Question

The enthalpy change for the transition of liquid water to steam is 30kilo joule / mole at 27 ° centigrade. The entropy change for the process would be what ? options are(1) 0.1 joule/kilomole (2) 100 joule / kilo mole (3) 10 joule / kilo mole (4) none of these

Answer 1

 

During phase transformation processes, such as vapourization of liquid water to water vapour, the entropy of the process can be found using the given formula:

ΔS = ΔHvapTwhere,ΔS = Change in entropy of the processΔHvap = Enthapy of vapourization = 40.8 kJ/mol = 40800 J/molT = Temperature at which phase transition occurs = (100+273) K = 373 KSubstituting these values, we get:ΔS = 40800 J/mol373 K=109.38 J/K/molΔS = ΔHvapTwhere,ΔS = Change in entropy of the processΔHvap = Enthapy of vapourization = 40.8 kJ/mol = 40800 J/molT = Temperature at which phase transition occurs = 100+273 K = 373 KSubstituting these values, we get:ΔS = 40800 J/mol373 K=109.38 J/K/mol