JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    Two identical containers each of volume \[{{V}_{0}}\] are joined by a pipe of negligible volume. The containers contain identical gases at temperature. To and pressure Po. One container is heated to temperature \[2{{T}_{0}}\] while maintaining the other at the same temperature. The common pressure of the gas is P & n is the number of moles of gas in container at temperature\[2{{T}_{0}}\]. Mark the correct expression for this situation

    A) \[P=2{{P}_{0}}\]                              

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

    C) \[2{{P}_{0}}{{V}_{0}}=3nR{{T}_{0}}\]      

    D) \[3{{P}_{0}}{{V}_{0}}=2nR{{T}_{0}}\]

    Correct Answer: C

    Solution :

    \[{{P}_{0}}{{V}_{0}}={{n}_{0}}R{{T}_{0}}\][Initially for both containers] For container having temperature \[2{{T}_{0}},P{{V}_{0}}=nR\times 2{{T}_{0}}\] For container having temperature \[{{T}_{0}},P{{V}_{0}}=(2{{n}_{0}}-n)R{{T}_{0}}\] As number of moles of gas is conserved, so\[2{{n}_{0}}=3n\]\[\Rightarrow \]\[2\frac{{{P}_{0}}{{V}_{0}}}{R{{T}_{0}}}=3\times \frac{P{{V}_{0}}}{2R{{T}_{0}}}=3n\]\[\Rightarrow \]\[P=\frac{4}{3}{{P}_{0}}\]

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