## Example 9.6

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

Ethan-Van To Dis2L
Posts: 50
Joined: Fri Sep 29, 2017 7:05 am

### Example 9.6

In example 9.6, it talks about the entropy of vaporization of acetone at 296 K, but I'm kind of confused on why you we would use the molar heat capacity for the first step, and then 4R for the heat capacity in the third step. Can someone explain to me?

jillian1k
Posts: 54
Joined: Sat Jul 22, 2017 3:00 am

### Re: Example 9.6

In step 1, acetone is in its liquid state, so you would use the molar heat capacity that corresponds to this phase. In step 2, liquid acetone in vaporized into acetone vapor at its boiling point. So, in step 3, acetone is now in the gas phase. Therefore, you have to use the corresponding Cp (since the system is at constant P) for a gas. Acetone vapor is a nonlinear molecule, so you use the 4R value for Cp in the chart on page 281.