Question

A welder using a tank of volume 7.90×10⁻² m3 fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.30×10⁵ Pa and temperature of 35.0 ∘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 21.3 ∘C, the gauge pressure of the oxygen in the tank is 1.60×10⁵ Pa .

Part A

Find the initial mass of oxygen. (kg)

Part B

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

Answer 1

Given that

Tank volume V1 = 7.90* 10^-2  m^3

as per question

P1 = Gauge pressure  + 1 atm = 3.30*10^5 + 1.013*10^5 = 4.313*10^5 Pa

Temperature T1 = 35+ 273.15 = 308.15 K

T2 = 21.3+273.15 = 294.45 K

P2 = 1.60*10^5 + 1.013*10^5 = 2.613*10^5 Pa

We know that ideal gas equation

PV = nRT

P1V1= n1RT1

n1 = P1V1/ RT1

where Gas constant R = 8.314 J/mol.K

n1 = (4.313*10^5)(7.90* 10⁻² ) / (8.314*308.15)

n1 = 13.29 mol

Now initial mass of oxygen will be

Mi = Mm*n1

Mi = 32* 13.29

Mi = 425.28 g -------------------------answer 1 part

part (2)

mass of oxygen that leaked out (M)

P2V2 = n2RT2

n2 = P2V2 / RT2

n2 = (2.613*10^5)( 7.90* 10^-2) / (8.314*294.45)

n2 = 8.43 mol

final mass of oxygen

Mf = Mm*n2

Mf = 32*8.43

Mf = 269.76 g

Now mass of oxygen that leaked out (M) = Mi - Mf

M = 425.28 - 269.76 =  155.52 g

M = 0.1555 kg

Good Luck !!