What mass, in grams, of oxygen gas (O2) is contained in a 10.5 liter tank at 27.3 degrees Celsius and 1.83 atmospheres? Show all of the work used to solve this problem.
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To solve this problem, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
We need to find the number of moles of oxygen gas (O2) first using the given conditions:
1. Convert the volume from liters to m^3:
10.5 L * (1 m^3 / 1000 L) = 0.0105 m^3
2. Convert the temperature from degrees Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 27.3 + 273.15 = 300.45 K
3. Convert the pressure from atmospheres to Pascals:
1 atm * (101325 Pa / 1 atm) = 101325 Pa
Now we can rearrange the ideal gas law equation to solve for the number of moles:
n = (PV) / (RT)
Substituting the given values:
n = (101325 Pa * 0.0105 m^3) / ((8.314 J/mol·K) * 300.45 K)
Simplifying the equation:
n ≈ 0.0409 moles
Next, we can calculate the mass of oxygen gas using the molar mass of oxygen:
1. Find the molar mass of oxygen gas (O2):
Molar mass of O2 = (2 * atomic mass of oxygen)
The atomic mass of oxygen is approximately 16.00 g/mol.
Therefore, the molar mass of oxygen gas (O2) is approximately 2 * 16.00 g/mol = 32.00 g/mol.
2. Calculate the mass of oxygen gas:
Mass = n × molar mass
Substituting the values:
Mass = 0.0409 moles × 32.00 g/mol
Mass ≈ 1.31 grams (rounded to two decimal places)
Therefore, there are approximately 1.31 grams of oxygen gas (O2) contained in the 10.5 liter tank at 27.3 degrees Celsius and 1.83 atmospheres.