Entropy Equation

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

Lindsay H 2B
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Entropy Equation

In my notes from the lecture a few weeks ago on wednesday 1/24 I have written that ∆S=-nRTln(V2/V1) but the solutions manual and the book both use that equation without the negative sign. Can anyone explain why that is?

Chem_Mod
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Re: ∆S=nRTln(Vf/Vi)

The equation you wrote down is actually the equation for work. Please see Dr. Lavelle's constants and equations sheet on his website for clarification. https://lavelle.chem.ucla.edu/wp-conten ... ations.pdf

Also note that there is no T variable in the equation for $\Delta S$ when there is a volume change.

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Re: ∆S=nRTln(Vf/Vi)

Your expression is indeed that for work, not change in entropy. But this is a matter of going through the derivation.
$\Delta U = q+w$

Under an isothermal process,
$\Delta U = 0$
, so that
$q=-w$

Work under an isothermal reaction is
$W = -nRT(\frac{V_{2}}{V_{1}})$

And change in entropy, with constant temperature, is
$\Delta S = \frac{q}{T}$

Substitute w for q in the above equation
$\Delta S = \frac{-w}{T}$

Substitute the expression of W under an isothermal process for
$\Delta S = nRln(\frac{V_{2}}{V_{1}})$

The final equation is how you would calculate the change in entropy for an isothermal process.