## HW:9.47 Isothermal Irreversible Free expansion

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

Yinhan_Liu_1D
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### HW:9.47 Isothermal Irreversible Free expansion

For part (b), If the expansion is irreversible and w=0 , how could we calculate the entropy change of the system?

Why does it equal to +3.84J/K, the entropy change in the (a)isothermal reversible expansion?

Chem_Mod
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### Re: HW:9.47 Isothermal Irreversible Free expansion

This is a good problem because it brings to light an otherwise easily forgotten aspect about entropy.

When isothermally and reversibly expanding to a larger volume, we can use the equation $\Delta S_{sys} = nRln(\frac{Vf}{Vi})$. As the volume expands, to keep the temperature the same (remember its isothermal) we can think of heat entering the system in an amount that corresponds with our $\Delta S_{sys}$. This heat can conversely be thought of as negatively entering the surroundings (-q) In the same way that we have $\Delta S_{sys}$ = +3.84 J/k, we have an identical but opposite decrease in the entropy of the surroundings(-3.84J/k from this -q. Adding them together we see that $\Delta S_{total} = 0$.

In the case of the irreversible free expansion, we know that no work is done because we cannot do work with "free expansion". For an isothermal process, the $\Delta U$ is always zero so we know that q + w = 0. If w = 0 then q must also = 0. So there is no way for us to calculate the entropy change of the surroundings because no heat was transferred in the first place from the surroundings.

Now here is where things get really interesting, when looking at the entropy change of only the system, because entropy is a state function, we don't actually need to know how we got from Vi to Vf. We simply need to know what the initial volume was and what the final volume is. In both cases of (a) and (b), they are the same initial and final state, so the $\Delta S$system = +3.84J/K in each scenario. Its only when we consider the $\Delta S$surroundings that things get different. For reversible expansion, the entropy goes down in the surroundings to compensate for the systems expansion, in the irreversible free expansion, there is no way for the surroundings entropy to decrease because it really hasn't communicated anything energetically with the system. In this way, we see $\Delta S_{surroundings}$ = 0 and hence $\Delta S_{total}$ = 0 + 3.84 J/k

Hope this explanation helps, good question!