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?

#### Solution 1

Let the volume of the round bottomed flask be *V.*

Then, the volume of air inside the flask at 27° C is *V.*

Now,

*V*_{1} = *V*

*T*_{1} = 27°C = 300 K

*V*_{2} =?

*T*_{2} = 477° C = 750 K

According to Charles’s law,

`V_1/T_1 = V_2/T_2`

`=> V_2 = (V_1T_2)/T_1`

`= 750V/300`

= 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`

#### Solution 2

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`