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NEET Sample Paper NEET Sample Test Paper-20

  • question_answer
    Two identical rigid containers joined by a small pipe initially contain the same gas at pressure \[{{P}_{o}}\]and absolute temperature \[{{T}_{o}}\]One container is now maintained at the same temperature while the other is heated to \[2{{T}_{o}}.\]The common pressure of the gases will be

    A) \[\frac{3}{2}{{P}_{0}}\]                             

    B) \[\frac{4}{3}{{P}_{0}}\]                 

    C)      \[\frac{5}{3}{{P}_{0}}\]                 

    D)      \[2{{P}_{0}}\]    

    Correct Answer: B

    Solution :

    Let \[{{V}_{0}}\]be the volume of each container is \[n=\frac{{{P}_{0}}{{V}_{0}}}{R{{T}_{0}}}\] Finally, number of moles in the two containers are \[{{n}_{1}}=\frac{{{P}_{0}}{{V}_{0}}}{R{{T}_{0}}}\]and \[{{n}_{2}}=\frac{P{{V}_{0}}}{R(2{{T}_{0}})}\] Total number of moles remain the same \[\Rightarrow \]\[{{n}_{1}}+{{n}_{2}}=n+n\] \[\Rightarrow \]\[\frac{P{{V}_{0}}}{R{{T}_{0}}}+\frac{P{{V}_{0}}}{2R{{T}_{0}}}=\frac{2{{P}_{0}}{{V}_{0}}}{R{{T}_{0}}}\]\[\therefore \]\[P=\frac{4}{3}{{P}_{0}}\] Hence, the correction option is (b).


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