Oxygen is filled in a closed metal jar of volume 1.0 × 10–3 m3 at a pressure of 1.5 × 105 Pa and temperature 400 K. The jar has a small leak in it. The atmospheric temperature is 300 K. Find the mass of the gas that leaks out by the time the pressure and the temperature inside the jar equalize with the surrounding.
We know ideal gas equation
PV=nRT
Where V= volume of gas
R=gas constant
T=temperature
n=number of moles of gas
P=pressure of gas
Given
Volume inside jar V1= 1.0 × 10–3 m3
Pressure inside jar P1 =1.5 × 105 Pa
Temperature inside jar T1=400K
Pressure of surrounding P2 =1atm=1.0Pa
Temperature of surrounding T2=300K
Let volume of oxygen at T2 and P2 =V2
When jar is in equilibrium with surrounding, temperature and pressure of oxygen gas inside jar will T2 and P2.
Number of moles will be same inside the jar, before and after equilibrium as no new oxygen gas has been added. Just temperature and pressure has been changed. Due to which volume will change.
Assuming there is no leak in jar, applying ideal gas equation before and after equilibrium, we get
Now if we consider leak,
Volume of gas leaked =V2-V1
= (1.125-1)m3
= 1.25m3
If n2 are number of moles leaked out, then
Mass of the gas leaked out =n2molar mass of oxygen molecule
Molar mass of oxygen molecule=32g/mol
Mass of gas leaked out=0.00532=0.16g
The mass of the gas that leaked out =0.16g.