Question

A welder using a tank of volume 7.30×10−2 m 3 fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.15×105 Pa and temperature of 38.1 ∘ C . Remember that the guage pressure of the gas in the tank is the difference between the absolute pressure of the gas inside the tank and the pressure outside the tank, in this case 1 atm. So the absolute pressure is equal to the guage pressure plus 1 atm. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 24.0 ∘ C , the gauge pressure of the oxygen in the tank is 1.80×105 Pa .

A) Find the initial mass of oxygen in kg.

B) Find the mass of oxygen that has leaked out in kg.

Answer 1

Given,

Volume of tank = 7.30 X 10^-2 m³

Initial gauge pressure, P₁ = 3.15 X 10⁵ Pa : 1 atm = 1.01325 X 10⁵ Pa

Absolute pressure = Gauge Pr. + Atmospheric Pr. = 3.15 X 10⁵ +1.01325 X 10⁵ = 4.16325 X 10⁵ Pa

Initial temperature = 38.1 ⁰C = (38.1 + 273.16) K = 311.26 K

Final gauge pressure = 1.8 X 10⁵ Pa ; Final absolute pressure = 2.81325 X 10⁵ Pa

Final temperature = 24⁰C = 24+273.16 K = 297.16 K

A) Using the ideal gas equation,

P₁V = n₁RT_1, where R=8.314 J/mol.K

4.16325 X 10⁵ X 7.3 X 10^-2 = n₁ X 8.314 X 311.26  

n₁ = 11.74 mol

Initial mass of oxygen = 11.74 X 32 = 375.68 g. = 0.37568 kg (Ans.)

B) Again using ideal gas equation,

P₂V = n₂ X R X T₂

2.81325 X 10⁵ X 7.3 X 10^-2 = n₂ X 8.314 X 297.16

n₂ = 8.31 mol ,  

Final mass of oxygen , m₂ = 8.31 X 32 = 266 g

Mass of oxygen that has leaked out = m₁ - m₂ = 375.68 - 266 = 109.68 g = 0.10968 kg. (Ans)