lung capacity was measured as 3.1 L at a body temperature of 37 ∘C and a pressure of 747 mmHg . How many moles of oxygen is in her lungs if air contains 21% oxygen?
Use PV = nRT and solve for n = number of mols air in the lungs.
Then n x 0.21 = mols oxygen.
.025
0.03 moles of O2
To find the number of moles of oxygen in the lungs, we need to use the ideal gas law equation:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of moles
R = Ideal gas constant
T = Temperature
First, we need to convert the lung capacity from liters to cubic meters, since the ideal gas law requires the volume to be in SI units. We know that 1 L is equal to 0.001 m³, so we can calculate:
V = 3.1 L * 0.001 m³/L
V = 0.0031 m³
Next, we need to convert the temperature from degrees Celsius to Kelvin. The Kelvin temperature scale is obtained by adding 273.15 to the Celsius value. So, we have:
T = 37°C + 273.15
T = 310.15 K
Now, let's calculate the partial pressure of oxygen using the information that air contains 21% oxygen. The total pressure of air is 747 mmHg, so:
Partial pressure of oxygen = 21% * Total pressure
Partial pressure of oxygen = 0.21 * 747 mmHg
Finally, we can rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Substituting the values we have:
n = (Partial pressure of oxygen * V) / (R * T)
Now, we'll substitute the values and solve:
n = (0.21 * 747 mmHg * 0.0031 m³) / (R * 310.15 K)
We need to use the appropriate value for the ideal gas constant. The correct value depends on the units of pressure and volume used. Since we have the pressure in mmHg and the volume in m³, we'll use the value for R in units of (mmHg * m³) / (K * mol):
R = 8.3145 (mmHg * m³) / (K * mol)
Now, we can calculate the number of moles of oxygen in the lungs.