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Thermodynamics

Question
CBSEENCH11006252

You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below:
        Liquid           straight b. straight p. space left parenthesis degree straight C right parenthesis space space space space space space space space space increment straight H subscript vap superscript 0 space end superscript space left parenthesis kJ space mol to the power of negative 1 end exponent right parenthesis
  space left parenthesis straight a right parenthesis space Ethanol with left parenthesis straight C subscript 2 straight H subscript 5 OH right parenthesis below space space space space space space space space space space space space space space space space space space 78.4 space space space space space space space space space 42.4
space left parenthesis straight b right parenthesis space Toluene with left parenthesis straight C subscript 2 straight H subscript 5 CH subscript 3 right parenthesis below space space space space space space space space space space space space space space space space space space 110.6 space space space space space space space 35.2 space space

Solution
left parenthesis straight a right parenthesis space increment straight S subscript vap space equals space fraction numerator increment straight H over denominator straight T subscript straight b end fraction space equals space fraction numerator 42.4 over denominator 273 plus 78.4 end fraction space equals space fraction numerator 42.4 over denominator 351.4 end fraction
space space space space space space space space space space space equals fraction numerator 42.4 over denominator 351.4 end fraction straight k space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent
therefore space space space space space space increment straight S subscript vap space equals space 0.1207 space kJ to the power of negative 1 end exponent mol to the power of negative 1 end exponent
Thus entropy of vapourisation of ethanol
                         = 120.7 JK-1 mol-1

left parenthesis straight b right parenthesis space increment straight S subscript vap space equals space fraction numerator increment straight H over denominator straight T subscript straight b end fraction space equals space fraction numerator 35.2 over denominator 273 plus 110.6 end fraction
space space space space space space space space space space space space space space space equals space 0.0918 space straight k thin space JK to the power of negative 1 end exponent mol to the power of negative 1 end exponent
space space space space space space space space space space space space space space space equals space 91.8 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent