J & K CET Medical J & K - CET Medical Solved Paper-2015

  • question_answer
    The energy per mole per degree of freedom of an ideal gas is

    A)  \[(3/2){{k}_{B}}T\]       

    B)  \[(1/2){{k}_{B}}T\]

    C)  \[(3/2)RT\]       

    D)  \[(1/2)RT\]

    Correct Answer: D

    Solution :

     The energy per mole per degree of freedom of an ideal gas is \[E=\left( \frac{1}{2}{{K}_{B}}T \right)({{N}_{A}})\] where, T = absolute temperature \[{{N}_{A}}=\]Avogadros number \[{{K}_{B}}=\]Boltzmanns constant \[\because \] \[R={{N}_{A}}{{K}_{B}}\] Therefore, \[E=\frac{1}{2}RT\]


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