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.


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