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JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • question_answer
    A gas molecule of mass M at the surface of the Earth has kinetic energy equivalent to \[0{}^\circ C\]. If it were to go up straight without colliding with any other molecules, how high it would rise? Assume that the height attained is much less than radius of the earth. (\[{{k}_{B}}\] is Boltzmann constant).     JEE Main Online Paper (Held On 19 April 2016)

    A) 0                                             

    B) \[\frac{273{{k}_{B}}}{2Mg}\]

    C) \[\frac{546{{k}_{B}}}{3Mg}\]                      

    D) \[\frac{819{{k}_{B}}}{2Mg}\]

    Correct Answer: D

    Solution :

    Kinetic energy of each molecule, \[K.E.=\frac{3}{2}{{K}_{B}}T\] In the given problem, Temperature, T = 0°C = 273 K Height attained by the gas molecule, h = ? \[K.E.=\frac{3}{2}{{K}_{B}}(273)=\frac{819{{K}_{B}}}{2}\] \[K.E.=P.E.\] \[\Rightarrow \]\[\frac{819{{K}_{B}}}{2}=Mgh\]or\[h=\frac{819{{K}_{B}}}{2Mg}\]


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