Entropy of free isothermal expansion of a perfect gas – why is it not 0?
Currently working on some thermodynamics (Atkins' Physical Chemistry) and I have a question about entropy changes. The textbook says that for the isothermal expansion of a perfect gas:
∆S = nR ln(Vf/Vi)
∆S(sur) = -nR ln(Vf/Vi)
Therefore ∆S(tot) = ∆S + ∆S(sur) = 0
Sure, I agree. But then when the expansion occurs freely (pressure = 0, so w = 0), it says that:
∆S = nR ln(Vf/Vi)
∆S(sur) = 0 (since ∆U = q + w = 0, hence q(sur) = 0)
Therefore ∆S(tot) = nR ln(Vf/Vi) > 0.
But I don't follow this. When w = 0, surely ∆S = q(rev)/T = -(w) / T = 0 too? Why do they use the other equation (nR ln (Vf/Vi))? Doesn't that apply only when w ≠ 0?