Question

A young male adult takes in about 4.49 x 10-4 m3 of fresh air during a normal breath. Fresh air contains approximately 21% oxygen. Assuming that the pressure in the lungs is 0.907 x 105 Pa and air is an ideal gas at a temperature of 310 K, find the number of oxygen molecules in a normal breath.

Answer 1

Volume of oxygen in a normal breath = V = 21 % of 4.49 x 10^-4 m³

or, V = 4.49 x 10^-4 m³ x 21 / 100

or, V ~ 9.43 x 10^-5 m³.

Pressure in lungs = P = 0.907 x 10⁵ Pa.

Temperature of oxygen gas = T = 310 K.

Universal gas constant = R = 8.314 J / ( mol . K ).

Hence, number of moles n, of oxygen gas in a normal breath, from the ideal gas equation, is given by :

PV = nRT

or, n = PV / RT = ( 0.907 x 10⁵ x 9.43 x 10^-5 ) / ( 8.314 x 310 )

or, n ~ 3.32 x 10^-3.

Now, 1 mole of oxygen = 6.022 x 10²³ oxygen molecules.

Hence, the number of oxygen molecules in a normal breath = 3.32 x 10^-3 x 6.022 x 10²³ ~ 2 x 10²¹.