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JCECE Medical JCECE Medical 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 \[{{\text{7}}^{\text{o}}}\text{C}\text{.}\] The gas is\[(R=8.3\,\text{J}\,\text{mo}{{\text{l}}^{-1}}{{K}^{-1}})\]

    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.


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