## 9.47

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

Hannah Guo 3D
Posts: 56
Joined: Fri Sep 29, 2017 7:06 am

### 9.47

9.47 Initially a sample of ideal gas at 323 K occupies 1.67 L at 4.95 atm. The gas is allowed to expand to 7.33 L by two pathways: (a) isothermal, reversible expansion; (b) isothermal, irreversible free expansion. Calculate Stot, S, and Ssurr for each pathway.

For (a), why since the process is reversible, delta Stot=0?
Thank you!

Christy Lee 2H
Posts: 73
Joined: Fri Sep 29, 2017 7:07 am
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### Re: 9.47

For a reversible process, the reaction is assumed to be at equilibrium which means deltaS of the universe is zero because there is no entropy change at equilibrium.