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BCECE Engineering BCECE Engineering Solved Paper-2001

  • question_answer
    At 300 K, the ratio of the average molecular kinetic energy of \[\text{U}{{\text{F}}_{\text{6}}}\]to that of \[{{\text{H}}_{2}}\]is:

    A) 2 : 7                       

    B)        7 : 2

    C) 6 : 2                       

    D)        1 : 1

    Correct Answer: D

    Solution :

    Key Idea: Average molecular kinetic energy \[=\frac{3}{2}kT\] Given, T = 300 K for both \[U{{F}_{2}}\]and \[{{H}_{2}}.\] \[\therefore \]Average molecular kinetic energy does not depend on molecular mass of substance. Average molecular kinetic energy of \[{{H}_{2}}\] \[=\frac{3}{2}K\times 300\]                                               ?(i) Average molecular kinetic energy of \[U{{F}_{2}}\] \[=\frac{3}{2}k\times 300\]                                                ?(ii) Dividing Eq. (i) by Eq. (ii) we get 1 : 1


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