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 the round bottomed flask be V.
Then, the volume of air inside the flask at 27° C is V.
V1 = V
T1 = 27°C = 300 K
T2 = 477° C = 750 K
According to Charles’s law,
`V_1/T_1 = V_2/T_2`
`=> V_2 = (V_1T_2)/T_1`
= 2.5 V
Therefore, volume of air expelled out = 2.5 V – V = 1.5 V
Hence, fraction of air expelled out = `(1.5 V)/(2.5 V) = 3/5`
Suppose volume of vessel = V `cm^3`
i.e volume of air in the flask at `27^@ C = V cm^3`
`V_1/T_1 = V_2/T_2 "i.e" V/300 = V_2/750 or V_2 =- 2.5 V`
:. Volume expelled = 2.5 V - V = 1.5 V
`:. "Fraction of air expelled" = (1.5V)/(2.5 V) = 3/5`
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- The Gas Laws - Charles’ Law (Temperature - Volume Relationship)