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VMMC VMMC Medical Solved Paper-2014

  • question_answer
    Two rigid boxes containing different ideal gases are placed on 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

    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 :

    The system of two rigid boxes is shown in figure. \[\underset{\text{a}\,\text{mole}\,{{\text{N}}_{\text{2}}}}{\mathop{}}\,\frac{{}}{\text{Thermal}\,\text{contact}}\underset{\text{a}\,\text{mole}\,\text{He}}{\mathop{}}\,\] We have,\[U=\frac{1}{2}nf({{T}_{f}}-{{T}_{i}})\] \[\Delta {{U}_{A}}=1\times \frac{5R}{2}({{T}_{f}}-{{T}_{0}})\] \[\Delta {{U}_{B}}=1\times \frac{3R}{2}\left( {{T}_{f}}-\frac{7}{3}{{T}_{0}} \right)\] Now,\[\Delta {{U}_{A}}+\Delta {{U}_{B}}=0\] \[\frac{5R}{2}({{T}_{f}}-{{T}_{0}})+\frac{3R}{2}\left( {{T}_{f}}-\frac{7{{T}_{0}}}{3} \right)=0\] \[5{{T}_{f}}-5{{T}_{0}}+3{{T}_{f}}-7\,{{T}_{0}}=0\] \[\Rightarrow \]\[8{{T}_{f}}=12{{T}_{0}}\]\[\Rightarrow \]\[{{T}_{f}}=\frac{12}{8}{{T}_{0}}=\frac{3}{2}{{T}_{0}}\]


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