## 2010 Midterm Question #3 Part B

$\Delta S = \frac{q_{rev}}{T}$

704578485
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### 2010 Midterm Question #3 Part B

Question 3 involves a thermodynamic process in which a system is kept at 300K and is free to expand. In Part A it asks for q, w, deltaU, and deltaS for the system assuming it undergoes isothermal, reversible expansion. Part B asks for the same values but now assuming it is an isothermal, free expansion process in a vacuum. So since deltaU, q, and w are all 0 here wouldn't deltaS also be 0? It says the deltaS would just be the same as part A but I do not see why.

Arsheeta Kumar 1B
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Joined: Fri Sep 25, 2015 3:00 am

### Re: 2010 Midterm Question #3 Part B

Even though the deltaU, q, and w are all zero, the deltaS of the system is not zero. Delta S of the surroundings would be zero, but to find the deltaS, you would use the formula dS=nRln(v2/v1) instead of q/T because the system is irreversible.