Haryana PMT Haryana PMT Solved Paper-2011

  • question_answer
    What is  when \[1.00\text{ }mole\]of liquid water vaporises at\[{{100}^{o}}C\]? The heat of vaporization. \[\Delta {{H}^{o}}_{vap}\] of water at \[{{100}^{o}}C\] is \[40.66\text{ }kJ\text{ }mo{{l}^{-1}}\].

    A)  \[36.73\text{ }kJ\text{ }mo{{l}^{-1}}\]

    B)  \[40.52\text{ }kJ\text{ }mo{{l}^{-1}}\]

    C)  \[37.56\text{ }kJ\text{ }mo{{l}^{-1}}\]

    D)  \[42.05\text{ }kJ\text{ }mo{{l}^{-1}}\]

    Correct Answer: C

    Solution :

                    \[{{H}_{2}}O(l){{H}_{2}}O(g)\] \[\Delta {{n}_{g}}=1-0=1\] \[\Delta {{H}^{o}}=\Delta {{U}^{o}}+\Delta {{n}_{g}}RT\] \[40.66=\Delta {{U}^{o}}+1\times 8.314\times {{10}^{-3}}\times 373\] \[40.66=\Delta {{U}^{o}}+3.10\] \[\Delta {{U}^{o}}=40.66-3.10=37.56\,kJ\,\,mo{{l}^{-1}}\]


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