I would calculate mols HNO3 produced by 532 g NO2 and substitute into PV = nRT and solve for L HNO3 at the temperature and pressure.
Then ppm = L HNO3/1,000,000 L air at 20C.
0.08E-6 = LHNO3 produced/x
Solve for x = L air
Check my thinking.
Nitrogen dioxide reacts with water and oxygen in the atmosphere to produce acid rain.
4NO2(g) + 2H2O(g) + O2(g) = 4HNO3(g)
Assuming that the temperature and the pressure remain constant, calculate the volume of air at 1 atm, 20 C, necessary to dilute the HNO3(g) to 0.080 ppm (by volume) when 532 g of NO2(g) is allowed to react.
I am wondering what should I do with the 0.080 ppm. I am pretty sure I know how to do everything else but, I am not sure how to use the ppm in to the equation of if it used after the pv=nRT equation.
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