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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is \[T{{V}^{x}}=\]constant, then x is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) \[\frac{2}{3}\]                                      

    B) \[\frac{3}{5}\]

    C) \[\frac{2}{5}\]              

    D)                  \[\frac{5}{3}\]

    Correct Answer: C

    Solution :

    For an adiabatic process, \[T{{V}^{\gamma -1}}=\]constant For a diatomic molecule,\[\gamma =\frac{{{C}_{P}}}{{{C}_{V}}}=\frac{7}{5}\] So, \[x=\gamma -1=\frac{7}{5}-1=\frac{2}{5}\]


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