Partition Function of 2 State System

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In a two-state system with energy levels of 0 and V, the partition function is correctly calculated as Z = 1 + e^(-V/kT). The average energy is not simply V; it is determined by the probabilities of occupying each state, leading to a more complex relationship. Entropy is not zero in this system; it reflects the uncertainty and distribution of states, even when one state has zero energy. The confusion arises from misinterpreting the relationship between energy, partition function, and entropy in statistical mechanics. Therefore, the entropy of a two-state system with one state at zero energy is not zero.
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If I have a 2 state system with energy levels of the 2 states to be 0 and V. I find the partition function to be Z = 1 + e^(-V/kT). Am I correct? If so, does that not mean the average energy is V? and thus the entropy is 0? This doesn't make sense, how is the entropy of a 2 state system (when 1 state is zero energy) 0?!

Thanks!
 
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Is the entropy of a 2 state system with 1 state with energy 0 equal to 0?
 

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