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JEE Main & Advanced Sample Paper JEE Main Sample Paper-36

  • question_answer
    1/2 mole of helium gas is contained in a container at S.T.P. The heat energy needed to double the pressure of the gas, keeping the volume constant (heat capacity of the gas \[=3\,J{{g}^{-1}}{{K}^{-1}}\]) is

    A)  \[3276\text{ }J\]                       

    B)  \[1638\text{ }J\]

    C)  \[819\text{ }J\]                         

    D)  \[409.5\text{ }J\]

    Correct Answer: B

    Solution :

     \[{{T}_{1}}=273K,\,\,{{T}_{2}}=2\times 273\] (on doubling P at const V, T doubles) \[\Delta T=273K,\,Q=n{{C}_{v}}dt,\] \[ms=3J/g/K,n{{C}_{v}}=3\]\[\Rightarrow \] \[\frac{1}{4}\times {{C}_{V}}=3\] \[\therefore \]    \[{{C}_{V}}=12\,J/mol/K,\,\,n'{{C}_{V}}=\frac{1}{2}\times 12=6\] \[Q=n'{{C}_{V}}dt=6\times 273=1638J\]


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