JEE Main & Advanced Chemistry States of Matter Question Bank Kinetic molecular theory of gases and Molecular collisions

  • question_answer The average K.E. of an ideal gas in calories per mole is approximately equal to    [EAMCET 1989]

    A)                 Three times the absolute temperature

    B)                 Absolute temperature 

    C)                 Two times the absolute temperature

    D)                 1.5 times the absolute temperature

    Correct Answer: A

    Solution :

               \[K.E.=\frac{3}{2}.RT=\frac{3}{2}.2.T\]         \[\because \,\,R\approx 2cal{{K}^{-1}}mo{{l}^{-1}}\]                                 \[K.E.=3T\]


You need to login to perform this action.
You will be redirected in 3 sec spinner