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JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    Initially, the root mean square (rms) velocity of \[{{N}_{2}}\]molecules at certain temperature is u. If this temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms, then the new rms velocity will be :   JEE Main Online Paper (Held On 10 April 2016)

    A) 2 u         

    B) 14 u

    C) u / 2                      

    D) 4 u

    Correct Answer: A

    Solution :

                    \[rms({{N}_{2}})=\sqrt{\frac{3RT}{{{M}_{{{N}_{2}}}}}}=\sqrt{\frac{3RT}{28}}=U\]                              ?(1) After dissociation \[rms(N)=\sqrt{\frac{3R(2T)}{{{M}_{N}}}}=\sqrt{\frac{3R(2T)}{14}}=2u\]                                


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