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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» 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` |
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