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### 9.47 Isothermal Irreversible Free Expansion

Posted: Wed Feb 14, 2018 11:42 am
When solving part b asking about the isothermal irreversible free expansion of an ideal gas, if the expansion is irreversible and w=0, how do we calculate the entropy change of the system? The solutions manual says it is equal to 3.84J/K (the entropy change in the isothermal reversible expansion). Can anyone explain why? Thanks!

### Re: 9.47 Isothermal Irreversible Free Expansion

Posted: Wed Feb 14, 2018 1:41 pm
You would use $\Delta S = nRln(\frac{v2}{v1})$

### Re: 9.47 Isothermal Irreversible Free Expansion

Posted: Wed Feb 14, 2018 2:47 pm
It would be the same as ΔSsys of part a, the isothermal and reversible process. ΔSsurr = 0 for free expansion so ΔStot = ΔSsys.

### Re: 9.47 Isothermal Irreversible Free Expansion

Posted: Wed Feb 14, 2018 9:12 pm
Since entropy is a state function, the pathway taken from final to initial doesn't matter. ∆Ssys for irreversible and reversible should be the same.