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NEET Sample Paper NEET Sample Test Paper-72

  • question_answer
    The K.E. of N molecules of \[{{O}_{2}}\] is x joules a\[-123{}^\circ C\]. Another sample of \[{{O}_{2}}\text{ }at\text{ }27{}^\circ C\] has a KE of 2x joules. The latter sample contains

    A) \[N\,\,molecules\,\,of\,\,{{O}_{2}}\]

    B) \[2/N\,\,molecules\,of\,\,{{O}_{2}}\]

    C) \[N/2\,\,molecules\,of\,\,{{O}_{2}}\]

    D) \[N/4\,\,molecules\,of\,\,{{O}_{2}}\]

    Correct Answer: A

    Solution :

    \[KE\,\,=\,\,\frac{3}{2}RT;\,\,T=-123+273=+150\,\,K\] \[\frac{3}{2}\times R\times 150=\frac{3}{2}\times 8.314\times 75=x\,J\] \[= 225 \times  8.314 =x\,J\] \[\operatorname{At}\,\,27{}^\circ C=27+223=300K\] \[KE\,\,joules\,\,=2x\,\,joules = \frac{3}{2}\,\,\times \, 8.314 \times  300\] N molecules \[\therefore ~~x joules =3\times 8.314\times 75\] In both the cases x joules correspond to N molecules.


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