## 6th edition 9.11

isochoric/isometric: $\Delta V = 0$
isothermal: $\Delta T = 0$
isobaric: $\Delta P = 0$

Julia Lung 1I
Posts: 30
Joined: Fri Jun 23, 2017 11:39 am

### 6th edition 9.11

Calculate the change in entropy when the pressure of 1.50 mol Ne(g) is decreased isothermally from 15.0 atm to 0.500 atm. Assume ideal behavior.
I got the right answer for this problem using ΔS=nRln(P1/P2) = ΔS=(1mol)(8.314 J/k*mol)ln(15.0atm/0.500atm) = 42.2J/K but I'm confused how the units cancel because the R constant that I used is J/K*mol that is not L*atm/K*mol even though the pressure is using atmospheres.

Emily Ng_4C
Posts: 65
Joined: Fri Sep 28, 2018 12:17 am

### Re: 6th edition 9.11

When you divide the pressures, the atm's cancel each other out. Your ending units should be J/ K.