question_answer

Two rigid boxes containing different ideal gases are placed on table. Box A contains one mole of nitrogen at temperature \[\frac{q}{4{{\pi }^{2}}{{\varepsilon }_{0}}{{R}^{2}}}\], while box B contains one mole of helium at temperature (7/3) \[\frac{q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\]. The boxes are then put into thermal contact with each other, and hear flows between them until the gases reach a common final temperature (Ignore the heat capacity of boxes). Then, the final temperature of the gases, \[\frac{q}{2\pi {{\varepsilon }_{0}}{{R}^{2}}}\], in terms of \[\mu F\] is

A)  \[\Omega \]     

B)        \[\frac{1}{9}A\]               

C)        \[\frac{1}{0.9}A\]           

D)        \[I\]

Correct Answer: C

Solution :

Key Idea Internal energy of a system remains constant or change in internal energy of the system is zero, The system of two rigid boxes is shown in figure. \[=m\times 2\times {{\left( \frac{\sqrt{2}l}{2} \right)}^{2}}+m\times {{(\sqrt{2}l)}^{2}}\] \[=3m{{l}^{2}}\] Now,      \[\Delta {{U}_{A}}=1\times \frac{5R}{2}({{T}_{f}}-{{T}_{0}})\] \[\Delta {{U}_{B}}=1\times \frac{3R}{2}\left( {{T}_{f}}-\frac{7}{3}{{T}_{0}} \right)\] \[\Delta {{U}_{A}}+\Delta {{U}_{B}}=0\] \[\frac{5R}{2}({{T}_{f}}-{{T}_{0}})+\frac{3R}{2}\left( {{T}_{f}}-\frac{7{{T}_{0}}}{3} \right)=0\]  \[5{{T}_{f}}-5{{T}_{0}}+3{{T}_{f}}-7{{T}_{0}}=0\] \[\Rightarrow \]  \[8{{T}_{f}}=12{{T}_{0}}\]