BCECE Engineering BCECE Engineering Solved Paper-2002

  • question_answer
    Find the temperature of hydrogen gas which has the same velocity as that of oxygen at a temperature of \[0{{\,}^{o}}C:\]

    A) \[273\div 16\text{ }K\]    

    B)        \[~273\times 8\,K\]

    C) \[273\div 32K\]     

    D)        \[273\times 4\,K\]

    Correct Answer: A

    Solution :

    Key Idea: \[{{v}_{rms}}=\sqrt{\frac{3RT}{M}}\] For \[{{O}_{2}}\]               \[T=273K,\,M=32\] \[\therefore \]  \[{{v}_{rms}}({{O}_{2}})=\sqrt{\frac{3R\times 273}{32}}\]                             ?(i) For \[{{H}_{2}}\]\[T=?\,M=2\] \[{{v}_{rms}}\,{{H}_{2}}=\sqrt{\frac{3R\times T}{2}}\] Given \[{{v}_{rms}}{{O}_{2}}={{v}_{rms}}\,{{H}_{2}}\] \[\therefore \]  \[\sqrt{\frac{3R\times 273}{32}}=\sqrt{\frac{3R\times T}{2}}\] or            \[\frac{273}{32}=\frac{T}{2}\] \[\therefore \]  \[T=\frac{273\times 2}{32}=\frac{273}{16}K\]


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