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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? - Chemistry

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?

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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,

V1 = V

T1 = 27°C = 300 K

V2 =?

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`

`= 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`

Concept: The Gas Laws - Charles’ Law (Temperature - Volume Relationship)
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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 5 States of Matter
Q 11 | Page 153
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