question_answer

'n' moles of an ideal gas undergoes a process \[A\to B\]as shown in the figure. The maximum temperature of the gas during the process will be : [JEE Main Solved Paper-2016 ]

A) \[\frac{9{{P}_{0}}{{V}_{0}}}{nR}\]                            

B) \[\frac{9{{P}_{0}}{{V}_{0}}}{4nR}\]

C) \[\frac{3{{P}_{0}}{{V}_{0}}}{2nR}\]                          

D) \[\frac{9{{P}_{0}}{{V}_{0}}}{2nR}\]

Correct Answer: B

Solution :

                T will be max where product of PV is max. equation of line\[P=\frac{-{{P}_{0}}}{{{V}_{0}}}V+3{{P}_{0}}\] \[PV=\frac{-{{P}_{0}}}{{{V}_{0}}}{{V}^{2}}+3{{P}_{0}}V=x\](says) \[\left. \begin{align}   & \frac{dx}{dV}=0\Rightarrow V=\frac{3{{V}_{0}}}{2} \\  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow P=\frac{3{{P}_{0}}}{2} \\ \end{align} \right\}\] here PV product is max. \[\Rightarrow \]\[T=\frac{PV}{nR}=\frac{9}{4}\frac{{{P}_{0}}{{V}_{0}}}{nR}\] Alternate Since initial and final temperature are equal hence maximum temperature is at middle of line. PV = nRT \[\frac{\left( \frac{3{{P}_{0}}}{2} \right)\left( \frac{3{{P}_{0}}}{2} \right)}{nR}={{T}_{\max }}.\Rightarrow \frac{9{{P}_{0}}{{V}_{0}}}{4nR}={{T}_{\max }}.\]