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A student forgot to add the reaction mixture to the round bottomed flask at $27^{\circ} \mathrm{C}$ 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 ${ }^{\circ} \mathrm{C}$. What fraction of air would have been expelled out?
Chemistry 101
Chemistry 102
Chapter 5
States of Matter
Gases
Temperature and the Kinetic Theory of Gases
Solids
Carleton College
Drexel University
Brown University
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Hello students in this question, a student forward to add the reaction mixture to the round bottom flask. At temperature T one equals to 27 degrees integrate. This is equal to 300 Calvin but instead he she placed the flask on the flame and after lapse of time he realized his mistake and the temperature T two has become affordable. Seven degrees integrated. And this temperature it is equal to 750 Calvin. Okay, so now we have to determine what fraction of the air for have been expelled out. So we have to determine the fraction of the volume that is we even by we too okay Or fraction 1-. We even might be too okay. So this is at the constant pressure because this is in the open flus. So pressure is here constant. So from the ideal situation T. V equals too N. RT So this pressure is constant. So we can write that volume will be directly proportional to temperature T. So even by Me Too, this is equal to temperature T even minus Tito. So the fraction this can be calculated by one minus we even by V two, this is equal to one minus Stephen mine by T two. So this fraction, this is the fraction. So supposed calling it a small F so one minus Stephen which is 300 Calvin and T two which is 7 50 Calvin. So from here this fraction Will be equal to the three x 5. So this fraction has been expelled out so this becomes the answer for this question. Okay, thank you.
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