Question

A student forget to add the reaction mixture to the round bottomed flask at 27°C but instead he placed the flask on the flame. After a lapse of time, he realized his mistake and using a pyrometer he found that the temperature of the flask was 477°C. What fraction of air would have been expelled out?

Answer 1

Let the volume of air in flask at 27°C = 300 K be V₁ and that of the same amount of the gas at 477°C (750 K) be V₂. According to the Charle's law

{V₂}/{T₂} = {V₁}/{T₁}    ....(i)

Volume of the gas expelled out = V₂ - V₁ then fraction of the gas expelled out

= {V₂ - V₁}/{V₂}

= 1 - {V₁}/{V₂}     ....(ii)

From Charle's law we can write

{V₁}/{V₂} = {T₁}/{T₂}

Fraction of the air expelled

= 1 - {T₁}/{T₂} - {T₂ - T₁}/{T₂}

= {750 K - 300 K}/{750 K}

= {450 K}/{750 K} = 3/5 = 0.6

Thus, three fifth (0.6) of the gas is expelled out.