JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Evening)

  • question_answer
    Two kg of a monoatomic gas is at a pressure of \[4\times {{10}^{4}}N/{{m}^{2}}\]. The density of the gas is \[8\text{ }kg/{{m}^{3}}\]. What is the order of energy of the gas due to its thermal motion? [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

    A) \[{{10}^{4}}\] J

    B) \[{{10}^{3}}J\]

    C) \[{{10}^{5}}\text{ }J\]

    D)                  \[{{10}^{6}}J\]

    Correct Answer: A

    Solution :

    \[P=\frac{\rho RT}{M}\] \[4\times {{10}^{4}}=\frac{8}{4\times 1.6\times {{10}^{-27}}}~\,\times 8.314\,T\] \[E\,\,=\,\,\frac{3}{2}\,PV\,=\,\frac{3}{2}\times 4\times {{10}^{4}}\,\times \frac{2}{8.4}\,=\,1.5\times {{10}^{4}}\]                                   


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