Midterm 2012, Q3A

Boltzmann Equation for Entropy:

Moderators: Chem_Mod, Chem_Admin

Timmy Gozali
Posts: 5
Joined: Fri Sep 25, 2015 3:00 am

Midterm 2012, Q3A

Postby Timmy Gozali » Sun Feb 07, 2016 3:23 pm

The question is telling us to determine whether BH2F or O2 has a higher residual entropy.

The solution then says that W for BH2F is 3 while W for O2 is 1. Can someone explain why that is?

Jake Ney lecture 1 discussion 1F
Posts: 26
Joined: Fri Sep 25, 2015 3:00 am

Re: Midterm 2012, Q3A

Postby Jake Ney lecture 1 discussion 1F » Sun Feb 07, 2016 3:46 pm

O2 has only one possible state due to its lewis structure O=O. This is why its degeneracy, which is the number of possible states to the number of particles in that are in each state is 1.

The lewis structure of BH2F has three possible states as there are three alternate positions for the fluorine and hydrogen atoms which is why its w value is 3.

H-B-H
/
F

F-B-H
/
H

H-B-F
/
H


Return to “Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy”

Who is online

Users browsing this forum: No registered users and 2 guests