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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2012

  • question_answer
    If a constant volume gas thermometer records a pressure of 20 kPa at tripple point of water and pressure of 14.3 kPa at the dry ice, then the temperature of dry ice will be

    A)  \[-77.85{}^\circ C\]    

    B)  \[+77.85{}^\circ C\]

    C)  \[-66.05{}^\circ C\]    

    D)  None of these

    Correct Answer: A

    Solution :

     Given,\[{{p}_{T}}=20.0\,kPa\]and\[p=14.3\text{ }kPa\] By the relation, \[T=(273.16K)\left( \frac{p}{{{p}_{T}}} \right)\] \[=(273.16K)\times \left( \frac{14.3kPa}{20.0kPa} \right)=195.3K\] On celsium scale \[\theta ({}^\circ C)=T(K)-273.15K\] \[=195.3-273.15\] \[=-77.83{}^\circ C\]


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