Respuesta :
There will be 0.040 mol of oxygen molecules containing at the end of strong inhalation.
Explanation:
Given
volume = 5.0 liter = 5.0 [tex]\times[/tex] 10^-3 m^3, temperature = 37 degree celsius = 310 K,
Pressure p = 1 atm = 1.013 [tex]\times[/tex] 10^5 Pa.
By using Boyle's law,
PV = nRT
where P represents the pressure,
V represents the volume,
R represents the gas constant,
T represents the temperature
n = PV / RT
= (1.013 [tex]\times[/tex] 10^5) (5.0 [tex]\times[/tex] 10^-3) / (9.31) (310)
n = 0.20 mol
N oxygen = 0.20 [tex]\times[/tex] 0.20 = 0.040 mol.
The number of mole of oxygen molecules contained in the lungs at the end of a strong inhalation is 0.04 mole
We'll begin by calculating the number of mole of air in the lungs.
Volume (V) = 5 L
Temperature (T) = 37 °C = 37 + 273 = 310 K
Pressure (P) = 1 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Number of mole (n) =?
PV = nRT
Divide both side by RT
n = PV / RT
n = (1 × 5) / (0.0821 × 310)
n = 0.196 mole
- Finally, we shall determine the number of mole of oxygen molecules.
Mole of air = 0.196 mole
Percentage of oxygen = 20% = 0.2
Mole of Oxygen molecules =?
Mole of Oxygen = percent × total mole
Mole of Oxygen = 0.04 mole
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