CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    The temperature of an ideal gas is reduced from\[927{}^\circ C\]to\[27{}^\circ C\]The rms velocity of the molecules becomes:

    A)  double the initial value

    B)  half of the initial value

    C)  four times the initial value

    D)  ten times the initial value

    E)  \[\sqrt{(927/27)}\]

    Correct Answer: B

    Solution :

    \[{{v}_{rms}}\propto \sqrt{T}\] \[\therefore \]  \[\frac{v_{rms}^{}}{{{v}_{rms}}}=\sqrt{\left( \frac{T}{T} \right)}=\sqrt{\frac{300}{1200}}=\frac{1}{2}\] \[\Rightarrow \]               \[v_{rms}^{}=\frac{{{v}_{rms}}}{2}\]


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