1. 0 = 0 (ground state, no internal energy), 1 2 = 2·1 3 = 3·1
Example:
Assume that you have 3 molecules. Each molecule has the following energy states
0 = 0 (ground state, no internal energy), 1
2 = 2·1 3 = 3·1
4 = 4·1 …
Assume that the total energy of the whole system is 31
N0, N1, N2, N3, … = number of molecules in state 0, 1, 2, 3, …
 three configurations to distribute energy: E
0=0 1 2 3 4
configuration
(ensemble state) N0 N1 N2 N3
tot.particles
k Nk
tot.energy
k Nkk I 2 1 3
31 II
III

2. Possible configuration, i.e. distributions to molecules A, B, C
ensemble state
mol. state I II
III 3 A B C 2 1
ABC 0 BC AC AB W 3 6 1
W = Number of possible configurations per given
ensemble state = statistical weight