## Help with Self Test 9.5A

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

Susanna
Posts: 11
Joined: Tue Nov 17, 2015 3:00 am

### Help with Self Test 9.5A

Calculate the change in entropy when the volume of 2.00 mol Ar(g) is decreased from 10.00 L to 5.00 L while the temperature decreases from 300 K to 100 K. Assume ideal behavior.

The answer is -38.9 J K^-1 but I got -57.34 J K^-1.
I followed the example above and first used delta S = nRlnV2/V1 and got -11.53 J K^-1 and then used delta S = CvlnT2/T1 and Cv=nCv,m and got -45.81 J K^-1. I then added both delta S' together and got -57.34, can someone tell me where I went wrong or if you approached the problem differently? Thanks.

Chem_Mod
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### Re: Help with Self Test 9.5A

Your value -11.53 J/K is correct. However, when you calculate for the nCvln(T2/T1) I actually get the value -27.4 J/K.
To calculate did you do $2 mol (\frac{3}2{})(8.314\frac{J}{mol\cdot K})ln(\frac{100K}{300K})$?

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