• 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 term of ${{T}_{0}}$ is A) ${{T}_{f}}=\frac{7}{3}{{T}_{0}}$ B) ${{T}_{f}}=\frac{3}{2}{{T}_{0}}$ C) ${{T}_{f}}=\frac{5}{2}{{T}_{0}}$ D) ${{T}_{f}}=\frac{3}{7}{{T}_{0}}$

 [b] When two gases are mixed together then Heat lost by the Helium gas = Heat gained by the Nitrogen gas ${{\mu }_{B}}\times {{({{C}_{v}})}_{He}}\times \left( \frac{7}{3}{{T}_{0}}-{{T}_{f}} \right)={{\mu }_{A}}\times {{({{C}_{v}})}_{N2}}\times ({{T}_{f}}-{{T}_{0}})$ $1\times \frac{3}{2}R\times \left( \frac{7}{3}{{T}_{0}}-{{T}_{f}} \right)=1\times \frac{5}{2}R\times ({{T}_{0}}-{{T}_{f}})$ By solving we get ${{T}_{f}}=\frac{3}{2}{{T}_{0}}$