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Haryana PMT Haryana PMT Solved Paper-2005

  • question_answer
    Two perfect gases at absolute temperatures \[{{T}_{1}}\] and \[{{T}_{2}}\]are mixed. There is no loss of energy. The temperature of mixture if masses of molecules are \[{{m}_{1}}\] and \[{{m}_{2}}\] and the number of molecules in the gases are \[{{n}_{1}}\] and \[{{n}_{2}}\] respectively, is :

    A) \[\frac{{{T}_{1}}+{{T}_{2}}}{2}\]                                

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

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

    D) \[\sqrt{{{T}_{1}}{{T}_{2}}/{{n}_{1}}{{n}_{2}}}\]

    Correct Answer: B

    Solution :

                    We know \[E=\frac{3}{2}nkT\]k = Boltzmams constant \[\therefore \]  \[\frac{3}{2}{{n}_{1}}k{{T}_{1}}+\frac{3}{2}{{n}_{2}}k{{T}_{2}}=\frac{3}{2}({{n}_{1}}+{{n}_{2}})kT\]                 \[T=\frac{{{n}_{1}}{{T}_{1}}+{{n}_{2}}{{T}_{2}}}{({{n}_{1}}+{{n}_{2}})}\]


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