Question

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.

Answer 1

Answer:

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