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JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    One mole of an ideal gas is expanded isothermally and reversibly to half of its initial pressure. \[\Delta S\]for the process in \[J\,{{K}^{-1}}\,mo{{l}^{-1}}\] is [\[\ell n2=0.693\]and\[R=8.314J/(mol\,K)\]].   JEE Main Online Paper (Held On 26-May-2012)  

    A) 6.76       

    B)                                        5.76

    C)                        10.76                                    

    D)                        8.03

    Correct Answer: B

    Solution :

                    For isothermal process\[(\Delta T=0)\] \[\Delta S=R\ell n\frac{{{P}_{1}}}{{{P}_{2}}}=8.314\ell n2\] \[=8.314\times 0.693\]


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