temperature and pressure. The ambient concentration of ozone in the atmosphere is much higher than inside the house. The amount of gone being taken out of the air when it is Inside the house is 65pp6" The air intake is 1500 L/mun, with 24.4L of air 1500 24.4 = 61.47 moles the average molar mass of air is 28.9659. The total influx of air mass per minute is 28.965x61.47 = 1780.69 The concentration of ozone lost is 65ppb or 65g/100 The mass of ozone lost per minute is 65x109x1780.6 = 0.0vollsg or 0.115 ing. Thus, the rate of ozone loss per minute is sollsinglman.