The density of air is 1.25 gram per liter determine the volume percentage of oxygen in the air at 1 atmospheric pressure and 27 degree celcius temprature? Assume air is a mixture of nitrogen and oxygen.
Nuts to this. If the volume is one liter, you have one liter of Nitrogen, and one liter of oxygen. Gases spread to fill the volume.
If you want the mass percentage, that is easy.
Assume a volume V.
MassV= 1.25V grams.
but massV=molemassAir*molesAir.
now, molesair = PV/RT
molemassAir=massV*RT/PV=1.25RT/P
so calculate that.
Now, molemassAir=32X+28(1-x)
and you can solve for X (percent of O2 by mass).
To determine the volume percentage of oxygen in air, we need to know the molar mass of nitrogen and oxygen.
The molar mass of nitrogen (N₂) is 28 grams per mole, and the molar mass of oxygen (O₂) is 32 grams per mole.
First, let's determine the molar mass of air, which is a mixture of nitrogen and oxygen. Since air is about 78% nitrogen and 21% oxygen, we can calculate the average molar mass as follows:
Average molar mass of air = (0.78 * Molar mass of nitrogen) + (0.21 * Molar mass of oxygen)
= (0.78 * 28 g/mol) + (0.21 * 32 g/mol)
≈ 28.98 g/mol
Next, we can calculate the density of air using its molar mass and the ideal gas law:
Density = (Molar mass * Pressure) / (Gas constant * Temperature)
Given:
Molar mass of air = 28.98 g/mol
Pressure = 1 atm
Temperature = 27°C = 27 + 273.15 = 300.15 K (converted to Kelvin)
Gas constant (R) = 0.0821 L·atm/(K·mol) (based on the ideal gas law)
Density = (28.98 g/mol * 1 atm) / (0.0821 L·atm/(K·mol) * 300.15 K)
≈ 1.16 g/L
Now, we can determine the volume percentage of oxygen in air at this density. Since the density of air is 1.25 g/L and the molar mass of oxygen is 32 g/mol, we can calculate the density of oxygen (DO₂) as follows:
DO₂ = (1.16 g/L * Volume percentage of oxygen) / 100%
Dividing the density of air by the density of oxygen, we get:
1.25 g/L / DO₂ = 1.16 g/L * 100% / 32 g/mol
Solving for DO₂, we find:
DO₂ ≈ 37.45 g/L
Now we substitute this value into the equation above to determine the volume percentage of oxygen:
37.45 g/L = (1.16 g/L * Volume percentage of oxygen) / 100%
Solving for the volume percentage of oxygen:
Volume percentage of oxygen ≈ (37.45 g/L * 100%) / 1.16 g/L
Volume percentage of oxygen ≈ 3226.72% / 1.16%
Volume percentage of oxygen ≈ 2786.9%
Therefore, at 1 atmospheric pressure and 27 degrees Celsius, the volume percentage of oxygen in air is approximately 2786.9%.