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

Correct Answer: B

Solution :

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