## Example 4I.3

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

Reina Robles 2B
Posts: 71
Joined: Fri Aug 09, 2019 12:16 am

### Example 4I.3

Q: Calculate ΔS, ΔSsurr, and ΔStot for a) isothermal, reversible expansion and b) isothermal, free expansion of 1.00 mol of ideal gas molecules from 8.00 L to 20.00 L at 292 K.

a) isothermal & reversible expansion:
ΔS = +7.6 J/K
ΔSsurr = -7.6 J/K
ΔStot = 0

b) isothermal, free expansion (and implied that it's irreversible too):
ΔSsurr = 0 (Free expansion means no external pressure, so no work is done because the gas isn't expanding against anything. Therefore, q = 0, and ΔS = 0 because ΔS = q/T).
ΔS = +7.6 J/K (The same as the reversible path, according to the book).
ΔStot = +7.6 J/K

I understand why ΔSsurr is 0, but I don't understand why ΔS is +7.6 J/K. Shouldn't it also be 0 because work still is 0? Thanks for the help!

Jacey Yang 1F
Posts: 101
Joined: Fri Aug 09, 2019 12:17 am

### Re: Example 4I.3

Entropy is a state function, so the change in entropy of the system is the same regardless of whether the path is reversible or irreversible.

Shutong Hou_1F
Posts: 117
Joined: Sat Sep 14, 2019 12:17 am

### Re: Example 4I.3

I agree with the peer above's explanation. In addition, the initial & final states are the same because the process is isothermal, so, according to ΔS = q(rev)/T, ΔS should be the same.

Reina Robles 2B
Posts: 71
Joined: Fri Aug 09, 2019 12:16 am

### Re: Example 4I.3

Jacey Yang 1F wrote:Entropy is a state function, so the change in entropy of the system is the same regardless of whether the path is reversible or irreversible.

Shutong Hou_1F wrote:I agree with the peer above's explanation. In addition, the initial & final states are the same because the process is isothermal, so, according to ΔS = q(rev)/T, ΔS should be the same.

I see, thank you!

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