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Chhattisgarh PMT Chhattisgarh 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\[\text{(R=8}\text{.3J}\,\text{mo}{{\text{l}}^{\text{-1}}}{{\text{K}}^{\text{01}}}\text{)}\]

    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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