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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.