Answer:
The answer to your question is: 131.2 g of oxygen
Explanation:
Data
Volume = V = 16.6 l
Temperature = T = 34°C = 307°K
Pressure = 6.22 atm
R = 0.08205 (atm l/mol °K)
Atomic mass of oxygen = 32 g
Formula
PV = nRT
n = PV / RT
Process
n = (6.22 x 16.6) / (0.08205 x 307)
n = 103.25 / 25.19
n = 4.09 moles
1 mol ------------------ 32 g of oxygen
4.09 ------------------- x
x = (4.09 x 32) / 1 = 131.2 g of oxygen
Taking into account the ideal gas law, the mass of oxygen gas in a 16.6 L container at 34.0°C and 6.22 atm is 131.2 grams.
In first place, you have to know that an ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:
P×V = n×R×T
In this case:
Replacing in the ideal gas law:
6.22 atm×16.6 L = n×0.082[tex]\frac{atm*L}{mol*K}[/tex]×307 K
Solving:
[tex]n\frac{6.22 atmx16.6 L}{0.082\frac{atm*L}{mol*K}x307 K}[/tex]
n=4.10 moles
Finally, you know that the molar mass of O₂ is 32 [tex]\frac{g}{mole}[/tex]. So to convert 4.10 moles to mass you use the following relationship:
[tex]mass=4.10 molesx32\frac{g}{mole}[/tex]
mass= 131.2 grams
In summary, the mass of oxygen gas in a 16.6 L container at 34.0°C and 6.22 atm is 131.2 grams.
Learn more about the ideal gas law:
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