BHU PMT BHU PMT Solved Paper-2003

  • question_answer
    When the temperature of an ideal gas is increased from\[27{}^\circ C\]to\[927{}^\circ C,\] the kinetic energy will be:

    A)  same                                   

    B)  eight times

    C)  four times         

    D)  twice

    Correct Answer: C

    Solution :

                     Key Idea: First write two different temperature from formula \[r\propto {{n}^{2}}\] \[KE=\frac{3}{2}RT\]and then directed them to find the change in KE \[{{T}_{1}}=27+273=300\text{ }K,\] \[{{T}_{2}}=927+273=1200\text{ }K\] \[K{{E}_{1}}=\frac{3}{2}R{{T}_{1}}=\frac{3}{2}\times R\times 300\]                 \[K{{E}_{2}}=\frac{3}{2}R{{T}_{2}}=\frac{3}{2}\times R\times 1200\]                 \[\frac{K{{E}_{2}}}{K{{E}_{1}}}=\frac{\frac{3}{2}R\times 1200}{\frac{3}{2}R\times 300}=4\] \[\therefore \]KE is increased 4 times.


You need to login to perform this action.
You will be redirected in 3 sec spinner