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.
[tex]PV = nRT[/tex]
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.
[tex]21.1 kPa \times \frac{1atm}{101.3kPa}= 0.208 atm[/tex]
V = 3.0 L
T = 295 K
R = 0.0821 L-atm/mol K
Let us rearrange the equation to solve for n.
[tex]n = \frac{PV}{RT}[/tex]
[tex]n = \frac{0.208atm\times 3.0L}{0.0821 L.atm/mol K\times 295 K}[/tex]
[tex]n = 0.026 mol[/tex]
0.026 moles of oxygen are in the lung
