## 9.13 [ENDORSED]

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

Matthew Lee 3L
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Joined: Fri Sep 29, 2017 7:07 am

### 9.13

For question 9.13, the question requires two separate calculations of entropy and then the sum of those is the answer. For the change in entropy using temperature, it uses the constant R in the equation instead of C like it does on the formula sheet. Is this because there is no constant pressure or constant volume?

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### Re: 9.13  [ENDORSED]

Please be aware that the solution manual (6th edition) contains an error in the solution for this problem.

https://lavelle.chem.ucla.edu/wp-conten ... rs_6Ed.pdf

More directly to your question, the value of R is not simply substituted for C. The derivation for the change in entropy at isothermal conditions is as follows:

$\textup{d}S = \frac{q_{\textup{rev}}}{T}$
$q = -w \; (\textup{for isothermal processes})$
$q = nRT\textup{ln}\frac{V_{\textup{f}}}{V_{\textup{i}}}$

Substituting for q above:

$\Delta S = \frac{q_{\textup{rev}}}{T}=\frac{nRT\textup{ln}\frac{V_{\textup{f}}}{V_{\textup{i}}}}{T}=nR\textup{ln}\frac{V_{\textup{f}}}{V_{\textup{i}}}$

Since entropy is a state function, it does not matter how we go from state 1 to state 2. Both temperature and volume change simultaneously in the problem description, so we can think of this as a 2-step process. In the first step we make the process isothermal, in the second we continue from that point with an isochoric (constant volume) change.

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