Question

A human lung at maximum capacity has a volume of 3.0 liters. If the partial pressure of oxygen in the air is 21.1 kilopascals and the air temperature is 295 K, how many moles of oxygen are in the lung?

Answer 1

Answer : 0.026 moles of oxygen are in the lung

Explanation :

We can solve the given question using ideal gas law.

The equation is given below.

$PV = nRT$

We have been given P = 21.1 kPa

Let us convert pressure from kPa to atm unit.

The conversion factor used here is 1 atm = 101.3 kPa.

$21.1 kPa \times \frac{1atm}{101.3kPa}= 0.208 atm$

V = 3.0 L

T = 295 K

R = 0.0821 L-atm/mol K

Let us rearrange the equation to solve for n.

$n = \frac{PV}{RT}$

$n = \frac{0.208atm\times 3.0L}{0.0821 L.atm/mol K\times 295 K}$

$n = 0.026 mol$

0.026 moles of oxygen are in the lung