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?

Asked by Pragya Singh | 1 year ago |  95

##### Solution :-

Let the volume of the container be V.

The volume of the air inside the container at 27°C is V.

Now,

V1 = V

T1 = 27°C= 300 K

V2 = ?

T2 = 477°C = 750 K

Acc to Charles’s law,

$$\dfrac{V_{1}}{T_{1}}​​ = \dfrac{V_{2}}{T_{2}}$$

$$V_2 = \dfrac{V_{ 1 }T_{ 2 }}{T_{ 1 }}$$

=$$\dfrac{750V }{300 }$$

= 2.5 V

Therefore, volume of air expelled out

= 2.5 V – V = 1.5 V

Hence, fraction of air expelled out

$$\dfrac{1.5V }{ 2.5V }$$

$$\dfrac{3 }{ 5 }$$

Answered by Abhisek | 1 year ago

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