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JEE Main & Advanced JEE Main Solved Paper-2016

  • question_answer
    Two closed bulbs of equal volume(V) containing an ideal gas initially at pressure \[{{p}_{i}}\] and temperature \[{{T}_{1}}\]are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to \[{{T}_{2}}\]The final pressure \[{{p}_{f}}\] is :- [JEE Main Solved Paper-2016 ]

    A) \[2{{p}_{i}}\left( \frac{{{T}_{1}}{{T}_{2}}}{{{T}_{1}}+{{T}_{2}}} \right)\]   

    B) \[{{p}_{i}}\left( \frac{{{T}_{1}}{{T}_{2}}}{{{T}_{1}}+{{T}_{2}}} \right)\]

    C) \[2{{p}_{i}}\left( \frac{{{T}_{1}}}{{{T}_{1}}+{{T}_{2}}} \right)\]                    

    D) \[2{{p}_{i}}\left( \frac{{{T}_{2}}}{{{T}_{1}}+{{T}_{2}}} \right)\]

    Correct Answer: D

    Solution :

                    Initial moles and final moles are equal \[{{({{n}_{T}})}_{i}}={{({{n}_{T}})}_{f}}\] \[\frac{{{P}_{i}}V}{R{{T}_{1}}}+\frac{{{P}_{i}}V}{R{{T}_{1}}}=\frac{{{P}_{f}}V}{R{{T}_{1}}}+\frac{{{P}_{f}}V}{R{{T}_{2}}}\] \[2\frac{{{P}_{i}}}{{{T}_{1}}}=\frac{{{P}_{f}}}{{{T}_{1}}}+\frac{{{P}_{f}}}{{{T}_{2}}}\]                          \[{{P}_{f}}=\frac{2{{P}_{i}}{{T}_{2}}}{{{T}_{1}}+{{T}_{2}}}\]


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