The temperature of an open room of volume 30 m^3 increases from 17^@C to 27^@C due to the sunshine. The atmospheric pressure in the room remains 1 times 10^5Pa. IF n_i and n_f are the number of molecules in the room before and after heating then n_f-n_i will be

The temperature of an open room of volume 30 m^3 increases from 17^@C

1.	The temperature of an open room of volume 30 m^3 increases from 17^@C to 27^@C due to the sunshine. The atmospheric pressure in the room remains 1 times 10^5Pa. IF n_i and n_f are the number of molecules in the room before and after heating then n_f-n_i will be
Answer» <html><body><p>`-1.61 <a href="https://interviewquestions.tuteehub.com/tag/times-1420471" style="font-weight:bold;" target="_blank" title="Click to know more about TIMES">TIMES</a> 10^23`<br/>`1.38 times 10^23`<br/>`2.5 times 10^25`<br/>`-2.5 times 10^25`</p>Solution :IF `n_1` and `n_2` are the <a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> of moles an <a href="https://interviewquestions.tuteehub.com/tag/initial-499269" style="font-weight:bold;" target="_blank" title="Click to know more about INITIAL">INITIAL</a> and final temperature inside the room <a href="https://interviewquestions.tuteehub.com/tag/respectively-1186938" style="font-weight:bold;" target="_blank" title="Click to know more about RESPECTIVELY">RESPECTIVELY</a> then <br/> `n_1=(p_1V_1)/(RT_1)=(10^5times30)/(8.31times(273+17))`<br/> `=1.245 times10^3` <br/>`n_2=(p_2V_2)/(RT_2)=(10^5times30)/(8.31times(273+27))`<br/> `=1.203 times10^3` <br/>Therefore `n_f-n_i=(n_2-n_1) times 6.023 times 10^23` <br/> `=-2.5 times 10^25`</body></html>