A young male adult takes in about 4.0 10-4 m3 of fresh air during a normal breath. Fresh air contains approximately 21% oxygen. Assuming that the pressure in the lungs is 1.0 105 Pa and air is an ideal gas at a temperature of 310 K, find the number of oxygen molecules in a normal breath.
I answered this yesterday.
The answer was posted at
http://www.jiskha.com/display.cgi?id=1292881652
To find the number of oxygen molecules in a normal breath, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P = Pressure in Pascals (Pa)
V = Volume in cubic meters (m³)
n = Number of moles of gas
R = Ideal gas constant (8.314 J/mol·K)
T = Temperature in Kelvin (K)
First, let's convert the volume of air from cubic meters (m³) to moles by using the ideal gas law equation:
n = PV / RT
The pressure in the lungs is given as 1.0 × 10^5 Pa. The volume of fresh air in a normal breath is 4.0 × 10^-4 m³. The temperature is given as 310 K. Plugging these values into the equation, we have:
n = (1.0 × 10^5 Pa) × (4.0 × 10^-4 m³) / (8.314 J/mol·K × 310 K)
Calculating the numerator:
(1.0 × 10^5 Pa) × (4.0 × 10^-4 m³) = 4.0 × 10¹ Pa·m³
Calculating the denominator:
8.314 J/mol·K × 310 K = 2574.14 J/mol
Dividing the numerator by the denominator, we get:
4.0 × 10¹ Pa·m³ / 2574.14 J/mol = 1.5532 × 10^-2 mol
Now that we have the number of moles (n), we can calculate the number of oxygen molecules using Avogadro's number (6.022 × 10²³ molecules/mol):
Number of oxygen molecules = n × Avogadro's number
Number of oxygen molecules = (1.5532 × 10^-2 mol) × (6.022 × 10²³ molecules/mol)
Calculating the product, we get:
(1.5532 × 10^-2 mol) × (6.022 × 10²³ molecules/mol) = 9.3534 × 10²¹ molecules
Therefore, there are approximately 9.3534 × 10²¹ oxygen molecules in a normal breath.