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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 ?

Answer» <html><body><p></p>Solution :Here vessel is not changed hence the pressure remain <a href="https://interviewquestions.tuteehub.com/tag/constant-930172" style="font-weight:bold;" target="_blank" title="Click to know more about CONSTANT">CONSTANT</a> as atmoshperic pressure.<br/>Initial <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> `= <a href="https://interviewquestions.tuteehub.com/tag/27-298706" style="font-weight:bold;" target="_blank" title="Click to know more about 27">27</a>^(@)C = (27+273)=300 K=T_(1)`<br/>Let Initial volume of gas (Vessel) `= <a href="https://interviewquestions.tuteehub.com/tag/v-722631" style="font-weight:bold;" target="_blank" title="Click to know more about V">V</a> dm^(<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>)`<br/>Tempe. at the end `= (477^(@)C+273^(@)C)=750 K = T_(2)`<br/>The volume of the gas in vessel `= V_(2)`<br/>As per Charle.s law `(V_(1))/(T_(1))=(V_(2))/(T_(2))`<br/>`therefore V_(2)=(V dm^(3)xx750K)/(300 K)=2.5V dm^(3)`<br/>The air expelled out `= (V_(2)-V_(1))=(2.5 V-V)dm^(3)`<br/>`= 1.5 V`<br/>Fraction of the air expelled out `(1.5V)/(2.5V)=(3)/(5)=0.6`</body></html>


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