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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 45

  • question_answer
    Figure shows a cylinder \[1\text{ }m\]long with a thin massless piston clamped in such a way that it divides the cylinder into two equal parts. The whole system is placed in a large water bath maintained at temperature\[300\text{ }K\]. The wall of the cylinders is highly conducting. The left side of the cylinder contain 1 mole of helium gas at pressure \[4\times {{10}^{5}}Pa\] and the right side also contain helium gas at pressure\[{{10}^{5}}Pa\]. Now the piston is released. The piston finally attains the equilibrium position. How much heat will be transmitted by the water bath in the process? Take \[R=8.3\text{ }J/mole\text{ }K,\]ln \[(1.6)=0.47\] and ln \[(0.4)=-0.916\]]

    A) \[200\,J\]          

    B)        \[400\,J\]

    C) \[600\,J\]          

    D)        \[800\,J\]

    Correct Answer: C

    Solution :

    [c] The process is isothermal. \[{{V}_{left}}=\frac{4{{V}_{0}}}{5}\] and \[{{V}_{right}}=\frac{{{V}_{0}}}{5}\] Total heat supplied = Work done by the gas \[\Delta W=\int{\frac{nRT}{V}\,dV={{n}_{left}}}RT\ell n\left( \frac{4{{V}_{0}}}{5\frac{{{V}_{0}}}{2}} \right)={{n}_{Right}}RT\ell n\left( \frac{{{V}_{0}}}{5\frac{{{V}_{0}}}{2}} \right)=600J\]


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