## 9.11

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

Posts: 59
Joined: Fri Sep 29, 2017 7:07 am

### 9.11

This homework problem says to

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 believe we uses the equation

delta S = n*R*ln(P2/P1)

But what R value do we use? The 8.314 J/K*mol value wouldnt cancel with the atm right?

904676178
Posts: 5
Joined: Fri Jun 17, 2016 11:28 am

### Re: 9.11

I believe yes we do use that r value (8.314J/K*MOL) and the atm should just cancel each other out when you divide (p2/p1). so you are left with JK-1 in answer.

Sarah Sharma 2J
Posts: 33
Joined: Fri Sep 29, 2017 7:05 am

### Re: 9.11

Since you are working with ideal gases you would use the R value of 8.314 J/K mol.