J & K CET Engineering J and K - CET Engineering Solved Paper-2008

  • question_answer
    For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molecular specific heat \[({{C}_{V}})\] and gas constant (R)

    A)  \[{{C}_{V}}=\frac{R}{2}\]

    B)  \[{{C}_{V}}=R\]

    C)  \[{{C}_{V}}=2R\]

    D)  \[{{C}_{V}}=3R\]

    Correct Answer: D

    Solution :

    From \[{{C}_{V}}=\frac{1}{2}fR=\frac{1}{2}\times 6R=3R\]


You need to login to perform this action.
You will be redirected in 3 sec spinner