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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume \[{{V}_{1}}\] and contains ideal gas at pressure \[{{P}_{1}}\], and temperature \[{{T}_{1}}\] The other chamber has volume \[{{V}_{2}}\] and contains ideal gas at pressure \[{{P}_{2}}\] and temperature \[{{T}_{2}}\] If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be  

    A) \[\frac{{{T}_{1}}{{T}_{2}}\left( {{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}} \right)}{{{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}}}\] 

    B) \[\frac{{{P}_{1}}{{V}_{1}}{{T}_{1}}+{{P}_{2}}{{V}_{2}}{{T}_{2}}}{{{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}}}\]

    C) \[\frac{{{P}_{1}}{{V}_{1}}{{T}_{2}}+{{P}_{2}}{{V}_{2}}{{T}_{1}}}{{{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}}}\]

    D)        \[\frac{{{T}_{1}}{{T}_{2}}\left( {{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}} \right)}{{{P}_{1}}{{V}_{1}}{{T}_{1}}+{{P}_{2}}{{V}_{2}}{{T}_{2}}}\]

    Correct Answer: A

    Solution :

    [a] Here \[Q=0\]and \[W=0.\] Therefore from first law of thermodynamics \[\Delta U=Q+W=0\] \[\therefore \]Internal energy of the system with partition = Internal energy of the system without partition. \[{{n}_{1}}{{C}_{v}}{{T}_{1}}+{{n}_{2}}{{C}_{v}}{{T}_{2}}=\left( {{n}_{1}}+{{n}_{2}} \right){{C}_{v}}T\] \[\therefore T=\frac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}}{{{n}_{1}}+{{n}_{2}}}\] \[\text{But }{{n}_{1}}=\frac{{{P}_{1}}{{V}_{1}}}{R{{T}_{1}}}\text{ and }{{n}_{2}}=\frac{{{P}_{2}}{{V}_{2}}}{R{{T}_{2}}}\] \[\therefore T=\frac{\frac{{{P}_{1}}{{V}_{1}}}{R{{T}_{1}}}\times {{T}_{1}}+\frac{{{P}_{2}}{{V}_{2}}}{R{{T}_{2}}}\times {{T}_{2}}}{\frac{{{P}_{1}}{{V}_{1}}}{R{{T}_{1}}}+\frac{{{P}_{2}}{{V}_{2}}}{R{{T}_{2}}}}=\frac{{{T}_{1}}{{T}_{2}}\left( {{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}} \right)}{{{P}_{1}}{{V}_{1}}{{T}_{2}}+{{P}_{2}}{{V}_{2}}{{T}_{1}}}\]


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