Entropy Change in Free Expansion: Why △S=ncᵥln(T_f/T_i)+nRln(V_f/V_i)?

[spideyjj]

In a free expansion, I know that we cannot use the equation dS=dQ/T...(1). Instead we use dS>dQ/T...(2).

The question is that why we can use △S=ncᵥln(T_f/T_i)+nRln(V_f/V_i) , which is derived from the equation(1), to calculate the entropy change? Shouldn’t it be a inequality too?

 

[Lord Jestocost]

Entropy is a state function and doesn’t depend on the path by which the system arrived at its present state. The entropy change on going from an initial state to a final state is thus independent of how the final state is arrived at. To calculate the entropy change dS for an irreversible process, one generally designs a reversible process - linking the same two endpoints - by means of which dS can be calculated using dS = δQ_rev/T.

 

[Chestermiller]

Entropy is a state function and doesn’t depend on the path by which the system arrived at its present state. The entropy change on going from an initial state to a final state is thus independent of how the final state is arrived at. To calculate the entropy change dS for an irreversible process, one generally designs a reversible process - linking the same two endpoints - by means of which dS can be calculated using dS = δQ_rev/T.

I would add that the designed reversible process does not need to bear any resemblance whatsoever to the actual irreversible process, except insofar as matching the initial and final thermodynamic equilibrium states.

 

[Herman Trivilino]

Don't the beginning and ending states have to be equilibrium states for the scheme described in Posts #2 and #3 to be valid ways of finding the change in entropy?

 

[Chestermiller]

Don't the beginning and ending states have to be equilibrium states for the scheme described in Posts #2 and #3 to be valid ways of finding the change in entropy?

Isn't that what I said?