December 22, 2024, 06:01:41 PM
Forum Rules: Read This Before Posting


Topic: Entropy Adiabatic Free Expansion  (Read 4395 times)

0 Members and 1 Guest are viewing this topic.

Offline gingi85

  • Regular Member
  • ***
  • Posts: 46
  • Mole Snacks: +0/-1
Entropy Adiabatic Free Expansion
« on: December 03, 2007, 04:44:05 PM »
I am having trouble with this problem. I seem to be getting a negative number ...

One mole of a monoatomic ideal gas at T1=300K and P1=10atm goes through free expansion at Pex=2 atm. The W done by the system is 228 calories. What is the change in entropy?

I divided it into two processes: 1) Isobaric from T1 to Tfinal, and 2) Isothermic from V1 to Vfinal.

V1 = nRT1/P1 = 2.49 m3

Pex(V2-V1)= W

582.7 m3 (!)

Tfinal = P2V2/R = 14042 K (?!?!)

(something tells me I've already made the mistake ...)



Process 1:

delta-S = Cpln(Tfinal/T1)

Cp = 3R/2

delta-S = (3*R/2)ln(14042/300) = 47.88 J/K

Process 2:

delta-S = nRln(Vf/V1) = 8.3ln(582/2.49) = 45.2698 J/K

delta-Stot = 93.15 J/K

Where did I make the mistake?


Offline Yggdrasil

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 3215
  • Mole Snacks: +485/-21
  • Gender: Male
  • Physical Biochemist
Re: Entropy Adiabatic Free Expansion
« Reply #1 on: December 03, 2007, 07:03:10 PM »
V1 = 2.49 dm3 since R = 0.08206 dm3 atm mol-1 K-1

(1 dm3 = 1 L)

Sponsored Links