Total lung capacity of a typical adult is approximately 5.0L. Approximately 20% of the air is oxygen, as air is 20% oxygen. At sea level and at an average body temperature of 37 C, how many moles of oxygen do the lungs contain at the end of an inflation?
n=PV/RT
V=5.0L=.005m^3
P=1.013e^5Pa
R=8.31J/mole*K
T=37c=310k
so n=124.85
then I times by .20 to get the moles of Oxygen...but I'm missing something?
Avogadro's number?
Wait!! You might know how to do it, but I still don't. Help :(
Well, it looks like you've got most of the information there, but let me help you fill in the missing piece. You're on the right track with using the ideal gas law equation, n = PV/RT, to calculate the number of moles of gas.
To find the moles of oxygen specifically, you're correct to multiply the total number of moles by the fraction representing the proportion of oxygen in the air, which in this case is 0.20.
So, if we plug the values into the equation:
n = (1.013e^5 Pa) * (0.005 m^3) / (8.31 J/mole*K) * (310 K)
After doing the math, you should get a value for n, which represents the total number of moles of air in the lungs. Then, to find the moles of oxygen, simply multiply n by 0.20.
And remember, laughter is the best medicine, so don't forget to give yourself a chuckle while doing your calculations!
To calculate the number of moles of oxygen in the lungs, you are on the right track by using the ideal gas law equation, n = PV/RT. However, there are a few corrections and additional steps needed to get the correct answer.
First, let's convert the given pressure from Pa (Pascals) to atm (atmospheres). 1 atm is equivalent to 1.013e^5 Pa. So, the pressure becomes:
P = 1.013e^5 Pa * (1 atm / 1.013e^5 Pa) = 1 atm
Now, let's convert the volume from liters to cubic meters, as the ideal gas constant R is given in J/(mol·K). Since 1 L = 0.001 m^3, the volume becomes:
V = 5.0 L * (0.001 m^3 / 1 L) = 0.005 m^3
Next, we need to convert the temperature from Celsius to Kelvin. The conversion from Celsius to Kelvin is K = °C + 273, so the temperature becomes:
T = 37 °C + 273 = 310 K
Now, plug the values of P, V, R, and T into the ideal gas law equation:
n = PV / RT
= (1 atm) * (0.005 m^3) / (8.31 J/(mol·K) * 310 K)
≈ 0.000805 mol
So, at the end of an inflation, you would have approximately 0.000805 moles of air in the lungs. However, you are interested in the moles of oxygen specifically. Since air is approximately 20% oxygen, you can multiply the total moles of air by 0.20 (20%) to get the moles of oxygen:
n_oxygen = 0.000805 mol * 0.20
= 0.000161 mol
Therefore, the lungs would contain approximately 0.000161 moles of oxygen at the end of an inflation.