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NEET Sample Paper NEET Sample Test Paper-73

  • question_answer
    Normal boiling point of water is 373 K (at 760 mm). Vapour pressure of water at 298 K is 23 mm. If  y   the enthalpy of evaporation is 40.656 kJ/mole, the boiling point of water at 23 mm pressure will be

    A) 250 K               

    B) 294 K

    C) 51.6 K              

    D) 12.5 K

    Correct Answer: B

    Solution :

    Applying Clausius-Clapeyron equation \[\log \frac{{{P}_{2}}}{{{P}_{1}}}=\frac{\Delta {{H}_{V}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}\times {{T}_{2}}} \right]\] \[\log \frac{760}{23}=\frac{40656}{2.303\times 8.314}\left[ \frac{373-{{T}_{1}}}{373T} \right]\] This gives \[{{T}_{1}}\,\,=\, 294.4\,\,K\]


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