BCECE Engineering BCECE Engineering Solved Paper-2010

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

    A) \[1092{}^\circ C\]   

    B)                        \[1492{}^\circ C\]

    C) \[273\text{ }K\]               

    D)        \[819{}^\circ C\]

    Correct Answer: D

    Solution :

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


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