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 ?
(i) The entropy of fusion. It may be defined as the entropy change when one mole of the solid substance changes into liquid form at its melting point. For example when ice melts as
where ∆Hfusion is the enthalpy of fusion and Tf is the fusion temperature. Since ∆Hfus is +ve, therefore ∆Sjfus is +ve, hence the process of fusion is accompanied by an increase of entropy.
(ii) The entropy of vapourisation. It may be defined as the entropy change when one mole of the liquid changes into vapour at its boiling point.
Since ∆Hvap is +ve, therefore ∆Svap is +ve, hence the process of vapourisation is accompanied by an increase of entropy.
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).
What are the two tendencies which determine the feasibility of process? How are the two related to each other?