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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is: Atomic masses: He = 4 u, O = 16 u)   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 300 K                    

    B) 600 K

    C) 1200 K                  

    D) 2400 K

    Correct Answer: D

    Solution :

                    \[{{V}_{rms}}=\sqrt{\frac{3RT}{M}}\] \[{{V}_{rms({{O}_{2}})}}={{V}_{rms(He)}}\] \[\sqrt{\frac{3R{{T}_{{{O}_{2}}}}}{{{M}_{{{O}_{2}}}}}}=\sqrt{\frac{3R{{T}_{He}}}{{{M}_{He}}}}\]or\[\frac{{{T}_{{{O}_{2}}}}}{{{M}_{{{O}_{2}}}}}=\frac{{{T}_{He}}}{{{M}_{He}}}\] \[\therefore \]\[{{T}_{{{O}_{2}}}}=\frac{300\times 32}{4}=2400K\]


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