Answer:
There are 0.1125 g of O₂ 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 atm·L·mol⁻¹·K⁻¹)
Now we calculate the number of moles of air in 1 L at sea level (this means with P=1atm):
1 atm * 1 L = n₁ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K
n₁=0.04092 moles
Now we calculate n₂, 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 = n₂ * 0.082 atm·L·mol⁻¹·K⁻¹ * 298 K
n₂=0.02414 moles
In order to calculate the difference in O₂, we substract n₂ from n₁:
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 O₂ in air (20.95%), and then use the molecular weight to calculate the grams of O₂ in 20.95% of 0.01678 moles: