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JEE Main & Advanced Sample Paper JEE Main Sample Paper-3

  • 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\[=3J{{g}^{-1}}{{K}^{-1}})\]is

    A)  3276 J                  

    B)  1638 J

    C)  819 J                    

    D)  409.5 J

    Correct Answer: B

    Solution :

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


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