Question

The estimated average concentration of NO2 in the air in theUS in 2006 was 0.016 ppm. (a) Calculate the partial pressure of the NO2 in a sample of this air when the atmospheric pressure is 755 torr (99.1kPa). (b) How many molecules of NO2 are present under these conditions at 20 C in a room that measures 15 x 14 x 8 ft? Please go in detail with question B. I'm not sure how to incorporate room measurements, Thanks!

Answer 1

(a) Calculate the partial pressure of the NO2 in a sample of this air when the atmospheric pressure is 755 torr (99.1kPa).

Partial pressure is related to mole fraction as

Pi = xi * P

where

Pi = partial pressure of a gas

xi = mole fraction of individula gas in a mixture

P = total pressure

1 ppm is 1 mg/L

1 L of air is

PV = nRT

755 x 1 = n x 62.363 x 293

n = 0.041 moles of air

1 mg of NO2 is 1 x 10^-3/46 g/mol = 2.17 x 10^-5 moles

0.016 ppm, 0.016 mg which is 2.17 x 10^-5 moles x 0.016 = 3.48 x 10^-7 moles

SO mole fraction is 3.48 x 10^-7 moles/0.041 moles = 8.49 x 10^-6 mole fraction

SO partial pressure = 8.49 x 10^-6 mole fraction x 755 torr = 0.00641 Torr

(b) How many molecules of NO2 are present under these conditions at 20 C in a room that measures 15 x 14 x 8 ft? Please go in detail with question B. I'm not sure how to incorporate room measurements, Thanks

15 x 14 x 18 = 3780 ft³

1 ft³ = 28.3 L

So 3780 ft³ = 3780 x 28.3 L = 106974 L

PV = nRT

0.00641 Torr x 106974 L = n x 62.363 x 293

n = 0.0375 moles

which will be 0.0375 x 6.023 x 10²³ molecules = 2.259 x 10²² molecules are present in the room