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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    At which temperature nitrogen under \[1.00\,\,atm\] pressure has the same root mean square speed as that of\[C{{O}_{2}}\]at\[STP\]

    A) \[{{0}^{o}}C\]                                   

    B)                        \[{{27}^{o}}C\]

    C)                        \[-{{99}^{o}}C\]

    D)                        \[-{{200}^{o}}C\]

    Correct Answer: C

    Solution :

                    Let RMS speed of nitrogen at\[T\,\,\,K\]be\[u\]and is equal to that of\[C{{O}_{2}}\]at\[STP\] \[u=\sqrt{\frac{3RT}{28}}=\sqrt{\frac{3R\times 273}{44}}\] \[T=\frac{273\times 28}{44}=173.73K=-{{99.27}^{o}}C\]


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