During an experiment, an idea/gas is found to obey an additional law VP^(2) = constant . The gas is initially at temperature T and volume V. What will be the temperature of the gas when it expands to a volume 2 V?

During an experiment, an idea/gas is found to obey an additional law V

1.	During an experiment, an idea/gas is found to obey an additional law VP^(2) = constant . The gas is initially at temperature T and volume V. What will be the temperature of the gas when it expands to a volume 2 V?
Answer» <html><body><p><<a href="https://interviewquestions.tuteehub.com/tag/p-588962" style="font-weight:bold;" target="_blank" title="Click to know more about P">P</a>></p>Solution :According to the given problem `VP^(2)` =<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>. constant. So the gas equation <a href="https://interviewquestions.tuteehub.com/tag/pv-593601" style="font-weight:bold;" target="_blank" title="Click to know more about PV">PV</a> = nRT in the light of above (eliminating P) <a href="https://interviewquestions.tuteehub.com/tag/yields-1465159" style="font-weight:bold;" target="_blank" title="Click to know more about YIELDS">YIELDS</a> <br/> `((K)/(sqrt(V))) V = nRt , `i.e., `sqrt(V)= (nR)/(K ) T ` <br/> `therefore sqrt((V_(1))/(V_(2))) = ((T_(1))/(T_(2)))," i.e., " sqrt((V)/(2V)) = (T)/(T^(1))` <br/> or `T^(1) = (sqrt(2))T `</body></html>