## Last Question on the Midterm [ENDORSED]

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

Gabriel Ordonez 2K
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Joined: Sat Jul 20, 2019 12:15 am

### Last Question on the Midterm

With the conclusion of the midterm, does anyone know how the final question should've been approached, calculated and solved? What were the types of concepts that were in play in this specific assessment?

Jocelyn Thorp 1A
Posts: 103
Joined: Wed Sep 18, 2019 12:20 am

### Re: Last Question on the Midterm

I'd like to know as well, so I'm commenting to easily navigate back.

Qiu Ya Wu 4I
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Joined: Fri Sep 20, 2019 12:17 am

### Re: Last Question on the Midterm

I'm not sure about the exact details of the question, but the answers for the midterms should be posted soon on Lavelle's website complete with step by step solutions and a point breakdown!

Zoya Mulji 1K
Posts: 105
Joined: Thu Jul 11, 2019 12:15 am

### Re: Last Question on the Midterm

I think the last question was about finding entropy in both an irreverisble and reversible expansion. You approach both using the entropy formula (i believe it was the change in volume formula) but this will only give you the answer to deltaS. To find deltaSsurrounding and deltaStotal you had to know that in an irreversible expansion, deltaSsurrounding is 0 and in an reversible expansion, deltaStotal is 0. Knowing these two properties would allow you to plug pieces into deltaStotal= deltaS + deltaSsurroundings and find the missing ones.

Junwei Sun 4I
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Joined: Wed Oct 02, 2019 12:16 am

### Re: Last Question on the Midterm

I think the last question is about finding entropy for reversible and irreversible pathway. For the reversible isothermal pathway deltaS total would be zero and therefore deltaS system plus deltaS of surrounding would be zero. I think for the question we can calculate deltaS for the system by using nRln(V2/V1) formula. And deltaS surrounding is just going to be the negative of that value. For an irreversible, free expansion since S is a state function deltaS system for the irreversible path is the same as deltaS system for the reversible path. Also for free expansion work done is zero. Since deltaU = q+w q here is also zero. Therefore deltaS surrounding for the irreversible pathway is zero.

Chem_Mod
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### Re: Last Question on the Midterm  [ENDORSED]

See the homework problems.

Q7, 16 points, is HW 4.43.

Q8, 14 points, is HW 4I.9.