RAJASTHAN ­ PET Rajasthan PET Solved Paper-2004

  • question_answer
    At normal temperature for the diatomic gas, the number of degree of freedom is

    A)  7               

    B)  6

    C)  5               

    D)  4

    Correct Answer: C

    Solution :

     \[\gamma =1+\frac{2}{f}\] For diatomic gas \[\gamma =\frac{{{C}_{P}}}{{{C}_{V}}}=\frac{7/2}{5/2}=\frac{7}{5}\] \[\frac{7}{5}=1+\frac{2}{f}\] \[f=5\]


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