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 ?

Explanation

Here vessel is not changed hence the pressure remain constant as atmoshperic pressure.
Initial temperature `= 27^(@)C = (27+273)=300 K=T_(1)`
Let Initial volume of gas (Vessel) `= V dm^(3)`
Tempe. at the end `= (477^(@)C+273^(@)C)=750 K = T_(2)`
The volume of the gas in vessel `= V_(2)`
As per Charle.s law `(V_(1))/(T_(1))=(V_(2))/(T_(2))`
`therefore V_(2)=(V dm^(3)xx750K)/(300 K)=2.5V dm^(3)`
The air expelled out `= (V_(2)-V_(1))=(2.5 V-V)dm^(3)`
`= 1.5 V`
Fraction of the air expelled out `(1.5V)/(2.5V)=(3)/(5)=0.6`