question_answer

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature \[{{T}_{0}},\] while box B contains one mole of helium at temperature (7/3)\[{{T}_{0}}\]. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases,\[{{T}_{f}}\]in terms of\[{{T}_{0}}\]is     AIEEE  Solved  Paper-2006

A) \[{{T}_{f}}=\frac{3}{7}{{T}_{0}}\]                              

B) \[{{T}_{f}}=\frac{7}{3}{{T}_{0}}\]              

C)        \[{{T}_{f}}=\frac{3}{2}{{T}_{0}}\]              

D)        \[{{T}_{f}}=\frac{5}{2}{{T}_{0}}\]

Correct Answer: C

Solution :

Here, change in internal energy of the system is zero. i.e., increase in internal energy of one is equal to decrease in internal energy of other. \[\Delta {{U}_{A}}=1\times \frac{5R}{2}({{T}_{f}}-{{T}_{0}})\] \[\Delta {{U}_{B}}=1\times \frac{3R}{2}\left( {{T}_{f}}-\frac{7}{3}{{T}_{0}} \right)\] Now,     \[\Delta {{U}_{A}}+\Delta {{U}_{B}}=0\] \[\Rightarrow \]\[\frac{5R}{2}({{T}_{f}}-{{T}_{0}})+\frac{3R}{2}\left( {{T}_{f}}-\frac{7{{T}_{0}}}{3} \right)=0\] \[\Rightarrow \]\[5{{T}_{f}}-5{{T}_{0}}+3{{T}_{f}}-7{{T}_{0}}=0\] \[\Rightarrow \]\[8{{T}_{f}}=12{{T}_{0}}\] \[\Rightarrow \]\[{{T}_{f}}=\frac{2}{8}{{T}_{0}}=\frac{3}{2}{{T}_{0}}\]