2. Solved Papers
  3. JIPMER

JIPMER Jipmer Medical Solved Paper-2009

  • question_answer
    The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by\[7{}^\circ C\]. The gas is \[\left( R=8.3\,J\,mo{{1}^{-1}}{{K}^{-1}} \right)\]

    A)  diatomic

    B)  triatomic

    C)  a mixture of monoatomic and diatomic

    D)  monoatomic

    Correct Answer: A

    Solution :

    For adiabatic process, \[dQ=0\] So,          \[dU=-\Delta W\] \[\Rightarrow \]               \[n{{C}_{V}}dT=+146\times {{10}^{3}}J\] \[\Rightarrow \]               \[\frac{nfR}{2}\times 7=146\times {{10}^{3}}\] [\[f\to \]Degree of freedom] \[\Rightarrow \]               \[\frac{{{10}^{3}}\times f\times 8.3\times 7}{2}=146\times {{10}^{3}}\] \[f=5.02\approx 5\] So, it is a diatomic gas.


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