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NEET NEET SOLVED PAPER 2017

  • question_answer
    A 20 litre container at 400 K contains \[C{{O}_{2}}(g)\]pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the containers is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of \[C{{O}_{2}}\]attains its maximum value, will be                                                                                                                                                                  (Given that:\[SrC{{O}_{3}}(s)=SrO(s)+C{{O}_{2}}(g).\]\[{{K}_{p}}=1.6\,\text{atm})\]

    A)  2 litre   

    B)         5 litre

    C)  10 litre

    D)         4 litre

    Correct Answer: B

    Solution :

                     Max. pressure of \[C{{O}_{2}}=\]Pressure of \[C{{O}_{2}}\] at equilibrium For reaction,\[SrC{{O}_{3}}(s)SrO(s)+C{{O}_{2}}\] \[{{K}_{p}}={{P}_{C{{O}_{2}}}}=1.6\,\text{atm =}\]maximum pressure of \[C{{O}_{2}}\]   Volume of container at this stage, \[V=\frac{nRT}{P}\]                                        ?(i) Since container is sealed and reaction was not earlier at equilibrium \[\therefore \]  \[n=\text{constant}\] \[n=\frac{PV}{RT}=\frac{0.4\times 20}{RT}\]                                                                                       ?(ii) Put equation (ii) in equation (i) \[V=\left[ \frac{0.4\times 20}{RT} \right]\frac{RT}{1.6}=5L\]


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