Prove that in a reversible process:
∆(system) + ∆S(surroundings) = 0
Or
Prove that there is no net change in entropy in a reversible process.
What do you understand by:
(i) The entropy of fusion?
(ii) The entropy of vapourisation ?
You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below:
Liquid
Calculate the entropy change of n-hexane when 1 mole of it evaporates at 341.7 K(∆Hvap = 29.0 kJ mol–1).
The thermochemical equation is,
From the above equation, it is clear that when 1 mol of H2O(l) is formed, 286 kJ of heat is released. The same amount of heat is absorbed by the surroundings.
What are the two tendencies which determine the feasibility of process? How are the two related to each other?