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Thermodynamics

Question
CBSEENCH11006254

When one mole of a solid ‘S’ (molecular weight 46) at its melting point is changed into the liquid at the same temperature, an entropy change of 26.4 JK–1 takes place. Calculate the melting point of the solid if its enthalpy of fusion is 0 .1 kJg–1.

Solution

 Here space space increment straight S space equals space 26.4 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent
space increment straight H subscript straight f space equals space 0.1 space cross times space 10 to the power of negative 3 end exponent space cross times space 46 space straight J space mol to the power of negative 1 end exponent
space space space space space space space space space space space equals 4600 space straight J space mol to the power of negative 1 end exponent
We space know comma space space space increment straight S subscript straight f space equals space fraction numerator increment straight H subscript straight f over denominator straight T end fraction
or space space space space space space space space space space straight T space equals space fraction numerator increment straight H subscript straight f over denominator increment straight S subscript straight f end fraction
or space space space space space space space space space space straight T space equals space fraction numerator 4600 space straight J space mol to the power of negative 1 end exponent over denominator 26.4 space JK to the power of negative 1 end exponent space mol to the power of negative 1 end exponent end fraction
space space space space space space space space space space space space space space space equals space 174.24 space straight K. space space space space space space space space space