JEE Main & Advanced Chemistry States of Matter / पदार्थ की अवस्थाएँ - गैस एवं द्रव Question Bank Molecular speeds

  • question_answer
    The temperature of the gas is raised from \[{{27}^{o}}C\] to \[{{927}^{o}}C\], the root mean square velocity is   [CBSE PMT 1994]

    A)                 \[\sqrt{927/27}\]times the earlier value

    B)                 Same as before

    C)                 Halved

    D)                 Doubled

    Correct Answer: D

    Solution :

               \[{{U}_{2}}={{U}_{1}}\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}={{U}_{1}}\sqrt{\frac{1200}{300}}={{U}_{1}}\times 2\]                                 r.m.s. velocity will be doubled.


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