Question

A flask containing 250 mg of air at `27^(@)C` is heated till 25.5% of air by mass is expelled from it. What is the final temperatuer of the flask ?

Answer 1

Let the volume of flask be V. Let final temeperature be T(K).
Mass of air expelled at `T(K)=(250x25)/(100)=62.5mg`
Mass of air contained in flask `=250-62.5=187.5mg`
now, volume of total air (250 mg) at higher temperature
`V_(2)=(Vxx250)/(187.5)mL`
Now `(V_(2))/(T)=(V)/(300)`
or `T=(V_(2)xx300)/(V)=(Vxx250xx300)/(187.5xxV)=400K` or `127^(@)C`