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  3. Chemistry
  4. Molar heat of vaporisation of a liquid is 6 kJ mole -1. If the entropy change is 16 J mole -1 K -1, the boiling point of the liquid is

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Q. Molar heat of vaporisation of a liquid is $6\, kJ\, mole ^{-1}$. If the entropy change is $16\, J\, mole ^{-1} K ^{-1}$, the boiling point of the liquid is

JIPMERJIPMER 2013Thermodynamics

Solution:

$\Delta S=16 \,J mol ^{-1} K ^{-1}, \Delta H_{v}=6 kJ mol ^{-1}$

$T_{ b . p .}=\frac{\Delta H_{\text {vapour }}}{\Delta S_{\text {vapour }}}=\frac{6 \times 1000}{16}=375 K$