JEE Main & Advanced Physics Kinetic Theory of Gases Question Bank Self Evaluation Test - Kinetic Theory

  • question_answer
    Two gases occupy two containers A and B the gas in A, of volume \[0.10{{m}^{3}}\], exerts a pressure of 1.40 MPa and that in B of volume \[0.15{{m}^{3}}\] exerts a pressure 0.7 MPa. The two containers are united by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the container will be (in MPa)

    A) 0.70

    B) 0.98   

    C) 1.40

    D) 210

    Correct Answer: B

    Solution :

    [b] We know that  \[{{P}_{A}}{{V}_{A}}=nRT,{{P}_{B}}{{V}_{B}}={{n}_{B}}RT\]and  \[{{\text{P}}_{f}}({{V}_{A}}+{{V}_{B}})=({{n}_{A}}+{{n}_{B}})\,RT\]       \[{{\text{P}}_{f}}({{V}_{A}}+{{V}_{B}})={{P}_{A}}{{V}_{A}}+{{P}_{B}}{{V}_{B}}\]   \[\therefore {{P}_{f}}=\left( \frac{{{P}_{A}}{{V}_{A}}+{{P}_{B}}{{V}_{B}}}{{{V}_{A}}+{{V}_{B}}} \right)\]  \[=\frac{1.4\times 0.1+0.7\times 0.15}{0.1+0.15}MPa\] \[=0.98MPa\]


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