Multiple choice

A student forgot to add the reaction mixture to the round-bottomed flask at $27^o$C but put it on the flame. After a lapse of time, he realized his mistake. Using a pyrometer, he found the temperature of the flask as $477^o$C. What fraction of air would have been expelled out?

  1. $5/3$
  2. $3/5$
  3. $2/5$
  4. $3/2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using Charles's Law at constant pressure, V1/T1 = V2/T2. Temperatures must be in Kelvin: T1 = 27+273 = 300K, T2 = 477+273 = 750K. V2 = V1 * (750/300) = 2.5 * V1. The volume of air expelled is V2 - V1 = 1.5 * V1. The fraction expelled is (1.5 * V1) / (2.5 * V1) = 1.5/2.5 = 3/5.

AI explanation

Applying Charles's law, the volume of a gas at constant pressure is directly proportional to its absolute temperature. Converting the initial and final temperatures from Celsius to Kelvin gives initial values of 300 K and final values of 750 K. The fraction of air remaining in the flask is the initial volume divided by the final volume, which is 300 divided by 750, equaling 2/5. The fraction of air expelled out is the total minus this remaining fraction, calculated as 1 minus 2/5, which equals 3/5. The fraction of air expelled out is 3/5.

Keep practicing — related questions