A welder using a tank of volume 7.20 x 10^-2 m^3 fills it with oxygen (with a molar mass of 32.0 g/mol) at a gauge pressure of 3.20 x 10^5 Pa and temperature of 39.0°C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 20.9 °C, the gauge pressure of the oxygen in the tank is 1.85 x 10^5 Pa.

Part A) Find the initial mass of oxygen.

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

a) the initial mass of O₂ is 373.92 gr

b) the mass leaked of O₂ is 104.26 gr

Explanation:

we can assume ideal gas behaviour of oxygen, then we can calculate the mass using the ideal gas equation

P*V = n*R*T,

where P = absolute pressure, V = volume occupied by the gas, n = number of moles of gas, R = ideal gas constant = 8.314 J/mol K, T = absolute temperature

Initially

P = Pg + Pa (1 atm) = 3.20 * 10⁵ Pa + 101325 Pa = 4.21*10⁵ Pa

where Pg = gauge pressure, Pa=atmospheric pressure

T = 39 °C = 312 K

V = 7.20 * 10⁻² m³

therefore

P*V = n*R*T → n = P*V / (R*T)

replacing values

n = P*V / (R*T) = 4.21*10⁵ Pa*7.20 * 10⁻² m³ / (8.314 J/mol K*312 K) = 11.685 mol

since

m = n*M, where m = mass, n = number of moles, M = molecular weight of oxygen

then

m = n*M = 11.685 mol * 32.0 g/mol = 373.92 gr of O₂

therefore the initial mass of O₂ is 373.92 gr

for the part B)

P₂ = Pg₂ + Pa (1 atm) = 1.85*10⁵ Pa + 101325 Pa = 2.86*10⁵ Pa

T₂ = 20.9 °C = 293.9 K

V = 7.20 * 10⁻² m³

therefore

n₂ = P₂*V / (R*T₂) = 2.86*10⁵ Pa*7.20 * 10⁻² m³ / (8.314 J/mol K*293.9 K) = 8.427 mol

m₂ = n*M = 8.427 mol*32.0 g/mol = 269.66 gr of O₂

thus the mass leaked of oxygen is

m leaked = m - m₂ = 373.92 gr - 269.66 gr = 104.26 gr