If â–³Hvaporisation of substance X(l) (molar mass : 30 g/mol) is 300 J/g at it's boiling point 300 K, then molar entropy change for reversible condensation process is
A
30J/molK−1
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B
−60J/molK−1
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C
−30J/molK−1
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D
None of these
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Solution
The correct option is C−30J/molK−1 Vapurisation and condensation are reversible process, Enthalpy of vapourisation = 300J/g Enthalp of condensation = −300J/g Molar enthalpy of condensation = −300×30g/mol Molar entropy, △S=△HcondensationT △S=−300×30300 =−30J/mol/K