JEE Main & Advanced Sample Paper JEE Main Sample Paper-16

  • question_answer
    At \[{{27}^{o}}C\] if temperature of 1 mol of gas is increased by \[{{50}^{o}}C\] the % change in kinetic energy of system is-

    A)  16.67                                   

    B)  18.57

    C)  85.18                                   

    D)  7.66

    Correct Answer: A

    Solution :

    \[K.E.\propto T\] \[\therefore \]\[\frac{{{\left( KE \right)}_{1}}}{{{\left( KE \right)}_{2}}}=\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{300}{350}\] \[\therefore \]\[{{(KE)}_{2}}=\frac{350}{300}{{(KE)}_{1}}\]% change in K.E. \[=\frac{{{\left( KE \right)}_{2}}-{{\left( KE \right)}_{1}}}{{{\left( KE \right)}_{1}}}\times 100=\frac{350-300}{300}\times 100=16.67%\]


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