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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 31

  • question_answer
    A box containing N molecules of a perfect gas at temperature \[{{T}_{1}}\] and pressure\[{{P}_{1}}\]. The number of molecules in the box is doubled keeping the total kinetic energy of the gas same as before. If the new pressure is \[{{P}_{2}}\] and temperature \[{{T}_{2}}\], then-

    A) \[{{P}_{2}}={{P}_{1}},{{T}_{2}}={{T}_{1}}\]   

    B)        \[{{P}_{2}}={{P}_{1}},{{T}_{2}}=\frac{{{T}_{1}}}{2}\]

    C) \[{{P}_{2}}=2{{P}_{1}},\text{ }{{T}_{2}}={{T}_{1}}\]

    D)        \[{{P}_{2}}=2{{P}_{1}},\text{ }{{T}_{2}}=\frac{{{T}_{1}}}{2}\]

    Correct Answer: B

    Solution :

    [b] \[{{E}_{total}}=\frac{f}{2}NkT\] \[\Rightarrow \text{ }T\propto \frac{1}{N}\]              \[\left[ \therefore \text{ }{{E}_{total}},\text{ }f,k=constant \right]\] \[\Rightarrow \frac{{{T}_{2}}}{{{T}_{1}}}=\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{1}{2}\Rightarrow {{T}_{2}}=\frac{{{T}_{1}}}{2}\] Also from \[PV=NkT\Rightarrow P\propto NT\] \[\Rightarrow \frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{N}_{1}}}{{{N}_{2}}}.\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{1}{2}\times \frac{2}{1}=\frac{1}{1}\]


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