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

1 Answer

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

Related Questions

Critical temperature for carbon dioxide and methane are 31.1°C and – 81.9°C respectively. Which of these has stronger intermolecular forces and why?

Class 11 Chemistry States of Matter View Answer

In terms of Charles’ law explain why -273°C is the lowest possible temperature.

Class 11 Chemistry States of Matter View Answer

A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

Class 11 Chemistry States of Matter View Answer