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NEET AIPMT SOLVED PAPER SCREENING 2012

  • question_answer
    Standard enthalpy of vaporisation\[{{\Delta }_{\text{vap}}}{{\text{H}}^{}}\]for water at 100°C .is 40.66 kJ \[\text{mo}{{\text{l}}^{\text{-1}}}\]. The internal energy of vaporisation .of water at 100°C (in kJ \[\text{mo}{{\text{l}}^{\text{-1}}}\]is (Assume water vapour to behave like an ideal gas).

    A) + 37.56                 

    B) - 43.76  

    C)        + 43.76                 

    D)        + 40.66

    Correct Answer: A

    Solution :

    \[{{H}_{2}}O(l)\xrightarrow[{}]{{{100}^{o}}C}{{H}_{2}}O(g)\] \[{{\Delta }_{vap}}{{H}^{\Theta }}={{\Delta }_{vap}}{{E}^{\Theta }}+\Delta {{n}_{g}}RT\] For the above reaction, \[\Delta {{n}_{g}}={{n}_{p}}-{{n}_{r}}=1-0=1\] \[\therefore \]\[40.66\,kJ\,mo{{l}^{-1}}\,={{\Delta }_{vap}}{{E}^{\Theta }}+1\times 8.314\] \[\times {{10}^{-3}}\times 373\] \[{{\Delta }_{vap}}{{E}^{\Theta }}=40.66kJ\,mo{{l}^{-1}}-3.1\,kJ\,mo{{l}^{-1}}\] \[=+37.56\,kJ\,mo{{l}^{-1}}\]


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