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## Homework Statement

Calculate the entropy for an irreversible adiabatic expansion of an ideal gas which goes from (Vi, Pi) = (9.01L, 3.00 bar) to (Vf, Pf) = (15.44L, 1.50 bar)

## Homework Equations

[tex]\Delta[/tex]S = [tex]\int dq_{reversible}/T[/tex]

## The Attempt at a Solution

I understand that entropy is a state function, and thus independent of path. So the above equation can be used to calculate the change in entropy for an irreversible adiabatic expansion. However, what I don't understand is how the entropy can possibly be non-zero. I mean, in a reversible adiabatic expansion which has the same (Vi,Pi) and (Vf,Pf) as the irreversible case, q = 0 by definition. So if [tex]dq_{reversible}[/tex] = 0 then how can [tex]\Delta[/tex]S be non-zero?

Any help as to what I'm missing here will be highly appreciated!

Edit: having written this question out, I think I might be on to something. I don't think it's possible for the reversible expansion to be adiabatic because U is a path function, and must be the same for reversible and irreversible case. In irreversible case, delta U = w. In the reversible case, w is different, so the remainder of the energy must be necessarily in form of heat, which is q(reversible). Am I on the right track?

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