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?
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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Why did you do V2 - V1 and not V1-V2 as v1 has more volume than v2???
Thank you