## 4I.9

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

Justin Vayakone 1C
Posts: 110
Joined: Sat Sep 07, 2019 12:19 am

### 4I.9

"Initially an 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 ΔStot, ΔS, and ΔSsurr for each pathway."

For part b, why is the ΔS of the irreversible process the same as the reversible process?

Lindsey Chheng 1E
Posts: 110
Joined: Fri Aug 30, 2019 12:16 am

### Re: 4I.9

Justin Vayakone 1C wrote:"Initially an 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 ΔStot, ΔS, and ΔSsurr for each pathway."

For part b, why is the ΔS of the irreversible process the same as the reversible process?

In the textbook, it says that deltaS of the system is the same value whether or not the process is reversible or irreversible. The deltaS of the surroundings is what changes depending on the type of process; therefore changing the value of deltaS total because the sum of deltaS (sys) and deltaS (surr) is equal to deltaS total.

805303639
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### Re: 4I.9

The ΔS of the irreversible process is the same as that of the reversible process because entropy is a state property. The pathway is irrelevant.

Anthony Hatashita 4H
Posts: 103
Joined: Wed Sep 18, 2019 12:21 am

### Re: 4I.9

ΔS is a state function, so they are the same either way. In irreversible however, ΔSsurr=0.