BHU PMT BHU PMT (Mains) Solved Paper-2011

  • question_answer
    The temperature at which the velocity of oxygen will be half of that of hydrogen at NTP is

    A) \[1092{}^\circ C\]            

    B)  \[1492{}^\circ C\]

    C) \[273K\]                              

    D)  \[819{}^\circ C\]

    Correct Answer: D

    Solution :

                                    \[{{v}_{{{O}_{2}}}}=\frac{1}{2}{{v}_{{{H}_{2}}}}\] \[\therefore \]  \[\sqrt{\frac{3RT}{32}}=\frac{1}{2}\sqrt{\frac{3R\times 273}{2}}\] Or           \[\frac{T}{32}=\frac{273}{8}\] \[\therefore \]  \[T=4\times 273\]                 \[T=1092K=1092-273\] \[T=819{}^\circ C\]


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