Answer:
There are 0.1125 g of Oā less in 1 L of air at 14,000 ft than in 1 L of air at sea level.
Explanation:
To solve this problem we use the ideal gas law:
PV=nRT
Where P is pressure (in atm), V is volume (in L), n is the number of moles, T is temperature (in K), and R is a constant (0.082 atmĀ·LĀ·molā»Ā¹Ā·Kā»Ā¹)
Now we calculate the number of moles of air in 1 L at sea level (this means with P=1atm):
1 atm * 1 L = nā * 0.082 atmĀ·LĀ·molā»Ā¹Ā·Kā»Ā¹ * 298 K
nā=0.04092 moles
Now we calculate nā, the number of moles of air in L at an 14,000 ft elevation, this means with P = 0.59 atm:
0.59 atm * 1 L = nā * 0.082 atmĀ·LĀ·molā»Ā¹Ā·Kā»Ā¹ * 298 K
nā=0.02414 moles
In order to calculate the difference in Oā, we substract nā from nā:
0.04092 mol - 0.02414 mol = 0.01678 mol
Keep in mind that these 0.01678 moles are of air, which means that we have to look up in literature the content of Oā in air (20.95%), and then use the molecular weight to calculate the grams of Oā in 20.95% of 0.01678 moles:
[tex]0.01678mol*\frac{20.95}{100} *32\frac{g}{mol} =0.1125 gO_{2}[/tex]
