Example 4I.3

Volume:
Temperature:

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

Example 4I.3

Postby Reina Robles 2B » Tue Feb 11, 2020 4:28 pm

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

Postby Jacey Yang 1F » Wed Feb 12, 2020 2:45 pm

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

Postby Shutong Hou_1F » Sun Feb 16, 2020 7:14 pm

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

Postby Reina Robles 2B » Fri Mar 13, 2020 10:14 pm

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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