2. Solved Papers
  3. RAJASTHAN PMT

RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • 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 \[(R=8.3\,J\,mo{{l}^{-1}}\,{{K}^{-1}})\]

    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