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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
    0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume \[10\text{ }{{m}^{3}}\] at 1000 K. Given R is the gas constant in \[J{{K}^{-1}}\,\,mo{{l}^{-1}}\], x is - [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) \[\frac{2R}{4-R}\]

    B) \[\frac{4+R}{2R}\]

    C) \[\frac{4-R}{2R}\]

    D) \[\frac{2R}{4+R}\]

    Correct Answer: C

    Solution :

     \[PV=nRT\] \[200\text{ }\times \text{ }10=\left( 0.5\text{ }+\text{ }x \right)\text{ }R\text{ }\times \text{ }1000\] \[0.5+x=\frac{2}{R}\] \[x\,\,=\,\,\frac{4\,\,-\,\,R}{2\,R}\] Answer should be 3 but NTA has given answer 2. Which is not correct.


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