Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
Solution : Normal boiling point of isopropanol = 82.5 0cel Heat of Condensation for isopropanol at 82.5 0cel = 39.85 kJ/mol Now, Equation for change in entropy is given by, \Delta S=\frac{q_{\text {rev }}}{T} Where, qrev = heat change = -n*Heat of condensation = 1 * 39.85 = -39.85 kJ (n = 1 mol) ( -ve Sign shows heat is liberate during condensation) T = Constant temperature = 82.5 + 273 = 355.5 K Put all values in equation \Delta S=\frac{-39.85}{355.5} \Delta S = - 0.112 kJ / K = -112 J/K Answer: Change in entropy = -112 J/K (-ve sign of entropy shows that entropy is decrease during condensation process) ...