[G O D I V A]

Isothermal expansion of a perfect gas from a Vi to Vf.  If it is irreversible expansion (such as free expansion) then for du = dq + dw --> dw and du = 0 therefore dq = 0.  Therefore ds_surr=0 so ΔS_total is always > 0.

I dont get that; i must of completely forgot some parts of the first law.

I remember that for free expansion dw = 0, but I cant remember why free expansion is irreversible or why for irreversible du = 0.

[Yggdrasil]

For an ideal gas, U depends only on T.  Therefore, if the temperature does not change, the internal energy of the gas cannot change.  So, it's the isothermal part that tells you ΔU = 0, not the irreversible part.

[renge ishyo]

I remember that for free expansion dw = 0, but I cant remember why free expansion is irreversible

Free expansion is irreversible because the probability that a gas will appear distributed across the entire volume of a container (call this volume "V2") is much much higher than that of a gas localizing itself in one small part of the container while leaving a vaccum in the other part (call this volume "V1"). This is described for the change in entropy of the system in the case of free expansion by:

 :delta: S_system = nRln(V2/V1)

If V1 is the volume the gas originally occupies in one small part of the container, and V2 is the volume of the entire container that the gas occupies after the expansion, then V2 is larger than V1. :delta: S_system will be positive in this situation (you can check by plugging in a larger number for V2 and taking the natural log...it will be positive). Now if the gas is allowed to freely expand from V1 to V2 in an isothermal situation without doing any work on surrounding molecules then for the surroundings, dw=0 for the work and dU = 0 since the temperature doesn't change (as Yggs mentioned, this is for the isothermal part of things), so that  :delta: S of the surroundings is ultimately equal to zero.

So now let's look at the situation of the entropy in total:

 :delta: S_total =  :delta: S_system +  :delta: S_surroundings

Since :delta: S_surroundings = 0:

 :delta: S_total = nRln(V2/V1)

And so for an expansion where V2 > V1, :delta: S_total is always positive.

While writing this out, I used the following website for reference: http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node40.html

[Yggdrasil]

It's also worth noting that for a process to be reversible the system must be in thermal equilibrium (T_sys = T_surr) and mechanical (P_sys = P_surr) equilibrium with the surroundings.  In the case of free expansion, P_surr = 0, so P_sys is not equal to P_surr.

[renge ishyo]

That's true; in fact, I think I have seen this same argument given with the equation for entropy of the system for free expansion involving pressures P2 and P1 where P2 > P1 as opposed to treating it by considering volumes V2 and V1, and the result came out the same (the ratio of the two pressures to each other and the two volumes to each other are the same in this case, and if you think about it, it is really the difference in pressure between the two areas of the container that is driving things forward).