NEET Physics Wave Mechanics NEET PYQ-Wave Mechanics

  • question_answer
    4.0 g of a gas occupies 22.4 L at NTP. The specific heat capacity of the gas at constant volume is \[5.0J{{K}^{-1}}mo{{l}^{-1}}\]. If the speed of sound in this gas at NTP is \[952\,m{{s}^{-1}}\] then the heat capacity at constant pressure is (Take gas constant\[R=8.3\,J{{K}^{-1}}mo{{l}^{-1}}\])      [NEET (Re) 2015]

    A)  \[8.0\,J{{K}^{-1}}mo{{l}^{-1}}\]

    B)       \[7.5\,\,J{{K}^{-1}}mo{{l}^{-1}}\]

    C)  \[7.0\,\,J{{K}^{-1}}mo{{l}^{-1}}\]    

    D)       \[8.5\,\,J{{K}^{-1}}mo{{l}^{-1}}\]

    Correct Answer: A

    Solution :

    Given, \[M=4\,gm,V=22.4\,L,{{C}_{v}}=5\,J{{K}^{-1}}mo{{l}^{-1}}\]
    \[{{V}_{\text{sound}}}=952m/s,{{C}_{p}}=?\]
    As. velocity of sound, \[{{V}_{sound}}=\sqrt{\frac{\gamma PV}{My}}\]
    \[\Rightarrow \]   \[\gamma =\frac{M}{pV}v_{sound}^{2}=\frac{{{C}_{p}}}{{{C}_{v}}}\]
    so heat capacity at constant pressure,
    \[{{C}_{p}}={{C}_{v}}\left[ \frac{M}{pV} \right]v_{sound}^{2}=5\left[ \frac{4\times {{10}^{-3}}}{{{10}^{5}}\times 22.4\times {{10}^{-3}}} \right]{{(952)}^{3}}\]            \[=\frac{20}{22.4}\times {{(952)}^{2}}\times {{10}^{-5}}\]
    \[=809.200\times {{10}^{-5}}\,=8.09\,\,J/mol\,K\]


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