## 9.35

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

Cam Bear 2F
Posts: 60
Joined: Thu Jul 27, 2017 3:01 am

### 9.35

On the solution manual errors page, why do they use .5 mol for B and C rather than 1 mol? And how do you know that 1 vibrational degree of freedom changes the Cv from 5R/2 to 3R?

Errors page explanation:
Gas B (0.5 mol diatomic, no vibrational degrees of freedom):
∆S = nCv ln(T2/T1) = (0.5 moles)(5R/2)ln(T2/T1) and because all undergo the same
temperature change, delta S is essentially 5R/4.

Gas C (0.5 mol diatomic, 1 vibrational degree of freedom):
∆S = nCv ln(T2/T1) = (0.5 moles)(3R)ln(T2/T1) and because all undergo the same
temperature change, delta S is essentially 3R/2.

Thuy-Anh Bui 1I
Posts: 56
Joined: Sat Jul 22, 2017 3:00 am
Been upvoted: 2 times

### Re: 9.35

0.5 mol is used for B and C because they are both described as diatomic molecules. This means that 1 mol of atoms is actually 0.5 mol of gas since the atoms pair up.

1 vibrational degree means that we are dealing with a "complex" molecule so you would use the corresponding Cv value of 3R. No vibrational degree means that we are dealing with a normal linear molecule and the Cv value is 5R/2.

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