Question

The enthalpy of vaporisation of a substance is 8400 J mol−1 and its boiling point is −173.15∘C. The entropy change for vaporisation is:

Answer 1

Explanation:
Given: Enthalpy of vaporisation of a substance is: 8400 J mol−1Temperature or boiling point of substance is: −173.15∘C We have temperature in Kelvin,= Celsius temperature +273.15∘C=−173.15∘C+273.15∘C=100 KNow, we know,Change in entropy, ΔS=qrevT=(8400100)=84 Jmol−1K−1 Hence, the correct answer is '84'.