Consider a room that is 14 ft x 20 ft with a 10 ft ceiling H

Consider a room that is 14 ft x 20 ft with a 10 ft ceiling. How many molecules of air are present in this room at 20°C and 750 ton? If a pollutant is present at 2.2 ppm, how many pollutant molecules are in this room? air molecules pollutant molecules

Solution

Volume of room = length * breadth * height = 14ft * 20ft * 10ft = 2800 ft^3

1 ft^3 = 28.3168 L

Volume in L = 79287.17 L

Using the ideal gas equation

P = 750 torr = 750/760 atm

PV = nRT

750/760 * 79287.17 = n * 0.0821 * (273+20)

n = 3252.66 moles

Number of air molecules = 6.023 * 10^(23) * 3252.66 = 1.9590 * 10^(27) molecules

Pollutant molecules = 2.2 * 10^(-6) * number of air molecules = 2.2 * 1.9590 * 10^(21) molecules

=> 4.3098 * 10^(21) molecules