Question

An adult inhales about 6.0 x 10⁻⁴ m³ of fresh air during a breath. Only 20% of fresh air is oxygen. Assume the pressure in the lungs is 1.0 x 10⁵ Pa and the air is at a temperature of 300 K.

How many oxygen molecules are in each breath?
A. 1.4 x 10²³
B. 2.9 x 10²¹
C. 4.9 x 10⁻⁴
D. 2.9 x 10²⁵

Answer 1

Answer: B.
`2.9* 10^(21)`

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number
`6.023* 10^(23)` of particles.

Given : Volume of fresh air =
`6.0* 10^(-4)m^3`

volume of oxygen =
`\frac{20}100}* 6.0* 10^(-4)m^3=1.2* 10^(-4)m^3`

According to the ideal gas equation:

`PV=nRT`

P = Pressure of the gas =
`1.0* 10^(5)Pa` = 0.98 atm

V= Volume of the gas =
`1.2* 10^(-4)m^3=0.12L`
`1m^3=1000L`

T= Temperature of the gas = 300 K

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= ?

`n=(PV)/(RT)=\frac{0.98atm* 0.12L}0.0821* 300}=4.8* 10^(-3)moles`

1 mole of oxygen contains =
`6.023* 10^(23)` molecules

Thus
`4.8* 10^(-3)moles` of oxygen contain=
`(6.023* 10^(23))/(1)* 4.8* 10^(-3)=2.9* 10^(21)` molecules of oxygen

Thus there are
`2.9* 10^(21)` oxygen molecules in each breath