9.11

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

Caitlin Mispagel 1D
Posts: 47
Joined: Tue Oct 10, 2017 7:13 am

9.11

For this question it asks "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 am using the equation deltaS= nRln(P2/P1) but I am getting the wrong answer. I found that the problem was that to get the right answer I need to use 15.0 as P2 and 0.500 as P1 but since 0.500 is the final and 15.0 is the initial why wouldn't it be the other way around?

Sabrina Dunbar 1I
Posts: 57
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

Re: 9.11

I got the correct answer taking the natural log of (.500 atm/15 atm) and then multiplying it by the constant value R and 1.5 as the number of moles of neon. I will include the substituted values that I used below. Maybe it was a simple calculation error?

∆S=(1.5 mol)(8.314 J/Kmol)(ln(.500 atm/15 atm))=42.4 J/K

Lena Nguyen 2H
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am
Been upvoted: 1 time

Re: 9.11

You're correct about the initial and final pressures, but the equation is deltaS = nRln(P1/P2) according to the solutions manual.

Mitch Mologne 1A
Posts: 74
Joined: Fri Sep 29, 2017 7:04 am

Re: 9.11

I believe it is because the starting equation is deltaS=nRTln(V2/V1) with volumes instead of pressure. If we apply Boyle's law that Pressure and Volume are inversely proportional (V1P1=V2P2) then we would be able to inverse the volumes to be pressures, thus deltaS=nRTln(P1/P2). To my knowledge, that is why you were able to get the right answer by having the pressures flipped.

Morgan Baxter 1E
Posts: 50
Joined: Thu Jul 27, 2017 3:00 am

Re: 9.11

You are correct in thinking that the initial pressure (P1) is 15 and the final pressure (P2) is .5. However, you are using an incorrect equation, which resulted in the wrong answer. The equation is change in entropy= nRln(P1/P2). The equations for change in volume and temperature have ln(V2/V1) and ln(T2/T1) respectively, however the pressure equation has ln(P1/P2) because according to Boyle's law, pressure is inversely proportional to volume. Therefore, if the volume equation is: change in entropy= nRln(V2/V1), then substituting pressure for volume according to boyles law, you get change in entropy= nRln(P1/P2). Once you use the correct formula, you will get the correct answer.