JEE Main & Advanced Physics Kinetic Theory of Gases Question Bank Self Evaluation Test - Kinetic Theory

  • question_answer
    The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K the root-mean-square velocity of the gas molecules is v, at 480 K it becomes

    A) 4v

    B) 2v   

    C) v/2 

    D) v/4

    Correct Answer: B

    Solution :

    [b] \[\frac{{{\left( {{V}_{rms}} \right)}_{1}}}{{{\left( {{V}_{rms}} \right)}_{2}}}=\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\Rightarrow \frac{V}{{{\left( {{V}_{rms}} \right)}^{2}}}=\sqrt{\frac{120}{480}}\] \[\Rightarrow \frac{V}{\left( {{V}_{rms}} \right){{ & }_{2}}}=\frac{1}{2}\Rightarrow {{\left( {{V}_{rms}} \right)}_{2}}=2V\]


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