Short Answer Type
Long Answer TypeProve 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 ?
Short Answer TypeYou 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).
Long Answer TypeWhat are the two tendencies which determine the feasibility of process? How are the two related to each other?
The following two tendencies are responsible for determining the feasibility of a process:
(i) The tendency of a system to acquire a state of minimum energy i.e. energy factor. It is expressed in terms of enthalpy change (∆H). A negative value of ∆H suggests that system has the tendency to proceed.
(ii) The tendency of a system to acquire a state of maximum randomness i.e. randomness factor. It is expressed by T∆S where T is the absolute temperature and ∆S is the change in entropy. A positive value of T∆S indicates an inherent tendency of a process to occur.
Relation between two tendencies:
The overall tendency or the driving force of a process is expressed in terms of free energy change (∆G). This is expressed as ∆G = ∆H – T∆S
This equation is called Gibb’s Helmholtz equation. For a spontaneous process, ∆G should have a negative value i.e. the system should undergo a decrease in its free energy.
what are the signs of 
?
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