A. The initial mass of oxygen in the tank is 480.96 g
B. The mass of oxygen that leaked out is 384.96 g
A. Determination of the initial mass of oxygen in the tank.
- We'll begin by calculating the initial number of mole of oxygen in the tank
Volume (V) = 0.1 m³
Pressure (P) = 4×10⁵ Pa
Temperature (T) = 47 °C = 47 + 273 = 320 K
Gas constant (R) = 8.314 Pa.m³/Kmol
number of mole (n) = ?
PV = nRT
0.1 × 4×10⁵ = n × 8.314 × 320
40000 = n × 2660.48
Divide both side by 2660.48
n = 40000 / 2660.48
n = 15.03 moles
- Finally, we shall determine the initial mass of oxygen.
mole = 15.03 moles
molar mass of oxygen gas = 32 g/mol
mass of oxygen gas = ?
Mass = mole × molar mass
Mass of oxygen gas = 15.03 × 32
Mass of oxygen gas = 480.96 g
Therefore, the initial mass of oxygen in the tank is 480.96 g
B. Determination of the mass of oxygen that leaked out of the tank.
- We'll begin by calculating the number of mole of oxygen that leaked out of the tank
Volume (V) = 0.1 m³
Pressure (P) = 3×10⁵ Pa
Temperature (T) = 27 °C = 27 + 273 = 300 K
Gas constant (R) = 8.314 Pa.m³/Kmol
number of mole (n) = ?
PV = nRT
0.1 × 3×10⁵ = n × 8.314 × 300
30000 = n × 2494.2
Divide both side by 2494.2
n = 30000 / 2660.48
n = 12.03 moles
- Finally, we shall determine the mass of oxygen that leaked out of tank.
mole = 12.03 moles
molar mass of oxygen gas = 32 g/mol
mass of oxygen gas = ?
Mass = mole × molar mass
Mass of oxygen gas = 12.03 × 32
Mass of oxygen gas = 384.96 g
Therefore, the mass of oxygen that leaked out of tank is 384.96 g
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