Question

The estimated average concentration of NO2 in air in the United States in 2006 was 0.016 ppm.

Part A

Calculate the partial pressure of the NO2 in a sample of this air when the atmospheric pressure is 746 torr (99.5 kPa ).

Part B

How many molecules of NO2 are present under these conditions at 20 ∘C in a room that measures 10 × 16 × 7 ft ?

Answer 1

Partial Pressure:

Pa = mole fraction of a x Total pressure

Pt = 746 torr

Mole Fraction NO2 = 0.016 ppm or 0.016 mol per 10^6 mol of air = 1.6*10^-8 mol of NO2/mol air

Pa = (1.6*10^-8) * 746 = 1.19*10^-5 torr

partial pressure of the NO2 = 1.19 x10^-5 torr

PART B

PV = nRT

n = PV/RT

P = 1.19 x10^-5 torr = 1.57 x 10^-8 atm

T = 20C = 293 K

V = 10 x 16 x 7 ft = 1120 ft3 = 31707.2 L

n = PV/RT

n = (1.57 x 10^-8)(31707.2) / (0.0821 x 293) = 2.07 x 10^-5 mol of NO2

1 mol = 6.022*10^23 molecules

2.07 x 10^-5 x 6.022*10^23 = 1.25 x 10^19 molecules of NO2

molecules of NO2 = 1.25 x 10^19