Question

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

Calculate the partial pressure of the NO2 in a sample of this air when the atmospheric pressure is 748 torr (99.7 kPa ).

How many molecules of NO2 are present under these conditions at 17 ∘C in a room that measures 16 × 11 × 6 ft ?

Answer 1

Concentration of a gas in ppm is also equal to µmol/mol of the gas in the atmosphere. Therefore, the partial pressure of NO₂ is given as

P(NO₂) = (ppm concentration)*10^-6*(atmospheric pressure) = (0.016)*10^-6*(99.7 kPa)*(1000 Pa/1 kPa) = 0.0015952 Pa = 1.5952*10^-3 Pa (ans).

Volume of the room = (16*11*6) ft³ = 1056 ft³ = (1056 cu.ft)*(28.3168 L/1 cu.ft) = 29902.5408 L (ans).

We know that 1 Pa = 9.8692*10^-6 atm.

Use the ideal gas law to calculate the number of moles of NO₂.

P*V = n*R*T

===> (1.5952*10^-3 Pa)*(9.8692*10^-6 atm/1 Pa)*29902.5408 L = n*(0.082 L-1tm/mol.K)*(273 + 17) K

===> n = 1.97967*10^-5 mol.

We know that 1 mole of an ideal gas = 6.023*10²³ molecules; therefore,

1.97967*10^-5 mol NO₂ = (6.023*10²³*1.96967*10^-5) molecules = 1.19235*10¹⁹ molecules ≈ 1.20*10¹⁹ molecules (ans).