## 9.47 Isothermal Irreversible Free Expansion

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

Cam Bear 2F
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Joined: Thu Jul 27, 2017 3:01 am

### 9.47 Isothermal Irreversible Free Expansion

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!

Clara Hu 1G
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### Re: 9.47 Isothermal Irreversible Free Expansion

You would use $\Delta S = nRln(\frac{v2}{v1})$

Tatiana Hage 2E
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### Re: 9.47 Isothermal Irreversible Free Expansion

It would be the same as ΔSsys of part a, the isothermal and reversible process. ΔSsurr = 0 for free expansion so ΔStot = ΔSsys.

Janine Chan 2K
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Joined: Fri Sep 29, 2017 7:04 am

### Re: 9.47 Isothermal Irreversible Free Expansion

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.