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Manipal Medical Manipal Medical Solved Paper-2009

  • question_answer
    An electron tube was sealed off during manufacture at a pressure of\[1.2\times {{10}^{-7}}\]mm of mercury at\[27{}^\circ C\]. Its volume is\[100\text{ }c{{m}^{3}}\]. The number of molecules that remain in the tube is

    A)  \[2\times {{10}^{16}}\]        

    B)  \[3\times {{10}^{15}}\]

    C)  \[3.86\times {{10}^{11}}\]      

    D)  \[5\times {{10}^{11}}\]

    Correct Answer: C

    Solution :

    From \[pV=nRT\] \[\Rightarrow \] \[(1.2\times {{10}^{-10}}\times 13.6\times {{10}^{3}}\times 9.8)\times {{10}^{-4}}\] \[=n\times 8.31\times (273+27)\] \[\therefore \]number of moles \[n=6.42\times {{10}^{-13}}\] Hence, number of molecules \[=6.42\times {{10}^{-13}}\times 6.02\times {{10}^{23}}\] \[=3.86\times {{10}^{11}}\]


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