## HOTDOG Question 5A

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

Lexie Baughman 2C
Posts: 30
Joined: Sat Oct 06, 2018 12:16 am

### HOTDOG Question 5A

Can someone explain why we don't use the Cv for a monatomic ideal gas (3/2R) when calculating the change in entropy that accounts for the temperature change of the system? I have delta S = (3.73 mol)(8.314 J/Kmol)ln(348/323) = 3.48 J/K, which includes R but not the 3/2.

Kevin Tang 4L
Posts: 83
Joined: Fri Sep 28, 2018 12:28 am

### Re: HOTDOG Question 5A

I think you do use Cv,m for this because we are using the equation DeltaS=Cv,m(lnT2/T1).

Thus, for DeltaS for the change in temperature we do (3.73mol)(3/2)(8.3145 J*mol-1*K-1)(ln 348K/323K)

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