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

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