Question 11

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

Answer

Let the volume of  air in the  flask at 27°C be V1 & that of the same amount of the gas at 477°C be V2.

According to charle’s law

V2/T2  =  V1/ T1   …………………(1)

NOW  volume of gas expelled out = V2 – V1, THEN

Fraction of the gas expelled out = (V2 – V1 ) / V2 = 1- (V1/ V2)     …………….(2)

Also from equation (1)  V1/ V2 = T1/ T2   …………..(3)

Substituting the values of (3) in (2), we get

Fraction of the air expelled  = 1- T1/ T2 = (T2 – T1)/ T2

 = 750- 300 /750 = 0.6

Hence, fraction of air expelled out is 0.6 or 3/5 th

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